CHEMISTRY
Raymond Chang
Williams College
Kenneth A. Goldsby
Florida State University
CHEMISTRY, TWELFTH EDITION
Published by McGraw-Hill Education, 2 Penn Plaza, New York, NY 10121. Copyright 2016 by McGraw-Hill Education. All rights
reserved. Printed in the United States of America. Previous editions © 2013, 2010, and 2007. No part of this publication may be
reproduced or distributed in any form or by any means, or stored in a database or retrieval system, without the prior written consent
of McGraw-Hill Education, including, but not limited to, in any network or other electronic storage or transmission, or broadcast for
distance learning.
Some ancillaries, including electronic and print components, may not be available to customers outside the United States.
This book is printed on acid-free paper.
1 2 3 4 5 6 7 8 9 0 DOW/DOW 1 0 9 8 7 6 5
ISBN 978–0–07–802151–0
MHID 0–07–802151–0
Senior Vice President, Products & Markets: Kurt L. Strand
Vice President, General Manager, Products & Markets: Marty Lange
Vice President, Content Design & Delivery: Kimberly Meriwether David
Managing Director: Thomas Timp
Director: David Spurgeon, PhD
Director, Product Development: Jodi Rhomberg
Marketing Director: Tamara L. Hodge
Director of Digital Content Development: Shirely Hino, PhD
Director, Content Design & Delivery: Linda Avenarius
Program Manager: Lora Neyens
Content Project Managers: Sandy Wille (core); Tammy Juran (assessment)
Buyer: Sandy Ludovissy
Design: David Hash
Content Licensing Specialists: John Leland (photo); DeAnna Dausner (text)
Cover Image: “Water desalination across nanoporous graphene”; © David Cohen-Tanugi and Professor Jeffrey C Grossman
at the Massachusetts Institute of Technology
Compositor: Aptara®, Inc.
Typeface: 10/12 Times Roman
Printer: R. R. Donnelley
All credits appearing on page or at the end of the book are considered to be an extension of the copyright page.
Library of Congress Cataloging-in-Publication Data
Chang, Raymond.
Chemistry.—Twelfth edition / Raymond Chang, Williams College, Kenneth A. Goldsby,
Florida State University.
pages cm
Includes index.
ISBN 978-0-07-802151-0 (0-07-802151-0 : alk. paper) 1. Chemistry—Textbooks.
I. Goldsby, Kenneth A. II. Title.
QD31.3.C38 2016
540—dc23
2014024893
The Internet addresses listed in the text were accurate at the time of publication. The inclusion of a website does not indicate an
endorsement by the authors or McGraw-Hill Education, and McGraw-Hill Education does not guarantee the accuracy of the
information presented at these sites.
www.mhhe.com
About the Authors
Raymond Chang
was born in Hong Kong and grew up in
Shanghai and Hong Kong. He received his B.Sc. degree in chemistry
from London University, and his Ph.D. in chemistry from Yale University.
After doing postdoctoral research at Washington University and teaching
for a year at Hunter College of the City University of New York, he
joined the chemistry department at Williams College.
Professor Chang has served on the American Chemical Society
Examination Committee, the National Chemistry Olympiad Examination,
and the Graduate Record Examination (GRE) Committee. He has written
books on physical chemistry, industrial chemistry, and physical science.
He has also coauthored books on the Chinese language, children’s picture books, and a novel for young readers.
For relaxation, Professor Chang does gardening, plays the harmonica, and practices the piano.
Ken Goldsby
was born and raised in Pensacola, Florida. He
received his B.A. in chemistry and mathematical science from Rice
University. After obtaining his Ph.D. in chemistry from the University
of North Carolina at Chapel Hill, Ken carried out postdoctoral research
at Ohio State University.
Since joining the Department of Chemistry and Biochemistry at
Florida State University in 1986, Ken has received several teaching and
advising awards, including the Cottrell Family Professorship for Teaching
in Chemistry. In 1998 he was selected as the Florida State University
Distinguished Teaching Professor. Ken also works with students in his
laboratory on a project to initiate collaborations between science departments and technical arts programs.
When he is not working, Ken enjoys hanging out with his family.
They especially like spending time together at the coast.
iii
Contents in Brief
1
2
3
4
5
6
7
8
9
10
Chemistry: The Study of Change 1
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
Intermolecular Forces and Liquids and Solids
Atoms, Molecules, and Ions
Mass Relationships in Chemical Reactions
Reactions in Aqueous Solutions
Gases
75
118
172
Thermochemistry 230
Quantum Theory and the Electronic Structure of Atoms
Periodic Relationships Among the Elements
Chemical Bonding I: Basic Concepts
274
326
368
Chemical Bonding II: Molecular Geometry and Hybridization
of Atomic Orbitals 412
465
Physical Properties of Solutions 518
Chemical Kinetics
562
Chemical Equilibrium
621
Acids and Bases 666
Acid-Base Equilibria and Solubility Equilibria
Entropy, Free Energy, and Equilibrium
Electrochemistry
Nuclear Chemistry
720
776
812
862
Chemistry in the Atmosphere
900
Metallurgy and the Chemistry of Metals
930
Nonmetallic Elements and Their Compounds
956
Transition Metals Chemistry and Coordination Compounds
Organic Chemistry 1025
Synthetic and Natural Organic Polymers 1058
Appendix
Appendix
Appendix
Appendix
iv
38
1
2
3
4
Derivation of the Names of Elements A-1
Units for the Gas Constant A-7
Thermodynamic Data at 1 atm and 25°C A-8
Mathematical Operations A-13
994
Contents
List of Applications xix
List of Animations xx
Preface xxi
Setting the Stage for Learning xxix
A Note to the Student xxxii
CHAPTER 1
Chemistry:
C
hem
The Study of Change 1
1.1
1.2
1.3
Chemistry: A Science for the Twenty-First Century 2
The Study of Chemistry 2
The Scientific Method 4
CHEMISTRY in Action
The Search for the Higgs Boson 6
1.4
1.5
1.6
1.7
Classifications of Matter 6
The Three States of Matter 9
Physical and Chemical Properties of Matter 10
Measurement 11
CHEMISTRY in Action
The Importance of Units 17
1.8 Handling Numbers 18
1.9 Dimensional Analysis in Solving Problems 23
1.10 Real-World Problem Solving: Information, Assumptions,
and Simplifications 27
Key Equations 28
Summary of Facts & Concepts 29
Key Words 29
Questions & Problems 29
CHEMICAL M YS TERY
The Disappearance of the Dinosaurs 36
v
vi
Contents
CHAPTER 2
Atoms,
A
tom Molecules, and Ions 38
2.1
2.2
2.3
2.4
The Atomic Theory 39
The Structure of the Atom 40
Atomic Number, Mass Number, and Isotopes 46
The Periodic Table 48
CHEMISTRY in Action
Distribution of Elements on Earth and in Living Systems 49
2.5
2.6
2.7
2.8
Molecules and Ions 50
Chemical Formulas 52
Naming Compounds 56
Introduction to Organic Compounds 65
Key Equation 67
Summary of Facts & Concepts 67
Key Words 67
Questions & Problems 68
CHAPTER 3
Mass
M
as Relationships in Chemical Reactions 75
3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.8
3.9
3.10
Atomic Mass 76
Avogadro’s Number and the Molar Mass of an Element 77
Molecular Mass 81
The Mass Spectrometer 83
Percent Composition of Compounds 85
Experimental Determination of Empirical Formulas 88
Chemical Reactions and Chemical Equations 90
Amounts of Reactants and Products 95
Limiting Reagents 99
Reaction Yield 103
CHEMISTRY in Action
Chemical Fertilizers 105
Key Equations 106
Summary of Facts & Concepts 106
Key Words 106
Questions & Problems 106
Contents
CHAPTER 4
Reactions in Aqueous Solutions 118
4.1
4.2
General Properties of Aqueous Solutions 119
Precipitation Reactions 121
CHEMISTRY in Action
An Undesirable Precipitation Reaction 126
4.3
4.4
Acid-Base Reactions 126
Oxidation-Reduction Reactions 132
CHEMISTRY in Action
Breathalyzer 144
4.5
4.6
4.7
4.8
Concentration of Solutions 145
Gravimetric Analysis 149
Acid-Base Titrations 151
Redox Titrations 155
CHEMISTRY in Action
Metal from the Sea 156
Key Equations 157
Summary of Facts & Concepts 158
Key Words 158
Questions & Problems 158
CHEMICAL M YS TERY
Who Killed Napoleon? 170
CHAPTER 5
Gases 172
5.1
5.2
5.3
5.4
5.5
5.6
Substances That Exist as Gases 173
Pressure of a Gas 174
The Gas Laws 178
The Ideal Gas Equation 184
Gas Stoichiometry 193
Dalton’s Law of Partial Pressures 195
CHEMISTRY in Action
Scuba Diving and the Gas Laws 200
5.7
The Kinetic Molecular Theory of Gases 202
CHEMISTRY in Action
Super Cold Atoms 208
5.8
Deviation from Ideal Behavior 210
Key Equations 213
Summary of Facts & Concepts 214
Key Words 214
Questions & Problems 215
CHEMICAL M YS TERY
Out of Oxygen 228
vii
viii
Contents
CHAPTER 6
Thermochemistry
T
her
230
6.1
6.2
6.3
The Nature of Energy and Types of Energy 231
Energy Changes in Chemical Reactions 232
Introduction to Thermodynamics 234
CHEMISTRY in Action
Making Snow and Inflating a Bicycle Tire 240
6.4
6.5
Enthalpy of Chemical Reactions 240
Calorimetry 246
CHEMISTRY in Action
White Fat Cells, Brown Fat Cells, and a Potential Cure for Obesity 250
6.6
Standard Enthalpy of Formation and Reaction 253
CHEMISTRY in Action
How a Bombardier Beetle Defends Itself 256
6.7
Heat of Solution and Dilution 258
Key Equations 261
Summary of Facts & Concepts 261
Key Words 262
Questions & Problems 262
CHEMICAL M YS TERY
The Exploding Tire 272
CHAPTER 7
Quantum
Q
uan
an
Theory and the
Electronic
Structure of Atoms 274
E
lec
7.1
7.2
7.3
7.4
From Classical Physics to Quantum Theory 275
The Photoelectric Effect 279
Bohr’s Theory of the Hydrogen Atom 282
The Dual Nature of the Electron 287
CHEMISTRY in Action
Laser—The Splendid Light 288
7.5
Quantum Mechanics 291
CHEMISTRY in Action
Electron Microscopy 292
7.6
7.7
7.8
Quantum Numbers 295
Atomic Orbitals 297
Electron Configuration 301
Contents
7.9
The Building-Up Principle 308
CHEMISTRY in Action
Quantum Dots 312
Key Equations 313
Summary of Facts & Concepts 314
Key Words 315
Questions & Problems 315
CHEMICAL M YS TERY
Discovery of Helium and the Rise and Fall of Coronium 324
CHAPTER 8
Periodic Relationships
Among the Elements 326
8.1
8.2
8.3
8.4
Development of the Periodic Table 327
Periodic Classification of the Elements 329
Periodic Variation in Physical Properties 333
Ionization Energy 340
CHEMISTRY in Action
The Third Liquid Element? 341
8.5
8.6
Electron Affinity 345
Variation in Chemical Properties
of the Representative Elements 347
CHEMISTRY in Action
Discovery of the Noble Gases 358
Key Equation 359
Summary of Facts & Concepts 359
Key Words 360
Questions & Problems 360
CHAPTER 9
Chemical Bonding I: Basic Concepts 368
9.1
9.2
9.3
Lewis Dot Symbols 369
The Ionic Bond 370
Lattice Energy of Ionic Compounds 372
CHEMISTRY in Action
Sodium Chloride—A Common and Important Ionic Compound 376
9.4
9.5
9.6
9.7
The Covalent Bond 377
Electronegativity 380
Writing Lewis Structures 384
Formal Charge and Lewis Structure 387
ix
x
Contents
9.8
9.9
The Concept of Resonance 390
Exceptions to the Octet Rule 392
CHEMISTRY in Action
Just Say NO 397
9.10 Bond Enthalpy 398
Key Equation 403
Summary of Facts & Concepts 403
Key Words 403
Questions & Problems 403
CHAPTER 10
Chemical
C
hem
Bonding II: Molecular Geometry
aand
nd Hybridization of Atomic Orbitals 412
10.1 Molecular Geometry 413
10.2 Dipole Moments 423
CHEMISTRY in Action
Microwave Ovens—Dipole Moments at Work 426
10.3 Valance Bond Theory 429
10.4 Hybridization of Atomic Orbitals 431
10.5 Hybridization in Molecules Containing Double
and Triple Bonds 440
10.6 Molecular Orbital Theory 443
10.7 Molecular Orbital Configurations 446
10.8 Delocalized Molecular Orbitals 452
CHEMISTRY in Action
Buckyball, Anyone? 454
Key Equations 456
Summary of Facts & Concepts 456
Key Words 456
Questions & Problems 457
CHAPTER 11
IIntermolecular
nter
Forces and Liquids
aand
nd Solids 465
11.1 The Kinetic Molecular Theory of Liquids and Solids 466
11.2 Intermolecular Forces 467
11.3 Properties of Liquids 473
CHEMISTRY in Action
A Very Slow Pitch 475
11.4 Crystal Structure 477
CHEMISTRY in Action
Why Do Lakes Freeze from the Top Down? 478
11.5 X-Ray Diffraction by Crystals 483
Contents
11.6 Types of Crystals 486
CHEMISTRY in Action
High-Temperature Superconductors 488
CHEMISTRY in Action
And All for the Want of a Button 492
11.7 Amorphous Solids 492
11.8 Phase Changes 493
11.9 Phase Diagrams 503
CHEMISTRY in Action
Hard-Boiling an Egg on a Mountaintop, Pressure Cookers,
and Ice Skating 505
CHEMISTRY in Action
Liquid Crystals 506
Key Equations 508
Summary of Facts & Concepts 508
Key Words 509
Questions & Problems 509
CHAPTER 12
Physical
P
hys
Properties of Solutions 518
12.1
12.2
12.3
12.4
12.5
Types of Solutions 519
A Molecular View of the Solution Process 520
Concentration Units 522
The Effect of Temperature on Solubility 527
The Effect of Pressure on the Solubility of Gases 529
CHEMISTRY in Action
The Killer Lake 531
12.6 Colligative Properties of Nonelectrolyte Solutions 532
12.7 Colligative Properties of Electrolyte Solutions 544
CHEMISTRY in Action
Dialysis 546
12.8 Colloids 546
Key Equations 549
Summary of Facts & Concepts 549
Key Words 550
Questions & Problems 550
CHEMICAL M YS TERY
The Wrong Knife 560
xi
xii
Contents
CHAPTER 13
Chemical
C
hem
Kinetics 562
13.1 The Rate of a Reaction 563
13.2 The Rate Law 571
13.3 The Relation Between Reactant Concentration and Time 575
CHEMISTRY in Action
Radiocarbon Dating 586
13.4 Activation Energy and Temperature Dependence
of Rate Constants 588
13.5 Reaction Mechanisms 594
13.6 Catalysis 599
CHEMISTRY in Action
Pharmacokinetics 606
Key Equations 608
Summary of Facts & Concepts 608
Key Words 609
Questions & Problems 609
CHAPTER 14
Chemical
C
hem
Equilibrium 621
14.1 The Concept of Equilibrium and
14.2
14.3
14.4
14.5
the Equilibrium Constant 622
Writing Equilibrium Constant Expressions 625
The Relationship Between Chemical Kinetics
and Chemical Equilibrium 637
What Does the Equilibrium Constant Tell Us? 638
Factors That Affect Chemical Equilibrium 644
CHEMISTRY in Action
Life at High Altitudes and Hemoglobin Production 651
CHEMISTRY in Action
The Haber Process 652
Key Equations 654
Summary of Facts & Concepts 654
Key Words 655
Questions & Problems 655
CHAPTER 15
Acids
A
cid and Bases 666
15.1
15.2
15.3
15.4
15.5
15.6
15.7
Brønsted Acids and Bases 667
The Acid-Base Properties of Water 668
pH—A Measure of Acidity 670
Strength of Acids and Bases 673
Weak Acids and Acid Ionization Constants 677
Weak Bases and Base Ionization Constants 685
The Relationship Between the Ionization Constants
of Acids and Their Conjugate Bases 687
Contents
15.8
15.9
15.10
15.11
15.12
xiii
Diprotic and Polyprotic Acids 688
Molecular Structure and the Strength of Acids 692
Acid-Base Properties of Salts 696
Acid-Base Properties of Oxides and Hydroxides 702
Lewis Acids and Bases 704
CHEMISTRY in Action
Antacids and the pH Balance in Your Stomach 706
Key Equations 708
Summary of Facts & Concepts 709
Key Words 709
Questions & Problems 709
CHEMICAL M YS TERY
Decaying Papers 718
CHAPTER 16
Acid-Base
A
cid
Equilibria and Solubility
Equilibria
720
E
qui
16.1
16.2
16.3
16.4
Homogeneous versus Heterogeneous Solution Equilibria 721
The Common Ion Effect 721
Buffer Solutions 724
Acid-Base Titrations 730
CHEMISTRY in Action
Maintaining the pH of Blood 732
16.5
16.6
16.7
16.8
16.9
16.10
Acid-Base Indicators 739
Solubility Equilibria 742
Separation of Ions by Fractional Precipitation 749
The Common Ion Effect and Solubility 751
pH and Solubility 753
Complex Ion Equilibria and Solubility 756
CHEMISTRY in Action
How an Eggshell Is Formed 760
16.11 Application of the Solubility Product Principle
to Qualitative Analysis 761
Key Equations 763
Summary of Facts & Concepts 764
Key Words 764
Questions & Problems 764
CHEMICAL M YS TERY
A Hard-Boiled Snack 774
xiv
Contents
CHAPTER 17
Entropy,
E
ntr
Free Energy, and Equilibrium 776
17.1
17.2
17.3
17.4
17.5
The Three Laws of Thermodynamics 777
Spontaneous Processes 777
Entropy 778
The Second Law of Thermodynamics 783
Gibbs Free Energy 789
CHEMISTRY in Action
The Efficiency of Heat Engines 790
17.6 Free Energy and Chemical Equilibrium 796
17.7 Thermodynamics in Living Systems 800
CHEMISTRY in Action
The Thermodynamics of a Rubber Band 801
Key Equations 803
Summary of Facts & Concepts 803
Key Words 803
Questions & Problems 804
CHAPTER 18
Electrochemistry
E
lec
812
18.1
18.2
18.3
18.4
18.5
18.6
Redox Reactions 813
Galvanic Cells 816
Standard Reduction Potentials 818
Thermodynamics of Redox Reactions 824
The Effect of Concentration of Cell Emf 827
Batteries 832
CHEMISTRY in Action
Bacteria Power 837
18.7 Corrosion 838
18.8 Electrolysis 841
CHEMISTRY in Action
Dental Filling Discomfort 846
Key Equations 848
Summary of Facts & Concepts 848
Key Words 849
Questions & Problems 849
CHEMICAL M YS TERY
Tainted Water 860
Contents
CHAPTER 19
Nuclear
N
ucl
Chemistry 862
19.1
19.2
19.3
19.4
19.5
The Nature of Nuclear Reactions 863
Nuclear Stability 865
Natural Radioactivity 870
Nuclear Transmutation 874
Nuclear Fission 877
CHEMISTRY in Action
Nature’s Own Fission Reactor 882
19.6 Nuclear Fusion 883
19.7 Uses of Isotopes 886
19.8 Biological Effects of Radiation 888
CHEMISTRY in Action
Food Irradiation 890
Key Equations 890
CHEMISTRY in Action
Boron Neutron Capture Therapy 891
Summary of Facts & Concepts 891
Key Words 892
Questions & Problems 892
CHEMICAL M YS TERY
The Art Forgery of the Twentieth Century 898
CHAPTER 20
Chemistry
C
hem
in the Atmosphere 900
20.1
20.2
20.3
20.4
20.5
20.6
20.7
20.8
Earth’s Atmosphere 901
Phenomena in the Outer Layers of the Atmosphere 905
Depletion of Ozone in the Stratosphere 907
Volcanoes 911
The Greenhouse Effect 912
Acid Rain 916
Photochemical Smog 919
Indoor Pollution 921
Summary of Facts & Concepts 924
Key Words 924
Questions & Problems 925
xv
xvi
Contents
CHAPTER 21
Metallurgy
M
eta
and the Chemistry of Metals 930
21.1
21.2
21.3
21.4
21.5
21.6
21.7
Occurrence of Metals 931
Metallurgical Processes 932
Band Theory of Electrical Conductivity 939
Periodic Trends in Metallic Properties 941
The Alkali Metals 942
The Alkaline Earth Metals 946
Aluminum 948
CHEMISTRY in Action
Recycling Aluminum 950
Summary of Facts & Concepts 952
Key Words 952
Questions & Problems 952
CHAPTER 22
Nonmetallic
N
on
Elements
aand
nd Their Compounds 956
22.1 General Properties of Nonmetals 957
22.2 Hydrogen 958
CHEMISTRY in Action
Metallic Hydrogen 962
22.3 Carbon 963
CHEMISTRY in Action
Synthetic Gas from Coal 966
22.4 Nitrogen and Phosphorus 967
CHEMISTRY in Action
Ammonium Nitrate—The Explosive Fertilizer 974
22.5 Oxygen and Sulfur 975
22.6 The Halogens 982
Summary of Facts & Concepts 989
Key Words 989
Questions & Problems 990
Contents
CHAPTER 23
Transition
T
ran
Metals Chemistry and
Coordination
Compounds 994
C
oor
23.1
23.2
23.3
23.4
23.5
Properties of the Transition Metals 995
Chemistry of Iron and Copper 998
Coordination Compounds 1000
Structure of Coordination Compounds 1005
Bonding in Coordination Compounds:
Crystal Field Theory 1009
23.6 Reactions of Coordination Compounds 1015
CHEMISTRY in Action
Coordination Compounds in Living Systems 1016
23.7 Applications of Coordination Compounds 1016
CHEMISTRY in Action
Cisplatin—The Anticancer Drug 1018
Key Equation 1020
Summary of Facts & Concepts 1020
Key Words 1020
Questions & Problems 1021
CHAPTER 24
Organic
O
rga
Chemistry 1025
24.1 Classes of Organic Compounds 1026
24.2 Aliphatic Hydrocarbons 1026
CHEMISTRY in Action
Ice That Burns 1038
24.3 Aromatic Hydrocarbons 1039
24.4 Chemistry of the Functional Groups 1042
CHEMISTRY in Action
The Petroleum Industry 1048
Summary of Facts & Concepts 1050
Key Words 1051
Questions & Problems 1051
CHEMICAL M YS TERY
The Disappearing Fingerprints 1056
xvii
xviii
Contents
CHAPTER 25
Synthetic
S
ynt
and Natural Organic
Polymers
1058
P
oly
25.1 Properties of Polymers 1059
25.2 Synthetic Organic Polymers 1059
25.3 Proteins 1065
CHEMISTRY in Action
Sickle Cell Anemia—A Molecular Disease 1072
25.4 Nucleic Acids 1073
CHEMISTRY in Action
DNA Fingerprinting 1076
Summary of Facts & Concepts 1077
Key Words 1077
Questions & Problems 1077
CHEMICAL M YS TERY
A Story That Will Curl Your Hair 1082
Appendix 1
Derivation of the Names of Elements A-1
Appendix 2
Units for the Gas Constant A-7
Appendix 3
Thermodynamic Data at 1 atm and 25°C A-8
Appendix 4
Mathematical Operations A-13
Glossary G-1
Answers to Even-Numbered Problems AP-1
Credits C-1
Index I-1
List of Applications
The opening sentence of this text is, “Chemistry
is an active, evolving science that has vital
importance to our world, in both the realm of nature and
the realm of society.” Throughout the text, Chemistry
in Action boxes and Chemical Mysteries give specific
examples of chemistry as active and evolving in all facets
of our lives.
Chemistry in Action
The Search for the Higgs Boson 6
The Importance of Units 17
Distribution of Elements on Earth
and in Living Systems 49
Chemical Fertilizers 105
An Undesirable Precipitation Reaction 126
Breathalyzer 144
Metal from the Sea 156
Scuba Diving and the Gas Laws 200
Super Cold Atoms 208
Making Snow and Inflating a Bicycle Tire 240
White Fat Cells, Brown Fat Cells, and a Potential Cure
for Obesity 250
How a Bombardier Beetle Defends Itself 256
Laser—The Splendid Light 288
Electron Microscopy 292
Quantum Dots 312
The Third Liquid Element? 341
Discovery of the Noble Gases 358
Sodium Chloride—A Common and Important
Ionic Compound 376
Just Say NO 397
Microwave Ovens—Dipole Moments at Work 426
Buckyball, Anyone? 454
A Very Slow Pitch 475
Why Do Lakes Freeze from the Top Down? 478
High-Temperature Superconductors 488
And All for the Want of a Button 492
Hard-Boiling an Egg on a Mountaintop, Pressure
Cookers, and Ice Skating 505
Liquid Crystals 506
The Killer Lake 531
Dialysis 546
Radiocarbon Dating 586
Pharmacokinetics 606
Life at High Altitudes and Hemoglobin Production 651
The Haber Process 652
Antacids and the pH Balance in Your Stomach 706
Maintaining the pH of Blood 732
How an Eggshell Is Formed 760
The Efficiency of Heat Engines 790
The Thermodynamics of a Rubber Band 801
Bacteria Power 837
Dental Filling Discomfort 846
Nature’s Own Fission Reactor 882
Food Irradiation 890
Boron Neutron Capture Therapy 891
Recycling Aluminum 950
Metallic Hydrogen 962
Synthetic Gas from Coal 966
Ammonium Nitrate—The Explosive Fertilizer 974
Coordination Compounds in Living Systems 1016
Cisplatin—The Anticancer Drug 1018
Ice That Burns 1038
The Petroleum Industry 1048
Sickle Cell Anemia—A Molecular Disease 1072
DNA Fingerprinting 1076
Chemical Mystery
The Disappearance of the Dinosaurs 36
Who Killed Napoleon? 170
Out of Oxygen 228
The Exploding Tire 272
Discovery of Helium and the Rise
and Fall of Coronium 324
The Wrong Knife 560
Decaying Papers 718
A Hard-Boiled Snack 774
Tainted Water 860
The Art Forgery of the Twentieth Century 898
The Disappearing Fingerprints 1056
A Story That Will Curl Your Hair 1081
xix
List of Animations
The animations below are correlated to Chemistry.
Within the chapter are icons letting the student and
instructor know that an animation is available for a specific topic. Animations can be found online in the Chang
Connect site.
Chang Animations
Absorption of Color (23.5)
Acid-Base Titrations (16.4)
Acid Ionization (15.5)
Activation Energy (13.4)
Alpha, Beta, and Gamma Rays (2.2)
α-Particle Scattering (2.2)
Atomic and Ionic Radius (8.3)
Base Ionization (15.6)
Buffer Solutions (16.3)
Catalysis (13.6)
Cathode Ray Tube (2.2)
Chemical Equilibrium (14.1)
Chirality (23.4, 24.2)
Collecting a Gas over Water (5.6)
Diffusion of Gases (5.7)
Dissolution of an Ionic and a Covalent Compound (12.2)
Electron Configurations (7.8)
Equilibrium Vapor Pressure (11.8)
Galvanic Cells (18.2)
The Gas Laws (5.3)
Heat Flow (6.2)
Hybridization (10.4)
Hydration (4.1)
Ionic vs. Covalent Bonding (9.4)
Le Chátelier’s Principle (14.5)
Limiting Reagent (3.9)
Line Spectra (7.3)
Making a Solution (4.5)
Millikan Oil Drop (2.2)
Nuclear Fission (19.5)
xx
Neutralization Reactions (4.3)
Orientation of Collision (13.4)
Osmosis (12.6)
Oxidation-Reduction Reactions (4.4)
Packing Spheres (11.4)
Polarity of Molecules (10.2)
Precipitation Reactions (4.2)
Preparing a Solution by Dilution (4.5)
Radioactive Decay (19.3)
Resonance (9.8)
Sigma and Pi Bonds (10.5)
Strong Electrolytes, Weak Electrolytes,
and Nonelectrolytes (4.1)
VSEPR (10.1)
More McGraw-Hill Education
Animations
Aluminum Production (21.7)
Atomic Line Spectra (7.3)
Cubic Unit Cells and Their Origins (11.4)
Cu/Zn Voltaic Cell (18.2)
Current Generation from a Voltaic Cell (18.2)
Dissociation of Strong and Weak Acids (15.4)
Emission Spectra (7.3)
Formation of Ag2S by Oxidation-Reduction (4.4)
Formation of an Ionic Compound (2.7)
Formation of a Covalent Bond (9.4)
Influence of Shape on Polarity (10.2)
Ionic and Covalent Bonding (9.4)
Molecular Shape and Orbital Hybridization (10.4)
Operation of a Voltaic Cell (18.2)
Phase Diagrams and the States of Matter (11.9)
Properties of Buffers (16.3)
Reaction of Cu with AgNO3 (4.4)
Reaction of Magnesium and Oxygen (4.4, 9.2)
Rutherford’s Experiment (2.2)
VSEPR Theory (10.1)
Preface
T
he twelfth edition continues the tradition by providing a firm foundation in chemical concepts and
principles and to instill in students an appreciation
of the vital part chemistry plays in our daily life. It is the
responsibility of the textbook authors to assist both instructors and their students in their pursuit of this objective by presenting a broad range of topics in a logical
manner. We try to strike a balance between theory and
application and to illustrate basic principles with everyday examples whenever possible.
As in previous editions, our goal is to create a text
that is clear in explaining abstract concepts, concise so
that it does not overburden students with unnecessary extraneous information, yet comprehensive enough so that
it prepares students to move on to the next level of learning. The encouraging feedback we have received from
instructors and students has convinced us that this
approach is effective.
The art program has been extensively revised in this
edition. Many of the laboratory apparatuses and scientific
instruments were redrawn to enhance the realism of the
components. Several of the drawings were updated to reflect advances in the science and applications described
in the text; see, for example, the lithium-ion battery depicted in Figure 18.10. Molecular structures were created
using ChemDraw, the gold standard in chemical drawing
software. Not only do these structures introduce students
to the convention used to represent chemical structures in
three dimensions that they will see in further coursework,
they also provide better continuity with the ChemDraw
application they will use in Connect, the online homework and practice system for our text.
In addition to revising the art program, over 100 new
photographs are added in this edition. These photos provide a striking look at processes that can be understood
by studying the underlying chemistry (see, for example,
Figure 19.15, which shows the latest attempt of using
lasers to induce nuclear fusion).
Problem Solving
The development of problem-solving skills has always
been a major objective of this text. The two major categories of learning are shown next.
Worked examples follow a proven step-by-step
strategy and solution.
• Problem statement is the reporting of the facts
needed to solve the problem based on the question
posed.
• Strategy is a carefully thought-out plan or method to
serve as an important function of learning.
•
•
•
Solution is the process of solving a problem given in
a stepwise manner.
Check enables the student to compare and verify
with the source information to make sure the answer
is reasonable.
Practice Exercise provides the opportunity to solve
a similar problem in order to become proficient in
this problem type. The Practice Exercises are available in the Connect electronic homework system.
The margin note lists additional similar problems to
work in the end-of-chapter problem section.
End-of-Chapter Problems are organized in various
ways. Each section under a topic heading begins with
Review Questions followed by Problems. The Additional
Problems section provides more problems not organized
by section, followed by the new problem type of
Interpreting, Modeling & Estimating.
Many of the examples and end-of-chapter problems present extra tidbits of knowledge and enable the
student to solve a chemical problem that a chemist
would solve. The examples and problems show students the real world of chemistry and applications to
everyday life situations.
Visualization
Graphs and Flow Charts are important in science. In
Chemistry, flow charts show the thought process of a concept and graphs present data to comprehend the concept.
A significant number of Problems and Review of
Concepts, including many new to this edition, include
graphical data.
Molecular art appears in various formats to serve
different needs. Molecular models help to visualize the
three-dimensional arrangement of atoms in a molecule.
Electrostatic potential maps illustrate the electron density
distribution in molecules. Finally, there is the macroscopic to microscopic art helping students understand
processes at the molecular level.
Photos are used to help students become familiar
with chemicals and understand how chemical reactions
appear in reality.
Figures of apparatus enable the student to visualize
the practical arrangement in a chemistry laboratory.
Study Aids
Setting the Stage
Each chapter starts with the Chapter Outline and A Look
Ahead.
xxi
xxii
Preface
Chapter Outline enables the student to see at a glance
the big picture and focus on the main ideas of the
chapter.
A Look Ahead provides the student with an overview of
concepts that will be presented in the chapter.
Tools to Use for Studying
Useful aids for studying are plentiful in Chemistry and
should be used constantly to reinforce the comprehension
of chemical concepts.
Marginal Notes are used to provide hints and feedback
to enhance the knowledge base for the student.
Worked Examples along with the accompanying
Practice Exercises are very important tools for
learning and mastering chemistry. The problemsolving steps guide the student through the critical
thinking necessary for succeeding in chemistry.
Using sketches helps student understand the inner
workings of a problem. (See Example 6.1 on
page 238.) A margin note lists similar problems in
the end-of-chapter problems section, enabling the
student to apply new skill to other problems of
the same type. Answers to the Practice Exercises
are listed at the end of the chapter problems.
Review of Concepts enables the student to evaluate
if they understand the concept presented in the
section.
Key Equations are highlighted within the chapter,
drawing the student’s eye to material that needs to
be understood and retained. The key equations are
also presented in the chapter summary materials for
easy access in review and study.
Summary of Facts and Concepts provides a quick
review of concepts presented and discussed in detail
within the chapter.
Key Words are a list of all important terms to help the
student understand the language of chemistry.
Testing Your Knowledge
Review of Concepts lets students pause and check to
see if they understand the concept presented and
discussed in the section occurred. Answers to the
Review of Concepts can be found in the Student
Solution Manual and online in the accompanying
Connect Chemistry companion website.
End-of-Chapter Problems enable the student to
practice critical thinking and problem-solving skills.
The problems are broken into various types:
• By chapter section. Starting with Review Questions to test basic conceptual understanding, followed by Problems to test the student’s skill in
solving problems for that particular section of the
chapter.
• Additional Problems uses knowledge gained from
the various sections and/or previous chapters to
solve the problem.
• Interpreting, Modeling & Estimating problems
teach students the art of formulating models and
estimating ballpark answers based on appropriate
assumptions.
Real-Life Relevance
Interesting examples of how chemistry applies to life are
used throughout the text. Analogies are used where
appropriate to help foster understanding of abstract
chemical concepts.
End-of-Chapter Problems pose many relevant
questions for the student to solve. Examples
include Why do swimming coaches sometimes
place a drop of alcohol in a swimmer’s ear to
draw out water? How does one estimate the
pressure in a carbonated soft drink bottle before
removing the cap?
Chemistry in Action boxes appear in every chapter on
a variety of topics, each with its own story of how
chemistry can affect a part of life. The student can
learn about the science of scuba diving and nuclear
medicine, among many other interesting cases.
Chemical Mystery poses a mystery case to the
student. A series of chemical questions provide
clues as to how the mystery could possibly be
solved. Chemical Mystery will foster a high
level of critical thinking using the basic problemsolving steps built up throughout the text.
Digital Resources
McGraw-Hill Education offers various tools and technology products to support Chemistry, 12th edition.
chemistry
McGraw-Hill ConnectPlus Chemistry provides online presentation, assignment, and assessment solutions. It connects your students with the tools and resources they’ll
need to achieve success. With ConnectPlus Chemistry, you
can deliver assignments, quizzes, and tests online. A robust
set of questions, problems, and interactives are presented
and aligned with the textbook’s learning goals. The integration of ChemDraw by PerkinElmer, the industry standard
in chemical drawing software, allows students to create
accurate chemical structures in their online homework
assignments. As an instructor, you can edit existing questions and author entirely new problems. Track individual
student performance—by question, assignment, or in relation to the class overall—with detailed grade reports.
Integrate grade reports easily with Learning Management
Systems (LMS), such as WebCT and Blackboard—and
much more. ConnectPlus Chemistry offers 24/7 online
access to an eBook. This media-rich version of the book
allows seamless integration of text, media, and assessment.
To learn more visit connect.mheducation.com
SmartBook is the first and only adaptive reading experience designed to change the way students read and
learn. It creates a personalized reading experience by
highlighting the most impactful concepts a student
needs to learn at that moment in time.
As a student engages with SmartBook,
the reading experience continuously
adapts by highlighting content based
on what the student knows and
doesn’t know. This ensures that the
focus is on the content he or she
needs to learn, while simultaneously
promoting long-term retention of material. Use SmartBook’s real-time reports to quickly identify the concepts
that require more attention from individual students—or the entire class.
The end result? Students are more
engaged with course content, can
better prioritize their time, and come
to class ready to participate.
Many questions within Connect Chemistry will
allow students a chemical drawing experience
that can be assessed directly inside of their
homework.
iii
xxiii
xxiv
Digital Resources
McGraw-Hill LearnSmart is available as a standalone
product or as an integrated feature of McGraw-Hill
Connect® Chemistry. It is an adaptive learning system
designed to help students learn faster, study more efficiently, and retain more knowledge for greater success.
LearnSmart assesses a student’s knowledge of course
content through a series of adaptive questions. It pinpoints concepts the student does not understand and maps
out a personalized study plan for success. This innovative
study tool also has features that allow instructors to see
exactly what students have accomplished and a built-in
assessment tool for graded assignments. Visit the following site for a demonstration. www.mhlearnsmart.com
Adaptive Probes
A student’s knowledge is intelligently probed by asking a series of questions. These questions dynamically
change both in the level of difficulty and in content
based on the student’s weak and strong areas. Each
practice session is based on the previous performance,
and LearnSmart uses sophisticated models for predicting what the student will forget and how to reinforce
that material typically forgotten. This saves students
study time and ensures that they have actual mastery
of the concepts.
Immediate Feedback
When a student incorrectly answers a probe, the correct
answer is provided, along with feedback.
Time Out
When LearnSmart has identified a specific subject area
where the student is struggling, he or she is given a “time
out” and directed to the textbook section or learning
objective for remediation. With ConnectPlus, students
are provided with a link to the specific page of the
eBook where they can study the material immediately.
Reporting
Dynamically generated reports document student progress and areas for additional reinforcement, offering
at-a-glance views of their strengths and weaknesses.
Reports Include:
• Most challenging learning objectives
• Tree of wisdom
• Test results
•
•
Current learning status
Metacognitive skills
•
•
Missed questions
Learning plan
Digital Resources
LearnSmart Labs for General Chemistry™
xxv
problem solving skills. And with ALEKS 360, your
student also has access to this text’s eBook. Learn
more at www.aleks.com/highered/ science
THE Virtual Lab Experience.
LearnSmart Labs is a must-see, outcomes-based lab
simulation. It assesses a student’s knowledge and adaptively corrects deficiencies, allowing the student to learn
faster and retain more knowledge with greater success.
First, a student’s knowledge is adaptively leveled on
core learning outcomes: Questioning reveals knowledge
deficiencies that are corrected by the delivery of content
that is conditional on a student’s response. Then, a simulated lab experience requires the student to think and act
like a scientist: Recording, interpreting, and analyzing
data using simulated equipment found in labs and clinics.
The student is allowed to make mistakes—a powerful
part of the learning experience! A virtual coach provides
subtle hints when needed; asks questions about the student’s choices; and allows the student to reflect upon and
correct those mistakes. Whether your need is to overcome
the logistical challenges of a traditional lab, provide better lab prep, improve student performance, or make your
online experience one that rivals the real world,
LearnSmart Labs accomplishes it all. Learn more at
www.mhlearnsmart.com
LearnSmart Prep is an adaptive tool that prepares students for the course they are about to take. It identifies the
prerequisite knowledge each student doesn’t know or
fully understand and provides learning resources to teach
essential concepts so he or she enters the classroom prepared to succeed.
ALEKS (Assessment and LEarning in Knowledge
Spaces) is a web-based system for individualized assessment and learning available 24/7 over the Internet.
ALEKS uses artificial intelligence to accurately determine a student’s knowledge and then guides her to the
material that she is most ready to learn. ALEKS offers
immediate feedback and access to ALEKSPedia—an
interactive text that contains concise entries on chemistry topics. ALEKS is also a full-featured course management system with rich reporting features that allow
instructors to monitor individual and class performance,
set student goals, assign/grade online quizzes, and
more. ALEKS allows instructors to spend more time on
concepts while ALEKS teaches students practical
McGraw-Hill Create™ is a self-service website that
allows you to create customized course materials using McGraw-Hill Education’s comprehensive, crossdisciplinary content and digital products. You can even
access third party content such as readings, articles,
cases, videos, and more. Arrange the content you’ve
selected to match the scope and sequence of your
course. Personalize your book with a cover design and
choose the best format for your students–eBook, color
print, or black-and-white print. And, when you are
done, you’ll receive a PDF review copy in just minutes!
www.mcgrawhillcreate.com
®
Tegrity Campus is a fully automated lecture capture solution used in traditional, hybrid, “flipped classes” and online courses to record lesson, lectures, and skills. Its
personalized learning features make study time incredibly efficient and its ability to affordably scale brings this
benefit to every student on campus. Patented search technology and real-time LMS integrations make Tegrity the
market-leading solution and service. Tegrity is available
as an integrated feature of McGraw-Hill Connect®
Chemistry and as a standalone.
Presentation Tools
Build instructional materials wherever, whenever, and
however you want! Access instructor tools from your text’s
Connect website to find photo’s, artwork, animations, and
other media that can be used to create customized lectures,
visually enhanced tests and quizzes, compelling course
websites, or attractive printed support materials. All assets
are copyrighted by McGraw-Hill Higher Education, but
can be used by instructors for classroom purposes. The
visual resources in this collection include:
• Art Full-color digital files of all illustrations in the
book.
• Photos The photo collection contains digital files of
photographs from the text.
• Tables Every table that appears in the text is available electronically.
• Animations Numerous full-color animations illustrating important processes are also provided.
xxvi
•
•
Digital Resources
PowerPoint Lecture Outlines Ready made presentations for each chapter of the text.
PowerPoint Slides All illustrations, photos, and
tables are pre-inserted by chapter into blank
PowerPoint slides.
and a summary of the corresponding text. Following
the summary are sample problems with detailed solutions. Each chapter has true–false questions and a
self-test, with all answers provided at the end of the
chapter.
Computerized Test Bank Online
Animations for MP3/iPod
A comprehensive bank of questions is provided within a
computerized test bank, enabling professors to prepare
and access tests or quizzes. Instructors can create or edit
questions, or drag-and-drop questions to prepare tests
quickly and easily. Tests can be published to their online
course, or printed for paper-based assignments.
A number of animations are available for download to
your MP3/iPod through the textbook’s Connect website.
Instructor’s Solution’s Manual
The Instructor’s Solution Manual, written by Raymond
Chang and Ken Goldsby, provides the solutions to most
end-of-chapter problems. The manual also provides the
difficulty level and category type for each problem. This
manual is available to instructors online in the text’s
Connect library tab.
Instructor’s Manual
The Instructor’s Manual provides a brief summary of the
contents of each chapter, along with the learning goals,
references to background concepts in earlier chapters,
and teaching tips. This manual can be found online for
instructors on the text’s Connect library tab.
For the Student
Students can order supplemental study materials by contacting their campus bookstore, calling 1-800-262-4729,
or online at http://shop.mheducation.com
Student Solutions Manual
ISBN 1-25-928622-3
The Student Solutions Manual is written by Raymond
Chang and Ken Goldsby. This supplement contains detailed solutions and explanations for even-numbered
problems in the main text. The manual also includes a
detailed discussion of different types of problems and approaches to solving chemical problems and tutorial solutions for many of the end-of-chapter problems in the text,
along with strategies for solving them. Note that solutions
to the problems listed under Interpreting, Modeling &
Estimating are not provided in the manual.
Student Study Guide ISBN 1-25-928623-1
This valuable ancillary contains material to help the
student practice problem-solving skills. For each section of a chapter, the author provides study objectives
Acknowledgments
We would like to thank the following reviewers and symposium participants, whose comments were of great help
to us in preparing this revision:
Kathryn S. Asala, University of North Carolina,
Charlotte
Mohd Asim Ansari, Fullerton College
Keith Baessler, Suffolk County Community College
Christian S. Bahn, Montana State University
Mary Fran Barber, Wayne State University
H. Laine Berghout, Weber State University
Feri Billiot, Texas A&M University Corpus Christi
John Blaha, Columbus State Community College
Marco Bonizzoni, University of Alabama–Tuscaloosa
Christopher Bowers, Ohio Northern University
Bryan Breyfogle, Missouri State University
Steve Burns, St. Thomas Aquinas College
Mark L. Campbell, United States Naval Academy
Tara Carpenter, University of Maryland
David Carter, Angelo State
Daesung Chong, Ball State University
Elzbieta Cook, Louisiana State University
Robert L. Cook, Louisiana State University
Colleen Craig, University of Washington
Brandon Cruickshank, Northern Arizona
University–Flagstaff
Elizabeth A. Clizbe, SUNY Buffalo
Mohammed Daoudi, University of Central Florida
Jay Deiner, New York City College of Technology
Dawn Del Carlo, University of Northern Iowa
Milagros Delgado, Florida International University,
Biscayne Bay Campus
Michael Denniston, Georgia Perimeter College
Stephanie R. Dillon, Florida State University
Anne Distler, Cuyahoga Community College
Bill Donovan, University of Akron
Mathilda D. Doorley, Southwest Tennessee Community
College
Digital Resources
Jack Eichler, University of California–Riverside
Bradley D. Fahlman, Central Michigan University
Lee Friedman, University of Maryland–College Park
Tiffany Gierasch, University of Maryland–
Baltimore County
Cameon Geyer, Olympic College
John Gorden, Auburn University
Tracy Hamilton, University of Alabama–Birmingham
Tony Hascall, Northern Arizona University
Lindsay M. Hinkle, Harvard University
Rebecca Hoenigman, Community College of Aurora
T. Keith Hollis, Mississippi State University
Byron Howell, Tyler Junior College
Michael R. Ivanov, Northeast Iowa Community College
David W. Johnson, University of Dayton
Steve Johnson, University of New England
Mohammad Karim, Tennessee State University–
Nashville
Jeremy Karr, College of Saint Mary
Vance Kennedy, Eastern Michigan University
Katrina Kline, University of Missouri
An-Phong Lee, Florida Southern College
Debbie Leedy, Glendale Community College
Willem R. Leenstra, University of Vermont
Barbara S. Lewis, Clemson University
Scott Luaders, Quincy University
Vicky Lykourinou, University of South Florida
Yinfa Ma, Missouri University of Science and
Technology
Sara-Kaye Madsen, South Dakota State University
Sharyl Majorski, Central Michigan University
Roy McClean, United States Naval Academy
Helene Maire-Afeli, University of South Carolina–Union
Tracy McGill, Emory University
David M. McGinnis, University of Arkansas–Fort Smith
Thomas McGrath, Baylor University
Deb Mlsna, Mississippi State University
Patricia Muisener, University of South Florida
Kim Myung, Georgia Perimeter College
Anne-Marie Nickel, Milwaukee School of Engineering
Krista Noren-Santmyer, Hillsborough Community
College–Brandon
Greg Oswald, North Dakota State University
John W. Overcash, University of Illinois–
Urbana–Champaign
Shadrick Paris, Ohio University
Manoj Patil, Western Iowa Technical College
John Pollard, University of Arizona
xxvii
Ramin Radfar, Wofford College
Betsy B. Ratcliff, West Virginia University
Mike Rennekamp, Columbus State Community College
Thomas G. Richmond, University of Utah
Steven Rowley, Middlesex County College
Joel W. Russell, Oakland University
Raymond Sadeghi, University of Texas–
San Antonio
Richard Schwenz, University of Northern Colorado
Allison S. Soult, University of Kentucky
Anne M. Spuches, East Carolina University
John Stubbs, University of New England
Katherine Stumpo, University of Tennessee–Martin
Jerry Suits, University of Northern Colorado
Charles Taylor, Florence Darlington
Technical College
Mark Thomson, Ferris State University
Eric M. Todd, University of Wisconsin–Stevens Point
Yijun Tang, University of Wisconsin–Oshkosh
Steve Theberge, Merrimack College
Lori Van Der Sluys, Penn State University
Lindsay B. Wheeler, University of Virginia
Gary D. White, Middle Tennessee State University
Stan Whittingham, Binghamton University
Troy Wolfskill, Stony Brook University
Anthony Wren, Butte College
Fadi Zaher, Gateway Technical College
Connect: Chemistry has been greatly enhanced by the
efforts of Yasmin Patell, Kansas State University;
MaryKay Orgill, University of Nevada–Las Vegas;
Mirela Krichten, The College of New Jersey; who did a
masterful job of authoring hints and feedback to augment
all of the system’s homework problems.
The following individuals helped write and review
learning goal–oriented content for LearnSmart for
General Chemistry:
Margaret Ruth Leslie, Kent State University
David G. Jones, Vistamar School
Erin Whitteck
Margaret Asirvatham, University of
Colorado–Boulder
Alexander J. Seed, Kent State University
Benjamin Martin, Texas State University–San Marcos
Claire Cohen, University of Toledo
Manoj Patil, Western Iowa Tech Community College
Adam I. Keller, Columbus State Community College
Peter de Lijser, California State University–Fullerton
Lisa Smith, North Hennepin Community College
xxviii
Digital Resources
We have benefited much from discussions with our
colleagues at Williams College and Florida State, and
correspondence with instructors here and abroad.
It is a pleasure to acknowledge the support given us
by the following members of McGraw-Hill Education’s
College Division: Tammy Ben, Thomas Timp, Marty Lange,
and Kurt Strand. In particular, we would like to mention
Sandy Wille for supervising the production, David Hash
for the book design, and John Leland, the content licensing
specialist, and Tami Hodge, the marketing manager, for
their suggestions and encouragement. We also thank our
sponsoring editor, David Spurgeon, for his advice and assistance. Our special thanks go to Jodi Rhomberg, the
product developer, for her supervision of the project at
every stage of the writing of this edition.
Finally, we would like to acknowledge Toni Michaels,
the photo researcher, for her resourcefulness in acquiring
the new images under a very tight schedule.
Setting the Stage for Learning
Real-Life Relevance
Interesting examples of how chemistry applies to life are used throughout the text. Analogies
are used where appropriate to help foster understanding of abstract chemical concepts.
cha21510_ch04_118-171.indd Page 144 31/05/14 1:11 PM f-w-196
/203/MH02197/cha21510_disk1of1/0078021510/cha21510_pagefiles
Chemistry in Action boxes appear in
every chapter on a variety of topics, each
with its own story of how chemistry can
affect a part of life. The student can learn
about the science of scuba diving and
nuclear
medicine, among many other
/203/MH02197/cha21510_disk1of1/0078021510/cha21510_pagefiles
interesting cases.
CHEMISTRY in Action
Breathalyzer
E
in the United
cha21510_ch03_075-117.indd Pagevery
99 year
7/19/14
11:02States
AMabout
f-49625,000 people are killed
and 500,000 more are injured as a result of drunk driving. In
spite of efforts to educate the public about the dangers of driving
while intoxicated and stiffer penalties for drunk driving offenses,
law enforcement agencies still have to devote a great deal of
work to removing drunk drivers from America’s roads.
The police often use a device called a breathalyzer to
test drivers suspected of being drunk. The chemical basis of
this device is a redox reaction. A sample of the driver’s
breath is drawn into the breathalyzer, where it is treated with
an acidic solution of potassium dichromate. The alcohol
(ethanol) in the breath is converted to acetic acid as shown in
the following equation:
3CH3CH2OH
ethanol
1
2K2Cr2O7
1
8H2SO4 ¡
potassium
dichromate
(orange yellow)
sulfuric
acid
3CH3COOH 1 2Cr2(SO4)3 1 2K2SO4 1 11H2O
acetic acid
chromium(III)
sulfate (green)
potassium
sulfate
In this reaction, the ethanol is oxidized to acetic acid and the
chromium(VI) in the orange-yellow dichromate ion is reduced
A driver being tested for blood alcohol content with a handheld breathalyzer.
to the green chromium(III) ion (see Figure 4.22). The driver’s
blood alcohol level can be determined readily by measuring the
degree of this color change (read from a calibrated meter on the
instrument). The current legal limit of blood alcohol content is
0.08 percent by mass. Anything higher constitutes intoxication.
Breath
Meter
Filter
Light
source
Chemical Mystery poses a mystery case
to the student. A series of chemical questions provide clues as to how the mystery
could possibly be solved. Chemical
Mystery will foster a high level of critical
thinking using the basic problem-solving
steps built up throughout the text.
Photocell
detector
Schematic diagram of a breathalyzer.
The alcohol in the driver’s breath is
reacted with a potassium dichromate
solution. The change in the absorption
of light due to the formation of
chromium(III) sulfate is registered by
the detector and shown on a meter,
which directly displays the alcohol
content in blood. The filter selects only
one wavelength of light for
measurement.
K2Cr2O7
solution
510_ch05_172-229.indd Page 211 31/05/14 1:15 PM f-w-196
/203/MH02197/cha21510_disk1of1/0078021510/cha21510_pagefiles
Before reaction has started
Visualization
CH4
H2
2.0
Molecular art
NH3
PV
1.0
RT
Ideal gas
0
200
400
Graphs and
Flow Charts
600 800 1000 1200
P (atm)
After reaction is complete
H2
CO
CH3OH
xxix
xxx
Setting the Stage for Learning
Key Equations
cha21510_ch02_038-074.indd Page 67 27/05/14 10:37 AM f-w-196
DU 5 q 1 w (6.1)
w 5 2PDV
cha21510_ch02_038-074.indd
Page 67(6.3)
27/05/14 10:37 AM f-w-196
H 5 U 1 PV (6.6)
DH 5 DU 1 PDV (6.8)
C 5 ms (6.11)
q 5 msDt (6.12)
q 5 CDt (6.13)
¢H°rxn 5 on¢H°f (products) 2 om¢H°f (reactants) (6.18)
DHsoln 5 U 1 DHhydr (6.20)
/203/MH02197/cha21510_disk1of1/0078021510/cha21510_pagefiles
Mathematical statement of the first law of thermodynamics.
Calculating work
done in gas expansion or gas compression.
/203/MH02197/cha21510_disk1of1/0078021510/cha21510_pagefiles
Definition of enthalpy.
Calculating enthalpy (or energy) change for a
constant-pressure process.
Definition of heat capacity.
Calculating heat change in terms of specific heat.
Calculating heat change in terms of heat capacity.
Calculating standard enthalpy of reaction.
Lattice energy and hydration contributions to heat of solution.
Summary of Facts & Concepts
Study Aids
Key Equations—material to
retain
Summary of Facts &
Concepts—quick review of
important concepts
Key Words—important terms
to understand
1. Modern chemistry began with Dalton’s atomic theory,
which states that all matter is composed of tiny, indivisible particles called atoms; that all atoms of the
same element are identical; that compounds contain
atoms of different elements combined in wholenumber ratios; and that atoms are neither created nor
destroyed in chemical reactions (the law of conservation of mass).
2. Atoms of constituent elements in a particular compound
are always combined in the same proportions by mass
(law of definite proportions). When two elements can
combine to form more than one type of compound, the
masses of one element that combine with a fixed mass
of the other element are in a ratio of small whole numbers (law of multiple proportions).
3. An atom consists of a very dense central nucleus
containing protons and neutrons, with electrons
moving about the nucleus at a relatively large distance from it.
4. Protons are positively charged, neutrons have no charge,
and electrons are negatively charged. Protons and neutrons have roughly the same mass, which is about 1840
times greater than the mass of an electron.
5. The atomic number of an element is the number of
protons in the nucleus of an atom of the element; it
6.
7.
8.
9.
10.
11.
determines the identity of an element. The mass
number is the sum of the number of protons and the
number of neutrons in the nucleus.
Isotopes are atoms of the same element with the same
number of protons but different numbers of neutrons.
Chemical formulas combine the symbols for the constituent elements with whole-number subscripts to
show the type and number of atoms contained in the
smallest unit of a compound.
The molecular formula conveys the specific number
and type of atoms combined in each molecule of a compound. The empirical formula shows the simplest ratios
of the atoms combined in a molecule.
Chemical compounds are either molecular compounds
(in which the smallest units are discrete, individual molecules) or ionic compounds, which are made of cations
and anions.
The names of many inorganic compounds can be
deduced from a set of simple rules. The formulas can be
written from the names of the compounds.
Organic compounds contain carbon and elements like
hydrogen, oxygen, and nitrogen. Hydrocarbon is the
simplest type of organic compound.
Key Words
Acid, p. 62
Alkali metals, p. 50
Alkaline earth metals, p. 50
Allotrope, p. 52
Alpha (α) particles, p. 43
Alpha (α) rays, p. 43
Anion, p. 51
Atom, p. 40
Atomic number (Z), p. 46
Base, p. 64
Beta (β) particles, p. 43
Beta (β) rays, p. 43
Binary compound, p. 56
Cation, p. 51
Chemical formula, p. 52
Diatomic molecule, p. 50
Electron, p. 41
Empirical formula, p. 53
Families, p. 48
Gamma (γ) rays, p. 43
Groups, p. 48
Halogens, p. 50
Hydrate, p. 64
Inorganic
compounds, p. 56
Ion, p. 50
Ionic compound, p. 51
Isotope, p. 46
Law of conservation of
mass, p. 40
Law of definite
proportions, p. 40
Law of multiple
proportions, p. 40
Mass number (A), p. 46
Metal, p. 48
Metalloid, p. 48
Molecular formula, p. 52
Molecule, p. 50
Monatomic ion, p. 51
Neutron, p. 45
Noble gases, p. 50
Nonmetal, p. 48
Nucleus, p. 44
Organic compound, p. 56
Oxoacid, p. 62
Oxoanion, p. 63
Periods, p. 48
Periodic table, p. 48
Polyatomic ion, p. 51
Polyatomic molecule, p. 50
Proton, p. 44
Radiation, p. 41
Radioactivity, p. 43
Structural formula, p. 53
Ternary compound, p. 57
xxxi
Setting the Stage for Learning
Chang Learning System
Review the section content by using this quick test for
acquired knowledge.
Review of Concepts
The diagrams here show three compounds AB2 (a), AC2 (b), and AD2 (c) dissolved
in water. Which is the strongest electrolyte and which is the weakest? (For
simplicity, water molecules are not shown.)
(a)
(b)
(c)
Example 4.6
Classify the following redox reactions and indicate changes in the oxidation numbers of
the elements:
(a) 2N2O(g) ¡ 2N2(g) 1 O2(g)
(b) 6Li(s) 1 N2(g) ¡ 2Li3N(s)
(c) Ni(s) 1 Pb(NO3)2(aq) ¡ Pb(s) 1 Ni(NO3)2(aq)
(d) 2NO2(g) 1 H2O(l) ¡ HNO2(aq) 1 HNO3(aq)
Strategy Review the definitions of combination reactions, decomposition reactions,
displacement reactions, and disproportionation reactions.
Solution
(a) This is a decomposition reaction because one reactant is converted to two different
products. The oxidation number of N changes from 11 to 0, while that of O
changes from 22 to 0.
(b) This is a combination reaction (two reactants form a single product). The oxidation
number of Li changes from 0 to 11 while that of N changes from 0 to 23.
(c) This is a metal displacement reaction. The Ni metal replaces (reduces) the Pb21 ion. The
oxidation number of Ni increases from 0 to 12 while that of Pb decreases from 12 to 0.
(d) The oxidation number of N is 14 in NO2 and it is 13 in HNO2 and 15 in HNO3.
Because the oxidation number of the same element both increases and decreases,
this is a disproportionation reaction.
Learn a problem-solving process of
strategizing, solving, and checking
your way to a solution.
Practice Exercise Identify the following redox reactions by type:
(a)
(b)
(c)
(d)
Fe 1 H2SO4 ¡ FeSO4 1 H2
S 1 3F2 ¡ SF6
2CuCl ¡ Cu 1 CuCl2
2Ag 1 PtCl2 ¡ 2AgCl 1 Pt
Use the problem-solving approach on real-world problems. Interpreting,
Modeling & Estimating problems provide students the opportunity to solve
problems like a chemist.
4.172 Potassium superoxide (KO2), a useful source of
oxygen employed in breathing equipment, reacts
with water to form potassium hydroxide, hydrogen peroxide, and oxygen. Furthermore, potassium superoxide also reacts with carbon dioxide to
form potassium carbonate and oxygen. (a) Write
equations for these two reactions and comment on
the effectiveness of potassium superoxide in this
application. (b) Focusing only on the reaction between KO2 and CO2, estimate the amount of KO2
needed to sustain a worker in a polluted environment for 30 min. See Problem 1.69 for useful
information.
•
A Note to the Student
G
eneral chemistry is commonly perceived to be
more difficult than most other subjects. There
is some justification for this perception. For
one thing, chemistry has a very specialized vocabulary.
At first, studying chemistry is like learning a new language. Furthermore, some of the concepts are abstract.
Nevertheless, with diligence you can complete this
course successfully, and you might even enjoy it. Here
are some suggestions to help you form good study
habits and master the material in this text.
• Attend classes regularly and take careful notes.
• If possible, always review the topics discussed in
class the same day they are covered in class. Use this
book to supplement your notes.
• Think critically. Ask yourself if you really understand the meaning of a term or the use of an equation.
A good way to test your understanding is to explain
a concept to a classmate or some other person.
• Do not hesitate to ask your instructor or your teaching assistant for help.
The twelfth edition tools for Chemistry are designed to
enable you to do well in your general chemistry course.
The following guide explains how to take full advantage
of the text, technology, and other tools.
• Before delving into the chapter, read the chapter outline and the chapter introduction to get a sense of the
important topics. Use the outline to organize your
note taking in class.
• At the end of each chapter you will find a summary of
facts and concepts, the key equations, and a list of key
words, all of which will help you review for exams.
xxxii
•
•
•
•
Definitions of the key words can be studied in context on the pages cited in the end-of-chapter list or in
the glossary at the back of the book.
Careful study of the worked-out examples in the
body of each chapter will improve your ability to
analyze problems and correctly carry out the calculations needed to solve them. Also take the time to
work through the practice exercise that follows each
example to be sure you understand how to solve the
type of problem illustrated in the example. The
answers to the practice exercises appear at the end of
the chapter, following the questions and problems.
For additional practice, you can turn to similar problems referred to in the margin next to the example.
The questions and problems at the end of the chapter
are organized by section.
The back inside cover shows a list of important
figures and tables with page references. This index
makes it convenient to quickly look up information
when you are solving problems or studying related
subjects in different chapters.
If you follow these suggestions and stay up-to-date
with your assignments, you should find that chemistry is
challenging, but less difficult and much more interesting
than you expected.
—Raymond Chang and Ken Goldsby
CHAPTER
1
Chemistry
The Study of Change
By applying electric fields to push DNA molecules
through pores created in graphene, scientists have
developed a technique that someday can be used for
fast sequencing the four chemical bases according to
their unique electrical properties.
CHAPTER OUTLINE
A LOOK AHEAD
1.1
Chemistry: A Science for
the Twenty-First Century
We begin with a brief introduction to the study of chemistry and describe its
role in our modern society. (1.1 and 1.2)
1.2
1.3
1.4
1.5
1.6
The Study of Chemistry
Next, we become familiar with the scientific method, which is a systematic
approach to research in all scientific disciplines. (1.3)
We define matter and note that a pure substance can either be an element or
a compound. We distinguish between a homogeneous mixture and a heterogeneous mixture. We also learn that, in principle, all matter can exist in one
of three states: solid, liquid, and gas. (1.4 and 1.5)
1.7
1.8
1.9
Measurement
To characterize a substance, we need to know its physical properties, which
can be observed without changing its identity and chemical properties,
which can be demonstrated only by chemical changes. (1.6)
Being an experimental science, chemistry involves measurements. We learn
the basic SI units and use the SI-derived units for quantities like volume and
density. We also become familiar with the three temperature scales: Celsius,
Fahrenheit, and Kelvin. (1.7)
1.10
Real-World Problem Solving:
Information, Assumptions,
and Simplifications
Chemical calculations often involve very large or very small numbers and a
convenient way to deal with these numbers is the scientific notation. In calculations or measurements, every quantity must show the proper number of
significant figures, which are the meaningful digits. (1.8)
We learn that dimensional analysis is useful in chemical calculations. By
carrying the units through the entire sequence of calculations, all the units
will cancel except the desired one. (1.9)
Solving real-world problems frequently involves making assumptions and
simplifications. (1.10)
The Scientific Method
Classifications of Matter
The Three States of Matter
Physical and Chemical
Properties of Matter
Handling Numbers
Dimensional Analysis
in Solving Problems
1
2
Chapter 1 ■ Chemistry: The Study of Change
C
hemistry is an active, evolving science that has vital importance to our world, in both the
realm of nature and the realm of society. Its roots are ancient, but as we will see, chemistry is every bit a modern science.
We will begin our study of chemistry at the macroscopic level, where we can see and
measure the materials of which our world is made. In this chapter, we will discuss the scientific
method, which provides the framework for research not only in chemistry but in all other
sciences as well. Next we will discover how scientists define and characterize matter. Then we
will spend some time learning how to handle numerical results of chemical measurements and
solve numerical problems. In Chapter 2, we will begin to explore the microscopic world of
atoms and molecules.
1.1 Chemistry: A Science for the Twenty-First Century
The Chinese characters for chemistry mean “The study of change.”
Chemistry is the study of matter and the changes it undergoes. Chemistry is often
called the central science, because a basic knowledge of chemistry is essential for
students of biology, physics, geology, ecology, and many other subjects. Indeed, it is
central to our way of life; without it, we would be living shorter lives in what we
would consider primitive conditions, without automobiles, electricity, computers, CDs,
and many other everyday conveniences.
Although chemistry is an ancient science, its modern foundation was laid in the
nineteenth century, when intellectual and technological advances enabled scientists to
break down substances into ever smaller components and consequently to explain
many of their physical and chemical characteristics. The rapid development of increasingly sophisticated technology throughout the twentieth century has given us even
greater means to study things that cannot be seen with the naked eye. Using computers and special microscopes, for example, chemists can analyze the structure of atoms
and molecules—the fundamental units on which the study of chemistry is based—and
design new substances with specific properties, such as drugs and environmentally
friendly consumer products.
It is fitting to ask what part the central science will have in the twenty-first
century. Almost certainly, chemistry will continue to play a pivotal role in all areas
of science and technology. Before plunging into the study of matter and its transformation, let us consider some of the frontiers that chemists are currently exploring
(Figure 1.1). Whatever your reasons for taking general chemistry, a good knowledge
of the subject will better enable you to appreciate its impact on society and on you
as an individual.
1.2 The Study of Chemistry
Compared with other subjects, chemistry is commonly believed to be more difficult,
at least at the introductory level. There is some justification for this perception; for
one thing, chemistry has a very specialized vocabulary. However, even if this is your
first course in chemistry, you already have more familiarity with the subject than you
may realize. In everyday conversations we hear words that have a chemical connection, although they may not be used in the scientifically correct sense. Examples are
“electronic,” “quantum leap,” “equilibrium,” “catalyst,” “chain reaction,” and “critical
mass.” Moreover, if you cook, then you are a practicing chemist! From experience
gained in the kitchen, you know that oil and water do not mix and that boiling water
left on the stove will evaporate. You apply chemical and physical principles when you
use baking soda to leaven bread, choose a pressure cooker to shorten the time it takes
to prepare soup, add meat tenderizer to a pot roast, squeeze lemon juice over sliced
1.2 The Study of Chemistry
(a)
(c)
(b)
(d)
Figure 1.1 (a) The output from an automated DNA sequencing machine. Each lane displays the
sequence (indicated by different colors) obtained with a separate DNA sample. (b) A graphene
supercapacitor. These materials provide some of the highest known energy-to-volume ratios and
response times. (c) Production of photovoltaic cells, used to convert light into electrical current. (d) The
leaf on the left was taken from a tobacco plant that was not genetically engineered but was exposed to
tobacco horn worms. The leaf on the right was genetically engineered and is barely attacked by the
worms. The same technique can be applied to protect the leaves of other types of plants.
pears to prevent them from turning brown or over fish to minimize its odor, and add
vinegar to the water in which you are going to poach eggs. Every day we observe
such changes without thinking about their chemical nature. The purpose of this
course is to make you think like a chemist, to look at the macroscopic world—the
things we can see, touch, and measure directly—and visualize the particles and events
of the microscopic world that we cannot experience without modern technology and
our imaginations.
At first some students find it confusing that their chemistry instructor and textbook
seem to be continually shifting back and forth between the macroscopic and microscopic worlds. Just keep in mind that the data for chemical investigations most often
come from observations of large-scale phenomena, but the explanations frequently lie
in the unseen and partially imagined microscopic world of atoms and molecules. In
other words, chemists often see one thing (in the macroscopic world) and think
another (in the microscopic world). Looking at the rusted nails in Figure 1.2, for
example, a chemist might think about the basic properties of individual atoms of
iron and how these units interact with other atoms and molecules to produce the
observed change.
3
4
Chapter 1 ■ Chemistry: The Study of Change
O2
88n
Fe2O3
Fe
Figure 1.2 A simplified molecular view of rust (Fe2O3) formation from iron (Fe) atoms and oxygen molecules (O2). In reality, the process
requires water and rust also contains water molecules.
1.3 The Scientific Method
All sciences, including the social sciences, employ variations of what is called the
scientific method, a systematic approach to research. For example, a psychologist
who wants to know how noise affects people’s ability to learn chemistry and a chemist interested in measuring the heat given off when hydrogen gas burns in air would
follow roughly the same procedure in carrying out their investigations. The first step
is to carefully define the problem. The next step includes performing experiments,
making careful observations, and recording information, or data, about the system—
the part of the universe that is under investigation. (In the examples just discussed,
the systems are the group of people the psychologist will study and a mixture of
hydrogen and air.)
The data obtained in a research study may be both qualitative, consisting of
general observations about the system, and quantitative, comprising numbers obtained
by various measurements of the system. Chemists generally use standardized symbols
and equations in recording their measurements and observations. This form of representation not only simplifies the process of keeping records, but also provides a common basis for communication with other chemists.
When the experiments have been completed and the data have been recorded, the
next step in the scientific method is interpretation, meaning that the scientist attempts
to explain the observed phenomenon. Based on the data that were gathered, the
researcher formulates a hypothesis, a tentative explanation for a set of observations.
Further experiments are devised to test the validity of the hypothesis in as many ways
as possible, and the process begins anew. Figure 1.3 summarizes the main steps of
the research process.
After a large amount of data has been collected, it is often desirable to summarize the information in a concise way, as a law. In science, a law is a concise
verbal or mathematical statement of a relationship between phenomena that is
always the same under the same conditions. For example, Sir Isaac Newton’s second law of motion, which you may remember from high school science, says that
force equals mass times acceleration (F 5 ma). What this law means is that an
1.3 The Scientific Method
Observation
Representation
Interpretation
Figure 1.3 The three levels of studying chemistry and their relationships. Observation deals
with events in the macroscopic world; atoms and molecules constitute the microscopic world.
Representation is a scientific shorthand for describing an experiment in symbols and chemical
equations. Chemists use their knowledge of atoms and molecules to explain an observed
phenomenon.
increase in the mass or in the acceleration of an object will always increase its
force proportionally, and a decrease in mass or acceleration will always decrease
the force.
Hypotheses that survive many experimental tests of their validity may evolve into
theories. A theory is a unifying principle that explains a body of facts and/or those
laws that are based on them. Theories, too, are constantly being tested. If a theory is
disproved by experiment, then it must be discarded or modified so that it becomes
consistent with experimental observations. Proving or disproving a theory can take
years, even centuries, in part because the necessary technology may not be available.
Atomic theory, which we will study in Chapter 2, is a case in point. It took more than
2000 years to work out this fundamental principle of chemistry proposed by Democritus, an ancient Greek philosopher. A more contemporary example is the search
for the Higgs boson discussed on page 6.
Scientific progress is seldom, if ever, made in a rigid, step-by-step fashion. Sometimes a law precedes a theory; sometimes it is the other way around. Two scientists
may start working on a project with exactly the same objective, but will end up taking drastically different approaches. Scientists are, after all, human beings, and their
modes of thinking and working are very much influenced by their background, training, and personalities.
The development of science has been irregular and sometimes even illogical.
Great discoveries are usually the result of the cumulative contributions and experience of many workers, even though the credit for formulating a theory or a law
is usually given to only one individual. There is, of course, an element of luck
involved in scientific discoveries, but it has been said that “chance favors the
prepared mind.” It takes an alert and well-trained person to recognize the significance of an accidental discovery and to take full advantage of it. More often than
not, the public learns only of spectacular scientific breakthroughs. For every success story, however, there are hundreds of cases in which scientists have spent
years working on projects that ultimately led to a dead end, and in which positive
achievements came only after many wrong turns and at such a slow pace that they
went unheralded. Yet even the dead ends contribute something to the continually
growing body of knowledge about the physical universe. It is the love of the
search that keeps many scientists in the laboratory.
Review of Concepts
Which of the following statements is true?
(a) A hypothesis always leads to the formulation of a law.
(b) The scientific method is a rigid sequence of steps in solving problems.
(c) A law summarizes a series of experimental observations; a theory provides
an explanation for the observations.
5
CHEMISTRY in Action
The Search for the Higgs Boson
I
n this chapter, we identify mass as a fundamental property of
matter, but have you ever wondered: Why does matter even
have mass? It might seem obvious that “everything” has mass,
but is that a requirement of nature? We will see later in our studies that light is composed of particles that do not have mass
when at rest, and physics tells us under different circumstances
the universe might not contain anything with mass. Yet we know
that our universe is made up of an uncountable number of particles with mass, and these building blocks are necessary to
form the elements that make up the people to ask such questions. The search for the answer to this question illustrates
nicely the process we call the scientific method.
Current theoretical models tell us that everything in the universe is based on two types of elementary particles: bosons and
fermions. We can distinguish the roles of these particles by considering the building blocks of matter to be constructed from fermions, while bosons are particles responsible for the force that holds
the fermions together. In 1964, three different research teams independently proposed mechanisms in which a field of energy permeates the universe, and the interaction of matter with this field is due
to a specific boson associated with the field. The greater the number
of these bosons, the greater the interaction will be with the field.
This interaction is the property we call mass, and the field and the
associated boson came to be named for Peter Higgs, one of the
original physicists to propose this mechanism.
This theory ignited a frantic search for the “Higgs boson”
that became one of the most heralded quests in modern science.
The Large Hadron Collider at CERN in Geneva, Switzerland
(described on p. 875) was constructed to carry out experiments
designed to find evidence for the Higgs boson. In these experiments, protons are accelerated to nearly the speed of light in
opposite directions in a circular 17-mile tunnel, and then allowed to collide, generating even more fundamental particles at
very high energies. The data are examined for evidence of an
excess of particles at an energy consistent with theoretical predictions for the Higgs boson. The ongoing process of theory
suggesting experiments that give results used to evaluate and
ultimately refine the theory, and so on, is the essence of the
scientific method.
Illustration of the data obtained from decay of the Higgs boson into other
particles following an 8-TeV collision at the Large Hadron Collider at CERN.
On July 4, 2012, scientists at CERN announced the discovery of the Higgs boson. It takes about 1 trillion proton-proton
collisions to produce one Higgs boson event, so it requires a
tremendous amount of data obtained from two independent sets
of experiments to confirm the findings. In science, the quest for
answers is never completely done. Our understanding can always
be improved or refined, and sometimes entire tenets of accepted
science are replaced by another theory that does a better job explaining the observations. For example, scientists are not sure if
the Higgs boson is the only particle that confers mass to matter, or
if it is only one of several such bosons predicted by other theories.
But over the long run, the scientific method has proven to
be our best way of understanding the physical world. It took
50 years for experimental science to validate the existence of the
Higgs boson. This discovery was greeted with great fanfare and
recognized the following year with a 2013 Nobel Prize in
Physics for Peter Higgs and François Englert, another one of the
six original scientists who first proposed the existence of a universal field that gives particles their mass. It is impossible to
imagine where science will take our understanding of the universe in the next 50 years, but we can be fairly certain that many
of the theories and experiments driving this scientific discovery
will be very different than the ones we use today.
1.4 Classifications of Matter
We defined chemistry in Section 1.1 as the study of matter and the changes it undergoes. Matter is anything that occupies space and has mass. Matter includes things
we can see and touch (such as water, earth, and trees), as well as things we cannot
(such as air). Thus, everything in the universe has a “chemical” connection.
6
1.4 Classifications of Matter
7
Chemists distinguish among several subcategories of matter based on composition and properties. The classifications of matter include substances, mixtures,
elements, and compounds, as well as atoms and molecules, which we will consider
in Chapter 2.
Substances and Mixtures
A substance is a form of matter that has a definite (constant) composition and distinct
properties. Examples are water, ammonia, table sugar (sucrose), gold, and oxygen.
Substances differ from one another in composition and can be identified by their
appearance, smell, taste, and other properties.
A mixture is a combination of two or more substances in which the substances
retain their distinct identities. Some familiar examples are air, soft drinks, milk, and
cement. Mixtures do not have constant composition. Therefore, samples of air collected in different cities would probably differ in composition because of differences
in altitude, pollution, and so on.
Mixtures are either homogeneous or heterogeneous. When a spoonful of sugar
dissolves in water we obtain a homogeneous mixture in which the composition of the
mixture is the same throughout. If sand is mixed with iron filings, however, the sand
grains and the iron filings remain separate (Figure 1.4). This type of mixture is called
a heterogeneous mixture because the composition is not uniform.
Any mixture, whether homogeneous or heterogeneous, can be created and then
separated by physical means into pure components without changing the identities
of the components. Thus, sugar can be recovered from a water solution by heating
the solution and evaporating it to dryness. Condensing the vapor will give us back
the water component. To separate the iron-sand mixture, we can use a magnet to
remove the iron filings from the sand, because sand is not attracted to the magnet
[see Figure 1.4(b)]. After separation, the components of the mixture will have the
same composition and properties as they did to start with.
Elements and Compounds
Substances can be either elements or compounds. An element is a substance that
cannot be separated into simpler substances by chemical means. To date, 118 elements have been positively identified. Most of them occur naturally on Earth. The
others have been created by scientists via nuclear processes, which are the subject of
Chapter 19 of this text.
Figure 1.4
(a) The mixture
contains iron filings and sand.
(b) A magnet separates the iron
filings from the mixture. The same
technique is used on a larger
scale to separate iron and steel
from nonmagnetic objects such as
aluminum, glass, and plastics.
(a)
(b)
8
Chapter 1 ■ Chemistry: The Study of Change
Table 1.1
Name
Some Common Elements and Their Symbols
Symbol
Aluminum
Arsenic
Barium
Bismuth
Bromine
Calcium
Carbon
Chlorine
Chromium
Cobalt
Copper
Al
As
Ba
Bi
Br
Ca
C
Cl
Cr
Co
Cu
Name
Fluorine
Gold
Hydrogen
Iodine
Iron
Lead
Magnesium
Manganese
Mercury
Nickel
Nitrogen
Symbol
F
Au
H
I
Fe
Pb
Mg
Mn
Hg
Ni
N
Name
Symbol
Oxygen
Phosphorus
Platinum
Potassium
Silicon
Silver
Sodium
Sulfur
Tin
Tungsten
Zinc
O
P
Pt
K
Si
Ag
Na
S
Sn
W
Zn
For convenience, chemists use symbols of one or two letters to represent the elements. The first letter of a symbol is always capitalized, but any following letters are
not. For example, Co is the symbol for the element cobalt, whereas CO is the formula
for the carbon monoxide molecule. Table 1.1 shows the names and symbols of some
of the more common elements; a complete list of the elements and their symbols
appears inside the front cover of this book. The symbols of some elements are derived
from their Latin names—for example, Au from aurum (gold), Fe from ferrum (iron),
and Na from natrium (sodium)—whereas most of them come from their English
names. Appendix 1 gives the origin of the names and lists the discoverers of most of
the elements.
Atoms of most elements can interact with one another to form compounds.
Hydrogen gas, for example, burns in oxygen gas to form water, which has properties that are distinctly different from those of the starting materials. Water is made
up of two parts hydrogen and one part oxygen. This composition does not change,
regardless of whether the water comes from a faucet in the United States, a lake in
Outer Mongolia, or the ice caps on Mars. Thus, water is a compound, a substance
composed of atoms of two or more elements chemically united in fixed proportions.
Unlike mixtures, compounds can be separated only by chemical means into their
pure components.
The relationships among elements, compounds, and other categories of matter are
summarized in Figure 1.5.
Review of Concepts
Which of the following diagrams represent elements and which represent
compounds? Each color sphere (or truncated sphere) represents an atom.
Different colored atoms indicate different elements.
(a)
(b)
(c)
(d)
9
1.5 The Three States of Matter
Matter
Separation by
physical methods
Mixtures
Heterogeneous
mixtures
Homogeneous
mixtures
Figure 1.5
Substances
Compounds
Separation by
chemical methods
Elements
Classification of matter.
1.5 The Three States of Matter
All substances, at least in principle, can exist in three states: solid, liquid, and
gas. As Figure 1.6 shows, gases differ from liquids and solids in the distances
between the molecules. In a solid, molecules are held close together in an orderly
fashion with little freedom of motion. Molecules in a liquid are close together but
are not held so rigidly in position and can move past one another. In a gas, the
molecules are separated by distances that are large compared with the size of the
molecules.
The three states of matter can be interconverted without changing the composition
of the substance. Upon heating, a solid (for example, ice) will melt to form a liquid
(water). (The temperature at which this transition occurs is called the melting point.)
Further heating will convert the liquid into a gas. (This conversion takes place at the
boiling point of the liquid.) On the other hand, cooling a gas will cause it to condense
into a liquid. When the liquid is cooled further, it will freeze into the solid form.
Figure 1.6 Microscopic views
of a solid, a liquid, and a gas.
Solid
Liquid
Gas
10
Chapter 1 ■ Chemistry: The Study of Change
Figure 1.7 The three states of
matter. A hot poker changes ice
into water and steam.
Figure 1.7 shows the three states of water. Note that the properties of water are unique
among common substances in that the molecules in the liquid state are more closely
packed than those in the solid state.
Review of Concepts
An ice cube is placed in a closed container. On heating, the ice cube first melts and
the water then boils to form steam. Which of the following statements is true?
(a) The physical appearance of the water is different at every stage of change.
(b) The mass of water is greatest for the ice cube and least for the steam.
1.6 Physical and Chemical Properties of Matter
Substances are identified by their properties as well as by their composition. Color,
melting point, and boiling point are physical properties. A physical property can be
measured and observed without changing the composition or identity of a substance.
For example, we can measure the melting point of ice by heating a block of ice and
recording the temperature at which the ice is converted to water. Water differs from
ice only in appearance, not in composition, so this is a physical change; we can freeze
1.7 Measurement
the water to recover the original ice. Therefore, the melting point of a substance is a
physical property. Similarly, when we say that helium gas is lighter than air, we are
referring to a physical property.
On the other hand, the statement “Hydrogen gas burns in oxygen gas to form
water” describes a chemical property of hydrogen, because to observe this property
we must carry out a chemical change, in this case burning. After the change, the
original chemical substance, the hydrogen gas, will have vanished, and all that will
be left is a different chemical substance—water. We cannot recover the hydrogen from
the water by means of a physical change, such as boiling or freezing.
Every time we hard-boil an egg, we bring about a chemical change. When subjected to a temperature of about 1008C, the yolk and the egg white undergo changes
that alter not only their physical appearance but their chemical makeup as well. When
eaten, the egg is changed again, by substances in our bodies called enzymes. This
digestive action is another example of a chemical change. What happens during digestion depends on the chemical properties of both the enzymes and the food.
All measurable properties of matter fall into one of two additional categories:
extensive properties and intensive properties. The measured value of an extensive
property depends on how much matter is being considered. Mass, which is the quantity of matter in a given sample of a substance, is an extensive property. More matter
means more mass. Values of the same extensive property can be added together. For
example, two copper pennies will have a combined mass that is the sum of the masses
of each penny, and the length of two tennis courts is the sum of the lengths of each
tennis court. Volume, defined as length cubed, is another extensive property. The value
of an extensive quantity depends on the amount of matter.
The measured value of an intensive property does not depend on how much matter is being considered. Density, defined as the mass of an object divided by its
volume, is an intensive property. So is temperature. Suppose that we have two beakers
of water at the same temperature. If we combine them to make a single quantity of
water in a larger beaker, the temperature of the larger quantity of water will be the
same as it was in two separate beakers. Unlike mass, length, and volume, temperature
and other intensive properties are not additive.
Review of Concepts
The diagram in (a) shows a compound made up of atoms of two elements
(represented by the green and red spheres) in the liquid state. Which of the
diagrams in (b)–(d) represents a physical change and which diagrams represent
a chemical change?
(a)
(b)
(c)
(d)
1.7 Measurement
The measurements chemists make are often used in calculations to obtain other related
quantities. Different instruments enable us to measure a substance’s properties: The
meterstick measures length or scale; the buret, the pipet, the graduated cylinder, and
Hydrogen burning in air to form
water.
11
12
Chapter 1 ■ Chemistry: The Study of Change
the volumetric flask measure volume (Figure 1.8); the balance measures
mass; the thermometer measures temperature. These instruments provide
measurements of macroscopic properties, which can be determined
directly. Microscopic properties, on the atomic or molecular scale, must
be determined by an indirect method, as we will see in Chapter 2.
A measured quantity is usually written as a number with an appropriate unit. To say that the distance between New York and San Francisco
by car along a certain route is 5166 is meaningless. We must specify that
the distance is 5166 kilometers. The same is true in
chemistry; units are essential to stating measurements correctly.
SI Units
For many years, scientists recorded measurements in metric units, which are related decimally, that is, by powers of 10. In 1960, however,
the General Conference of Weights and Measures, the international authority on units, proposed a revised metric system called the
International System of Units (abbreviated SI,
from the French Système Internationale d’Unites).
Table 1.2 shows the seven SI base units. All other
units of measurement can be derived from these
base units. Like metric units, SI units are modified in decimal fashion by a series of prefixes, as
shown in Table 1.3. We will use both metric and
SI units in this book.
Figure 1.8 Some common measuring devices
found in a chemistry laboratory. These devices
are not drawn to scale relative to one another.
We will discuss the uses of these measuring
devices in Chapter 4.
Volumetric flask
Graduated cylinder
Pipet
Buret
1.7 Measurement
Table 1.2
13
Note that a metric prefix simply represents
a number:
1 mm 5 1 3 1023 m
SI Base Units
Base Quantity
Name of Unit
Length
Mass
Time
Electrical current
Temperature
Amount of substance
Luminous intensity
meter
kilogram
second
ampere
kelvin
mole
candela
Symbol
m
kg
s
A
K
mol
cd
An astronaut jumping on the
surface of the moon.
Table 1.3
Prefixes Used with SI Units
Prefix
Symbol
teragigamegakilodecicentimillimicronanopicofemtoatto-
T
G
M
k
d
c
m
μ
n
p
f
a
Meaning
Example
1,000,000,000,000, or 1012
1,000,000,000, or 109
1,000,000, or 106
1,000, or 103
1/10, or 1021
1/100, or 1022
1/1,000, or 1023
1/1,000,000, or 1026
1/1,000,000,000, or 1029
1/1,000,000,000,000, or 10212
1/1,000,000,000,000,000, or 10215
1/1,000,000,000,000,000,000 or 10218
1
1
1
1
1
1
1
1
1
1
1
1
terameter (Tm) 5 1 3 1012 m
gigameter (Gm) 5 1 3 109 m
megameter (Mm) 5 1 3 106 m
kilometer (km) 5 1 3 103 m
decimeter (dm) 5 0.1 m
centimeter (cm) 5 0.01 m
millimeter (mm) 5 0.001 m
micrometer (μm) 5 1 3 1026 m
nanometer (nm) 5 1 3 1029 m
picometer (pm) 5 1 3 10212 m
femtometer (fm) 5 1 3 10215 m
attometer (am) 5 1 3 10218 m
Measurements that we will utilize frequently in our study of chemistry include
time, mass, volume, density, and temperature.
Mass and Weight
The terms “mass” and “weight” are often used interchangeably, although, strictly
speaking, they are different quantities. Whereas mass is a measure of the amount of
matter in an object, weight, technically speaking, is the force that gravity exerts on
an object. An apple that falls from a tree is pulled downward by Earth’s gravity. The
mass of the apple is constant and does not depend on its location, but its weight does.
For example, on the surface of the moon the apple would weigh only one-sixth what
it does on Earth, because the moon’s gravity is only one-sixth that of Earth. The
moon’s smaller gravity enabled astronauts to jump about rather freely on its surface
despite their bulky suits and equipment. Chemists are interested primarily in mass,
which can be determined readily with a balance; the process of measuring mass, oddly,
is called weighing.
The SI unit of mass is the kilogram (kg). Unlike the units of length and time,
which are based on natural processes that can be repeated by scientists anywhere, the
kilogram is defined in terms of a particular object (Figure 1.9). In chemistry, however,
the smaller gram (g) is more convenient:
1 kg 5 1000 g 5 1 3 103 g
Figure 1.9 The prototype
kilogram is made of a platinumiridium alloy. It is kept in a vault
at the International Bureau of
Weights and Measures in Sèvres,
France. In 2007 it was discovered
that the alloy has mysteriously lost
about 50 μg!
14
Chapter 1 ■ Chemistry: The Study of Change
Volume: 1000 cm3;
1000 mL;
1 dm3;
1L
Volume
The SI unit of length is the meter (m), and the SI-derived unit for volume is the
cubic meter (m3). Generally, however, chemists work with much smaller volumes,
such as the cubic centimeter (cm3) and the cubic decimeter (dm3):
1 cm3 5 (1 3 1022 m) 3 5 1 3 1026 m3
1 dm3 5 (1 3 1021 m) 3 5 1 3 1023 m3
Another common unit of volume is the liter (L). A liter is the volume occupied
by one cubic decimeter. One liter of volume is equal to 1000 milliliters (mL) or
1000 cm3:
1 cm
1 L 5 1000 mL
5 1000 cm3
5 1 dm3
10 cm = 1 dm
Volume: 1 cm3;
1 mL
and one milliliter is equal to one cubic centimeter:
1 cm
Figure 1.10 Comparison of two
1 mL 5 1 cm3
volumes, 1 mL and 1000 mL.
Figure 1.10 compares the relative sizes of two volumes. Even though the liter is not
an SI unit, volumes are usually expressed in liters and milliliters.
Density
The equation for density is
density 5
mass
volume
Table 1.4
Densities of Some
Substances at 25°C
Substance
Density
(g/cm3)
Air*
Ethanol
Water
Graphite
Table salt
Aluminum
Diamond
Iron
Lead
Mercury
Gold
Osmium†
0.001
0.79
1.00
2.2
2.2
2.70
3.5
7.9
11.3
13.6
19.3
22.6
*Measured at 1 atmosphere.
†
Osmium (Os) is the densest element
known.
or
d5
m
V
(1.1)
where d, m, and V denote density, mass, and volume, respectively. Because density is
an intensive property and does not depend on the quantity of mass present, for a given
substance the ratio of mass to volume always remains the same; in other words, V
increases as m does. Density usually decreases with temperature.
The SI-derived unit for density is the kilogram per cubic meter (kg/m3). This unit
is awkwardly large for most chemical applications. Therefore, grams per cubic centimeter (g/cm3) and its equivalent, grams per milliliter (g/mL), are more commonly used for solid and liquid densities. Because gas densities are often very low,
we express them in units of grams per liter (g/L):
1 g/cm3 5 1 g/mL 5 1000 kg/m3
1 g/L 5 0.001 g/mL
Table 1.4 lists the densities of several substances.
1.7 Measurement
15
Examples 1.1 and 1.2 show density calculations.
Example 1.1
Gold is a precious metal that is chemically unreactive. It is used mainly in jewelry,
dentistry, and electronic devices. A piece of gold ingot with a mass of 301 g has a
volume of 15.6 cm3. Calculate the density of gold.
Solution We are given the mass and volume and asked to calculate the density.
Therefore, from Equation (1.1), we write
d5
5
m
V
301 g
15.6 cm3
5 19.3 g/cm3
Practice Exercise A piece of platinum metal with a density of 21.5 g/cm3 has a
volume of 4.49 cm3. What is its mass?
Example 1.2
The density of mercury, the only metal that is a liquid at room temperature, is 13.6 g/mL.
Calculate the mass of 5.50 mL of the liquid.
Gold bars and the solid-state
arrangement of the gold atoms.
Similar problems: 1.21, 1.22.
Solution We are given the density and volume of a liquid and asked to calculate the
mass of the liquid. We rearrange Equation (1.1) to give
m5d3V
g
5 13.6
3 5.50 mL
mL
5 74.8 g
Practice Exercise The density of sulfuric acid in a certain car battery is 1.41 g/mL.
Calculate the mass of 242 mL of the liquid.
Temperature Scales
Three temperature scales are currently in use. Their units are 8F (degrees Fahrenheit), 8C (degrees Celsius), and K (kelvin). The Fahrenheit scale, which is the most
commonly used scale in the United States outside the laboratory, defines the normal
freezing and boiling points of water to be exactly 328F and 2128F, respectively.
The Celsius scale divides the range between the freezing point (08C) and boiling
point (1008C) of water into 100 degrees. As Table 1.2 shows, the kelvin is the SI
base unit of temperature: It is the absolute temperature scale. By absolute we mean
that the zero on the Kelvin scale, denoted by 0 K, is the lowest temperature that
can be attained theoretically. On the other hand, 08F and 08C are based on the
behavior of an arbitrarily chosen substance, water. Figure 1.11 compares the three
temperature scales.
The size of a degree on the Fahrenheit scale is only 100/180, or 5/9, of a degree
on the Celsius scale. To convert degrees Fahrenheit to degrees Celsius, we write
?°C 5 (°F 2 32°F) 3
5°C
9°F
(1.2)
Mercury.
Similar problems: 1.21, 1.22.
Note that the Kelvin scale does not have
the degree sign. Also, temperatures
expressed in kelvins can never be
negative.
16
Chapter 1 ■ Chemistry: The Study of Change
Figure 1.11 Comparison of the
three temperature scales: Celsius,
and Fahrenheit, and the absolute
(Kelvin) scales. Note that there are
100 divisions, or 100 degrees,
between the freezing point and
the boiling point of water on the
Celsius scale, and there are 180
divisions, or 180 degrees, between
the same two temperature limits
on the Fahrenheit scale. The
Celsius scale was formerly called
the centigrade scale.
373 K
100°C
310 K
37°C
298 K
25°C
Room
temperature
77°F
273 K
0°C
Freezing point
of water
32°F
Kelvin
Boiling point
of water
212°F
Body
temperature
Celsius
98.6°F
Fahrenheit
The following equation is used to convert degrees Celsius to degrees Fahrenheit:
?°F 5
9°F
3 (°C) 1 32°F
5°C
(1.3)
Both the Celsius and the Kelvin scales have units of equal magnitude; that is,
one degree Celsius is equivalent to one kelvin. Experimental studies have shown that
absolute zero on the Kelvin scale is equivalent to 2273.158C on the Celsius scale.
Thus, we can use the following equation to convert degrees Celsius to kelvin:
? K 5 (°C 1 273.15°C)
1K
1°C
(1.4)
We will frequently find it necessary to convert between degrees Celsius and
degrees Fahrenheit and between degrees Celsius and kelvin. Example 1.3 illustrates
these conversions.
The Chemistry in Action essay on page 17 shows why we must be careful with
units in scientific work.
Example 1.3
(a) Below the transition temperature of 21418C, a certain substance becomes a
superconductor; that is, it can conduct electricity with no resistance. What is the temperature
in degrees Fahrenheit? (b) Helium has the lowest boiling point of all the elements at
24528F. Convert this temperature to degrees Celsius. (c) Mercury, the only metal that exists
as a liquid at room temperature, melts at 238.98C. Convert its melting point to kelvins.
Magnet suspended above
superconductor cooled below
its transition temperature by
liquid nitrogen.
Solution These three parts require that we carry out temperature conversions, so we
need Equations (1.2), (1.3), and (1.4). Keep in mind that the lowest temperature on the
Kelvin scale is zero (0 K); therefore, it can never be negative.
(a) This conversion is carried out by writing
9°F
3 (2141°C) 1 32°F 5 2222°F
5°C
(Continued)
CHEMISTRY in Action
The Importance of Units
I
n December 1998, NASA launched the 125-million dollar
Mars Climate Orbiter, intended as the red planet’s first weather
satellite. After a 416-million mi journey, the spacecraft was supposed to go into Mars’ orbit on September 23, 1999. Instead, it
entered Mars’ atmosphere about 100 km (62 mi) lower than
planned and was destroyed by heat. The mission controllers said
the loss of the spacecraft was due to the failure to convert
English measurement units into metric units in the navigation
software.
Engineers at Lockheed Martin Corporation who built the
spacecraft specified its thrust in pounds, which is an English
unit. Scientists at NASA’s Jet Propulsion Laboratory, on the
other hand, had assumed that thrust data they received were
expressed in metric units, as newtons. Normally, pound is the
unit for mass. Expressed as a unit for force, however, 1 lb is
the force due to gravitational attraction on an object of that
mass. To carry out the conversion between pound and newton,
we start with 1 lb 5 0.4536 kg and from Newton’s second law
of motion,
scientist said: “This is going to be the cautionary tale that will
be embedded into introduction to the metric system in elementary school, high school, and college science courses till the end
of time.”
force 5 mass 3 acceleration
5 0.4536 kg 3 9.81 m/s2
5 4.45 kg m/s2
5 4.45 N
because 1 newton (N) 5 1 kg m/s2. Therefore, instead of
converting 1 lb of force to 4.45 N, the scientists treated it
as 1 N.
The considerably smaller engine thrust expressed in newtons resulted in a lower orbit and the ultimate destruction of the
spacecraft. Commenting on the failure of the Mars mission, one
Artist’s conception of the Martian Climate Orbiter.
(b) Here we have
(2452°F 2 32°F) 3
5°C
5 2269°C
9°F
(c) The melting point of mercury in kelvins is given by
(238.9°C 1 273.15°C) 3
1K
5 234.3 K
1°C
Similar problems: 1.24, 1.25, 1.26.
Practice Exercise Convert (a) 327.58C (the melting point of lead) to degrees
Fahrenheit; (b) 172.98F (the boiling point of ethanol) to degrees Celsius; and (c) 77 K,
the boiling point of liquid nitrogen, to degrees Celsius.
17
18
Chapter 1 ■ Chemistry: The Study of Change
Review of Concepts
The density of copper is 8.94 g/cm3 at 208C and 8.91 g/cm3 at 608C. This density
decrease is the result of which of the following?
(a) The metal expands.
(b) The metal contracts.
(c) The mass of the metal increases.
(d) The mass of the metal decreases.
1.8 Handling Numbers
Having surveyed some of the units used in chemistry, we now turn to techniques for handling numbers associated with measurements: scientific notation and significant figures.
Scientific Notation
Chemists often deal with numbers that are either extremely large or extremely small.
For example, in 1 g of the element hydrogen there are roughly
602,200,000,000,000,000,000,000
hydrogen atoms. Each hydrogen atom has a mass of only
0.00000000000000000000000166 g
These numbers are cumbersome to handle, and it is easy to make mistakes when using
them in arithmetic computations. Consider the following multiplication:
0.0000000056 3 0.00000000048 5 0.000000000000000002688
It would be easy for us to miss one zero or add one more zero after the decimal point.
Consequently, when working with very large and very small numbers, we use a system
called scientific notation. Regardless of their magnitude, all numbers can be expressed
in the form
N 3 10n
where N is a number between 1 and 10 and n, the exponent, is a positive or negative
integer (whole number). Any number expressed in this way is said to be written in
scientific notation.
Suppose that we are given a certain number and asked to express it in scientific
notation. Basically, this assignment calls for us to find n. We count the number of
places that the decimal point must be moved to give the number N (which is between
1 and 10). If the decimal point has to be moved to the left, then n is a positive integer; if it has to be moved to the right, n is a negative integer. The following examples
illustrate the use of scientific notation:
(1) Express 568.762 in scientific notation:
568.762 5 5.68762 3 102
Note that the decimal point is moved to the left by two places and n 5 2.
(2) Express 0.00000772 in scientific notation:
0.00000772 5 7.72 3 1026
Here the decimal point is moved to the right by six places and n 5 26.
1.8 Handling Numbers
Keep in mind the following two points. First, n 5 0 is used for numbers that are
not expressed in scientific notation. For example, 74.6 3 100 (n 5 0) is equivalent
to 74.6. Second, the usual practice is to omit the superscript when n 5 1. Thus, the
scientific notation for 74.6 is 7.46 3 10 and not 7.46 3 101.
Next, we consider how scientific notation is handled in arithmetic operations.
Addition and Subtraction
To add or subtract using scientific notation, we first write each quantity—say, N1 and
N2—with the same exponent n. Then we combine N1 and N2; the exponents remain
the same. Consider the following examples:
(7.4 3 103 ) 1 (2.1 3 103 ) 5 9.5 3 103
(4.31 3 104 ) 1 (3.9 3 103 ) 5 (4.31 3 104 ) 1 (0.39 3 104 )
5 4.70 3 104
22
23
(2.22 3 10 ) 2 (4.10 3 10 ) 5 (2.22 3 1022 ) 2 (0.41 3 1022 )
5 1.81 3 1022
Multiplication and Division
To multiply numbers expressed in scientific notation, we multiply N1 and N2 in the
usual way, but add the exponents together. To divide using scientific notation, we
divide N1 and N2 as usual and subtract the exponents. The following examples show
how these operations are performed:
(8.0 3 104 ) 3 (5.0 3 102 ) 5 (8.0 3 5.0)(10412 )
5 40 3 106
5 4.0 3 107
(4.0 3 1025 ) 3 (7.0 3 103 ) 5 (4.0 3 7.0)(102513 )
5 28 3 1022
5 2.8 3 1021
7
6.9 3 10
6.9
5
3 1072 (25)
25
3.0
3.0 3 10
5 2.3 3 1012
4
8.5 3 10
8.5
5
3 10429
9
5.0
5.0 3 10
5 1.7 3 1025
Significant Figures
Except when all the numbers involved are integers (for example, in counting the
number of students in a class), it is often impossible to obtain the exact value of
the quantity under investigation. For this reason, it is important to indicate the
margin of error in a measurement by clearly indicating the number of significant
figures, which are the meaningful digits in a measured or calculated quantity. When
significant figures are used, the last digit is understood to be uncertain. For example,
we might measure the volume of a given amount of liquid using a graduated cylinder with a scale that gives an uncertainty of 1 mL in the measurement. If the
volume is found to be 6 mL, then the actual volume is in the range of 5 mL to
7 mL. We represent the volume of the liquid as (6 6 1) mL. In this case, there is
only one significant figure (the digit 6) that is uncertain by either plus or minus
1 mL. For greater accuracy, we might use a graduated cylinder that has finer divisions, so that the volume we measure is now uncertain by only 0.1 mL. If the
volume of the liquid is now found to be 6.0 mL, we may express the quantity as
(6.0 6 0.1) mL, and the actual value is somewhere between 5.9 mL and 6.1 mL.
19
Any number raised to the power zero is
equal to one.
20
Chapter 1 ■ Chemistry: The Study of Change
We can further improve the measuring device and obtain more significant figures,
but in every case, the last digit is always uncertain; the amount of this uncertainty
depends on the particular measuring device we use.
Figure 1.12 shows a modern balance. Balances such as this one are available
in many general chemistry laboratories; they readily measure the mass of objects
to four decimal places. Therefore, the measured mass typically will have four
significant figures (for example, 0.8642 g) or more (for example, 3.9745 g).
Keeping track of the number of significant figures in a measurement such as mass
ensures that calculations involving the data will reflect the precision of the
measurement.
Figure 1.12 A Fisher Scientific
A-200DS Digital Recorder
Precision Balance.
Guidelines for Using Significant Figures
We must always be careful in scientific work to write the proper number of significant
figures. In general, it is fairly easy to determine how many significant figures a number has by following these rules:
1. Any digit that is not zero is significant. Thus, 845 cm has three significant figures,
1.234 kg has four significant figures, and so on.
2. Zeros between nonzero digits are significant. Thus, 606 m contains three significant figures, 40,501 kg contains five significant figures, and so on.
3. Zeros to the left of the first nonzero digit are not significant. Their purpose is
to indicate the placement of the decimal point. For example, 0.08 L contains
one significant figure, 0.0000349 g contains three significant figures, and
so on.
4. If a number is greater than 1, then all the zeros written to the right of the
decimal point count as significant figures. Thus, 2.0 mg has two significant
figures, 40.062 mL has five significant figures, and 3.040 dm has four significant figures. If a number is less than 1, then only the zeros that are at the end
of the number and the zeros that are between nonzero digits are significant. This
means that 0.090 kg has two significant figures, 0.3005 L has four significant
figures, 0.00420 min has three significant figures, and so on.
5. For numbers that do not contain decimal points, the trailing zeros (that is, zeros
after the last nonzero digit) may or may not be significant. Thus, 400 cm may
have one significant figure (the digit 4), two significant figures (40), or three
significant figures (400). We cannot know which is correct without more
information. By using scientific notation, however, we avoid this ambiguity.
In this particular case, we can express the number 400 as 4 3 102 for one
significant figure, 4.0 3 102 for two significant figures, or 4.00 3 102 for
three significant figures.
Example 1.4 shows the determination of significant figures.
Example 1.4
Determine the number of significant figures in the following measurements: (a) 394 cm,
(b) 5.03 g, (c) 0.714 m, (d) 0.052 kg, (e) 2.720 3 1022 atoms, (f ) 3000 mL.
Solution (a) Three , because each digit is a nonzero digit. (b) Three , because zeros
between nonzero digits are significant. (c) Three , because zeros to the left of the first
nonzero digit do not count as significant figures. (d) Two . Same reason as in (c).
(e) Four . Because the number is greater than one, all the zeros written to the right of
the decimal point count as significant figures. (f) This is an ambiguous case. The
number of significant figures may be four (3.000 3 103), three (3.00 3 103), two
(Continued)
1.8 Handling Numbers
(3.0 3 103), or one (3 3 103). This example illustrates why scientific notation must be
used to show the proper number of significant figures.
Practice Exercise Determine the number of significant figures in each of the following
measurements: (a) 35 mL, (b) 2008 g, (c) 0.0580 m3, (d) 7.2 3 104 molecules, (e) 830 kg.
A second set of rules specifies how to handle significant figures in calculations.
1. In addition and subtraction, the answer cannot have more digits to the right of
the decimal point than either of the original numbers. Consider these examples:
89.332
1 1.1 ←— one digit after the decimal point
90.432 ←— round off to 90.4
2.097
2 0.12 ←— two digits after the decimal point
1.977 ←— round off to 1.98
The rounding-off procedure is as follows. To round off a number at a certain point
we simply drop the digits that follow if the first of them is less than 5. Thus, 8.724
rounds off to 8.72 if we want only two digits after the decimal point. If the first
digit following the point of rounding off is equal to or greater than 5, we add 1 to
the preceding digit. Thus, 8.727 rounds off to 8.73, and 0.425 rounds off to 0.43.
2. In multiplication and division, the number of significant figures in the final product or quotient is determined by the original number that has the smallest number
of significant figures. The following examples illustrate this rule:
2.8 3 4.5039 5 12.61092 — round off to 13
6.85
5 0.0611388789 — round off to 0.0611
112.04
3. Keep in mind that exact numbers obtained from definitions or by counting numbers of objects can be considered to have an infinite number of significant figures.
For example, the inch is defined to be exactly 2.54 centimeters; that is,
1 in 5 2.54 cm
Thus, the “2.54” in the equation should not be interpreted as a measured number with
three significant figures. In calculations involving conversion between “in” and “cm,”
we treat both “1” and “2.54” as having an infinite number of significant figures.
Similarly, if an object has a mass of 5.0 g, then the mass of nine such objects is
5.0 g 3 9 5 45 g
The answer has two significant figures because 5.0 g has two significant figures.
The number 9 is exact and does not determine the number of significant figures.
Example 1.5 shows how significant figures are handled in arithmetic operations.
Example 1.5
Carry out the following arithmetic operations to the correct number of significant
figures: (a) 12,343.2 g 1 0.1893 g, (b) 55.67 L 2 2.386 L, (c) 7.52 m 3 6.9232,
(d) 0.0239 kg 4 46.5 mL, (e) 5.21 3 103 cm 1 2.92 3 102 cm.
(Continued)
Similar problems: 1.33, 1.34.
21
22
Chapter 1 ■ Chemistry: The Study of Change
Solution In addition and subtraction, the number of decimal places in the answer is
determined by the number having the lowest number of decimal places. In multiplication
and division, the significant number of the answer is determined by the number having
the smallest number of significant figures.
(a)
(b)
12,343.2 g
1
0.1893 g
12,343.3893 g ←— round off to 12,343.4 g
55.67 L
2 2.386 L
53.284 L ←— round off to 53.28 L
(c) 7.52 m 3 6.9232 5 52.06246 m ←— round off to 52.1 m
0.0239 kg
5 0.0005139784946 kg/mL ←— round off to 0.000514 kg/mL
46.5 mL
or 5.14 3 1024 kg/mL
2
(e) First we change 2.92 3 10 cm to 0.292 3 103 cm and then carry out the addition
(5.21 cm 1 0.292 cm) 3 103. Following the procedure in (a), we find the answer
is 5.50 3 103 cm.
(d)
Similar problems: 1.35, 1.36.
Practice Exercise Carry out the following arithmetic operations and round off the
answers to the appropriate number of significant figures: (a) 26.5862 L 1 0.17 L,
(b) 9.1 g 2 4.682 g, (c) 7.1 3 104 dm 3 2.2654 3 102 dm, (d) 6.54 g 4 86.5542 mL,
(e) (7.55 3 104 m) 2 (8.62 3 103 m).
The preceding rounding-off procedure applies to one-step calculations. In
chain calculations, that is, calculations involving more than one step, we can get
a different answer depending on how we round off. Consider the following twostep calculations:
First step:
Second step:
A3B5C
C3D5E
Let’s suppose that A 5 3.66, B 5 8.45, and D 5 2.11. Depending on whether we
round off C to three or four significant figures, we obtain a different number for E:
Method 1
3.66 3 8.45 5 30.9
30.9 3 2.11 5 65.2
Method 2
3.66 3 8.45 5 30.93
30.93 3 2.11 5 65.3
However, if we had carried out the calculation as 3.66 3 8.45 3 2.11 on a calculator without rounding off the intermediate answer, we would have obtained 65.3 as
the answer for E. Although retaining an additional digit past the number of significant figures for intermediate steps helps to eliminate errors from rounding, this
procedure is not necessary for most calculations because the difference between the
answers is usually quite small. Therefore, for most examples and end-of-chapter
problems where intermediate answers are reported, all answers, intermediate and
final, will be rounded.
Accuracy and Precision
In discussing measurements and significant figures, it is useful to distinguish between
accuracy and precision. Accuracy tells us how close a measurement is to the true
value of the quantity that was measured. Precision refers to how closely two or more
measurements of the same quantity agree with one another (Figure 1.13).
1.9 Dimensional Analysis in Solving Problems
23
Figure 1.13 The distribution of
holes formed by darts on a dart
board shows the difference
between precise and accurate.
(a) Good accuracy and good
precision. (b) Poor accuracy and
good precision. (c) Poor accuracy
and poor precision.
(a)
(b)
(c)
The difference between accuracy and precision is a subtle but important one.
Suppose, for example, that three students are asked to determine the mass of a piece
of copper wire. The results of two successive weighings by each student are
Average value
Student A
1.964 g
1.978 g
1.971 g
Student B
1.972 g
1.968 g
1.970 g
Student C
2.000 g
2.002 g
2.001 g
The true mass of the wire is 2.000 g. Therefore, Student B’s results are more precise
than those of Student A (1.972 g and 1.968 g deviate less from 1.970 g than 1.964 g
and 1.978 g from 1.971 g), but neither set of results is very accurate. Student C’s
results are not only the most precise, but also the most accurate, because the average
value is closest to the true value. Highly accurate measurements are usually precise
too. On the other hand, highly precise measurements do not necessarily guarantee
accurate results. For example, an improperly calibrated meterstick or a faulty balance
may give precise readings that are in error.
Review of Concepts
Give the length of the pencil with proper significant figures according to which
ruler you use for the measurement.
1.9 Dimensional Analysis in Solving Problems
Careful measurements and the proper use of significant figures, along with correct
calculations, will yield accurate numerical results. But to be meaningful, the answers
also must be expressed in the desired units. The procedure we use to convert between
units in solving chemistry problems is called dimensional analysis (also called the
factor-label method). A simple technique requiring little memorization, dimensional
analysis is based on the relationship between different units that express the same
24
Chapter 1 ■ Chemistry: The Study of Change
physical quantity. For example, by definition 1 in 5 2.54 cm (exactly). This equivalence enables us to write a conversion factor as follows:
1 in
2.54 cm
Because both the numerator and the denominator express the same length, this fraction
is equal to 1. Similarly, we can write the conversion factor as
2.54 cm
1 in
which is also equal to 1. Conversion factors are useful for changing units. Thus, if
we wish to convert a length expressed in inches to centimeters, we multiply the length
by the appropriate conversion factor.
12.00 in 3
2.54 cm
5 30.48 cm
1 in
We choose the conversion factor that cancels the unit inches and produces the desired
unit, centimeters. Note that the result is expressed in four significant figures because
2.54 is an exact number.
Next let us consider the conversion of 57.8 meters to centimeters. This problem
can be expressed as
? cm 5 57.8 m
By definition,
1 cm 5 1 3 10 22 m
Because we are converting “m” to “cm,” we choose the conversion factor that has
meters in the denominator,
1 cm
1 3 1022 m
and write the conversion as
? cm 5 57.8 m 3
1 cm
1 3 1022 m
5 5780 cm
5 5.78 3 103 cm
Note that scientific notation is used to indicate that the answer has three significant
figures. Again, the conversion factor 1 cm/1 3 1022 m contains exact numbers; therefore, it does not affect the number of significant figures.
In general, to apply dimensional analysis we use the relationship
given quantity 3 conversion factor 5 desired quantity
and the units cancel as follows:
Remember that the unit we want appears
in the numerator and the unit we want to
cancel appears in the denominator.
given unit 3
desired unit
5 desired unit
given unit
In dimensional analysis, the units are carried through the entire sequence of calculations. Therefore, if the equation is set up correctly, then all the units will cancel except
the desired one. If this is not the case, then an error must have been made somewhere,
and it can usually be spotted by reviewing the solution.
1.9 Dimensional Analysis in Solving Problems
25
A Note on Problem Solving
At this point you have been introduced to scientific notation, significant figures, and
dimensional analysis, which will help you in solving numerical problems. Chemistry
is an experimental science and many of the problems are quantitative in nature. The
key to success in problem solving is practice. Just as a marathon runner cannot prepare
for a race by simply reading books on running and a pianist cannot give a successful
concert by only memorizing the musical score, you cannot be sure of your understanding of chemistry without solving problems. The following steps will help to improve
your skill at solving numerical problems.
1. Read the question carefully. Understand the information that is given and what
you are asked to solve. Frequently it is helpful to make a sketch that will help
you to visualize the situation.
2. Find the appropriate equation that relates the given information and the
unknown quantity. Sometimes solving a problem will involve more than one
step, and you may be expected to look up quantities in tables that are not
provided in the problem. Dimensional analysis is often needed to carry out
conversions.
3. Check your answer for the correct sign, units, and significant figures.
4. A very important part of problem solving is being able to judge whether the
answer is reasonable. It is relatively easy to spot a wrong sign or incorrect units.
But if a number (say, 9) is incorrectly placed in the denominator instead of in
the numerator, the answer would be too small even if the sign and units of the
calculated quantity were correct.
5. One quick way to check the answer is to round off the numbers in the calculation in
such a way so as to simplify the arithmetic. The answer you get will not be exact,
but it will be close to the correct one.
Example 1.6
A person’s average daily intake of glucose (a form of sugar) is 0.0833 pound (lb). What
is this mass in milligrams (mg)? (1 lb 5 453.6 g.)
Strategy The problem can be stated as
? mg 5 0.0833 lb
The relationship between pounds and grams is given in the problem. This relationship
will enable conversion from pounds to grams. A metric conversion is then needed to
convert grams to milligrams (1 mg 5 1 3 1023 g). Arrange the appropriate
conversion factors so that pounds and grams cancel and the unit milligrams is
obtained in your answer.
Glucose tablets can provide
diabetics with a quick method for
raising their blood sugar levels.
Solution The sequence of conversions is
pounds ¡ grams ¡ milligrams
Conversion factors for some of the
English system units commonly used in
the United States for nonscientific
measurements (for example, pounds and
inches) are provided inside the back
cover of this book.
Using the following conversion factors
453.6 g
1 lb
and
1 mg
1 3 1023 g
(Continued)
26
Chapter 1 ■ Chemistry: The Study of Change
we obtain the answer in one step:
? mg 5 0.0833 lb 3
Similar problem: 1.45.
453.6 g
1 mg
3
5 3.78 3 104 mg
1 lb
1 3 1023 g
Check As an estimate, we note that 1 lb is roughly 500 g and that 1 g 5 1000 mg.
Therefore, 1 lb is roughly 5 3 105 mg. Rounding off 0.0833 lb to 0.1 lb, we get
5 3 104 mg, which is close to the preceding quantity.
Practice Exercise A roll of aluminum foil has a mass of 1.07 kg. What is its mass
in pounds?
As Examples 1.7 and 1.8 illustrate, conversion factors can be squared or cubed
in dimensional analysis.
Example 1.7
A liquid helium storage tank has a volume of 275 L. What is the volume in m3?
Strategy The problem can be stated as
? m3 5 275 L
How many conversion factors are needed for this problem? Recall that 1 L 5 1000 cm3
and 1 cm 5 1 3 1022 m.
Solution We need two conversion factors here: one to convert liters to cm3 and one to
convert centimeters to meters:
1000 cm3
1L
and
1 3 1022 m
1 cm
Because the second conversion deals with length (cm and m) and we want volume here,
it must therefore be cubed to give
A cryogenic storage tank for
liquid helium.
1 3 1022 m
1 3 1022 m
1 3 1022 m
1 3 1022 m 3
3
3
5a
b
1 cm
1 cm
1 cm
1 cm
This means that 1 cm3 5 1 3 1026 m3. Now we can write
Remember that when a unit is raised to a
power, any conversion factor you use
must also be raised to that power.
? m3 5 275 L 3
1000 cm3
1 3 1022 m 3
3a
b 5 0.275 m3
1L
1 cm
Check From the preceding conversion factors you can show that 1 L 5 1 3 1023 m3.
Similar problem: 1.50(d).
Therefore, a 275-L storage tank would be equal to 275 3 1023 m3 or 0.275 m3, which
is the answer.
Practice Exercise The volume of a room is 1.08 3 108 dm3. What is the volume in m3?
Example 1.8
Liquid nitrogen is obtained from liquefied air and is used to prepare frozen goods and in
low-temperature research. The density of the liquid at its boiling point (21968C or 77 K)
is 0.808 g/cm3. Convert the density to units of kg/m3.
(Continued)
1.10 Real-World Problem Solving: Information, Assumptions, and Simplifications
27
Strategy The problem can be stated as
? kg/m3 5 0.808 g/cm3
Two separate conversions are required for this problem: g ¡ kg and cm3 ¡ m3.
Recall that 1 kg 5 1000 g and 1 cm 5 1 3 1022 m.
Solution In Example 1.7 we saw that 1 cm3 5 1 3 1026 m3. The conversion factors are
1 kg
1000 g
and
1 cm3
1 3 1026 m3
Finally,
3
? kg/m 5
0.808 g
1 cm3
Liquid nitrogen is used for frozen
foods and low-temperature
research.
1 kg
1 cm3
5 808 kg/m3
3
3
1000 g
1 3 1026 m3
Check Because 1 m3 5 1 3 106 cm3, we would expect much more mass in 1 m3 than
in 1 cm3. Therefore, the answer is reasonable.
Similar problem: 1.51.
2
3
Practice Exercise The density of the lightest metal, lithium (Li), is 5.34 3 10 kg/m .
Convert the density to g/cm3.
Review of Concepts
The Food and Drug Administration recommends no more than 65 g of daily
intake of fat. What is this mass in pounds? (1 lb 5 453.6 g.)
1.10
Real-World Problem Solving:
Information, Assumptions, and Simplifications
In chemistry, as in other scientific disciplines, it is not always possible to solve a
numerical problem exactly. There are many reasons why this is the case. For example,
our understanding of a situation is not complete or data are not fully available. In
these cases, we must learn to make an intelligent guess. This approach is sometimes
called “ball-park estimates,” which are simple, quick calculations that can be done on
the “back of an envelope.” As you can imagine, in many cases the answers are only
order-of-magnitude estimates.†
In most of the example problems that you have seen so far, as well as the questions
given at the end of this and subsequent chapters, the necessary information is provided;
however, in order to solve important real-world problems such as those related to medicine, energy, and agriculture, you must be able to determine what information is needed
and where to find it. Much of the information you might need can be found in the
various tables located throughout the text, and a list of tables and important figures is
given on the inside back cover. In many cases, however, you will need to go to outside
sources to find the information you need. Although the Internet is a fast way to find
information, you must take care that the source is reliable and well referenced. One
excellent source is the National Institute of Standards and Technology (NIST).
In order to know what information you need, you will first have to formulate a plan
for solving the problem. In addition to the limitations of the theories used in science,
typically assumptions are made in setting up and solving the problems based on those
theories. These assumptions come at a price, however, as the accuracy of the answer is
reduced with increasing simplifications of the problem, as illustrated in Example 1.9.
†
An order of magnitude is a factor of 10.
28
Chapter 1 ■ Chemistry: The Study of Change
Example 1.9
A modern pencil “lead” is actually composed primarily of graphite, a form of carbon.
Estimate the mass of the graphite core in a standard No. 2 pencil before it is sharpened.
Strategy Assume that the pencil lead can be approximated as a cylinder. Measurement
of a typical unsharpened pencil gives a length of about 18 cm (subtracting the length of
the eraser head) and a diameter of roughly 2 mm for the lead. The volume of a cylinder
V is given by V 5 πr2l, where r is the radius and l is the length. Assuming that the
lead is pure graphite, you can calculate the mass of the lead from the volume using the
density of graphite given in Table 1.4.
Solution Converting the diameter of the lead to units of cm gives
2 mm 3
1 cm
5 0.2 cm
10 mm
which, along with the length of the lead, gives
0.2 cm 2
b 3 18 cm
2
5 0.57 cm3
V5πa
Rearranging Equation (1.1) gives
m5d3V
g
3 0.57 cm3
5 2.2
cm3
5 1g
Check Rounding off the values used to calculate the volume of the lead gives
3 3 (0.1 cm)2 3 20 cm 5 0.6 cm3. Multiplying that volume by roughly 2 g/cm3 gives
around 1 g, which agrees with the value just calculated.
Similar problems: 1.105, 1.106, 1.114.
Practice Exercise Estimate the mass of air in a ping pong ball.
Considering Example 1.9, even if the dimensions of the pencil lead were measured with greater precision, the accuracy of the final answer would be limited by the
assumptions made in modeling this problem. The pencil lead is actually a mixture of
graphite and clay, where the relative amounts of the two materials determine the softness of the lead, so the density of the material is likely to be different than 2.2 g/cm3.
You could probably find a better value for the density of the mixture used to make
No. 2 pencils, but it is not worth the effort in this case.
Key Equations
d5
m
(1.1)
V
?°C 5 (°F 2 32°F) 3
Equation for density
5°C
(1.2)
9°F
9°F
3 (°C) 1 32°F (1.3)
5°C
1K
(1.4)
? K 5 (°C 1 273.15°C)
1°C
?°F 5
Converting °F to °C
Converting °C to °F
Converting °C to K
Questions & Problems
29
Summary of Facts & Concepts
1. The study of chemistry involves three basic steps: observation, representation, and interpretation. Observation refers to measurements in the macroscopic world;
representation involves the use of shorthand notation
symbols and equations for communication; interpretations are based on atoms and molecules, which belong
to the microscopic world.
2. The scientific method is a systematic approach to research that begins with the gathering of information
through observation and measurements. In the process,
hypotheses, laws, and theories are devised and tested.
3. Chemists study matter and the changes it undergoes.
The substances that make up matter have unique physical properties that can be observed without changing
their identity and unique chemical properties that, when
they are demonstrated, do change the identity of the
4.
5.
6.
7.
substances. Mixtures, whether homogeneous or heterogeneous, can be separated into pure components by
physical means.
The simplest substances in chemistry are elements.
Compounds are formed by the chemical combination of
atoms of different elements in fixed proportions.
All substances, in principle, can exist in three states:
solid, liquid, and gas. The interconversion between
these states can be effected by changing the temperature.
SI units are used to express physical quantities in all
sciences, including chemistry.
Numbers expressed in scientific notation have the form
N 3 10n, where N is between 1 and 10, and n is a positive or negative integer. Scientific notation helps us
handle very large and very small quantities.
Key Words
Accuracy, p. 22
Chemical property, p. 11
Chemistry, p. 2
Compound, p. 8
Density, p. 11
Element, p. 7
Extensive property, p. 11
Heterogeneous mixture, p. 7
Homogeneous mixture, p. 7
Hypothesis, p. 4
Intensive property, p. 11
International System of Units
(SI), p. 12
Kelvin, p. 15
Law, p. 4
Liter, p. 14
Macroscopic property, p. 12
Mass, p. 11
Matter, p. 6
Microscopic property, p. 12
Mixture, p. 7
Physical property, p. 10
Precision, p. 22
Qualitative, p. 4
Quantitative, p. 4
Scientific method, p. 4
Significant figures, p. 19
Substance, p. 7
Theory, p. 5
Volume, p. 11
Weight, p. 13
Questions & Problems
to music would have been much greater if he had
married. (b) An autumn leaf gravitates toward the
ground because there is an attractive force between
the leaf and Earth. (c) All matter is composed of
very small particles called atoms.
• Problems available in Connect Plus
Red numbered problems solved in Student Solutions Manual
The Scientific Method
Review Questions
1.1
1.2
Explain what is meant by the scientific method.
What is the difference between qualitative data and
quantitative data?
Problems
1.3
• 1.4
Classify the following as qualitative or quantitative
statements, giving your reasons. (a) The sun is approximately 93 million mi from Earth. (b) Leonardo
da Vinci was a better painter than Michelangelo. (c)
Ice is less dense than water. (d) Butter tastes better
than margarine. (e) A stitch in time saves nine.
Classify each of the following statements as a hypothesis, a law, or a theory. (a) Beethoven’s contribution
Classification and Properties of Matter
Review Questions
1.5
• 1.6
1.7
1.8
Give an example for each of the following terms:
(a) matter, (b) substance, (c) mixture.
Give an example of a homogeneous mixture and an
example of a heterogeneous mixture.
Using examples, explain the difference between a
physical property and a chemical property.
How does an intensive property differ from an extensive property? Which of the following properties
are intensive and which are extensive? (a) length,
(b) volume, (c) temperature, (d) mass.
30
1.9
1.10
Chapter 1 ■ Chemistry: The Study of Change
Give an example of an element and a compound.
How do elements and compounds differ?
What is the number of known elements?
Problems
• 1.11
• 1.12
• 1.13
1.14
1.15
• 1.16
Do the following statements describe chemical or
physical properties? (a) Oxygen gas supports
combustion. (b) Fertilizers help to increase agricultural production. (c) Water boils below 1008C on top
of a mountain. (d) Lead is denser than aluminum.
(e) Uranium is a radioactive element.
Does each of the following describe a physical
change or a chemical change? (a) The helium gas
inside a balloon tends to leak out after a few hours.
(b) A flashlight beam slowly gets dimmer and finally goes out. (c) Frozen orange juice is reconstituted by adding water to it. (d) The growth of plants
depends on the sun’s energy in a process called photosynthesis. (e) A spoonful of table salt dissolves in
a bowl of soup.
Give the names of the elements represented by
the chemical symbols Li, F, P, Cu, As, Zn, Cl, Pt,
Mg, U, Al, Si, Ne. (See Table 1.1 and the inside
front cover.)
Give the chemical symbols for the following elements:
(a) cesium, (b) germanium, (c) gallium, (d) strontium,
(e) uranium, (f) selenium, (g) neon, (h) cadmium. (See
Table 1.1 and the inside front cover.)
Classify each of the following substances as an element or a compound: (a) hydrogen, (b) water, (c) gold,
(d) sugar.
Classify each of the following as an element, a
compound, a homogeneous mixture, or a heterogeneous mixture: (a) water from a well, (b) argon gas,
(c) sucrose, (d) a bottle of red wine, (e) chicken
noodle soup, (f ) blood flowing in a capillary,
(g) ozone.
Measurement
Problems
1.21
• 1.22
• 1.23
• 1.24
• 1.25
1.26
Handling Numbers
Review Questions
1.27
1.28
Review Questions
1.17
• 1.18
1.19
1.20
Name the SI base units that are important in chemistry. Give the SI units for expressing the following:
(a) length, (b) volume, (c) mass, (d) time, (e) energy,
(f ) temperature.
Write the numbers represented by the following prefixes: (a) mega-, (b) kilo-, (c) deci-, (d) centi-, (e) milli-,
(f) micro-, (g) nano-, (h) pico-.
What units do chemists normally use for density of
liquids and solids? For gas density? Explain the
differences.
Describe the three temperature scales used in the
laboratory and in everyday life: the Fahrenheit scale,
the Celsius scale, and the Kelvin scale.
Bromine is a reddish-brown liquid. Calculate its density
(in g/mL) if 586 g of the substance occupies 188 mL.
The density of methanol, a colorless organic liquid
used as solvent, is 0.7918 g/mL. Calculate the mass
of 89.9 mL of the liquid.
Convert the following temperatures to degrees
Celsius or Fahrenheit: (a) 958F, the temperature on a
hot summer day; (b) 128F, the temperature on a cold
winter day; (c) a 1028F fever; (d) a furnace operating
at 18528F; (e) 2273.158C (theoretically the lowest
attainable temperature).
(a) Normally the human body can endure a temperature of 1058F for only short periods of time without
permanent damage to the brain and other vital organs. What is this temperature in degrees Celsius?
(b) Ethylene glycol is a liquid organic compound
that is used as an antifreeze in car radiators. It
freezes at 211.58C. Calculate its freezing temperature in degrees Fahrenheit. (c) The temperature on
the surface of the sun is about 63008C. What is this
temperature in degrees Fahrenheit? (d) The ignition
temperature of paper is 4518F. What is the temperature in degrees Celsius?
Convert the following temperatures to kelvin:
(a) 1138C, the melting point of sulfur, (b) 378C, the
normal body temperature, (c) 3578C, the boiling
point of mercury.
Convert the following temperatures to degrees Celsius: (a) 77 K, the boiling point of liquid nitrogen,
(b) 4.2 K, the boiling point of liquid helium, (c) 601 K,
the melting point of lead.
What is the advantage of using scientific notation
over decimal notation?
Define significant figure. Discuss the importance of
using the proper number of significant figures in
measurements and calculations.
Problems
• 1.29
1.30
• 1.31
Express the following numbers in scientific notation:
(a) 0.000000027, (b) 356, (c) 47,764, (d) 0.096.
Express the following numbers as decimals:
(a) 1.52 3 1022, (b) 7.78 3 1028.
Express the answers to the following calculations in
scientific notation:
(a) 145.75 1 (2.3 3 1021)
(b) 79,500 4 (2.5 3 102)
(c) (7.0 3 1023) 2 (8.0 3 1024)
(d) (1.0 3 104) 3 (9.9 3 106)
Questions & Problems
1.32
• 1.33
1.34
• 1.35
• 1.36
1.37
• 1.38
Express the answers to the following calculations in
scientific notation:
(a) 0.0095 1 (8.5 3 1023)
(b) 653 4 (5.75 3 1028)
(c) 850,000 2 (9.0 3 105)
(d) (3.6 3 1024) 3 (3.6 3 106)
What is the number of significant figures in each of
the following measurements?
(a) 4867 mi
(b) 56 mL
(c) 60,104 tons
(d) 2900 g
(e) 40.2 g/cm3
(f) 0.0000003 cm
(g) 0.7 min
(h) 4.6 3 1019 atoms
How many significant figures are there in each of
the following? (a) 0.006 L, (b) 0.0605 dm,
(c) 60.5 mg, (d) 605.5 cm2, (e) 960 3 1023 g,
(f) 6 kg, (g) 60 m.
Carry out the following operations as if they were
calculations of experimental results, and express
each answer in the correct units with the correct
number of significant figures:
(a) 5.6792 m 1 0.6 m 1 4.33 m
(b) 3.70 g 2 2.9133 g
(c) 4.51 cm 3 3.6666 cm
(d) (3 3 104 g 1 6.827 g)/(0.043 cm3 2 0.021 cm3)
Carry out the following operations as if they were
calculations of experimental results, and express
each answer in the correct units with the correct
number of significant figures:
(a) 7.310 km 4 5.70 km
(b) (3.26 3 1023 mg) 2 (7.88 3 1025 mg)
(c) (4.02 3 106 dm) 1 (7.74 3 107 dm)
(d) (7.8 m 2 0.34 m)/(1.15 s 1 0.82 s)
Three students (A, B, and C) are asked to determine the volume of a sample of ethanol. Each student measures the volume three times with a
graduated cylinder. The results in milliliters are:
A (87.1, 88.2, 87.6); B (86.9, 87.1, 87.2); C (87.6,
87.8, 87.9). The true volume is 87.0 mL. Comment on the precision and the accuracy of each
student’s results.
Three apprentice tailors (X, Y, and Z) are assigned
the task of measuring the seam of a pair of trousers.
Each one makes three measurements. The results in
inches are X (31.5, 31.6, 31.4); Y (32.8, 32.3, 32.7);
Z (31.9, 32.2, 32.1). The true length is 32.0 in. Comment on the precision and the accuracy of each tailor’s measurements.
31
Dimensional Analysis
Problems
• 1.39
1.40
• 1.41
1.42
1.43
• 1.44
• 1.45
1.46
1.47
• 1.48
• 1.49
1.50
• 1.51
• 1.52
Carry out the following conversions: (a) 22.6 m to
decimeters, (b) 25.4 mg to kilograms, (c) 556 mL to
liters, (d) 10.6 kg/m3 to g/cm3.
Carry out the following conversions: (a) 242 lb to
milligrams, (b) 68.3 cm3 to cubic meters, (c) 7.2 m3
to liters, (d) 28.3 μg to pounds.
The average speed of helium at 258C is 1255 m/s.
Convert this speed to miles per hour (mph).
How many seconds are there in a solar year
(365.24 days)?
How many minutes does it take light from the
sun to reach Earth? (The distance from the sun
to Earth is 93 million mi; the speed of light 5
3.00 3 10 8 m/s.)
A jogger runs a mile in 8.92 min. Calculate the speed
in (a) in/s, (b) m/min, (c) km/h. (1 mi 5 1609 m;
1 in 5 2.54 cm.)
A 6.0-ft person weighs 168 lb. Express this person’s
height in meters and weight in kilograms. (1 lb 5
453.6 g; 1 m 5 3.28 ft.)
The speed limit on parts of the German autobahn
was once set at 286 kilometers per hour (km/h). Calculate the speed limit in miles per hour (mph).
For a fighter jet to take off from the deck of an aircraft carrier, it must reach a speed of 62 m/s. Calculate the speed in miles per hour (mph).
The “normal” lead content in human blood is about
0.40 part per million (that is, 0.40 g of lead per million grams of blood). A value of 0.80 part per million (ppm) is considered to be dangerous. How
many grams of lead are contained in 6.0 3 103 g of
blood (the amount in an average adult) if the lead
content is 0.62 ppm?
Carry out the following conversions: (a) 1.42 lightyears to miles (a light-year is an astronomical measure
of distance—the distance traveled by light in a year, or
365 days; the speed of light is 3.00 3 108 m/s).
(b) 32.4 yd to centimeters. (c) 3.0 3 1010 cm/s to ft/s.
Carry out the following conversions: (a) 70 kg, the
average weight of a male adult, to pounds. (b) 14 billion years (roughly the age of the universe) to seconds. (Assume there are 365 days in a year.) (c) 7 ft
6 in, the height of the basketball player Yao Ming, to
meters. (d) 88.6 m3 to liters.
Aluminum is a lightweight metal (density 5 2.70 g/
cm3) used in aircraft construction, high-voltage
transmission lines, beverage cans, and foils. What is
its density in kg/m3?
Ammonia gas is used as a refrigerant in large-scale cooling systems. The density of ammonia gas under certain
conditions is 0.625 g/L. Calculate its density in g/cm3.
32
Chapter 1 ■ Chemistry: The Study of Change
Additional Problems
1.53
• 1.54
1.55
• 1.56
• 1.57
• 1.58
• 1.59
1.60
• 1.61
1.62
• 1.63
Give one qualitative and one quantitative statement
about each of the following: (a) water, (b) carbon,
(c) iron, (d) hydrogen gas, (e) sucrose (cane sugar),
(f) table salt (sodium chloride), (g) mercury,
(h) gold, (i) air.
Which of the following statements describe physical properties and which describe chemical properties? (a) Iron has a tendency to rust. (b) Rainwater
in industrialized regions tends to be acidic. (c) Hemoglobin molecules have a red color. (d) When a
glass of water is left out in the sun, the water gradually disappears. (e) Carbon dioxide in air is converted to more complex molecules by plants during
photosynthesis.
In 2008, about 95.0 billion lb of sulfuric acid were
produced in the United States. Convert this quantity
to tons.
In determining the density of a rectangular metal
bar, a student made the following measurements:
length, 8.53 cm; width, 2.4 cm; height, 1.0 cm;
mass, 52.7064 g. Calculate the density of the metal
to the correct number of significant figures.
Calculate the mass of each of the following: (a) a
sphere of gold with a radius of 10.0 cm [the volume
of a sphere with a radius r is V 5 (4/3)πr3; the density of gold 5 19.3 g/cm3], (b) a cube of platinum of
edge length 0.040 mm (the density of platinum 5
21.4 g/cm3), (c) 50.0 mL of ethanol (the density of
ethanol 5 0.798 g/mL).
A cylindrical glass bottle 21.5 cm in length is filled
with cooking oil of density 0.953 g/mL. If the mass
of the oil needed to fill the bottle is 1360 g, calculate
the inner diameter of the bottle.
The following procedure was used to determine the
volume of a flask. The flask was weighed dry and
then filled with water. If the masses of the empty
flask and filled flask were 56.12 g and 87.39 g, respectively, and the density of water is 0.9976 g/cm3,
calculate the volume of the flask in cm3.
The speed of sound in air at room temperature is
about 343 m/s. Calculate this speed in miles per
hour. (1 mi 5 1609 m.)
A piece of silver (Ag) metal weighing 194.3 g is
placed in a graduated cylinder containing 242.0
mL of water. The volume of water now reads
260.5 mL. From these data calculate the density
of silver.
The experiment described in Problem 1.61 is a crude
but convenient way to determine the density of some
solids. Describe a similar experiment that would enable you to measure the density of ice. Specifically,
what would be the requirements for the liquid used
in your experiment?
A lead sphere of diameter 48.6 cm has a mass of
6.852 3 105 g. Calculate the density of lead.
• 1.64
• 1.65
• 1.66
• 1.67
• 1.68
• 1.69
• 1.70
• 1.71
Lithium is the least dense metal known (density:
0.53 g/cm3). What is the volume occupied by 1.20 3
103 g of lithium?
The medicinal thermometer commonly used in
homes can be read 60.18F, whereas those in the
doctor’s office may be accurate to 60.18C. In degrees Celsius, express the percent error expected
from each of these thermometers in measuring a
person’s body temperature of 38.98C.
Vanillin (used to flavor vanilla ice cream and other
foods) is the substance whose aroma the human
nose detects in the smallest amount. The threshold
limit is 2.0 3 10211 g per liter of air. If the current
price of 50 g of vanillin is $112, determine the cost
to supply enough vanillin so that the aroma could be
detected in a large aircraft hangar with a volume of
5.0 3 107 ft3.
At what temperature does the numerical reading on
a Celsius thermometer equal that on a Fahrenheit
thermometer?
Suppose that a new temperature scale has been devised
on which the melting point of ethanol (2117.38C) and
the boiling point of ethanol (78.38C) are taken as 08S
and 1008S, respectively, where S is the symbol for the
new temperature scale. Derive an equation relating a
reading on this scale to a reading on the Celsius scale.
What would this thermometer read at 258C?
A resting adult requires about 240 mL of pure
oxygen/min and breathes about 12 times every minute. If inhaled air contains 20 percent oxygen by
volume and exhaled air 16 percent, what is the volume of air per breath? (Assume that the volume of
inhaled air is equal to that of exhaled air.)
(a) Referring to Problem 1.69, calculate the total
volume (in liters) of air an adult breathes in a day.
(b) In a city with heavy traffic, the air contains 2.1 3
1026 L of carbon monoxide (a poisonous gas) per
liter. Calculate the average daily intake of carbon
monoxide in liters by a person.
Three different 25.0-g samples of solid pellets are
added to 20.0 mL of water in three different measuring cylinders. The results are shown here. Given the
densities of the three metals used, identify the cylinder that contains each sample of solid pellets: A
(2.9 g/cm3), B (8.3 g/cm3), and C (3.3 g/cm3).
30
30
30
20
20
20
(a)
(b)
(c)
Questions & Problems
1.72
• 1.73
• 1.74
• 1.75
• 1.76
• 1.77
The circumference of an NBA-approved basketball
is 29.6 in. Given that the radius of Earth is about
6400 km, how many basketballs would it take to
circle around the equator with the basketballs touching one another? Round off your answer to an integer with three significant figures.
A student is given a crucible and asked to prove
whether it is made of pure platinum. She first weighs
the crucible in air and then weighs it suspended in
water (density 5 0.9986 g/mL). The readings are
860.2 g and 820.2 g, respectively. Based on these
measurements and given that the density of platinum is 21.45 g/cm3, what should her conclusion be?
(Hint: An object suspended in a fluid is buoyed up
by the mass of the fluid displaced by the object. Neglect the buoyance of air.)
The surface area and average depth of the Pacific
Ocean are 1.8 3 108 km2 and 3.9 3 103 m, respectively. Calculate the volume of water in the ocean in
liters.
The unit “troy ounce” is often used for precious
metals such as gold (Au) and platinum (Pt). (1 troy
ounce 5 31.103 g.) (a) A gold coin weighs 2.41
troy ounces. Calculate its mass in grams. (b) Is a
troy ounce heavier or lighter than an ounce? (1 lb 5
16 oz; 1 lb 5 453.6 g.)
Osmium (Os) is the densest element known (density 5
22.57 g/cm3). Calculate the mass in pounds and in
kilograms of an Os sphere 15 cm in diameter (about
the size of a grapefruit). See Problem 1.57 for volume of a sphere.
Percent error is often expressed as the absolute value
of the difference between the true value and the experimental value, divided by the true value:
percent error 5
• 1.78
• 1.79
• 1.80
Ztrue value 2 experimental valueZ
Ztrue valueZ
• 1.81
1.82
• 1.83
1.84
• 1.85
• 1.86
3 100%
The vertical lines indicate absolute value. Calculate
the percent error for the following measurements:
(a) The density of alcohol (ethanol) is found to be
0.802 g/mL. (True value: 0.798 g/mL.) (b) The mass
of gold in an earring is analyzed to be 0.837 g. (True
value: 0.864 g.)
The natural abundances of elements in the human
body, expressed as percent by mass, are: oxygen
(O), 65 percent; carbon (C), 18 percent; hydrogen (H), 10 percent; nitrogen (N), 3 percent; calcium (Ca), 1.6 percent; phosphorus (P), 1.2
percent; all other elements, 1.2 percent. Calculate the mass in grams of each element in the
body of a 62-kg person.
The men’s world record for running a mile outdoors
(as of 1999) is 3 min 43.13 s. At this rate, how long
would it take to run a 1500-m race? (1 mi 5 1609 m.)
Venus, the second closest planet to the sun, has a
surface temperature of 7.3 3 102 K. Convert this
temperature to 8C and 8F.
• 1.87
• 1.88
• 1.89
• 1.90
33
Chalcopyrite, the principal ore of copper (Cu),
contains 34.63 percent Cu by mass. How many
grams of Cu can be obtained from 5.11 3 103 kg of
the ore?
It has been estimated that 8.0 3 104 tons of gold
(Au) have been mined. Assume gold costs $948 per
ounce. What is the total worth of this quantity of
gold?
A 1.0-mL volume of seawater contains about 4.0 3
10212 g of gold. The total volume of ocean water is
1.5 3 1021 L. Calculate the total amount of gold (in
grams) that is present in seawater, and the worth of
the gold in dollars (see Problem 1.82). With so much
gold out there, why hasn’t someone become rich by
mining gold from the ocean?
Measurements show that 1.0 g of iron (Fe) contains
1.1 3 1022 Fe atoms. How many Fe atoms are in 4.9 g
of Fe, which is the total amount of iron in the body
of an average adult?
The thin outer layer of Earth, called the crust,
contains only 0.50 percent of Earth’s total mass
and yet is the source of almost all the elements
(the atmosphere provides elements such as oxygen, nitrogen, and a few other gases). Silicon
(Si) is the second most abundant element in
Earth’s crust (27.2 percent by mass). Calculate
the mass of silicon in kilograms in Earth’s crust.
(The mass of Earth is 5.9 3 1021 tons. 1 ton 5
2000 lb; 1 lb 5 453.6 g.)
The radius of a copper (Cu) atom is roughly 1.3 3
10210 m. How many times can you divide evenly a
piece of 10-cm copper wire until it is reduced to two
separate copper atoms? (Assume there are appropriate tools for this procedure and that copper atoms
are lined up in a straight line, in contact with each
other. Round off your answer to an integer.)
One gallon of gasoline in an automobile’s engine
produces on the average 9.5 kg of carbon dioxide,
which is a greenhouse gas, that is, it promotes the
warming of Earth’s atmosphere. Calculate the annual production of carbon dioxide in kilograms if
there are 250 million cars in the United States and
each car covers a distance of 5000 mi at a consumption rate of 20 miles per gallon.
A sheet of aluminum (Al) foil has a total area
of 1.000 ft2 and a mass of 3.636 g. What is the
thickness of the foil in millimeters? (Density of
Al 5 2.699 g/cm3.)
Comment on whether each of the following is a
homogeneous mixture or a heterogeneous mixture: (a) air in a closed bottle and (b) air over New
York City.
Chlorine is used to disinfect swimming pools. The
accepted concentration for this purpose is 1 ppm
chlorine, or 1 g of chlorine per million grams of
water. Calculate the volume of a chlorine solution
(in milliliters) a homeowner should add to her
34
• 1.91
• 1.92
Chapter 1 ■ Chemistry: The Study of Change
swimming pool if the solution contains 6.0 percent
chlorine by mass and there are 2.0 3 104 gallons of
water in the pool. (1 gallon 5 3.79 L; density of
liquids 5 1.0 g/mL.)
An aluminum cylinder is 10.0 cm in length and has
a radius of 0.25 cm. If the mass of a single Al atom
is 4.48 3 10223g, calculate the number of Al atoms
present in the cylinder. The density of aluminum is
2.70 g/cm3.
A pycnometer is a device for measuring the density
of liquids. It is a glass flask with a close-fitting
ground glass stopper having a capillary hole
through it. (a) The volume of the pycnometer is
determined by using distilled water at 208C with a
known density of 0.99820 g/mL. First, the water is
filled to the rim. With the stopper in place, the fine
hole allows the excess liquid to escape. The pycnometer is then carefully dried with filter paper.
Given that the masses of the empty pycnometer
and the same one filled with water are 32.0764 g
and 43.1195 g, respectively, calculate the volume
of the pycnometer. (b) If the mass of the pycnometer filled with ethanol at 208C is 40.8051 g, calculate the density of ethanol. (c) Pycnometers can
also be used to measure the density of solids. First,
small zinc granules weighing 22.8476 g are placed
in the pycnometer, which is then filled with water.
If the combined mass of the pycnometer plus the
zinc granules and water is 62.7728 g, what is the
density of zinc?
• 1.95
• 1.96
1.97
• 1.98
• 1.93
• 1.94
In 1849 a gold prospector in California collected a
bag of gold nuggets plus sand. Given that the density of
gold and sand are 19.3 g/cm3 and 2.95 g/cm3,
respectively, and that the density of the mixture is
4.17 g/cm3, calculate the percent by mass of gold in
the mixture.
The average time it takes for a molecule to diffuse a
distance of x cm is given by
t5
1.99
• 1.100
x2
2D
where t is the time in seconds and D is the diffusion
coefficient. Given that the diffusion coefficient of
glucose is 5.7 3 1027 cm2/s, calculate the time it
would take for a glucose molecule to diffuse 10 μm,
which is roughly the size of a cell.
1.101
A human brain weighs about 1 kg and contains
about 1011 cells. Assuming that each cell is completely filled with water (density 5 1 g/mL), calculate the length of one side of such a cell if it were a
cube. If the cells are spread out in a thin layer that
is a single cell thick, what is the surface area in
square meters?
(a) Carbon monoxide (CO) is a poisonous gas because it binds very strongly to the oxygen carrier
hemoglobin in blood. A concentration of 8.00 3 102
ppm by volume of carbon monoxide is considered
lethal to humans. Calculate the volume in liters occupied by carbon monoxide in a room that measures
17.6 m long, 8.80 m wide, and 2.64 m high at this
concentration. (b) Prolonged exposure to mercury
(Hg) vapor can cause neurological disorders and respiratory problems. For safe air quality control, the
concentration of mercury vapor must be under 0.050
mg/m3. Convert this number to g/L. (c) The general
test for type II diabetes is that the blood sugar (glucose) level should be below 120 mg per deciliter
(mg/dL). Convert this number to micrograms per
milliliter (μg/mL).
A bank teller is asked to assemble “one-dollar” sets
of coins for his clients. Each set is made of three
quarters, one nickel, and two dimes. The masses of
the coins are: quarter: 5.645 g; nickel: 4.967 g;
dime: 2.316 g. What is the maximum number of
sets that can be assembled from 33.871 kg of quarters, 10.432 kg of nickels, and 7.990 kg of dimes?
What is the total mass (in g) of the assembled sets
of coins?
A graduated cylinder is filled to the 40.00-mL mark
with a mineral oil. The masses of the cylinder before
and after the addition of the mineral oil are 124.966 g
and 159.446 g, respectively. In a separate experiment, a metal ball bearing of mass 18.713 g is placed
in the cylinder and the cylinder is again filled to the
40.00-mL mark with the mineral oil. The combined
mass of the ball bearing and mineral oil is 50.952 g.
Calculate the density and radius of the ball bearing.
[The volume of a sphere of radius r is (4/3)πr3.]
A chemist in the nineteenth century prepared an unknown substance. In general, do you think it would
be more difficult to prove that it is an element or a
compound? Explain.
Bronze is an alloy made of copper (Cu) and tin (Sn)
used in applications that require low metal-on-metal
friction. Calculate the mass of a bronze cylinder of
radius 6.44 cm and length 44.37 cm. The composition of the bronze is 79.42 percent Cu and 20.58 percent Sn and the densities of Cu and Sn are 8.94 g/cm3
and 7.31 g/cm3, respectively. What assumption
should you make in this calculation?
You are given a liquid. Briefly describe steps you
would take to show whether it is a pure substance or
a homogeneous mixture.
Answers to Practice Exercises
1.102
• 1.103
A chemist mixes two liquids A and B to form a homogeneous mixture. The densities of the liquids are
2.0514 g/mL for A and 2.6678 g/mL for B. When
she drops a small object into the mixture, she finds
that the object becomes suspended in the liquid; that
is, it neither sinks nor floats. If the mixture is made
of 41.37 percent A and 58.63 percent B by volume,
what is the density of the metal? Can this procedure
be used in general to determine the densities of solids? What assumptions must be made in applying
this method?
Tums is a popular remedy for acid indigestion. A
typical Tums tablet contains calcium carbonate plus
some inert substances. When ingested, it reacts with
1.104
35
the gastric juice (hydrochloric acid) in the stomach
to give off carbon dioxide gas. When a 1.328-g tablet reacted with 40.00 mL of hydrochloric acid (density: 1.140 g/mL), carbon dioxide gas was given off
and the resulting solution weighed 46.699 g. Calculate the number of liters of carbon dioxide gas released if its density is 1.81 g/L.
A 250-mL glass bottle was filled with 242 mL of
water at 208C and tightly capped. It was then left
outdoors overnight, where the average temperature
was 258C. Predict what would happen. The density
of water at 208C is 0.998 g/cm3 and that of ice at
258C is 0.916 g/cm3.
Interpreting, Modeling & Estimating
1.105
1.106
1.107
1.108
1.109
What is the mass of one mole of ants? (Useful information: A mole is the unit used for atomic and subatomic particles. It is approximately 6 3 1023. A
1-cm-long ant weighs about 3 mg.)
How much time (in years) does an 80-year-old person spend sleeping during his or her life span?
Estimate the daily amount of water (in gallons) used
indoors by a family of four in the United States.
Public bowling alleys generally stock bowling balls
from 8 to 16 lb, where the mass is given in whole
numbers. Given that regulation bowling balls have a
diameter of 8.6 in, which (if any) of these bowling
balls would you expect to float in water?
Fusing “nanofibers” with
diameters of 100–300 nm
gives junctures with very
small volumes that would
potentially allow the study
of reactions involving
1 μm
only a few molecules. Estimate the volume in liters
of the junction formed between two such fibers with
internal diameters of 200 nm. The scale reads 1 μm.
1.110
1.111
1.112
1.113
1.114
1.115
Estimate the annual consumption of gasoline by
passenger cars in the United States.
Estimate the total amount of ocean water in liters.
Estimate the volume of blood in an adult in liters.
How far (in feet) does light travel in one nanosecond?
Estimate the distance (in miles) covered by an NBA
player in a professional basketball game.
In water conservation, chemists spread a thin film of
a certain inert material over the surface of water to
cut down on the rate of evaporation of water in reservoirs. This technique was pioneered by Benjamin
Franklin three centuries ago. Franklin found that
0.10 mL of oil could spread over the surface of water about 40 m2 in area. Assuming that the oil forms
a monolayer, that is, a layer that is only one molecule thick, estimate the length of each oil molecule
in nanometers. (1 nm 5 1 3 1029 m.)
Answers to Practice Exercises
1.1 96.5 g. 1.2 341 g. 1.3 (a) 621.58F, (b) 78.38C,
(c) 21968C. 1.4 (a) Two, (b) four, (c) three, (d) two,
(e) three or two. 1.5 (a) 26.76 L, (b) 4.4 g, (c) 1.6 3 107 dm2,
(d) 0.0756 g/mL, (e) 6.69 3 104 m. 1.6 2.36 lb. 1.7 1.08 3
105 m3. 1.8 0.534 g/cm3. 1.9 Roughly 0.03 g.
CHEMICAL M YS TERY
The Disappearance of the Dinosaurs
D
inosaurs dominated life on Earth for millions of years and then disappeared very
suddenly. To solve the mystery, paleontologists studied fossils and skeletons found in
rocks in various layers of Earth’s crust. Their findings enabled them to map out which
species existed on Earth during specific geologic periods. They also revealed no dinosaur
skeletons in rocks formed immediately after the Cretaceous period, which dates back some
36
65 million years. It is therefore assumed that the dinosaurs became extinct about 65 million years ago.
Among the many hypotheses put forward to account for their disappearance were
disruptions of the food chain and a dramatic change in climate caused by violent volcanic
eruptions. However, there was no convincing evidence for any one hypothesis until 1977.
It was then that a group of paleontologists working in Italy obtained some very puzzling
data at a site near Gubbio. The chemical analysis of a layer of clay deposited above
sediments formed during the Cretaceous period (and therefore a layer that records events
occurring after the Cretaceous period) showed a surprisingly high content of the element
iridium (Ir). Iridium is very rare in Earth’s crust but is comparatively abundant in asteroids.
This investigation led to the hypothesis that the extinction of dinosaurs occurred as
follows. To account for the quantity of iridium found, scientists suggested that a large
asteroid several miles in diameter hit Earth about the time the dinosaurs disappeared. The
impact of the asteroid on Earth’s surface must have been so tremendous that it literally
vaporized a large quantity of surrounding rocks, soils, and other objects. The resulting dust
and debris floated through the air and blocked the sunlight for months or perhaps years.
Without ample sunlight most plants could not grow, and the fossil record confirms that
many types of plants did indeed die out at this time. Consequently, of course, many planteating animals perished, and then, in turn, meat-eating animals began to starve. Dwindling
food sources would obviously affect large animals needing great amounts of food more
quickly and more severely than small animals. Therefore, the huge dinosaurs, the largest
of which might have weighed as much as 30 tons, vanished due to lack of food.
Chemical Clues
1.
How does the study of dinosaur extinction illustrate the scientific method?
2.
Suggest two ways that would enable you to test the asteroid collision hypothesis.
3.
In your opinion, is it justifiable to refer to the asteroid explanation as the theory of
dinosaur extinction?
4.
Available evidence suggests that about 20 percent of the asteroid’s mass turned to
dust and spread uniformly over Earth after settling out of the upper atmosphere. This
dust amounted to about 0.02 g/cm2 of Earth’s surface. The asteroid very likely had a
density of about 2 g/cm3. Calculate the mass (in kilograms and tons) of the asteroid
and its radius in meters, assuming that it was a sphere. (The area of Earth is 5.1 3
1014 m2; 1 lb 5 453.6 g.) (Source: Consider a Spherical Cow—A Course in Environmental Problem Solving by J. Harte, University Science Books, Mill Valley, CA 1988.
Used with permission.)
37
CHAPTER
2
Atoms, Molecules,
and Ions
Illustration depicting Marie and Pierre Curie at work in
their laboratory. The Curies studied and identified
many radioactive elements.
CHAPTER OUTLINE
A LOOK AHEAD
2.1
2.2
2.3
The Atomic Theory
We begin with a historical perspective of the search for the fundamental
units of matter. The modern version of atomic theory was laid by John Dalton
in the nineteenth century, who postulated that elements are composed of
extremely small particles, called atoms. All atoms of a given element are
identical, but they are different from atoms of all other elements. (2.1)
2.4
2.5
2.6
2.7
2.8
The Periodic Table
We note that, through experimentation, scientists have learned that an atom
is composed of three elementary particles: proton, electron, and neutron.
The proton has a positive charge, the electron has a negative charge, and the
neutron has no charge. Protons and neutrons are located in a small region at
the center of the atom, called the nucleus, while electrons are spread out
about the nucleus at some distance from it. (2.2)
We will learn the following ways to identify atoms. Atomic number is the
number of protons in a nucleus; atoms of different elements have different
atomic numbers. Isotopes are atoms of the same element having a different
number of neutrons. Mass number is the sum of the number of protons and
neutrons in an atom. Because an atom is electrically neutral, the number of
protons is equal to the number of electrons in it. (2.3)
Next we will see how elements can be grouped together according to their
chemical and physical properties in a chart called the periodic table. The
periodic table enables us to classify elements (as metals, metalloids, and
nonmetals) and correlate their properties in a systematic way. (2.4)
We will see that atoms of most elements interact to form compounds, which
are classified as molecules or ionic compounds made of positive (cations)
and negative (anions) ions. (2.5)
We learn to use chemical formulas (molecular and empirical) to represent
molecules and ionic compounds and models to represent molecules. (2.6)
We learn a set of rules that help us name the inorganic compounds. (2.7)
38
The Structure of the Atom
Atomic Number, Mass
Number, and Isotopes
Molecules and Ions
Chemical Formulas
Naming Compounds
Introduction to Organic
Compounds
Finally, we will briefly explore the organic world to which we will return in
a later chapter. (2.8)
2.1 The Atomic Theory
39
S
ince ancient times humans have pondered the nature of matter. Our modern ideas of the
structure of matter began to take shape in the early nineteenth century with Dalton’s atomic
theory. We now know that all matter is made of atoms, molecules, and ions. All of chemistry
is concerned in one way or another with these species.
2.1 The Atomic Theory
In the fifth century b.c. the Greek philosopher Democritus expressed the belief that
all matter consists of very small, indivisible particles, which he named atomos
(meaning uncuttable or indivisible). Although Democritus’ idea was not accepted
by many of his contemporaries (notably Plato and Aristotle), somehow it endured.
Experimental evidence from early scientific investigations provided support for the
notion of “atomism” and gradually gave rise to the modern definitions of elements
and compounds. In 1808 an English scientist and school teacher, John Dalton,†
formulated a precise definition of the indivisible building blocks of matter that we
call atoms.
Dalton’s work marked the beginning of the modern era of chemistry. The hypotheses about the nature of matter on which Dalton’s atomic theory is based can be
summarized as follows:
1. Elements are composed of extremely small particles called atoms.
2. All atoms of a given element are identical, having the same size, mass, and
chemical properties. The atoms of one element are different from the atoms of
all other elements.
3. Compounds are composed of atoms of more than one element. In any compound,
the ratio of the numbers of atoms of any two of the elements present is either an
integer or a simple fraction.
4. A chemical reaction involves only the separation, combination, or rearrangement
of atoms; it does not result in their creation or destruction.
Figure 2.1 is a schematic representation of the last three hypotheses.
Dalton’s concept of an atom was far more detailed and specific than Democritus’.
The second hypothesis states that atoms of one element are different from atoms of
all other elements. Dalton made no attempt to describe the structure or composition
of atoms—he had no idea what an atom is really like. But he did realize that the
†
John Dalton (1766–1844). English chemist, mathematician, and philosopher. In addition to the atomic
theory, he also formulated several gas laws and gave the first detailed description of color blindness, from
which he suffered. Dalton was described as an indifferent experimenter, and singularly wanting in the
language and power of illustration. His only recreation was lawn bowling on Thursday afternoons. Perhaps
it was the sight of those wooden balls that provided him with the idea of the atomic theory.
Atoms of element Y
Atoms of element X
(a)
Compounds of elements X and Y
(b)
Figure 2.1 (a) According to
Dalton’s atomic theory, atoms of
the same element are identical,
but atoms of one element are
different from atoms of other
elements. (b) Compound formed
from atoms of elements X and Y.
In this case, the ratio of the atoms
of element X to the atoms of
element Y is 2:1. Note that a
chemical reaction results only in
the rearrangement of atoms, not
in their destruction or creation.
40
Chapter 2
■
Atoms, Molecules, and Ions
Carbon monoxide
1
O
± 5 ±±± 5 ±
1
C
Carbon dioxide
2
O
± 5 ±±±±±±± 5 ±
1
C
Ratio of oxygen in
carbon monoxide to
oxygen in carbon dioxide: 1:2
Figure 2.2 An illustration of the
law of multiple proportions.
different properties shown by elements such as hydrogen and oxygen can be explained
by assuming that hydrogen atoms are not the same as oxygen atoms.
The third hypothesis suggests that, to form a certain compound, we need not only
atoms of the right kinds of elements, but specific numbers of these atoms as well.
This idea is an extension of a law published in 1799 by Joseph Proust,† a French
chemist. Proust’s law of definite proportions states that different samples of the same
compound always contain its constituent elements in the same proportion by mass.
Thus, if we were to analyze samples of carbon dioxide gas obtained from different
sources, we would find in each sample the same ratio by mass of carbon to oxygen.
It stands to reason, then, that if the ratio of the masses of different elements in a given
compound is fixed, the ratio of the atoms of these elements in the compound also
must be constant.
Dalton’s third hypothesis supports another important law, the law of multiple
proportions. According to the law, if two elements can combine to form more than
one compound, the masses of one element that combine with a fixed mass of the
other element are in ratios of small whole numbers. Dalton’s theory explains the law
of multiple proportions quite simply: Different compounds made up of the same
elements differ in the number of atoms of each kind that combine. For example,
carbon forms two stable compounds with oxygen, namely, carbon monoxide and
carbon dioxide. Modern measurement techniques indicate that one atom of carbon
combines with one atom of oxygen in carbon monoxide and with two atoms of
oxygen in carbon dioxide. Thus, the ratio of oxygen in carbon monoxide to oxygen
in carbon dioxide is 1:2. This result is consistent with the law of multiple proportions (Figure 2.2).
Dalton’s fourth hypothesis is another way of stating the law of conservation of
mass,‡ which is that matter can be neither created nor destroyed. Because matter is
made of atoms that are unchanged in a chemical reaction, it follows that mass must
be conserved as well. Dalton’s brilliant insight into the nature of matter was the main
stimulus for the rapid progress of chemistry during the nineteenth century.
Review of Concepts
The atoms of elements A (blue) and B (orange) form two compounds shown here.
Do these compounds obey the law of multiple proportions?
2.2 The Structure of the Atom
On the basis of Dalton’s atomic theory, we can define an atom as the basic unit of
an element that can enter into chemical combination. Dalton imagined an atom that
was both extremely small and indivisible. However, a series of investigations that
began in the 1850s and extended into the twentieth century clearly demonstrated
that atoms actually possess internal structure; that is, they are made up of even
smaller particles, which are called subatomic particles. This research led to the
discovery of three such particles—electrons, protons, and neutrons.
†
Joseph Louis Proust (1754–1826). French chemist. Proust was the first person to isolate sugar from grapes.
‡
According to Albert Einstein, mass and energy are alternate aspects of a single entity called mass-energy.
Chemical reactions usually involve a gain or loss of heat and other forms of energy. Thus, when energy is lost
in a reaction, for example, mass is also lost. Except for nuclear reactions (see Chapter 19), however, changes of
mass in chemical reactions are too small to detect. Therefore, for all practical purposes mass is conserved.
2.2 The Structure of the Atom
A
(−)
Cathode ray
Anode
(+)
Cathode
(−)
B
C
41
Figure 2.3 A cathode ray tube
with an electric field perpendicular
to the direction of the cathode
rays and an external magnetic
field. The symbols N and S denote
the north and south poles of the
magnet. The cathode rays will
strike the end of the tube at A in
the presence of a magnetic field,
at C in the presence of an electric
field, and at B when there are no
external fields present or when
the effects of the electric field and
magnetic field cancel each other.
Evacuated tube
(+)
Fluorescent screen
Magnet
The Electron
In the 1890s, many scientists became caught up in the study of radiation, the emission
and transmission of energy through space in the form of waves. Information gained
from this research contributed greatly to our understanding of atomic structure. One
device used to investigate this phenomenon was a cathode ray tube, the forerunner of
the television tube (Figure 2.3). It is a glass tube from which most of the air has been
evacuated. When the two metal plates are connected to a high-voltage source, the
negatively charged plate, called the cathode, emits an invisible ray. The cathode ray
is drawn to the positively charged plate, called the anode, where it passes through a
hole and continues traveling to the other end of the tube. When the ray strikes the
specially coated surface, it produces a strong fluorescence, or bright light.
In some experiments, two electrically charged plates and a magnet were added to the
outside of the cathode ray tube (see Figure 2.3). When the magnetic field is on and the
electric field is off, the cathode ray strikes point A. When only the electric field is on,
the ray strikes point C. When both the magnetic and the electric fields are off or when they
are both on but balanced so that they cancel each other’s influence, the ray strikes point B.
According to electromagnetic theory, a moving charged body behaves like a magnet and can
interact with electric and magnetic fields through which it passes. Because the cathode ray
is attracted by the plate bearing positive charges and repelled by the plate bearing negative
charges, it must consist of negatively charged particles. We know these negatively charged
particles as electrons. Figure 2.4 shows the effect of a bar magnet on the cathode ray.
(a)
(b)
Animation
Cathode Ray Tube
Electrons are normally associated with
atoms. However, they can also be studied
individually.
(c)
Figure 2.4 (a) A cathode ray produced in a discharge tube traveling from the cathode (left) to the anode (right). The ray itself is invisible,
but the fluorescence of a zinc sulfide coating on the glass causes it to appear green. (b) The cathode ray is bent downward when a bar
magnet is brought toward it. (c) When the polarity of the magnet is reversed, the ray bends in the opposite direction.
42
Chapter 2
■
Atoms, Molecules, and Ions
Figure 2.5 Schematic diagram
of Millikan’s oil drop experiment.
Atomizer
Fine mist of
oil particles
Electrically
charged plates
X ray source to produce
charge on oil droplet
(+)
Viewing microscope
(–)
Animation
Millikan Oil Drop
An English physicist, J. J. Thomson,† used a cathode ray tube and his knowledge
of electromagnetic theory to determine the ratio of electric charge to the mass of an
individual electron. The number he came up with was 21.76 3 108 C/g, where C
stands for coulomb, which is the unit of electric charge. Thereafter, in a series of
experiments carried out between 1908 and 1917, R. A. Millikan‡ succeeded in measuring the charge of the electron with great precision. His work proved that the charge
on each electron was exactly the same. In his experiment, Millikan examined the
motion of single tiny drops of oil that picked up static charge from ions in the air.
He suspended the charged drops in air by applying an electric field and followed their
motions through a microscope (Figure 2.5). Using his knowledge of electrostatics,
Millikan found the charge of an electron to be 21.6022 3 10219 C. From these data
he calculated the mass of an electron:
charge
charge/mass
21.6022 3 10219 C
5
21.76 3 108 C/g
5 9.10 3 10228 g
mass of an electron 5
This is an exceedingly small mass.
Radioactivity
In 1895 the German physicist Wilhelm Röntgen§ noticed that cathode rays caused
glass and metals to emit very unusual rays. This highly energetic radiation penetrated
matter, darkened covered photographic plates, and caused a variety of substances to
†
Joseph John Thomson (1856–1940). British physicist who received the Nobel Prize in Physics in 1906 for
discovering the electron.
‡
Robert Andrews Millikan (1868–1953). American physicist who was awarded the Nobel Prize in Physics
in 1923 for determining the charge of the electron.
§
Wilhelm Konrad Röntgen (1845–1923). German physicist who received the Nobel Prize in Physics in 1901
for the discovery of X rays.
43
2.2 The Structure of the Atom
–
α
Lead block
γ
Figure 2.6 Three types of rays
emitted by radioactive elements.
β rays consist of negatively
charged particles (electrons) and
are therefore attracted by the
positively charged plate. The
opposite holds true for α rays—
they are positively charged and
are drawn to the negatively
charged plate. Because γ rays
have no charges, their path is
unaffected by an external
electric field.
β
+
Radioactive substance
fluoresce. Because these rays could not be deflected by a magnet, they could not
contain charged particles as cathode rays do. Röntgen called them X rays because
their nature was not known.
Not long after Röntgen’s discovery, Antoine Becquerel,† a professor of physics
in Paris, began to study the fluorescent properties of substances. Purely by accident,
he found that exposing thickly wrapped photographic plates to a certain uranium
compound caused them to darken, even without the stimulation of cathode rays. Like
X rays, the rays from the uranium compound were highly energetic and could not be
deflected by a magnet, but they differed from X rays because they arose spontaneously. One of Becquerel’s students, Marie Curie,‡ suggested the name radioactivity to
describe this spontaneous emission of particles and/or radiation. Since then, any
element that spontaneously emits radiation is said to be radioactive.
Three types of rays are produced by the decay, or breakdown, of radioactive
substances such as uranium. Two of the three are deflected by oppositely charged
metal plates (Figure 2.6). Alpha (α) rays consist of positively charged particles, called
α particles, and therefore are deflected by the positively charged plate. Beta (β) rays,
or β particles, are electrons and are deflected by the negatively charged plate. The
third type of radioactive radiation consists of high-energy rays called gamma (γ) rays.
Like X rays, γ rays have no charge and are not affected by an external field.
Animation
Alpha, Beta, and Gamma Rays
Positive charge spread
over the entire sphere
The Proton and the Nucleus
By the early 1900s, two features of atoms had become clear: They contain electrons,
and they are electrically neutral. To maintain electric neutrality, an atom must contain
an equal number of positive and negative charges. Therefore, Thomson proposed that
an atom could be thought of as a uniform, positive sphere of matter in which electrons
are embedded like raisins in a cake (Figure 2.7). This so-called “plum-pudding” model
was the accepted theory for a number of years.
†
Antoine Henri Becquerel (1852–1908). French physicist who was awarded the Nobel Prize in Physics in
1903 for discovering radioactivity in uranium.
‡
Marie (Marya Sklodowska) Curie (1867–1934). Polish-born chemist and physicist. In 1903 she and her
French husband, Pierre Curie, were awarded the Nobel Prize in Physics for their work on radioactivity. In
1911, she again received the Nobel prize, this time in chemistry, for her work on the radioactive elements
radium and polonium. She is one of only three people to have received two Nobel prizes in science. Despite
her great contribution to science, her nomination to the French Academy of Sciences in 1911 was rejected
by one vote because she was a woman! Her daughter Irene, and son-in-law Frederic Joliot-Curie, shared
the Nobel Prize in Chemistry in 1935.
–
–
–
–
–
–
–
Figure 2.7 Thomson’s model of
the atom, sometimes described as
the “plum-pudding” model, after a
traditional English dessert
containing raisins. The electrons
are embedded in a uniform,
positively charged sphere.
44
Chapter 2
■
Atoms, Molecules, and Ions
Figure 2.8 (a) Rutherford’s
experimental design for
measuring the scattering of α
particles by a piece of gold foil.
Most of the α particles passed
through the gold foil with little or
no deflection. A few were
deflected at wide angles.
Occasionally an α particle was
turned back. (b) Magnified view of
α particles passing through and
being deflected by nuclei.
Gold foil
a –Particle
emitter
Slit
Detecting screen
(a)
Animation
α-Particle Scattering
Animation
Rutherford’s Experiment
A common non-SI unit for atomic length
is the angstrom (Å; 1 Å 5 100 pm).
(b)
In 1910 the New Zealand physicist Ernest Rutherford,† who had studied with
Thomson at Cambridge University, decided to use α particles to probe the structure
of atoms. Together with his associate Hans Geiger‡ and an undergraduate named
Ernest Marsden,§ Rutherford carried out a series of experiments using very thin foils
of gold and other metals as targets for α particles from a radioactive source (Figure
2.8). They observed that the majority of particles penetrated the foil either undeflected or with only a slight deflection. But every now and then an α particle was
scattered (or deflected) at a large angle. In some instances, an α particle actually
bounced back in the direction from which it had come! This was a most surprising
finding, for in Thomson’s model the positive charge of the atom was so diffuse that
the positive α particles should have passed through the foil with very little deflection.
To quote Rutherford’s initial reaction when told of this discovery: “It was as incredible as if you had fired a 15-inch shell at a piece of tissue paper and it came back
and hit you.”
Rutherford was later able to explain the results of the α-scattering experiment in
terms of a new model for the atom. According to Rutherford, most of the atom must
be empty space. This explains why the majority of α particles passed through the gold
foil with little or no deflection. The atom’s positive charges, Rutherford proposed, are
all concentrated in the nucleus, which is a dense central core within the atom. Whenever an α particle came close to a nucleus in the scattering experiment, it experienced
a large repulsive force and therefore a large deflection. Moreover, an α particle traveling directly toward a nucleus would be completely repelled and its direction would
be reversed.
The positively charged particles in the nucleus are called protons. In separate
experiments, it was found that each proton carries the same quantity of charge as an
electron and has a mass of 1.67262 3 10224 g—about 1840 times the mass of the
oppositely charged electron.
At this stage of investigation, scientists perceived the atom as follows: The mass
of a nucleus constitutes most of the mass of the entire atom, but the nucleus occupies
only about 1/1013 of the volume of the atom. We express atomic (and molecular)
dimensions in terms of the SI unit called the picometer (pm), where
1 pm 5 1 3 10212 m
†
Ernest Rutherford (1871–1937). New Zealand physicist. Rutherford did most of his work in England
(Manchester and Cambridge Universities). He received the Nobel Prize in Chemistry in 1908 for his investigations into the structure of the atomic nucleus. His often-quoted comment to his students was that “all
science is either physics or stamp-collecting.”
‡
Johannes Hans Wilhelm Geiger (1882–1945). German physicist. Geiger’s work focused on the structure
of the atomic nucleus and on radioactivity. He invented a device for measuring radiation that is now commonly called the Geiger counter.
§
Ernest Marsden (1889–1970). English physicist. It is gratifying to know that at times an undergraduate
can assist in winning a Nobel prize. Marsden went on to contribute significantly to the development of
science in New Zealand.
2.2 The Structure of the Atom
A typical atomic radius is about 100 pm, whereas the radius of an atomic nucleus is
only about 5 3 1023 pm. You can appreciate the relative sizes of an atom and its
nucleus by imagining that if an atom were the size of a sports stadium, the volume
of its nucleus would be comparable to that of a small marble. Although the protons
are confined to the nucleus of the atom, the electrons are conceived of as being spread
out about the nucleus at some distance from it.
The concept of atomic radius is useful experimentally, but we should not infer
that atoms have well-defined boundaries or surfaces. We will learn later that the outer
regions of atoms are relatively “fuzzy.”
The Neutron
45
If the size of an atom were
expanded to that of this sports
stadium, the size of the nucleus
would be that of a marble.
Rutherford’s model of atomic structure left one major problem unsolved. It was
known that hydrogen, the simplest atom, contains only one proton and that the
helium atom contains two protons. Therefore, the ratio of the mass of a helium
atom to that of a hydrogen atom should be 2:1. (Because electrons are much
lighter than protons, their contribution to atomic mass can be ignored.) In reality,
however, the ratio is 4:1. Rutherford and others postulated that there must be
another type of subatomic particle in the atomic nucleus; the proof was provided
by another English physicist, James Chadwick,† in 1932. When Chadwick bombarded a thin sheet of beryllium with α particles, a very high-energy radiation
similar to γ rays was emitted by the metal. Later experiments showed that the
rays actually consisted of a third type of subatomic particles, which Chadwick
named neutrons, because they proved to be electrically neutral particles having
a mass slightly greater than that of protons. The mystery of the mass ratio could
now be explained. In the helium nucleus there are two protons and two neutrons,
but in the hydrogen nucleus there is only one proton and no neutrons; therefore,
the ratio is 4:1.
Figure 2.9 shows the location of the elementary particles (protons, neutrons,
and electrons) in an atom. There are other subatomic particles, but the electron, the
†
James Chadwick (1891–1972). British physicist. In 1935 he received the Nobel Prize in Physics for
proving the existence of neutrons.
Figure 2.9 The protons and
neutrons of an atom are packed in
an extremely small nucleus.
Electrons are shown as “clouds”
around the nucleus.
Proton
Neutron
46
Chapter 2
■
Atoms, Molecules, and Ions
Table 2.1
Mass and Charge of Subatomic Particles
Charge
Particle
Electron*
Proton
Neutron
Mass (g)
Coulomb
228
Charge Unit
219
9.10938 3 10
1.67262 3 10224
1.67493 3 10224
21.6022 3 10
11.6022 3 10219
0
21
11
0
*More refined measurements have given us a more accurate value of an electron’s mass than Millikan’s.
proton, and the neutron are the three fundamental components of the atom that are
important in chemistry. Table 2.1 shows the masses and charges of these three
elementary particles.
2.3 Atomic Number, Mass Number, and Isotopes
Protons and neutrons are collectively
called nucleons.
All atoms can be identified by the number of protons and neutrons they contain. The
atomic number (Z) is the number of protons in the nucleus of each atom of an element. In a neutral atom the number of protons is equal to the number of electrons,
so the atomic number also indicates the number of electrons present in the atom. The
chemical identity of an atom can be determined solely from its atomic number. For
example, the atomic number of fluorine is 9. This means that each fluorine atom has
9 protons and 9 electrons. Or, viewed another way, every atom in the universe that
contains 9 protons is correctly named “fluorine.”
The mass number (A) is the total number of neutrons and protons present in the
nucleus of an atom of an element. Except for the most common form of hydrogen,
which has one proton and no neutrons, all atomic nuclei contain both protons and
neutrons. In general, the mass number is given by
mass number 5 number of protons 1 number of neutrons
5 atomic number 1 number of neutrons
(2.1)
The number of neutrons in an atom is equal to the difference between the mass number
and the atomic number, or (A 2 Z). For example, if the mass number of a particular
boron atom is 12 and the atomic number is 5 (indicating 5 protons in the nucleus), then
the number of neutrons is 12 2 5 5 7. Note that all three quantities (atomic number,
number of neutrons, and mass number) must be positive integers, or whole numbers.
Atoms of a given element do not all have the same mass. Most elements have two
or more isotopes, atoms that have the same atomic number but different mass numbers.
For example, there are three isotopes of hydrogen. One, simply known as hydrogen, has
one proton and no neutrons. The deuterium isotope contains one proton and one neutron,
and tritium has one proton and two neutrons. The accepted way to denote the atomic
number and mass number of an atom of an element (X) is as follows:
mass number 8n
atomic number 8n
A
ZX
Thus, for the isotopes of hydrogen, we write
1
1H
1
1H
2
1H
3
1H
2
1H
3
1H
hydrogen deuterium tritium
2.3 Atomic Number, Mass Number, and Isotopes
As another example, consider two common isotopes of uranium with mass numbers
of 235 and 238, respectively:
235
92U
238
92U
The first isotope is used in nuclear reactors and atomic bombs, whereas the second
isotope lacks the properties necessary for these applications. With the exception of
hydrogen, which has different names for each of its isotopes, isotopes of elements
are identified by their mass numbers. Thus, the preceding two isotopes are called
uranium-235 (pronounced “uranium two thirty-five”) and uranium-238 (pronounced
“uranium two thirty-eight”).
The chemical properties of an element are determined primarily by the protons
and electrons in its atoms; neutrons do not take part in chemical changes under normal conditions. Therefore, isotopes of the same element have similar chemistries,
forming the same types of compounds and displaying similar reactivities.
Example 2.1 shows how to calculate the number of protons, neutrons, and electrons using atomic numbers and mass numbers.
Example 2.1
Give the number of protons, neutrons, and electrons in each of the following species:
22
17
(a) 20
O, and (d) carbon-14.
11Na, (b) 11Na, (c)
Strategy Recall that the superscript denotes the mass number (A) and the subscript
denotes the atomic number (Z). Mass number is always greater than atomic number.
(The only exception is 11H, where the mass number is equal to the atomic number.) In
a case where no subscript is shown, as in parts (c) and (d), the atomic number can be
deduced from the element symbol or name. To determine the number of electrons,
remember that because atoms are electrically neutral, the number of electrons is equal
to the number of protons.
Solution
(a) The atomic number is 11, so there are 11 protons. The mass number is 20, so
the number of neutrons is 20 2 11 5 9. The number of electrons is the same as the
number of protons; that is, 11.
(b) The atomic number is the same as that in (a), or 11. The mass number is 22, so
the number of neutrons is 22 2 11 5 11. The number of electrons is 11. Note that
the species in (a) and (b) are chemically similar isotopes of sodium.
(c) The atomic number of O (oxygen) is 8, so there are 8 protons. The mass number is
17, so there are 17 2 8 5 9 neutrons. There are 8 electrons.
(d) Carbon-14 can also be represented as 14C. The atomic number of carbon is 6, so
there are 14 2 6 5 8 neutrons. The number of electrons is 6.
Practice Exercise How many protons, neutrons, and electrons are in the following
isotope of copper:
63
Cu?
Review of Concepts
(a) What is the atomic number of an element if one of its isotopes has
117 neutrons and a mass number of 195?
(b) Which of the following two symbols provides more information?
17
O or 8O.
Similar problems: 2.15, 2.16.
47
48
Chapter 2
■
Atoms, Molecules, and Ions
2.4 The Periodic Table
More than half of the elements known today were discovered between 1800 and 1900.
During this period, chemists noted that many elements show strong similarities to one
another. Recognition of periodic regularities in physical and chemical behavior and
the need to organize the large volume of available information about the structure and
properties of elemental substances led to the development of the periodic table, a
chart in which elements having similar chemical and physical properties are grouped
together. Figure 2.10 shows the modern periodic table in which the elements are
arranged by atomic number (shown above the element symbol) in horizontal rows
called periods and in vertical columns known as groups or families, according to
similarities in their chemical properties. Note that elements 113–118 have recently
been synthesized, although they have not yet been named.
The elements can be divided into three categories—metals, nonmetals, and
metalloids. A metal is a good conductor of heat and electricity while a nonmetal is usually a poor conductor of heat and electricity. A metalloid has properties that are intermediate between those of metals and nonmetals. Figure 2.10
shows that the majority of known elements are metals; only 17 elements are
nonmetals, and 8 elements are metalloids. From left to right across any period,
the physical and chemical properties of the elements change gradually from metallic to nonmetallic.
1
1A
1
H
18
8A
2
2A
13
3A
14
4A
15
5A
16
6A
17
7A
2
He
3
4
5
6
7
8
9
10
Li
Be
B
C
N
O
F
Ne
13
14
15
16
17
18
Al
Si
P
S
Cl
Ar
11
12
Na
Mg
3
3B
4
4B
5
5B
6
6B
7
7B
8
9
8B
10
11
1B
12
2B
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
K
Ca
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
Ga
Ge
As
Se
Br
Kr
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
Rb
Sr
Y
Zr
Nb
Mo
Tc
Ru
Rh
Pd
Ag
Cd
In
Sn
Sb
Te
I
Xe
55
56
57
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
Cs
Ba
La
Hf
Ta
W
Re
Os
Ir
Pt
Au
Hg
Tl
Pb
Bi
Po
At
Rn
113
114
115
116
117
118
87
88
89
104
105
106
107
108
109
110
111
112
Fr
Ra
Ac
Rf
Db
Sg
Bh
Hs
Mt
Ds
Rg
Cn
Fl
Lv
Metals
58
59
60
61
62
63
64
65
66
67
68
69
70
71
Ce
Pr
Nd
Pm
Sm
Eu
Gd
Tb
Dy
Ho
Er
Tm
Yb
Lu
Metalloids
90
91
92
93
94
95
96
97
98
99
100
101
102
103
Th
Pa
U
Np
Pu
Am
Cm
Bk
Cf
Es
Fm
Md
No
Lr
Nonmetals
Figure 2.10 The modern periodic table. The elements are arranged according to the atomic numbers above their symbols. With the
exception of hydrogen (H), nonmetals appear at the far right of the table. The two rows of metals beneath the main body of the table are
conventionally set apart to keep the table from being too wide. Actually, cerium (Ce) should follow lanthanum (La), and thorium (Th) should
come right after actinium (Ac). The 1–18 group designation has been recommended by the International Union of Pure and Applied
Chemistry (IUPAC) but is not yet in wide use. In this text, we use the standard U.S. notation for group numbers (1A–8A and 1B–8B). No
names have yet been assigned to elements 113, 115, 117, and 118.
CHEMISTRY in Action
Distribution of Elements on Earth and in Living Systems
T
elements, we should keep in mind that (1) the elements are not
evenly distributed throughout Earth’s crust, and (2) most elements occur in combined forms. These facts provide the basis
for most methods of obtaining pure elements from their compounds, as we will see in later chapters.
The accompanying table lists the essential elements in the
human body. Of special interest are the trace elements, such as
iron (Fe), copper (Cu), zinc (Zn), iodine (I), and cobalt (Co),
which together make up about 0.1 percent of the body’s mass.
These elements are necessary for biological functions such as
growth, transport of oxygen for metabolism, and defense
against disease. There is a delicate balance in the amounts of
these elements in our bodies. Too much or too little over an
extended period of time can lead to serious illness, retardation,
or even death.
he majority of elements are naturally occurring. How are
these elements distributed on Earth, and which are essential
to living systems?
Earth’s crust extends from the surface to a depth of about
40 km (about 25 mi). Because of technical difficulties, scientists
have not been able to study the inner portions of Earth as easily
as the crust. Nevertheless, it is believed that there is a solid core
consisting mostly of iron at the center of Earth. Surrounding the
core is a layer called the mantle, which consists of hot fluid containing iron, carbon, silicon, and sulfur.
Of the 83 elements that are found in nature, 12 make up
99.7 percent of Earth’s crust by mass. They are, in decreasing
order of natural abundance, oxygen (O), silicon (Si), aluminum
(Al), iron (Fe), calcium (Ca), magnesium (Mg), sodium (Na),
potassium (K), titanium (Ti), hydrogen (H), phosphorus (P),
and manganese (Mn). In discussing the natural abundance of the
Essential Elements in the Human Body
Mantle
Crust
Element
Percent by Mass*
Oxygen
Carbon
Hydrogen
Nitrogen
Calcium
Phosphorus
Potassium
Sulfur
Chlorine
Core
2900 km 3480 km
Structure of Earth’s interior.
Element
65
18
10
3
1.6
1.2
0.2
0.2
0.2
Percent by Mass*
Sodium
Magnesium
Iron
Cobalt
Copper
Zinc
Iodine
Selenium
Fluorine
0.1
0.05
,0.05
,0.05
,0.05
,0.05
,0.05
,0.01
,0.01
*Percent by mass gives the mass of the element in grams present in a 100-g sample.
All others 5.3%
Magnesium 2.8%
Calcium 4.7%
Oxygen
45.5%
Iron 6.2%
Silicon
27.2%
(a)
Aluminum 8.3%
Oxygen
65%
Carbon
18%
All others 1.2%
Phosphorus 1.2%
Calcium 1.6%
Nitrogen 3%
Hydrogen 10%
(b)
(a) Natural abundance of the elements in percent by mass. For example, oxygen’s abundance is 45.5 percent.
This means that in a 100-g sample of Earth’s crust there are, on the average, 45.5 g of the element oxygen.
(b) Abundance of elements in the human body in percent by mass.
49
50
Chapter 2
■
Atoms, Molecules, and Ions
Elements are often referred to collectively by their periodic table group number
(Group 1A, Group 2A, and so on). However, for convenience, some element groups
have been given special names. The Group 1A elements (Li, Na, K, Rb, Cs, and Fr)
are called alkali metals, and the Group 2A elements (Be, Mg, Ca, Sr, Ba, and Ra)
are called alkaline earth metals. Elements in Group 7A (F, Cl, Br, I, and At) are
known as halogens, and elements in Group 8A (He, Ne, Ar, Kr, Xe, and Rn) are called
noble gases, or rare gases.
The periodic table is a handy tool that correlates the properties of the elements
in a systematic way and helps us to make predictions about chemical behavior. We
will take a closer look at this keystone of chemistry in Chapter 8.
The Chemistry in Action essay on p. 49 describes the distribution of the elements
on Earth and in the human body.
Review of Concepts
In viewing the periodic table, do chemical properties change more markedly
across a period or down a group?
2.5 Molecules and Ions
Of all the elements, only the six noble gases in Group 8A of the periodic table (He,
Ne, Ar, Kr, Xe, and Rn) exist in nature as single atoms. For this reason, they are
called monatomic (meaning a single atom) gases. Most matter is composed of molecules or ions formed by atoms.
Molecules
We will discuss the nature of chemical
bonds in Chapters 9 and 10.
1A
H 2A
8A
3A 4A 5A 6A 7A
N O F
Cl
Br
I
Elements that exist as diatomic
molecules.
A molecule is an aggregate of at least two atoms in a definite arrangement held
together by chemical forces (also called chemical bonds). A molecule may contain
atoms of the same element or atoms of two or more elements joined in a fixed ratio,
in accordance with the law of definite proportions stated in Section 2.1. Thus, a molecule is not necessarily a compound, which, by definition, is made up of two or more
elements (see Section 1.4). Hydrogen gas, for example, is a pure element, but it
consists of molecules made up of two H atoms each. Water, on the other hand, is a
molecular compound that contains hydrogen and oxygen in a ratio of two H atoms
and one O atom. Like atoms, molecules are electrically neutral.
The hydrogen molecule, symbolized as H2, is called a diatomic molecule because
it contains only two atoms. Other elements that normally exist as diatomic molecules
are nitrogen (N2) and oxygen (O2), as well as the Group 7A elements—fluorine (F2),
chlorine (Cl2), bromine (Br2), and iodine (I2). Of course, a diatomic molecule can
contain atoms of different elements. Examples are hydrogen chloride (HCl) and carbon monoxide (CO).
The vast majority of molecules contain more than two atoms. They can be atoms
of the same element, as in ozone (O3), which is made up of three atoms of oxygen,
or they can be combinations of two or more different elements. Molecules containing
more than two atoms are called polyatomic molecules. Like ozone, water (H2O) and
ammonia (NH3) are polyatomic molecules.
Ions
An ion is an atom or a group of atoms that has a net positive or negative charge.
The number of positively charged protons in the nucleus of an atom remains the same
during ordinary chemical changes (called chemical reactions), but negatively charged
51
2.5 Molecules and Ions
electrons may be lost or gained. The loss of one or more electrons from a neutral
atom results in a cation, an ion with a net positive charge. For example, a sodium
atom (Na) can readily lose an electron to become a sodium cation, which is represented by Na1:
In Chapter 8 we will see why atoms of
different elements gain (or lose) a
specific number of electrons.
Na1 Ion
11 protons
10 electrons
Na Atom
11 protons
11 electrons
On the other hand, an anion is an ion whose net charge is negative due to an increase
in the number of electrons. A chlorine atom (Cl), for instance, can gain an electron
to become the chloride ion Cl2:
Cl2 Ion
17 protons
18 electrons
Cl Atom
17 protons
17 electrons
Sodium chloride (NaCl), ordinary table salt, is called an ionic compound because it
is formed from cations and anions.
An atom can lose or gain more than one electron. Examples of ions formed by
the loss or gain of more than one electron are Mg21, Fe31, S22, and N32. These ions,
as well as Na1 and Cl2, are called monatomic ions because they contain only one
atom. Figure 2.11 shows the charges of a number of monatomic ions. With very few
exceptions, metals tend to form cations and nonmetals form anions.
In addition, two or more atoms can combine to form an ion that has a net
positive or net negative charge. Polyatomic ions such as OH2 (hydroxide ion),
CN2 (cyanide ion), and NH14 (ammonium ion) are ions containing more than
one atom.
Review of Concepts
(a) What does S8 signify? How does it differ from 8S?
(b) Determine the number of protons and electrons for the following ions:
(a) P32 and (b) Ti41.
1
1A
18
8A
2
2A
13
3A
Li+
17
7A
C4–
N3–
O2–
F–
P3–
S2–
Cl–
Se2–
Br–
Te2–
I–
8
9
8B
10
11
1B
12
2B
Cr 2+
Cr 3+
Mn2+
Mn3+
Fe2+
Fe3+
Co2+
Co3+
Ni2+
Ni3+
Cu+
Cu2+
Zn2+
Sr2+
Ag+
Cd2+
Sn2+
Sn4+
Ba2+
Au+
Au3+
Hg2+
2
Hg2+
Pb2+
Pb4+
K+
Ca2+
Rb+
Cs+
5
5B
16
6A
7
7B
Mg2+
4
4B
15
5A
6
6B
Na+
3
3B
14
4A
Al3+
Figure 2.11 Common monatomic ions arranged according to their positions in the periodic table. Note that the Hg221 ion contains two atoms.
52
Chapter 2
■
Atoms, Molecules, and Ions
2.6 Chemical Formulas
Chemists use chemical formulas to express the composition of molecules and
ionic compounds in terms of chemical symbols. By composition we mean not
only the elements present but also the ratios in which the atoms are combined.
Here we are concerned with two types of formulas: molecular formulas and
empirical formulas.
Molecular Formulas
A molecular formula shows the exact number of atoms of each element in the smallest unit of a substance. In our discussion of molecules, each example was given with
its molecular formula in parentheses. Thus, H2 is the molecular formula for hydrogen,
O2 is oxygen, O3 is ozone, and H2O is water. The subscript numeral indicates the
number of atoms of an element present. There is no subscript for O in H2O because
there is only one atom of oxygen in a molecule of water, and so the number “one”
is omitted from the formula. Note that oxygen (O2) and ozone (O3) are allotropes of
oxygen. An allotrope is one of two or more distinct forms of an element. Two allotropic forms of the element carbon—diamond and graphite—are dramatically different
not only in properties but also in their relative cost.
Molecular Models
Molecules are too small for us to observe directly. An effective means of visualizing
them is by the use of molecular models. Two standard types of molecular models are
currently in use: ball-and-stick models and space-filling models (Figure 2.12). In balland-stick model kits, the atoms are wooden or plastic balls with holes in them. Sticks
or springs are used to represent chemical bonds. The angles they form between atoms
approximate the bond angles in actual molecules. With the exception of the H atom,
the balls are all the same size and each type of atom is represented by a specific color.
In space-filling models, atoms are represented by truncated balls held together by snap
See back endpaper for color codes
for atoms.
Molecular
formula
Structural
formula
Hydrogen
Water
Ammonia
Methane
H2
H2O
NH3
CH4
H±N±H
W
H
H
W
H±C±H
W
H
H±H
H±O±H
Ball-and-stick
model
Space-filling
model
Figure 2.12 Molecular and structural formulas and molecular models of four common molecules.
53
2.6 Chemical Formulas
fasteners, so that the bonds are not visible. The balls are proportional in size to atoms.
The first step toward building a molecular model is writing the structural formula,
which shows how atoms are bonded to one another in a molecule. For example, it is
known that each of the two H atoms is bonded to an O atom in the water molecule.
Therefore, the structural formula of water is H¬O¬H. A line connecting the two
atomic symbols represents a chemical bond.
Ball-and-stick models show the three-dimensional arrangement of atoms clearly,
and they are fairly easy to construct. However, the balls are not proportional to the
size of atoms. Furthermore, the sticks greatly exaggerate the space between atoms in
a molecule. Space-filling models are more accurate because they show the variation
in atomic size. Their drawbacks are that they are time-consuming to put together and
they do not show the three-dimensional positions of atoms very well. Molecular modeling software can also be used to create ball-and-stick and space-filling models. We
will use both models extensively in this text.
Empirical Formulas
The molecular formula of hydrogen peroxide, a substance used as an antiseptic and
as a bleaching agent for textiles and hair, is H2O2. This formula indicates that each
hydrogen peroxide molecule consists of two hydrogen atoms and two oxygen atoms.
The ratio of hydrogen to oxygen atoms in this molecule is 2:2 or 1:1. The empirical
formula of hydrogen peroxide is HO. Thus, the empirical formula tells us which elements are present and the simplest whole-number ratio of their atoms, but not necessarily the actual number of atoms in a given molecule. As another example, consider
the compound hydrazine (N2H4), which is used as a rocket fuel. The empirical formula of hydrazine is NH2. Although the ratio of nitrogen to hydrogen is 1:2 in both
the molecular formula (N2H4) and the empirical formula (NH2), only the molecular
formula tells us the actual number of N atoms (two) and H atoms (four) present in a
hydrazine molecule.
Empirical formulas are the simplest chemical formulas; they are written by reducing the subscripts in the molecular formulas to the smallest possible whole numbers.
Molecular formulas are the true formulas of molecules. If we know the molecular
formula, we also know the empirical formula, but the reverse is not true. Why, then,
do chemists bother with empirical formulas? As we will see in Chapter 3, when chemists analyze an unknown compound, the first step is usually the determination of the
compound’s empirical formula. With additional information, it is possible to deduce
the molecular formula.
For many molecules, the molecular formula and the empirical formula are one
and the same. Some examples are water (H2O), ammonia (NH3), carbon dioxide
(CO2), and methane (CH4).
Examples 2.2 and 2.3 deal with writing molecular formulas from molecular models and writing empirical formulas from molecular formulas.
H2O2
The word “empirical” means “derived from
experiment.” As we will see in Chapter 3,
empirical formulas are determined
experimentally.
C
N
H
Methylamine
Cl
Example 2.2
Write the molecular formula of methylamine, a colorless gas used in the production of
pharmaceuticals and pesticides, from its ball-and-stick model, shown in the margin.
Solution Refer to the labels (also see back end papers). There are five H atoms, one
C atom, and one N atom. Therefore, the molecular formula is CH5N. However, the
standard way of writing the molecular formula for methylamine is CH3NH2 because it
shows how the atoms are joined in the molecule.
Practice Exercise Write the molecular formula of chloroform, which is used as a
solvent and a cleaning agent. The ball-and-stick model of chloroform is shown in
the margin.
H
C
Chloroform
Similar problems: 2.47, 2.48.
54
Chapter 2
■
Atoms, Molecules, and Ions
Example 2.3
Write the empirical formulas for the following molecules: (a) diborane (B2H6), used in
rocket propellants; (b) dimethyl fumarate (C8H12O4), a substance used to treat psoriasis,
a skin disease; and (c) vanillin (C8H8O3), a flavoring agent used in foods and beverages.
Strategy Recall that to write the empirical formula, the subscripts in the molecular
formula must be converted to the smallest possible whole numbers.
Solution
Similar problems: 2.45, 2.46.
(a) There are two boron atoms and six hydrogen atoms in diborane. Dividing the
subscripts by 2, we obtain the empirical formula BH3.
(b) In dimethyl fumarate there are 8 carbon atoms, 12 hydrogen atoms, and 4 oxygen
atoms. Dividing the subscripts by 4, we obtain the empirical formula C2H3O. Note
that if we had divided the subscripts by 2, we would have obtained the formula
C4H6O2. Although the ratio of carbon to hydrogen to oxygen atoms in C4H6O2 is the
same as that in C2H3O (2:3:1), C4H6O2 is not the simplest formula because its
subscripts are not in the smallest whole-number ratio.
(c) Because the subscripts in C8H8O3 are already the smallest possible whole numbers,
the empirical formula for vanillin is the same as its molecular formula.
Practice Exercise Write the empirical formula for caffeine (C8H10N4O2), a stimulant
found in tea and coffee.
Formula of Ionic Compounds
Sodium metal reacting with chlorine
gas to form sodium chloride.
(a)
The formulas of ionic compounds are usually the same as their empirical formulas
because ionic compounds do not consist of discrete molecular units. For example, a
solid sample of sodium chloride (NaCl) consists of equal numbers of Na1 and
Cl2 ions arranged in a three-dimensional network (Figure 2.13). In such a compound
there is a 1:1 ratio of cations to anions so that the compound is electrically neutral.
As you can see in Figure 2.13, no Na1 ion in NaCl is associated with just one particular Cl2 ion. In fact, each Na1 ion is equally held by six surrounding Cl2 ions and
vice versa. Thus, NaCl is the empirical formula for sodium chloride. In other ionic
compounds, the actual structure may be different, but the arrangement of cations and
anions is such that the compounds are all electrically neutral. Note that the charges
on the cation and anion are not shown in the formula for an ionic compound.
For ionic compounds to be electrically neutral, the sum of the charges on the
cation and anion in each formula unit must be zero. If the charges on the cation and
anion are numerically different, we apply the following rule to make the formula
(b)
(c)
Figure 2.13 (a) Structure of solid NaCl. (b) In reality, the cations are in contact with the anions. In both (a) and (b), the smaller spheres
represent Na1 ions and the larger spheres, Cl2 ions. (c) Crystals of NaCl.
2.6 Chemical Formulas
electrically neutral: The subscript of the cation is numerically equal to the charge on
the anion, and the subscript of the anion is numerically equal to the charge on the
cation. If the charges are numerically equal, then no subscripts are necessary. This
rule follows from the fact that because the formulas of ionic compounds are usually
empirical formulas, the subscripts must always be reduced to the smallest ratios. Let
us consider some examples.
• Potassium Bromide. The potassium cation K1 and the bromine anion Br2 combine to form the ionic compound potassium bromide. The sum of the charges is
11 1 (21) 5 0, so no subscripts are necessary. The formula is KBr.
• Zinc Iodide. The zinc cation Zn21 and the iodine anion I2 combine to form zinc
iodide. The sum of the charges of one Zn21 ion and one I2 ion is 12 1 (21) 5
11. To make the charges add up to zero we multiply the 21 charge of the anion
by 2 and add the subscript “2” to the symbol for iodine. Therefore the formula
for zinc iodide is ZnI2.
• Aluminum Oxide. The cation is Al31 and the oxygen anion is O22. The following diagram helps us determine the subscripts for the compound formed by the
cation and the anion:
Al 3 1
55
Refer to Figure 2.11 for charges of cations
and anions.
O22
Al2 O3
The sum of the charges is 2(13) 1 3(22) 5 0. Thus, the formula for aluminum
oxide is Al2O3.
Note that in each of the three examples,
the subscripts are in the smallest ratios.
Example 2.4
Magnesium nitride is used to prepare Borazon, a very hard compound employed in
cutting tools and machine parts. Write the formula of magnesium nitride, containing the
Mg21 and N32 ions.
Strategy Our guide for writing formulas for ionic compounds is electrical neutrality;
that is, the total charge on the cation(s) must be equal to the total charge on the
anion(s). Because the charges on the Mg21 and N32 ions are not equal, we know the
formula cannot be MgN. Instead, we write the formula as MgxNy, where x and y are
subscripts to be determined.
Solution To satisfy electrical neutrality, the following relationship must hold:
(12)x 1 (23)y 5 0
When magnesium burns in air, it
forms both magnesium oxide and
magnesium nitride.
Solving, we obtain xyy 5 3y2. Setting x 5 3 and y 5 2, we write
Mg 2 1
N
3 2
Mg3 N2
Check The subscripts are reduced to the smallest whole-number ratio of the atoms
because the chemical formula of an ionic compound is usually its empirical formula.
Practice Exercise Write the formulas of the following ionic compounds: (a) chromium
sulfate (containing the Cr31 and SO 422 ions) and (b) titanium oxide (containing the Ti41
and O22 ions).
Similar problems: 2.43, 2.44.
56
Chapter 2
■
Atoms, Molecules, and Ions
Review of Concepts
Match each of the diagrams shown here with the following ionic compounds:
Al2O3, LiH, Na2S, Mg(NO3)2. (Green spheres represent cations and red spheres
represent anions.)
(a)
(b)
(c)
(d)
2.7 Naming Compounds
For names and symbols of the elements,
see front endpapers.
When chemistry was a young science and the number of known compounds was
small, it was possible to memorize their names. Many of the names were derived
from their physical appearance, properties, origin, or application—for example,
milk of magnesia, laughing gas, limestone, caustic soda, lye, washing soda, and
baking soda.
Today the number of known compounds is well over 66 million. Fortunately, it
is not necessary to memorize their names. Over the years chemists have devised a
clear system for naming chemical substances. The rules are accepted worldwide,
facilitating communication among chemists and providing a useful way of labeling
an overwhelming variety of substances. Mastering these rules now will prove beneficial almost immediately as we proceed with our study of chemistry.
To begin our discussion of chemical nomenclature, the naming of chemical compounds, we must first distinguish between inorganic and organic compounds. Organic
compounds contain carbon, usually in combination with elements such as hydrogen,
oxygen, nitrogen, and sulfur. All other compounds are classified as inorganic compounds. For convenience, some carbon-containing compounds, such as carbon monoxide (CO), carbon dioxide (CO2), carbon disulfide (CS2), compounds containing the
cyanide group (CN2), and carbonate (CO322) and bicarbonate (HCO32) groups are
considered to be inorganic compounds. Section 2.8 gives a brief introduction to
organic compounds.
To organize and simplify our venture into naming compounds, we can divide
inorganic compounds into four categories: ionic compounds, molecular compounds,
acids and bases, and hydrates.
Ionic Compounds
1A
8A
2A
Li
Na Mg
K Ca
Rb Sr
Cs Ba
3A 4A 5A 6A 7A
N O F
Al
S Cl
Br
I
The most reactive metals (green)
and the most reactive nonmetals
(blue) combine to form ionic
compounds.
Animation
Formation of an Ionic Compound
In Section 2.5 we learned that ionic compounds are made up of cations (positive ions)
and anions (negative ions). With the important exception of the ammonium ion, NH1
4,
all cations of interest to us are derived from metal atoms. Metal cations take their
names from the elements. For example,
Element
Na
sodium
K
potassium
Mg
magnesium
Al
aluminum
Name of Cation
1
Na
K1
Mg21
Al31
sodium ion (or sodium cation)
potassium ion (or potassium cation)
magnesium ion (or magnesium cation)
aluminum ion (or aluminum cation)
Many ionic compounds are binary compounds, or compounds formed from just
two elements. For binary compounds, the first element named is the metal cation,
followed by the nonmetallic anion. Thus, NaCl is sodium chloride. The anion is
named by taking the first part of the element name (chlorine) and adding “-ide.”
57
2.7 Naming Compounds
Table 2.2
The “-ide” Nomenclature of Some Common Monatomic Anions
According to Their Positions in the Periodic Table
Group 4A
Group 5A
42
C carbide (C )*
Si silicide (Si42)
Group 6A
32
N nitride (N )
P phosphide (P32)
Group 7A
22
O oxide (O )
S sulfide (S22)
Se selenide (Se22)
Te telluride (Te22)
F fluoride (F2)
Cl chloride (Cl2)
Br bromide (Br2)
I iodide (I2)
*The word “carbide” is also used for the anion C22
2 .
Potassium bromide (KBr), zinc iodide (ZnI2), and aluminum oxide (Al2O3) are also
binary compounds. Table 2.2 shows the “-ide” nomenclature of some common monatomic anions according to their positions in the periodic table.
The “-ide” ending is also used for certain anion groups containing different elements, such as hydroxide (OH2) and cyanide (CN2). Thus, the compounds LiOH and
KCN are named lithium hydroxide and potassium cyanide, respectively. These and a
number of other such ionic substances are called ternary compounds, meaning compounds consisting of three elements. Table 2.3 lists alphabetically the names of a
number of common cations and anions.
Certain metals, especially the transition metals, can form more than one type of
cation. Take iron as an example. Iron can form two cations: Fe21 and Fe31. An older
nomenclature system that is still in limited use assigns the ending “-ous” to the
cation with fewer positive charges and the ending “-ic” to the cation with more
positive charges:
Fe21
Fe31
ferrous ion
ferric ion
3B 4B 5B 6B 7B
8B
1B 2B
The transition metals are the
elements in Groups 1B and 3B–8B
(see Figure 2.10).
The names of the compounds that these iron ions form with chlorine would thus be
FeCl2
FeCl3
ferrous chloride
ferric chloride
This method of naming ions has some distinct limitations. First, the “-ous” and “-ic”
suffixes do not provide information regarding the actual charges of the two cations
involved. Thus, the ferric ion is Fe31, but the cation of copper named cupric has the
formula Cu21. In addition, the “-ous” and “-ic” designations provide names for only
two different elemental cations. Some metallic elements can assume three or more
different positive charges in compounds. Therefore, it has become increasingly common to designate different cations with Roman numerals. This is called the Stock†
system. In this system, the Roman numeral I indicates one positive charge, II means
two positive charges, and so on. For example, manganese (Mn) atoms can assume
several different positive charges:
Mn21: MnO
Mn31: Mn2O3
Mn41: MnO2
manganese(II) oxide
manganese(III) oxide
manganese(IV) oxide
These names are pronounced “manganese-two oxide,” “manganese-three oxide,”
and “manganese-four oxide.” Using the Stock system, we denote the ferrous ion
†
Alfred E. Stock (1876–1946). German chemist. Stock did most of his research in the synthesis and characterization of boron, beryllium, and silicon compounds. He was the first scientist to explore the dangers
of mercury poisoning.
FeCl2 (left) and FeCl3 (right).
Keep in mind that the Roman numerals
refer to the charges on the metal cations.
58
Chapter 2
■
Atoms, Molecules, and Ions
Table 2.3
Names and Formulas of Some Common Inorganic Cations
and Anions
Cation
Anion
31
aluminum (Al )
ammonium (NH14)
barium (Ba21)
cadmium (Cd21)
calcium (Ca21)
cesium (Cs1)
chromium(III) or chromic (Cr31)
cobalt(II) or cobaltous (Co21)
copper(I) or cuprous (Cu1)
copper(II) or cupric (Cu21)
hydrogen (H1)
iron(II) or ferrous (Fe21)
iron(III) or ferric (Fe31)
lead(II) or plumbous (Pb21)
lithium (Li1)
magnesium (Mg21)
manganese(II) or manganous (Mn21)
mercury(I) or mercurous (Hg221)*
mercury(II) or mercuric (Hg21)
potassium (K1)
rubidium (Rb1)
silver (Ag1)
sodium (Na1)
strontium (Sr21)
tin(II) or stannous (Sn21)
zinc (Zn21)
bromide (Br2)
carbonate (CO22
3 )
chlorate (ClO2
3)
chloride (Cl2)
chromate (CrO22
4 )
2
cyanide (CN )
dichromate (Cr2O22
7 )
dihydrogen phosphate (H2PO2
4)
fluoride (F2)
hydride (H2)
hydrogen carbonate or bicarbonate (HCO2
3)
hydrogen phosphate (HPO422)
hydrogen sulfate or bisulfate (HSO2
4)
2
hydroxide (OH )
iodide (I2)
nitrate (NO2
3)
nitride (N32)
nitrite (NO2
2)
oxide (O22)
permanganate (MnO2
4)
peroxide (O222)
phosphate (PO432)
sulfate (SO22
4 )
sulfide (S22)
sulfite (SO22
3 )
thiocyanate (SCN2)
*Mercury(I) exists as a pair as shown.
Nontransition metals such as tin (Sn) and
lead (Pb) can also form more than one
type of cations.
and the ferric ion as iron(II) and iron(III), respectively; ferrous chloride becomes
iron(II) chloride, and ferric chloride is called iron(III) chloride. In keeping with
modern practice, we will favor the Stock system of naming compounds in
this textbook.
Examples 2.5 and 2.6 illustrate how to name ionic compounds and write formulas for ionic compounds based on the information given in Figure 2.11 and Tables 2.2
and 2.3.
Example 2.5
Name the following compounds: (a) Fe(NO3)2, (b) Na2HPO4, and (c) (NH4)2SO3.
Strategy Our reference for the names of cations and anions is Table 2.3. Keep in mind
that if a metal can form cations of different charges (see Figure 2.11), we need to use
the Stock system.
(Continued)
2.7 Naming Compounds
59
Solution
(a) The nitrate ion (NO2
3 ) bears one negative charge, so the iron ion must have two
positive charges. Because iron forms both Fe21 and Fe31 ions, we need to use the
Stock system and call the compound iron(II) nitrate.
(b) The cation is Na1 and the anion is HPO22
4 (hydrogen phosphate). Because sodium
only forms one type of ion (Na1), there is no need to use sodium(I) in the name.
The compound is sodium hydrogen phosphate.
22
(c) The cation is NH1
4 (ammonium ion) and the anion is SO3 (sulfite ion). The
compound is ammonium sulfite.
Similar problems: 2.57(b), (e), (f).
Practice Exercise Name the following compounds: (a) PbO and (b) LiClO3.
Example 2.6
Write chemical formulas for the following compounds: (a) mercury(I) nitrate, (b) cesium
oxide, and (c) strontium nitride.
Strategy We refer to Table 2.3 for the formulas of cations and anions. Recall that the
Roman numerals in the Stock system provide useful information about the charges of
the cation.
Solution
(a) The Roman numeral shows that the mercury ion bears a 11 charge. According to
Table 2.3, however, the mercury(I) ion is diatomic (that is, Hg221) and the nitrate ion
is NO2
3 . Therefore, the formula is Hg2(NO3)2.
(b) Each oxide ion bears two negative charges, and each cesium ion bears one
positive charge (cesium is in Group 1A, as is sodium). Therefore, the formula
is Cs2O.
(c) Each strontium ion (Sr21) bears two positive charges, and each nitride ion (N32)
bears three negative charges. To make the sum of the charges equal zero, we must
adjust the numbers of cations and anions:
Note that the subscripts of this ionic
compound are not reduced to the
smallest ratio because the Hg(I) ion
exists as a pair or dimer.
3(12) 1 2(23) 5 0
Thus, the formula is Sr3N2.
Similar problems: 2.59(a), (b), (d), (h), (i).
Practice Exercise Write formulas for the following ionic compounds: (a) rubidium
sulfate and (b) barium hydride.
Molecular Compounds
Unlike ionic compounds, molecular compounds contain discrete molecular units. They
are usually composed of nonmetallic elements (see Figure 2.10). Many molecular
compounds are binary compounds. Naming binary molecular compounds is similar to
naming binary ionic compounds. We place the name of the first element in the formula first, and the second element is named by adding -ide to the root of the element
name. Some examples are
HCl
HBr
SiC
hydrogen chloride
hydrogen bromide
silicon carbide
60
Chapter 2
■
Atoms, Molecules, and Ions
It is quite common for one pair of elements to form several different compounds.
In these cases, confusion in naming the compounds is avoided by the use of Greek
prefixes to denote the number of atoms of each element present (Table 2.4). Consider
the following examples:
Table 2.4
Greek Prefixes Used
in Naming Molecular
Compounds
Prefix
Meaning
monoditritetrapentahexaheptaoctanonadeca-
1
2
3
4
5
6
7
8
9
10
CO
CO2
SO2
SO3
NO2
N2O4
carbon monoxide
carbon dioxide
sulfur dioxide
sulfur trioxide
nitrogen dioxide
dinitrogen tetroxide
The following guidelines are helpful in naming compounds with prefixes:
• The prefix “mono-” may be omitted for the first element. For example, PCl3 is
named phosphorus trichloride, not monophosphorus trichloride. Thus, the absence
of a prefix for the first element usually means there is only one atom of that
element present in the molecule.
• For oxides, the ending “a” in the prefix is sometimes omitted. For example, N2O4
may be called dinitrogen tetroxide rather than dinitrogen tetraoxide.
Exceptions to the use of Greek prefixes are molecular compounds containing
hydrogen. Traditionally, many of these compounds are called either by their common,
nonsystematic names or by names that do not specifically indicate the number of H
atoms present:
B2H6
CH4
SiH4
NH3
PH3
H2O
H2S
Binary compounds containing carbon
and hydrogen are organic compounds;
they do not follow the same naming
conventions. We will discuss the naming
of organic compounds in Chapter 24.
diborane
methane
silane
ammonia
phosphine
water
hydrogen sulfide
Note that even the order of writing the elements in the formulas for hydrogen compounds is irregular. In water and hydrogen sulfide, H is written first, whereas it
appears last in the other compounds.
Writing formulas for molecular compounds is usually straightforward. Thus, the
name arsenic trifluoride means that there are three F atoms and one As atom in each
molecule, and the molecular formula is AsF3. Note that the order of elements in the
formula is the same as in its name.
Example 2.7
Name the following molecular compounds: (a) PBr5 and (b) As2O5.
Strategy We refer to Table 2.4 for the prefixes used in naming molecular compounds.
Solution
Similar problems: 2.57(c), (i), ( j).
(a) Because there are five bromine atoms present, the compound is phosphorus
pentabromide.
(b) There are two arsenic atoms and five oxygen atoms present, so the compound is
diarsenic pentoxide. Note that the “a” is omitted in “penta.”
Practice Exercise Name the following molecular compounds: (a) NF3 and (b) Cl2O7.
2.7 Naming Compounds
Example 2.8
Write chemical formulas for the following molecular compounds: (a) bromine trifluoride
and (b) diboron trioxide.
Strategy We refer to Table 2.4 for the prefixes used in naming molecular compounds.
Solution
(a) Because there are three fluorine atoms and one bromine atom present, the formula
is BrF3.
(b) There are two boron atoms and three oxygen atoms present, so the formula is B2O3.
Similar problems: 2.59(g), ( j).
Practice Exercise Write chemical formulas for the following molecular compounds:
(a) sulfur tetrafluoride and (b) dinitrogen pentoxide.
Figure 2.14 summarizes the steps for naming ionic and binary molecular
compounds.
Review of Concepts
Why is it that the name for SeCl2, selenium dichloride, contains a prefix, but the
name for SrCl2, strontium chloride, does not?
Compound
Ionic
Molecular
Cation: metal or NH+4
Anion: monatomic or
polyatomic
• Binary compounds
of nonmetals
Naming
Cation has
only one charge
• Alkali metal cations
• Alkaline earth metal cations
• Ag+, Al3+, Cd2+, Zn2+
Cation has more
than one charge
• Other metal cations
Naming
Naming
• Name metal first
• If monatomic anion,
add “–ide” to the
root of the element
name
• If polyatomic anion,
use name of anion
(see Table 2.3)
• Name metal first
• Specify charge of
metal cation with
Roman numeral
in parentheses
• If monatomic anion,
add “–ide” to the
root of the element
name
• If polyatomic anion,
use name of anion
(see Table 2.3)
Figure 2.14 Steps for naming ionic and binary molecular compounds.
• Use prefixes for
both elements present
(Prefix “mono–”
usually omitted for
the first element)
• Add “–ide” to the
root of the second
element
61
62
Chapter 2
■
Atoms, Molecules, and Ions
Acids and Bases
HCl
Naming Acids
An acid can be described as a substance that yields hydrogen ions (H1) when dissolved
in water. (H1 is equivalent to one proton, and is often referred to that way.) Formulas
for acids contain one or more hydrogen atoms as well as an anionic group. Anions
whose names end in “-ide” form acids with a “hydro-” prefix and an “-ic” ending, as
shown in Table 2.5. In some cases two different names seem to be assigned to the
same chemical formula.
H3O+
HCl
HCl
hydrogen chloride
hydrochloric acid
Cl–
When dissolved in water, the HCl
molecule is converted to the
H1 and Cl– ions. The H1 ion is
associated with one or more
water molecules, and is usually
represented as H3O1.
The name assigned to the compound depends on its physical state. In the gaseous or
pure liquid state, HCl is a molecular compound called hydrogen chloride. When it is
dissolved in water, the molecules break up into H1 and Cl2 ions; in this state, the
substance is called hydrochloric acid.
Oxoacids are acids that contain hydrogen, oxygen, and another element (the central element). The formulas of oxoacids are usually written with the H first, followed
by the central element and then O. We use the following five common acids as our
references in naming oxoacids:
H2CO3
HClO3
HNO3
H3PO4
H2SO4
H
O
C
carbonic acid
chloric acid
nitric acid
phosphoric acid
sulfuric acid
Often two or more oxoacids have the same central atom but a different number of O
atoms. Starting with our reference oxoacids whose names all end with “-ic,” we use
the following rules to name these compounds.
H2CO3
H
O
N
1. Addition of one O atom to the “-ic” acid: The acid is called “per . . . -ic” acid.
Thus, adding an O atom to HClO3 changes chloric acid to perchloric acid,
HClO4.
2. Removal of one O atom from the “-ic” acid: The acid is called “-ous” acid. Thus,
nitric acid, HNO3, becomes nitrous acid, HNO2.
3. Removal of two O atoms from the “-ic” acid: The acid is called “hypo . . . -ous”
acid. Thus, when HBrO3 is converted to HBrO, the acid is called hypobromous
acid.
HNO3
Table 2.5
Note that these acids all exist as molecular
compounds in the gas phase.
Some Simple Acids
Acid
Corresponding Anion
HF (hydrofluoric acid)
HCl (hydrochloric acid)
HBr (hydrobromic acid)
HI (hydroiodic acid)
HCN (hydrocyanic acid)
H2S (hydrosulfuric acid)
F2 (fluoride)
Cl2 (chloride)
Br2 (bromide)
I2 (iodide)
CN2 (cyanide)
S22 (sulfide)
63
2.7 Naming Compounds
Removal of
Oxoacid
Oxoanion
all H+ ions
per– –ic acid
Figure 2.15 Naming oxoacids
and oxoanions.
per– –ate
+[O]
–ate
Reference “–ic” acid
–[O]
–ite
“–ous” acid
–[O]
hypo– –ous acid
hypo– –ite
The rules for naming oxoanions, anions of oxoacids, are as follows:
1. When all the H ions are removed from the “-ic” acid, the anion’s name ends with
“-ate.” For example, the anion CO322 derived from H2CO3 is called carbonate.
2. When all the H ions are removed from the “-ous” acid, the anion’s name ends
with “-ite.” Thus, the anion ClO2
2 derived from HClO2 is called chlorite.
3. The names of anions in which one or more but not all the hydrogen ions have
been removed must indicate the number of H ions present. For example, consider
the anions derived from phosphoric acid:
H3PO4
H2PO42
HPO422
PO432
phosphoric acid
dihydrogen phosphate
hydrogen phosphate
phosphate
Note that we usually omit the prefix “mono-” when there is only one H in the anion.
Figure 2.15 summarizes the nomenclature for the oxoacids and oxoanions, and Table 2.6
gives the names of the oxoacids and oxoanions that contain chlorine.
Example 2.9 deals with the nomenclature for an oxoacid and an oxoanion.
Table 2.6
Names of Oxoacids and Oxoanions That Contain Chlorine
Acid
Corresponding Anion
HClO4 (perchloric acid)
HClO3 (chloric acid)
HClO2 (chlorous acid)
HClO (hypochlorous acid)
ClO2
4
ClO2
3
ClO2
2
ClO2
(perchlorate)
(chlorate)
(chlorite)
(hypochlorite)
O
P
H3PO4
H
64
Chapter 2
■
Atoms, Molecules, and Ions
Example 2.9
Name the following oxoacid and oxoanions: (a) H2SO3, a very unstable acid formed
when SO2(g) reacts with water, (b) H2AsO2
4, once used to control ticks and lice on
livestock, and (c) SeO232, used to manufacture colorless glass. H3AsO4 is arsenic acid,
and H2SeO4 is selenic acid.
Strategy We refer to Figure 2.15 and Table 2.6 for the conventions used in naming
oxoacids and oxoanions.
Solution
Similar problems: 2.58(f).
(a) We start with our reference acid, sulfuric acid (H2SO4). Because H2SO3 has one
fewer O atom, it is called sulfurous acid.
(b) Because H3AsO4 is arsenic acid, the AsO432 ion is named arsenate. The H2AsO2
4 anion is
formed by adding two H1 ions to AsO432, so H2AsO2
is
called
dihydrogen
arsenate.
4
(c) The parent acid is H2SeO3. Because the acid has one fewer O atom than selenic
acid (H2SeO4), it is called selenous acid. Therefore, the SeO322 anion derived from
H2SeO3 is called selenite.
Practice Exercise Name the following oxoacid and oxoanion: (a) HBrO and (b) HSO24 .
Naming Bases
A base can be described as a substance that yields hydroxide ions (OH2) when dissolved in water. Some examples are
NaOH
KOH
Ba(OH)2
sodium hydroxide
potassium hydroxide
barium hydroxide
Ammonia (NH3), a molecular compound in the gaseous or pure liquid state, is
also classified as a common base. At first glance this may seem to be an exception
to the definition of a base. But note that as long as a substance yields hydroxide ions
when dissolved in water, it need not contain hydroxide ions in its structure to be
considered a base. In fact, when ammonia dissolves in water, NH3 reacts partially
with water to yield NH41 and OH2 ions. Thus, it is properly classified as a base.
Review of Concepts
Why is the following question ambiguous: What is the name of HF? What
additional information is needed to answer the question?
Hydrates
Hydrates are compounds that have a specific number of water molecules attached to
them. For example, in its normal state, each unit of copper(II) sulfate has five water
molecules associated with it. The systematic name for this compound is copper(II)
sulfate pentahydrate, and its formula is written as CuSO4 ? 5H2O. The water molecules
can be driven off by heating. When this occurs, the resulting compound is CuSO4, which
is sometimes called anhydrous copper(II) sulfate; “anhydrous” means that the compound
no longer has water molecules associated with it (Figure 2.16). Some other hydrates are
BaCl2 ? 2H2O
LiCl ? H2O
MgSO4 ? 7H2O
Sr(NO3)2 ? 4H2O
barium chloride dihydrate
lithium chloride monohydrate
magnesium sulfate heptahydrate
strontium nitrate tetrahydrate
2.8 Introduction to Organic Compounds
65
Figure 2.16 CuSO4 ? 5H2O (left)
is blue; CuSO4 (right) is white.
Table 2.7
Common and Systematic Names of Some Compounds
Formula
Common Name
Systematic Name
H2O
NH3
CO2
NaCl
N2O
CaCO3
CaO
Ca(OH)2
NaHCO3
Na2CO3 ? 10H2O
MgSO4 ? 7H2O
Mg(OH)2
CaSO4 ? 2H2O
Water
Ammonia
Dry ice
Table salt
Laughing gas
Marble, chalk, limestone
Quicklime
Slaked lime
Baking soda
Washing soda
Epsom salt
Milk of magnesia
Gypsum
Dihydrogen monoxide
Trihydrogen nitride
Solid carbon dioxide
Sodium chloride
Dinitrogen monoxide
Calcium carbonate
Calcium oxide
Calcium hydroxide
Sodium hydrogen carbonate
Sodium carbonate decahydrate
Magnesium sulfate heptahydrate
Magnesium hydroxide
Calcium sulfate dihydrate
Familiar Inorganic Compounds
Some compounds are better known by their common names than by their systematic
chemical names. Familiar examples are listed in Table 2.7.
CH3OH
2.8 Introduction to Organic Compounds
The simplest type of organic compounds is the hydrocarbons, which contain only
carbon and hydrogen atoms. The hydrocarbons are used as fuels for domestic and
industrial heating, for generating electricity and powering internal combustion engines,
and as starting materials for the chemical industry. One class of hydrocarbons is called
the alkanes. Table 2.8 shows the names, formulas, and molecular models of the first
10 straight-chain alkanes, in which the carbon chains have no branches. Note that all
the names end with -ane. Starting with C5H12, we use the Greek prefixes in Table 2.4
to indicate the number of carbon atoms present.
The chemistry of organic compounds is largely determined by the functional
groups, which consist of one or a few atoms bonded in a specific way. For example,
when an H atom in methane is replaced by a hydroxyl group (¬OH), an amino group
(¬NH2), and a carboxyl group (¬COOH), the following molecules are generated:
H
H
H
C
OH
H
Methanol
H
C
NH2
H
Methylamine
H
H
O
C
C
CH3NH2
OH
H
Acetic acid
CH3COOH
66
Chapter 2
■
Atoms, Molecules, and Ions
Table 2.8
The First Ten Straight-Chain Alkanes
Name
Formula
Methane
CH4
Ethane
C2H6
Propane
C3H8
Butane
C4H10
Pentane
C5H12
Hexane
C6H14
Heptane
C7H16
Octane
C8H18
Nonane
C9H20
Decane
C10H22
Molecular Model
The chemical properties of these molecules can be predicted based on the reactivity
of the functional groups. Although the nomenclature of the major classes of organic
compounds and their properties in terms of the functional groups will not be discussed
until Chapter 24, we will frequently use organic compounds as examples to illustrate
chemical bonding, acid-base reactions, and other properties throughout the book.
Review of Concepts
How many different molecules can you generate by replacing one H atom with a
hydroxyl group (¬OH) in butane (see Table 2.8)?
Key Words
67
Key Equation
mass number 5 number of protons 1 number of neutrons
5 atomic number 1 number of neutrons (2.1)
Summary of Facts & Concepts
1. Modern chemistry began with Dalton’s atomic theory,
which states that all matter is composed of tiny, indivisible particles called atoms; that all atoms of the
same element are identical; that compounds contain
atoms of different elements combined in wholenumber ratios; and that atoms are neither created nor
destroyed in chemical reactions (the law of conservation of mass).
2. Atoms of constituent elements in a particular compound
are always combined in the same proportions by mass
(law of definite proportions). When two elements can
combine to form more than one type of compound, the
masses of one element that combine with a fixed mass
of the other element are in a ratio of small whole numbers (law of multiple proportions).
3. An atom consists of a very dense central nucleus
containing protons and neutrons, with electrons
moving about the nucleus at a relatively large distance from it.
4. Protons are positively charged, neutrons have no charge,
and electrons are negatively charged. Protons and neutrons have roughly the same mass, which is about 1840
times greater than the mass of an electron.
5. The atomic number of an element is the number of
protons in the nucleus of an atom of the element; it
6.
7.
8.
9.
10.
11.
determines the identity of an element. The mass
number is the sum of the number of protons and the
number of neutrons in the nucleus.
Isotopes are atoms of the same element with the same
number of protons but different numbers of neutrons.
Chemical formulas combine the symbols for the constituent elements with whole-number subscripts to
show the type and number of atoms contained in the
smallest unit of a compound.
The molecular formula conveys the specific number
and type of atoms combined in each molecule of a compound. The empirical formula shows the simplest ratios
of the atoms combined in a molecule.
Chemical compounds are either molecular compounds
(in which the smallest units are discrete, individual molecules) or ionic compounds, which are made of cations
and anions.
The names of many inorganic compounds can be
deduced from a set of simple rules. The formulas can be
written from the names of the compounds.
Organic compounds contain carbon and elements like
hydrogen, oxygen, and nitrogen. Hydrocarbon is the
simplest type of organic compound.
Key Words
Acid, p. 62
Alkali metals, p. 50
Alkaline earth metals, p. 50
Allotrope, p. 52
Alpha (α) particles, p. 43
Alpha (α) rays, p. 43
Anion, p. 51
Atom, p. 40
Atomic number (Z), p. 46
Base, p. 64
Beta (β) particles, p. 43
Beta (β) rays, p. 43
Binary compound, p. 56
Cation, p. 51
Chemical formula, p. 52
Diatomic molecule, p. 50
Electron, p. 41
Empirical formula, p. 53
Families, p. 48
Gamma (γ) rays, p. 43
Groups, p. 48
Halogens, p. 50
Hydrate, p. 64
Inorganic
compounds, p. 56
Ion, p. 50
Ionic compound, p. 51
Isotope, p. 46
Law of conservation of
mass, p. 40
Law of definite
proportions, p. 40
Law of multiple
proportions, p. 40
Mass number (A), p. 46
Metal, p. 48
Metalloid, p. 48
Molecular formula, p. 52
Molecule, p. 50
Monatomic ion, p. 51
Neutron, p. 45
Noble gases, p. 50
Nonmetal, p. 48
Nucleus, p. 44
Organic compound, p. 56
Oxoacid, p. 62
Oxoanion, p. 63
Periods, p. 48
Periodic table, p. 48
Polyatomic ion, p. 51
Polyatomic molecule, p. 50
Proton, p. 44
Radiation, p. 41
Radioactivity, p. 43
Structural formula, p. 53
Ternary compound, p. 57
68
Chapter 2
■
Atoms, Molecules, and Ions
Questions & Problems
• Problems available in Connect Plus
• 2.16
Red numbered problems solved in Student Solutions Manual
Structure of the Atom
Review Questions
2.1
2.2
2.3
2.4
2.5
2.6
Define the following terms: (a) α particle, (b) β particle, (c) γ ray, (d) X ray.
Name the types of radiation known to be emitted by
radioactive elements.
Compare the properties of the following: α particles, cathode rays, protons, neutrons, electrons.
What is meant by the term “fundamental particle”?
Describe the contributions of the following scientists
to our knowledge of atomic structure: J. J. Thomson,
R. A. Millikan, Ernest Rutherford, James Chadwick.
Describe the experimental basis for believing that
the nucleus occupies a very small fraction of the
volume of the atom.
Problems
• 2.7
2.8
The diameter of a helium atom is about 1 3 102 pm.
Suppose that we could line up helium atoms side by
side in contact with one another. Approximately
how many atoms would it take to make the distance
from end to end 1 cm?
Roughly speaking, the radius of an atom is about
10,000 times greater than that of its nucleus. If an
atom were magnified so that the radius of its nucleus
became 2.0 cm, about the size of a marble, what would
be the radius of the atom in miles? (1 mi 5 1609 m.)
15
33
63
84
130
186
202
7N, 16S, 29Cu, 38Sr, 56Ba, 74 W, 80Hg
• 2.17
• 2.18
2.9
2.10
2.11
2.12
Use the helium-4 isotope to define atomic number
and mass number. Why does a knowledge of atomic
number enable us to deduce the number of electrons
present in an atom?
Why do all atoms of an element have the same
atomic number, although they may have different
mass numbers?
What do we call atoms of the same elements with
different mass numbers?
Explain the meaning of each term in the symbol ZAX.
Problems
• 2.13
• 2.14
• 2.15
Write the appropriate symbol for each of the following isotopes: (a) Z 5 11, A 5 23; (b) Z 5 28,
A 5 64.
Write the appropriate symbol for each of the following isotopes: (a) Z 5 74, A 5 186; (b) Z 5 80,
A 5 201.
The Periodic Table
Review Questions
2.19
2.20
2.21
• 2.22
What is the periodic table, and what is its significance in the study of chemistry?
State two differences between a metal and a
nonmetal.
Write the names and symbols for four elements
in each of the following categories: (a) nonmetal,
(b) metal, (c) metalloid.
Define, with two examples, the following terms:
(a) alkali metals, (b) alkaline earth metals, (c) halogens, (d) noble gases.
Problems
2.23
2.24
Atomic Number, Mass Number, and Isotopes
Review Questions
Indicate the number of protons, neutrons, and electrons in each of the following species:
2.25
• 2.26
Elements whose names end with -ium are usually
metals; sodium is one example. Identify a nonmetal
whose name also ends with -ium.
Describe the changes in properties (from metals to
nonmetals or from nonmetals to metals) as we move
(a) down a periodic group and (b) across the periodic table from left to right.
Consult a handbook of chemical and physical data
(ask your instructor where you can locate a copy
of the handbook) to find (a) two metals less dense
than water, (b) two metals more dense than mercury, (c) the densest known solid metallic element, (d) the densest known solid nonmetallic
element.
Group the following elements in pairs that you
would expect to show similar chemical properties:
K, F, P, Na, Cl, and N.
Molecules and Ions
Review Questions
What is the mass number of an iron atom that has 28
neutrons?
Calculate the number of neutrons in 239Pu.
For each of the following species, determine the number
of protons and the number of neutrons in the nucleus:
25
48
79
195
3
4
24
2He, 2He, 12Mg, 12Mg, 22Ti, 35Br, 78Pt
2.27
2.28
2.29
What is the difference between an atom and a
molecule?
What are allotropes? Give an example. How are
allotropes different from isotopes?
Describe the two commonly used molecular
models.
69
Questions & Problems
2.30
Give an example of each of the following: (a) a monatomic cation, (b) a monatomic anion, (c) a polyatomic
cation, (d) a polyatomic anion.
Chemical Formulas
Review Questions
2.37
Problems
• 2.31
Which of the following diagrams represent diatomic
molecules, polyatomic molecules, molecules that
are not compounds, molecules that are compounds,
or an elemental form of the substance?
2.38
2.39
2.40
2.41
2.42
What does a chemical formula represent? What is
the ratio of the atoms in the following molecular
formulas? (a) NO, (b) NCl3, (c) N2O4, (d) P4O6
Define molecular formula and empirical formula.
What are the similarities and differences between
the empirical formula and molecular formula of a
compound?
Give an example of a case in which two molecules
have different molecular formulas but the same empirical formula.
What does P4 signify? How does it differ from 4P?
What is an ionic compound? How is electrical neutrality maintained in an ionic compound?
Explain why the chemical formulas of ionic compounds are usually the same as their empirical
formulas.
Problems
(a)
• 2.32
(b)
(c)
Which of the following diagrams represent diatomic
molecules, polyatomic molecules, molecules that
are not compounds, molecules that are compounds,
or an elemental form of the substance?
• 2.43
• 2.44
• 2.45
• 2.46
(a)
• 2.33
• 2.34
• 2.35
• 2.36
(b)
(c)
Identify the following as elements or compounds:
NH3, N2, S8, NO, CO, CO2, H2, SO2.
Give two examples of each of the following: (a) a
diatomic molecule containing atoms of the same
element, (b) a diatomic molecule containing atoms of different elements, (c) a polyatomic molecule containing atoms of the same element, (d) a
polyatomic molecule containing atoms of different elements.
Give the number of protons and electrons in each of
the following common ions: Na1, Ca21, Al31, Fe21,
I2, F2, S22, O22, and N32.
Give the number of protons and electrons in each of
the following common ions: K1, Mg21, Fe31, Br2,
Mn21, C42, Cu21.
• 2.47
Write the formulas for the following ionic compounds: (a) sodium oxide, (b) iron sulfide (containing the Fe21 ion), (c) cobalt sulfate (containing the
Co31 and SO422 ions), and (d) barium fluoride.
(Hint: See Figure 2.11.)
Write the formulas for the following ionic compounds: (a) copper bromide (containing the Cu1
ion), (b) manganese oxide (containing the Mn31
ion), (c) mercury iodide (containing the Hg221 ion),
and (d) magnesium phosphate (containing the PO432
ion). (Hint: See Figure 2.11.)
What are the empirical formulas of the following
compounds? (a) C2N2, (b) C6H6, (c) C9H20, (d) P4O10,
(e) B2H6
What are the empirical formulas of the following
compounds? (a) Al2Br6, (b) Na2S2O4, (c) N2O5,
(d) K2Cr2O7
Write the molecular formula of glycine, an amino
acid present in proteins. The color codes are: black
(carbon), blue (nitrogen), red (oxygen), and gray
(hydrogen).
H
O
C
N
70
• 2.48
Chapter 2
■
Atoms, Molecules, and Ions
Write the molecular formula of ethanol. The color
codes are: black (carbon), red (oxygen), and gray
(hydrogen).
H
O
2.61
Sulfur (S) and fluorine (F) form several different
compounds. One of them, SF6, contains 3.55 g of F
for every gram of S. Use the law of multiple proportions to determine n, which represents the number of
F atoms in SFn, given that it contains 2.37 g of F for
every gram of S.
Name the following compounds.
2.62
C
O
N
• 2.49
• 2.50
Which of the following compounds are likely to be
ionic? Which are likely to be molecular? SiCl4, LiF,
BaCl2, B2H6, KCl, C2H4
Which of the following compounds are likely to be
ionic? Which are likely to be molecular? CH4, NaBr,
BaF2, CCl4, ICl, CsCl, NF3
Naming Inorganic Compounds
2.52
2.53
2.54
2.55
2.56
What is the difference between inorganic compounds and organic compounds?
What are the four major categories of inorganic
compounds?
Give an example each for a binary compound and a
ternary compound.
What is the Stock system? What are its advantages
over the older system of naming cations?
Explain why the formula HCl can represent two different chemical systems.
Define the following terms: acids, bases, oxoacids,
oxoanions, and hydrates.
2.63
Pair the following species that contain the same
number of electrons: Ar, Sn41, F2, Fe31, P32, V,
Ag1, N32.
Write the correct symbols for the atoms that contain:
(a) 25 protons, 25 electrons, and 27 neutrons; (b) 10
protons, 10 electrons, and 12 neutrons; (c) 47 protons, 47 electrons, and 60 neutrons; (d) 53 protons,
53 electrons, and 74 neutrons; (e) 94 protons,
94 electrons, and 145 neutrons.
2.64
Additional Problems
2.65
A sample of a uranium compound is found to be losing mass gradually. Explain what is happening to
the sample.
In which one of the following pairs do the two species resemble each other most closely in chemical
properties? Explain. (a) 11H and 11H1, (b) 147N and
14 32
12
13
7N , (c) 6C and 6C.
One isotope of a metallic element has mass number
65 and 35 neutrons in the nucleus. The cation derived from the isotope has 28 electrons. Write the
symbol for this cation.
One isotope of a nonmetallic element has mass
number 127 and 74 neutrons in the nucleus. The
anion derived from the isotope has 54 electrons.
Write the symbol for this anion.
Determine the molecular and empirical formulas of
the compounds shown here. (Black spheres are
carbon and gray spheres are hydrogen.)
• 2.66
• 2.67
Problems
• 2.57
• 2.58
• 2.59
• 2.60
Name these compounds: (a) Na2CrO4, (b) K2HPO4,
(c) HBr (gas), (d) HBr (in water), (e) Li2CO3,
(f) K2Cr2O7, (g) NH4NO2, (h) PF3, (i) PF5, (j) P4O6,
(k) CdI2, (l) SrSO4, (m) Al(OH)3, (n) Na2CO3 ? 10H2O.
Name these compounds: (a) KClO, (b) Ag2CO3,
(c) FeCl2, (d) KMnO4, (e) CsClO3, (f) HIO, (g) FeO,
(h) Fe2O3, (i) TiCl4, ( j) NaH, (k) Li3N, (l) Na2O,
(m) Na2O2, (n) FeCl3 ? 6H2O.
Write the formulas for the following compounds:
(a) rubidium nitrite, (b) potassium sulfide, (c) sodium
hydrogen sulfide, (d) magnesium phosphate, (e) calcium hydrogen phosphate, (f) potassium dihydrogen
phosphate, (g) iodine heptafluoride, (h) ammonium
sulfate, (i) silver perchlorate, (j) boron trichloride.
Write the formulas for the following compounds:
(a) copper(I) cyanide, (b) strontium chlorite, (c) perbromic acid, (d) hydroiodic acid, (e) disodium ammonium
phosphate, (f ) lead(II) carbonate, (g) tin(II) fluoride,
(h) tetraphosphorus decasulfide, (i) mercury(II) oxide,
( j) mercury(I) iodide, (k) selenium hexafluoride.
Br
B
Review Questions
2.51
Al
F
2.68
• 2.69
(a)
2.70
(b)
(c)
(d)
What is wrong with or ambiguous about the phrase
“four molecules of NaCl”?
71
Questions & Problems
2.71
2.72
• 2.73
The following phosphorus sulfides are known: P4S3,
P4S7, and P4S10. Do these compounds obey the law
of multiple proportions?
Which of the following are elements, which are
molecules but not compounds, which are compounds but not molecules, and which are both
compounds and molecules? (a) SO 2, (b) S 8,
(c) Cs, (d) N2O5, (e) O, (f) O2, (g) O3, (h) CH4,
(i) KBr, (j) S, (k) P4, (l) LiF
The following table gives numbers of electrons,
protons, and neutrons in atoms or ions of a number of elements. Answer the following: (a) Which
of the species are neutral? (b) Which are negatively charged? (c) Which are positively charged?
(d) What are the conventional symbols for all the
species?
Atom or Ion
of Element
A
B
C
D
E
F
G
Number of electrons
Number of protons
Number of neutrons
5
5
5
10
7
7
18
19
20
28
30
36
36
35
46
5
5
6
9
9
10
2.74
2.75
2.76
2.77
2.78
Identify the elements represented by the following
symbols and give the number of protons and neu63
107
182
trons in each case: (a) 20
10X, (b) 29X, (c) 47X, (d) 74X,
234
X,
(f)
X.
(e) 203
84
94
Each of the following pairs of elements will react to
form an ionic compound. Write the formulas and
name these compounds: (a) barium and oxygen,
(b) calcium and phosphorus, (c) aluminum and sulfur, (d) lithium and nitrogen.
Match the descriptions [(a)–(h)] with each of the
following elements: P, Cu, Kr, Sb, Cs, Al, Sr, Cl.
(a) A transition metal, (b) a nonmetal that forms a
23 ion, (c) a noble gas, (d) an alkali metal, (e) a
metal that forms a 13 ion, (f) a metalloid, (g) an
element that exists as a diatomic gas molecule,
(h) an alkaline earth metal.
Explain why anions are always larger than the atoms
from which they are derived, whereas cations are
always smaller than the atoms from which they are
derived. (Hint: Consider the electrostatic attraction
between protons and electrons.)
(a) Describe Rutherford’s experiment and how it led
to the structure of the atom. How was he able to
estimate the number of protons in a nucleus from the
scattering of the α particles? (b) Consider the 23Na
atom. Given that the radius and mass of the nucleus
are 3.04 3 10215 m and 3.82 3 10223 g, respectively,
calculate the density of the nucleus in g/cm3. The
radius of a 23Na atom is 186 pm. Calculate the density of the space occupied by the electrons in the
sodium atom. Do your results support Rutherford’s
model of an atom? [The volume of a sphere of
radius r is (4/3)πr3.]
2.79
Caffeine, shown here, is a psychoactive stimulant
drug. Write the molecular formula and empirical
formula of the compound.
H
O
N
C
2.80
Acetaminophen, shown here, is the active ingredient
in Tylenol. Write the molecular formula and empirical formula of the compound.
O
H
N
C
2.81
2.82
• 2.83
What is wrong with the chemical formula for each
of the following compounds: (a) magnesium
iodate [Mg(IO4)2], (b) phosphoric acid (H3PO3),
(c) barium sulfite (BaS), (d) ammonium bicarbonate (NH3HCO3)?
What is wrong with the names (in parentheses) for
each of the following compounds: SnCl4 (tin chloride), (b) Cu2O [copper(II) oxide], (c) Co(NO3)2
(cobalt nitrate), (d) Na2Cr2O7 (sodium chromate)?
Fill in the blanks in the following table.
54
21
26Fe
Symbol
Protons
5
Neutrons
6
Electrons
5
Net charge
2.84
• 2.85
79
86
16
117
136
18
79
23
0
(a) Which elements are most likely to form ionic
compounds? (b) Which metallic elements are
most likely to form cations with different
charges?
Write the formula of the common ion derived from
each of the following: (a) Li, (b) S, (c) I, (d) N,
(e) Al, (f) Cs, (g) Mg
72
2.86
• 2.87
2.88
• 2.89
• 2.90
2.91
2.92
• 2.93
• 2.94
2.95
2.96
2.97
• 2.98
• 2.99
Chapter 2
■
Atoms, Molecules, and Ions
Which of the following symbols provides more information about the atom: 23Na or 11Na? Explain.
Write the chemical formulas and names of binary acids and oxoacids that contain Group 7A elements. Do
the same for elements in Groups 3A, 4A, 5A, and 6A.
Of the 118 elements known, only two are liquids at
room temperature (25°C). What are they? (Hint:
One element is a familiar metal and the other element is in Group 7A.)
For the noble gases (the Group 8A elements), 42He,
20
40
84
132
10Ne, 18Ar, 36Kr, and 54Xe, (a) determine the number of protons and neutrons in the nucleus of each
atom, and (b) determine the ratio of neutrons to protons in the nucleus of each atom. Describe any general trend you discover in the way this ratio changes
with increasing atomic number.
List the elements that exist as gases at room temperature. (Hint: Most of these elements can be found
in Groups 5A, 6A, 7A, and 8A.)
The Group 1B metals, Cu, Ag, and Au, are called
coinage metals. What chemical properties make
them specially suitable for making coins and
jewelry?
The elements in Group 8A of the periodic table are
called noble gases. Can you suggest what “noble”
means in this context?
The formula for calcium oxide is CaO. What are
the formulas for magnesium oxide and strontium
oxide?
A common mineral of barium is barytes, or barium sulfate (BaSO4). Because elements in the
same periodic group have similar chemical properties, we might expect to find some radium sulfate (RaSO4) mixed with barytes since radium is
the last member of Group 2A. However, the only
source of radium compounds in nature is in uranium minerals. Why?
List five elements each that are (a) named after
places, (b) named after people, (c) named after a
color. (Hint: See Appendix 1.)
One isotope of a nonmetallic element has mass
number 77 and 43 neutrons in the nucleus. The
anion derived from the isotope has 36 electrons.
Write the symbol for this anion.
Fluorine reacts with hydrogen (H) and deuterium
(D) to form hydrogen fluoride (HF) and deuterium
fluoride (DF), where deuterium (21H) is an isotope
of hydrogen. Would a given amount of fluorine react with different masses of the two hydrogen isotopes? Does this violate the law of definite
proportion? Explain.
Predict the formula and name of a binary compound
formed from the following elements: (a) Na and H,
(b) B and O, (c) Na and S, (d) Al and F, (e) F and O,
(f ) Sr and Cl.
Identify each of the following elements: (a) a halogen whose anion contains 36 electrons, (b) a radioactive noble gas with 86 protons, (c) a Group 6A
• 2.100
element whose anion contains 36 electrons, (d) an
alkali metal cation that contains 36 electrons, (e) a
Group 4A cation that contains 80 electrons.
Write the molecular formulas for and names of the
following compounds.
S
N
F
P
Br
Cl
2.101
Show the locations of (a) alkali metals, (b) alkaline
earth metals, (c) the halogens, and (d) the noble
gases in the following outline of a periodic table.
Also draw dividing lines between metals and metalloids and between metalloids and nonmetals.
1A
8A
2A
• 2.102
3A 4A 5A 6A 7A
Fill the blanks in the following table.
Cation
Anion
Formula
Name
Magnesium bicarbonate
SrCl2
Fe31
NO2
2
Manganese(II) chlorate
SnBr4
Co21
Hg221
PO32
4
I2
Cu2CO3
Lithium nitride
31
Al
• 2.103
22
S
Some compounds are better known by their common names than by their systematic chemical
names. Give the chemical formulas of the following
substances: (a) dry ice, (b) table salt, (c) laughing
gas, (d) marble (chalk, limestone), (e) quicklime,
(f) slaked lime, (g) baking soda, (h) washing soda,
(i) gypsum, (j) milk of magnesia.
73
Questions & Problems
• 2.104
• 2.105
2.106
• 2.107
• 2.108
• 2.109
On p. 40 it was pointed out that mass and energy are
alternate aspects of a single entity called massenergy. The relationship between these two physical
quantities is Einstein’s famous equation, E 5 mc2,
where E is energy, m is mass, and c is the speed of
light. In a combustion experiment, it was found that
12.096 g of hydrogen molecules combined with
96.000 g of oxygen molecules to form water and
released 1.715 3 103 kJ of heat. Calculate the corresponding mass change in this process and comment on whether the law of conservation of mass
holds for ordinary chemical processes. (Hint: The
Einstein equation can be used to calculate the
change in mass as a result of the change in energy.
1 J 5 1 kg m2/s2 and c 5 3.00 3 108 m/s.)
Draw all possible structural formulas of the following
hydrocarbons: CH4, C2H6, C3H8, C4H10, and C5H12.
(a) Assuming nuclei are spherical in shape, show
that its radius r is proportional to the cube root of
mass number (A). (b) In general, the radius of a
nucleus is given by r 5 r0 A1/3, where r0 is a proportionality constant given by 1.2 3 10215 m. Calculate
the volume of the 73Li nucleus. (c) Given that the
radius of a Li atom is 152 pm, calculate the fraction
of the atom’s volume occupied by the nucleus. Does
your result support Rutherford’s model of an atom?
Draw two different structural formulas based on the
molecular formula C2H6O. Is the fact that you can have
more than one compound with the same molecular formula consistent with Dalton’s atomic theory?
Ethane and acetylene are two gaseous hydrocarbons. Chemical analyses show that in one sample of
ethane, 2.65 g of carbon are combined with 0.665 g
of hydrogen, and in one sample of acetylene, 4.56 g
of carbon are combined with 0.383 g of hydrogen.
(a) Are these results consistent with the law of multiple proportions? (b) Write reasonable molecular
formulas for these compounds.
A cube made of platinum (Pt) has an edge length of
1.0 cm. (a) Calculate the number of Pt atoms in the
cube. (b) Atoms are spherical in shape. Therefore,
the Pt atoms in the cube cannot fill all of the available space. If only 74 percent of the space inside
the cube is taken up by Pt atoms, calculate the
radius in picometers of a Pt atom. The density of
• 2.110
• 2.111
Pt is 21.45 g/cm3 and the mass of a single Pt atom
is 3.240 3 10222 g. [The volume of a sphere of
radius r is (4/3)πr3.]
A monatomic ion has a charge of 12. The nucleus
of the parent atom has a mass number of 55. If the
number of neutrons in the nucleus is 1.2 times that
of the number of protons, what is the name and symbol of the element?
In the following 2 3 2 crossword, each letter must
be correct four ways: horizontally, vertically, diagonally, and by itself. When the puzzle is complete, the
four spaces will contain the overlapping symbols of
10 elements. Use capital letters for each square.
There is only one correct solution.*
2
3
4
Horizontal
1–2:
3–4:
Two-letter symbol for a metal used in ancient times
Two-letter symbol for a metal that burns in air and is
found in Group 5A
Vertical
1–3:
2–4:
Two-letter symbol for a metalloid
Two-letter symbol for a metal used in U.S. coins
Single Squares
1:
2:
3:
4:
A colorful nonmetal
A colorless gaseous nonmetal
An element that makes fireworks green
An element that has medicinal uses
Diagonal
1–4:
2–3:
• 2.112
Two-letter symbol for an element used in electronics
Two-letter symbol for a metal used with Zr to make
wires for superconducting magnets
Name the following acids.
H
S
N
Cl
1
C
O
*Reproduced with permission of S. J. Cyvin of the University of Trondheim (Norway). This puzzle appeared in Chemical & Engineering News,
December 14, 1987 (p. 86) and in Chem Matters, October 1988.
74
2.113
2.114
Chapter 2
■
Atoms, Molecules, and Ions
Calculate the density of the nucleus of a 56
26 Fe atom,
given that the nuclear mass is 9.229 3 10223 g. From
your result, comment on the fact that any nucleus
containing more than one proton must have neutrons
present as well. (Hint: See Problem 2.106.)
Element X reacts with element Y to form an ionic
compound containing X41 and Y22 ions. Write a
formula for the compound and suggest in which
periodic groups these elements are likely to be
found. Name a representative compound.
2.115
Methane, ethane, and propane are shown in Table 2.8.
Show that the following data are consistent with the
law of multiple proportions.
Methane
Ethane
Propane
Mass of Carbon
in 1 g Sample
Mass of Hydrogen
in 1 g Sample
0.749 g
0.799 g
0.817 g
0.251 g
0.201 g
0.183 g
Interpreting, Modeling & Estimating
2.116
2.117
2.118
2.119
2.120
In the Rutherford scattering experiment, an α particle is heading directly toward a gold nucleus. The
particle will come to a halt when its kinetic energy is
completely converted to electrical potential energy.
When this happens, how close will the α particle
with a kinetic energy of 6.0 3 10214 J be from the
nucleus? [According to Coulomb’s law, the electrical potential energy between two charged particles
is E 5 kQ1Q2/r, where Q1 and Q2 are the charges (in
coulombs) of the α particle and the gold nucleus, r
is the distance of separation in meters, and k is a
constant equal to 9.0 3 109 kg ? m3/s2 ? C2. Joule (J)
is the unit of energy where 1 J 5 1 kg ? m2/s2.]
Estimate the relative sizes of the following species:
Li, Li1, Li2.
Compare the atomic size of the following two magnesium isotopes: 24Mg and 26Mg.
Using visible light, we humans cannot see any
object smaller than 2 3 1025 cm with an unaided
eye. Roughly how many silver atoms must be lined
up for us to see the atoms?
If the size of the nucleus of an atom were that of a
pea, how far would the electrons be (on average)
from the nucleus in meters?
2.121
2.122
Sodium and potassium are roughly equal in natural
abundance in Earth’s crust and most of their compounds are soluble. However, the composition of
seawater is much higher in sodium than potassium.
Explain.
One technique proposed for recycling plastic grocery bags is to heat them at 700°C and high pressure
to form carbon microspheres that can be used in a
number of applications. Electron microscopy shows
some representative carbon microspheres obtained
in this manner, where the scale is given in the bottom right corner of the figure. Determine the number of carbon atoms in a typical carbon microsphere.
5 μm
Answers to Practice Exercises
2.1 29 protons, 34 neutrons, and 29 electrons. 2.2 CHCl3.
2.3 C4H5N2O. 2.4 (a) Cr2(SO4)3, (b) TiO2. 2.5 (a) Lead(II)
oxide, (b) lithium chlorate. 2.6 (a) Rb2SO4, (b) BaH2.
2.7 (a) Nitrogen trifluoride, (b) dichlorine heptoxide.
2.8 (a) SF4, (b) N2O5. 2.9 (a) Hypobromous acid,
(b) hydrogen sulfate ion.
CHAPTER
3
Mass Relationships in
Chemical Reactions
Fireworks are chemical reactions noted for the
spectacular colors rather than the energy or useful
substances they produce.
CHAPTER OUTLINE
A LOOK AHEAD
3.1
3.2
Atomic Mass
3.3
3.4
3.5
Molecular Mass
We begin by studying the mass of an atom, which is based on the carbon-12
isotope scale. An atom of the carbon-12 isotope is assigned a mass of exactly
12 atomic mass units (amu). To work with the more convenient scale of
grams, we use the molar mass. The molar mass of carbon-12 has a mass
of exactly 12 grams and contains an Avogadro’s number (6.022 3 1023) of
atoms. The molar masses of other elements are also expressed in grams and
contain the same number of atoms. (3.1 and 3.2)
3.6
Experimental Determination
of Empirical Formulas
Our discussion of atomic mass leads to molecular mass, which is the sum of
the masses of the constituent atoms present. We learn that the most direct
way to determine atomic and molecular mass is by the use of a mass spectrometer. (3.3 and 3.4)
3.7
Chemical Reactions and
Chemical Equations
To continue our study of molecules and ionic compounds, we learn how to
calculate the percent composition of these species from their chemical
formulas. (3.5)
3.8
Amounts of Reactants
and Products
We will see how the empirical and molecular formulas of a compound are
determined by experiment. (3.6)
3.9
3.10
Limiting Reagents
Next, we learn how to write a chemical equation to describe the outcome of
a chemical reaction. A chemical equation must be balanced so that we have
the same number and type of atoms for the reactants, the starting materials,
and the products, the substances formed at the end of the reaction. (3.7)
Building on our knowledge of chemical equations, we then proceed to study
the mass relationships of chemical reactions. A chemical equation enables
us to use the mole method to predict the amount of product(s) formed,
knowing how much the reactant(s) was used. We will see that a reaction’s
yield depends on the amount of limiting reagent (a reactant that is used up
first) present. (3.8 and 3.9)
We will learn that the actual yield of a reaction is almost always less than
that predicted from the equation, called the theoretical yield, because of
various complications. (3.10)
Avogadro’s Number and the
Molar Mass of an Element
The Mass Spectrometer
Percent Composition
of Compounds
Reaction Yield
75
76
Chapter 3 ■ Mass Relationships in Chemical Reactions
I
n this chapter, we will consider the masses of atoms and molecules and what happens to
them when chemical changes occur. Our guide for this discussion will be the law of conservation of mass.
3.1 Atomic Mass
Section 3.4 describes a method for
determining atomic mass.
One atomic mass unit is also called
one dalton.
In this chapter, we will use what we have learned about chemical structure and formulas in studying the mass relationships of atoms and molecules. These relationships
in turn will help us to explain the composition of compounds and the ways in which
composition changes.
The mass of an atom depends on the number of electrons, protons, and neutrons
it contains. Knowledge of an atom’s mass is important in laboratory work. But atoms
are extremely small particles—even the smallest speck of dust that our unaided eyes
can detect contains as many as 1 3 1016 atoms! Clearly we cannot weigh a single
atom, but it is possible to determine the mass of one atom relative to another experimentally. The first step is to assign a value to the mass of one atom of a given element
so that it can be used as a standard.
By international agreement, atomic mass (sometimes called atomic weight) is the
mass of the atom in atomic mass units (amu). One atomic mass unit is defined as a mass
exactly equal to one-twelfth the mass of one carbon-12 atom. Carbon-12 is the carbon
isotope that has six protons and six neutrons. Setting the atomic mass of carbon-12 at
12 amu provides the standard for measuring the atomic mass of the other elements. For
example, experiments have shown that, on average, a hydrogen atom is only 8.400 percent as massive as the carbon-12 atom. Thus, if the mass of one carbon-12 atom is exactly
12 amu, the atomic mass of hydrogen must be 0.08400 3 12 amu or 1.008 amu. Similar calculations show that the atomic mass of oxygen is 16.00 amu and that of iron is
55.85 amu. Thus, although we do not know just how much an average iron atom’s mass
is, we know that it is approximately 56 times as massive as a hydrogen atom.
Average Atomic Mass
Atomic
number
6
C
12.01
12
C
98.90%
Atomic
mass
13
C
1.10%
Natural abundances of C-12 and
C-13 isotopes.
When you look up the atomic mass of carbon in a table such as the one on the inside
front cover of this book, you will find that its value is not 12.00 amu but 12.01 amu.
The reason for the difference is that most naturally occurring elements (including
carbon) have more than one isotope. This means that when we measure the atomic
mass of an element, we must generally settle for the average mass of the naturally
occurring mixture of isotopes. For example, the natural abundances of carbon-12 and
carbon-13 are 98.90 percent and 1.10 percent, respectively. The atomic mass of
carbon-13 has been determined to be 13.00335 amu. Thus, the average atomic mass
of carbon can be calculated as follows:
average atomic mass
of natural carbon 5 (0.9890)(12 amu) 1 (0.0110)(13.00335 amu)
5 12.01 amu
Note that in calculations involving percentages, we need to convert percentages to
fractions. For example, 98.90 percent becomes 98.90/100, or 0.9890. Because there
are many more carbon-12 atoms than carbon-13 atoms in naturally occurring carbon,
the average atomic mass is much closer to 12 amu than to 13 amu.
It is important to understand that when we say that the atomic mass of carbon is
12.01 amu, we are referring to the average value. If carbon atoms could be examined
individually, we would find either an atom of atomic mass exactly 12 amu or one of
13.00335 amu, but never one of 12.01 amu. Example 3.1 shows how to calculate the
average atomic mass of an element.
3.2 Avogadro’s Number and the Molar Mass of an Element
77
Example 3.1
Boron is used in the manufacture of ceramics and polymers such as fiberglass. The
atomic masses of its two stable isotopes, 105B (19.80 percent) and 115B (80.20 percent),
are 10.0129 amu and 11.0093 amu, respectively. The boron-10 isotope is also
important as a neutron-capturing agent in nuclear reactors. Calculate the average
atomic mass of boron.
Strategy Each isotope contributes to the average atomic mass based on its relative
abundance. Multiplying the mass of an isotope by its fractional abundance (not percent)
will give the contribution to the average atomic mass of that particular isotope.
Solution First the percent abundances are converted to fractions: 19.80 percent to
19.80/100 or 0.1980 and 80.20 percent to 80.20/100 or 0.8020. We find the contribution
to the average atomic mass for each isotope, and then add the contributions together to
obtain the average atomic mass.
(0.1980) (10.0129 amu) 1 (0.8020) (11.0093 amu) 5 10.8129 amu
Check The average atomic mass should be between the two isotopic masses; therefore,
the answer is reasonable. Note that because there are more 115B isotopes than 105B
isotopes, the average atomic mass is closer to 11.0093 amu than to 10.0129 amu.
63
Practice Exercise The atomic masses of the two stable isotopes of copper, 29
Cu
(69.17 percent) and 65
Cu
(30.83
percent),
are
62.9296
amu
and
64.9278
amu,
respectively.
29
Calculate the average atomic mass of copper.
The atomic masses of many elements have been accurately determined to five or
six significant figures. However, for our purposes we will normally use atomic masses
accurate only to four significant figures (see table of atomic masses inside the front
cover). For simplicity, we will omit the word “average” when we discuss the atomic
masses of the elements.
Boron and the solid-state structure
of boron.
Review of Concepts
There are two stable isotopes of iridium: 191Ir (190.96 amu) and 193Ir (192.96 amu).
If you were to randomly pick an iridium atom from a large collection of iridium
atoms, which isotope are you more likely to select?
Similar problems: 3.5, 3.6.
3.2 Avogadro’s Number and the Molar Mass
of an Element
Atomic mass units provide a relative scale for the masses of the elements. But because
atoms have such small masses, no usable scale can be devised to weigh them in
calibrated units of atomic mass units. In any real situation, we deal with macroscopic
samples containing enormous numbers of atoms. Therefore, it is convenient to have
a special unit to describe a very large number of atoms. The idea of a unit to denote
a particular number of objects is not new. For example, the pair (2 items), the dozen
(12 items), and the gross (144 items) are all familiar units. Chemists measure atoms
and molecules in moles.
In the SI system the mole (mol) is the amount of a substance that contains as
many elementary entities (atoms, molecules, or other particles) as there are atoms in
exactly 12 g (or 0.012 kg) of the carbon-12 isotope. The actual number of atoms in
12 g of carbon-12 is determined experimentally. This number is called Avogadro’s
The adjective formed from the noun
“mole” is “molar.”
78
Chapter 3 ■ Mass Relationships in Chemical Reactions
Figure 3.1 One mole each
of several common elements.
Carbon (black charcoal powder),
sulfur (yellow powder), iron
(as nails), copper wires, and
mercury (shiny liquid metal).
number (NA), in honor of the Italian scientist Amedeo Avogadro.† The currently
accepted value is
NA 5 6.0221413 3 1023
In calculations, the units of molar mass
are g/mol or kg/mol.
The molar masses of the elements are
given on the inside front cover of the book.
Generally, we round Avogadro’s number to 6.022 3 1023. Thus, just as 1 dozen
oranges contains 12 oranges, 1 mole of hydrogen atoms contains 6.022 3 1023 H
atoms. Figure 3.1 shows samples containing 1 mole each of several common elements.
The enormity of Avogadro’s number is difficult to imagine. For example, spreading
6.022 3 1023 oranges over the entire surface of Earth would produce a layer 9 mi into
space! Because atoms (and molecules) are so tiny, we need a huge number to study
them in manageable quantities.
We have seen that 1 mole of carbon-12 atoms has a mass of exactly 12 g and
contains 6.022 3 1023 atoms. This mass of carbon-12 is its molar mass (m), defined
as the mass (in grams or kilograms) of 1 mole of units (such as atoms or molecules)
of a substance. Note that the molar mass of carbon-12 (in grams) is numerically equal
to its atomic mass in amu. Likewise, the atomic mass of sodium (Na) is 22.99 amu
and its molar mass is 22.99 g; the atomic mass of phosphorus is 30.97 amu and its
molar mass is 30.97 g; and so on. If we know the atomic mass of an element, we
also know its molar mass.
Knowing the molar mass and Avogadro’s number, we can calculate the mass of
a single atom in grams. For example, we know the molar mass of carbon-12 is 12 g
and there are 6.022 3 1023 carbon-12 atoms in 1 mole of the substance; therefore,
the mass of one carbon-12 atom is given by
12 g carbon-12 atoms
6.022 3 1023 carbon-12 atoms
†
5 1.993 3 10223 g
Lorenzo Romano Amedeo Carlo Avogadro di Quaregua e di Cerreto (1776–1856). Italian mathematical
physicist. He practiced law for many years before he became interested in science. His most famous work,
now known as Avogadro’s law (see Chapter 5), was largely ignored during his lifetime, although it became
the basis for determining atomic masses in the late nineteenth century.
3.2 Avogadro’s Number and the Molar Mass of an Element
Mass of
element (m)
Number of moles
of element (n)
m /}
n}
nNA
N/NA
79
Number of atoms
of element (N)
Figure 3.2 The relationships between mass (m in grams) of an element and number of moles of an element (n) and between
number of moles of an element and number of atoms (N) of an element. m is the molar mass (g/mol) of the element and NA is
Avogadro’s number.
We can use the preceding result to determine the relationship between atomic
mass units and grams. Because the mass of every carbon-12 atom is exactly 12 amu,
the number of atomic mass units equivalent to 1 gram is
12 amu
amu
1 carbon-12 atom
5
3
gram
1 carbon-12 atom
1.993 3 10223 g
5 6.022 3 1023 amu/g
Thus,
1 g 5 6.022 3 1023 amu
and
1 amu 5 1.661 3 10224 g
This example shows that Avogadro’s number can be used to convert from the atomic
mass units to mass in grams and vice versa.
The notions of Avogadro’s number and molar mass enable us to carry out conversions between mass and moles of atoms and between moles and number of atoms
(Figure 3.2). We will employ the following conversion factors in the calculations:
1 mol X
molar mass of X
and
After some practice, you can use the
equations in Figure 3.2 in calculations:
n 5 m/m and N 5 nNA.
1 mol X
6.022 3 1023 X atoms
where X represents the symbol of an element. Using the proper conversion factors
we can convert one quantity to another, as Examples 3.2–3.4 show.
Example 3.2
Helium (He) is a valuable gas used in industry, low-temperature research, deep-sea
diving tanks, and balloons. How many moles of He atoms are in 6.46 g of He?
Strategy We are given grams of helium and asked to solve for moles of helium.
What conversion factor do we need to convert between grams and moles? Arrange the
appropriate conversion factor so that grams cancel and the unit moles is obtained for
your answer.
Solution The conversion factor needed to convert between grams and moles is the
molar mass. In the periodic table (see inside front cover) we see that the molar mass
of He is 4.003 g. This can be expressed as
1 mol He 5 4.003 g He
From this equality, we can write two conversion factors
1 mol He
4.003 g He
and
A scientific research helium balloon.
4.003 g He
1 mol He
(Continued)
80
Chapter 3 ■ Mass Relationships in Chemical Reactions
The conversion factor on the left is the correct one. Grams will cancel, leaving the unit
mol for the answer, that is,
6.46 g He 3
1 mol He
5 1.61 mol He
4.003 g He
Thus, there are 1.61 moles of He atoms in 6.46 g of He.
Similar problem: 3.15.
Check Because the given mass (6.46 g) is larger than the molar mass of He, we expect
to have more than 1 mole of He.
Practice Exercise How many moles of magnesium (Mg) are there in 87.3 g of Mg?
Example 3.3
Zinc (Zn) is a silvery metal that is used in making brass (with copper) and in plating
iron to prevent corrosion. How many grams of Zn are in 0.356 mole of Zn?
Strategy We are trying to solve for grams of zinc. What conversion factor do we need
to convert between moles and grams? Arrange the appropriate conversion factor so that
moles cancel and the unit grams are obtained for your answer.
Zinc.
Solution The conversion factor needed to convert between moles and grams is the
molar mass. In the periodic table (see inside front cover) we see the molar mass of Zn
is 65.39 g. This can be expressed as
1 mol Zn 5 65.39 g Zn
From this equality, we can write two conversion factors
1 mol Zn
65.39 g Zn
and
65.39 g Zn
1 mol Zn
The conversion factor on the right is the correct one. Moles will cancel, leaving unit of
grams for the answer. The number of grams of Zn is
0.356 mol Zn 3
65.39 g Zn
5 23.3 g Zn
1 mol Zn
Thus, there are 23.3 g of Zn in 0.356 mole of Zn.
Similar problem: 3.16.
Check Does a mass of 23.3 g for 0.356 mole of Zn seem reasonable? What is the
mass of 1 mole of Zn?
Practice Exercise Calculate the number of grams of lead (Pb) in 12.4 moles of lead.
Example 3.4
The C60 molecule is called buckminsterfullerene because its shape resembles the
geodesic domes designed by the visionary architect R. Buckminster Fuller. What is the
mass (in grams) of one C60 molecule?
Strategy The question asks for the mass of one C60 molecule. Determine the moles of
C atoms in one C60 molecule, and then use the molar mass of C to calculate the mass
of one molecule in grams.
Buckminsterfullerene (C60) or
“buckyball.”
(Continued)
3.3 Molecular Mass
Solution Because one C60 molecule contains 60 C atoms, and 1 mole of C contains
6.022 3 1023 C atoms and has a mass of 12.011 g, we can calculate the mass of one
C60 molecule as follows:
1 C60 molecule 3
12.01 g
1 mol C
60 C atoms
3
5 1.197 3 10221 g
3
1 mol C
1 C60 molecule
6.022 3 1023 C atoms
Check Because 6.022 3 1023 atoms of C have a mass 12.01 g, a molecule containing
only 60 carbon atoms should have a significantly smaller mass.
Similar problems: 3.20, 3.21.
Practice Exercise Gold atoms form small clusters containing a fixed number of
atoms. What is the mass (in grams) of one Au31 cluster?
Review of Concepts
Referring to the periodic table in the inside front cover and Figure 3.2, determine
which of the following contains the largest number of atoms: (a) 7.68 g of He,
(b) 112 g of Fe, and (c) 389 g of Hg.
3.3 Molecular Mass
If we know the atomic masses of the component atoms, we can calculate the mass of
a molecule. The molecular mass (sometimes called molecular weight) is the sum of the
atomic masses (in amu) in the molecule. For example, the molecular mass of H2O is
2(atomic mass of H) 1 atomic mass of O
or
2(1.008 amu) 1 16.00 amu 5 18.02 amu
In general, we need to multiply the atomic mass of each element by the number of
atoms of that element present in the molecule and sum over all the elements. Example 3.5 illustrates this approach.
Example 3.5
Calculate the molecular masses (in amu) of the following compounds: (a) sulfur dioxide
(SO2), a gas that is responsible for acid rain, and (b) caffeine (C8H10N4O2), a stimulant
present in tea, coffee, and cola beverages.
Strategy How do atomic masses of different elements combine to give the molecular
SO2
mass of a compound?
Solution To calculate molecular mass, we need to sum all the atomic masses in the
molecule. For each element, we multiply the atomic mass of the element by the number
of atoms of that element in the molecule. We find atomic masses in the periodic table
(inside front cover).
(a) There are two O atoms and one S atom in SO2, so that
molecular mass of SO2 5 32.07 amu 1 2(16.00 amu)
5 64.07 amu
(Continued)
81
82
Chapter 3 ■ Mass Relationships in Chemical Reactions
(b) There are eight C atoms, ten H atoms, four N atoms, and two O atoms in caffeine,
so the molecular mass of C8H10N4O2 is given by
Similar problems: 3.23, 3.24.
8(12.01 amu) 1 10(1.008 amu) 1 4(14.01 amu) 1 2(16.00 amu) 5 194.20 amu
Practice Exercise What is the molecular mass of methanol (CH4O)?
From the molecular mass we can determine the molar mass of a molecule or
compound. The molar mass of a compound (in grams) is numerically equal to its
molecular mass (in amu). For example, the molecular mass of water is 18.02 amu, so
its molar mass is 18.02 g. Note that 1 mole of water weighs 18.02 g and contains
6.022 3 1023 H2O molecules, just as 1 mole of elemental carbon contains 6.022 3 1023
carbon atoms.
As Examples 3.6 and 3.7 show, a knowledge of the molar mass enables us to
calculate the numbers of moles and individual atoms in a given quantity of a
compound.
Example 3.6
Methane (CH4) is the principal component of natural gas. How many moles of CH4 are
present in 6.07 g of CH4?
CH4
Strategy We are given grams of CH4 and asked to solve for moles of CH4. What
conversion factor do we need to convert between grams and moles? Arrange the
appropriate conversion factor so that grams cancel and the unit moles are obtained for
your answer.
Solution The conversion factor needed to convert between grams and moles is the
molar mass. First we need to calculate the molar mass of CH4, following the procedure
in Example 3.5:
molar mass of CH4 5 12.01 g 1 4(1.008 g)
5 16.04 g
Methane gas burning on a cooking
range.
Because
1 mol CH4 5 16.04 g CH4
the conversion factor we need should have grams in the denominator so that the unit g
will cancel, leaving the unit mol in the numerator:
1 mol CH4
16.04 g CH4
We now write
6.07 g CH4 3
1 mol CH4
5 0.378 mol CH4
16.04 g CH4
Thus, there is 0.378 mole of CH4 in 6.07 g of CH4.
Check Should 6.07 g of CH4 equal less than 1 mole of CH4? What is the mass of
Similar problem: 3.26.
1 mole of CH4?
Practice Exercise Calculate the number of moles of chloroform (CHCl3) in 198 g of
chloroform.
3.4 The Mass Spectrometer
83
Example 3.7
How many hydrogen atoms are present in 25.6 g of urea [(NH2)2CO], which is used as
a fertilizer, in animal feed, and in the manufacture of polymers? The molar mass of urea
is 60.06 g.
Strategy We are asked to solve for atoms of hydrogen in 25.6 g of urea. We cannot convert
directly from grams of urea to atoms of hydrogen. How should molar mass and Avogadro’s
number be used in this calculation? How many moles of H are in 1 mole of urea?
Solution To calculate the number of H atoms, we first must convert grams of urea to
moles of urea using the molar mass of urea. This part is similar to Example 3.2. The
molecular formula of urea shows there are four moles of H atoms in one mole of urea
molecule, so the mole ratio is 4:1. Finally, knowing the number of moles of H atoms,
we can calculate the number of H atoms using Avogadro’s number. We need two
conversion factors: molar mass and Avogadro’s number. We can combine these
conversions
Urea.
grams of urea ¡ moles of urea ¡ moles of H ¡ atoms of H
into one step:
25.6 g (NH2 ) 2CO 3
1 mol (NH2 ) 2CO
6.022 3 1023 H atoms
4 mol H
3
3
1 mol (NH2 ) 2CO
1 mol H
60.06 g (NH2 ) 2CO
5 1.03 3 1024 H atoms
Check Does the answer look reasonable? How many atoms of H would 60.06 g of
urea contain?
Similar problems: 3.27, 3.28.
Practice Exercise How many H atoms are in 72.5 g of isopropanol (rubbing alcohol),
C3H8O?
Finally, note that for ionic compounds like NaCl and MgO that do not contain
discrete molecular units, we use the term formula mass instead. The formula unit of
NaCl consists of one Na1 ion and one Cl2 ion. Thus, the formula mass of NaCl is
the mass of one formula unit:
formula mass of NaCl 5 22.99 amu 1 35.45 amu
5 58.44 amu
and its molar mass is 58.44 g.
Review of Concepts
Determine the molecular mass and the molar mass of citric acid, H3C6H5O7.
3.4 The Mass Spectrometer
The most direct and most accurate method for determining atomic and molecular
masses is mass spectrometry, which is depicted in Figure 3.3. In one type of a mass
spectrometer, a gaseous sample is bombarded by a stream of high-energy electrons.
Collisions between the electrons and the gaseous atoms (or molecules) produce
positive ions by dislodging an electron from each atom or molecule. These positive
Note that the combined mass of a Na1 ion
and a Cl2 ion is equal to the combined
mass of a Na atom and a Cl atom.
84
Chapter 3 ■ Mass Relationships in Chemical Reactions
Detecting
screen
Accelerating
plates
Electron
beam
Magnet
Ion beam
Sample
gas
Filament
Figure 3.3 Schematic diagram of one type of mass spectrometer.
Note that it is possible to determine the
molar mass of a compound without
knowing its chemical formula.
ions (of mass m and charge e) are accelerated by two oppositely charged plates as
they pass through the plates. The emerging ions are deflected into a circular path
by a magnet. The radius of the path depends on the charge-to-mass ratio (that is,
e/m). Ions of smaller e/m ratio trace a wider curve than those having a larger e/m
ratio, so that ions with equal charges but different masses are separated from one
another. The mass of each ion (and hence its parent atom or molecule) is determined
from the magnitude of its deflection. Eventually the ions arrive at the detector,
which registers a current for each type of ion. The amount of current generated is
directly proportional to the number of ions, so it enables us to determine the relative
abundance of isotopes.
The first mass spectrometer, developed in the 1920s by the English physicist
F. W. Aston,† was crude by today’s standards. Nevertheless, it provided indisputable
evidence of the existence of isotopes—neon-20 (atomic mass 19.9924 amu and natural abundance 90.92 percent) and neon-22 (atomic mass 21.9914 amu and natural
abundance 8.82 percent). When more sophisticated and sensitive mass spectrometers
became available, scientists were surprised to discover that neon has a third stable
isotope with an atomic mass of 20.9940 amu and natural abundance 0.257 percent
(Figure 3.4). This example illustrates how very important experimental accuracy is to
a quantitative science like chemistry. Early experiments failed to detect neon-21
because its natural abundance is just 0.257 percent. In other words, only 26 in 10,000
Ne atoms are neon-21. The masses of molecules can be determined in a similar manner by the mass spectrometer.
Review of Concepts
Explain how the mass spectrometer enables chemists to determine the average
atomic mass of chlorine, which has two stable isotopes (35Cl and 37Cl).
†
Francis William Aston (1877–1945). English chemist and physicist. He was awarded the Nobel Prize in
Chemistry in 1922 for developing the mass spectrometer.
3.5 Percent Composition of Compounds
Intensity of peaks
20
10 Ne(90.92%)
21
10 Ne(0.26%)
19
20
22
10 Ne(8.82%)
21
22
Atomic mass (amu)
23
Figure 3.4 The mass spectrum of the three isotopes of neon.
3.5 Percent Composition of Compounds
As we have seen, the formula of a compound tells us the numbers of atoms of each
element in a unit of the compound. However, suppose we needed to verify the purity
of a compound for use in a laboratory experiment. From the formula we could calculate what percent of the total mass of the compound is contributed by each element.
Then, by comparing the result to the percent composition obtained experimentally for
our sample, we could determine the purity of the sample.
The percent composition by mass is the percent by mass of each element in a
compound. Percent composition is obtained by dividing the mass of each element in
1 mole of the compound by the molar mass of the compound and multiplying by 100
percent. Mathematically, the percent composition of an element in a compound is
expressed as
percent composition of an element 5
n 3 molar mass of element
3 100%
molar mass of compound
(3.1)
where n is the number of moles of the element in 1 mole of the compound. For
example, in 1 mole of hydrogen peroxide (H2O2) there are 2 moles of H atoms and
2 moles of O atoms. The molar masses of H2O2, H, and O are 34.02 g, 1.008 g,
and 16.00 g, respectively. Therefore, the percent composition of H2O2 is calculated
as follows:
2 3 1.008 g H
3 100% 5 5.926%
34.02 g H2O2
2 3 16.00 g O
%O 5
3 100% 5 94.06%
34.02 g H2O2
%H 5
The sum of the percentages is 5.926% 1 94.06% 5 99.99%. The small discrepancy from 100 percent is due to the way we rounded off the molar masses of
the elements. If we had used the empirical formula HO for the calculation, we
H2O2
85
86
Chapter 3 ■ Mass Relationships in Chemical Reactions
would have obtained the same percentages. This is so because both the molecular
formula and empirical formula tell us the percent composition by mass of the
compound.
Example 3.8
Phosphoric acid (H3PO4) is a colorless, syrupy liquid used in detergents, fertilizers,
toothpastes, and in carbonated beverages for a “tangy” flavor. Calculate the percent
composition by mass of H, P, and O in this compound.
Strategy Recall the procedure for calculating a percentage. Assume that we have
H3PO4
1 mole of H3PO4. The percent by mass of each element (H, P, and O) is given by the
combined molar mass of the atoms of the element in 1 mole of H3PO4 divided by the
molar mass of H3PO4, then multiplied by 100 percent.
Solution The molar mass of H3PO4 is 97.99 g. The percent by mass of each of the
elements in H3PO4 is calculated as follows:
3(1.008 g) H
3 100% 5 3.086%
97.99 g H3PO4
30.97 g P
%P 5
3 100% 5 31.61%
97.99 g H3PO4
4(16.00 g) O
3 100% 5 65.31%
%O 5
97.99 g H3PO4
%H 5
Similar problem: 3.40.
Check Do the percentages add to 100 percent? The sum of the percentages is
(3.086% 1 31.61% 1 65.31%) 5 100.01%. The small discrepancy from 100 percent
is due to the way we rounded off.
Practice Exercise Calculate the percent composition by mass of each of the elements
in sulfuric acid (H2SO4).
Mass
percent
Convert to grams and
divide by molar mass
Moles of
each element
Divide by the smallest
number of moles
Mole ratios
of elements
Change to
integer subscripts
Empirical
formula
Figure 3.5 Procedure for
calculating the empirical formula
of a compound from its percent
compositions.
The procedure used in the example can be reversed if necessary. Given the percent
composition by mass of a compound, we can determine the empirical formula of the
compound (Figure 3.5). Because we are dealing with percentages and the sum of all
the percentages is 100 percent, it is convenient to assume that we started with 100 g
of a compound, as Example 3.9 shows.
Example 3.9
Ascorbic acid (vitamin C) cures scurvy. It is composed of 40.92 percent carbon (C),
4.58 percent hydrogen (H), and 54.50 percent oxygen (O) by mass. Determine its
empirical formula.
Strategy In a chemical formula, the subscripts represent the ratio of the number of
moles of each element that combine to form one mole of the compound. How can we
convert from mass percent to moles? If we assume an exactly 100-g sample of the
compound, do we know the mass of each element in the compound? How do we then
convert from grams to moles?
Solution If we have 100 g of ascorbic acid, then each percentage can be converted
directly to grams. In this sample, there will be 40.92 g of C, 4.58 g of H, and 54.50 g
of O. Because the subscripts in the formula represent a mole ratio, we need to
convert the grams of each element to moles. The conversion factor needed is the
(Continued)
3.5 Percent Composition of Compounds
87
molar mass of each element. Let n represent the number of moles of each element
so that
1 mol C
5 3.407 mol C
12.01 g C
1 mol H
5 4.54 mol H
nH 5 4.58 g H 3
1.008 g H
1 mol O
5 3.406 mol O
nO 5 54.50 g O 3
16.00 g O
nC 5 40.92 g C 3
Thus, we arrive at the formula C3.407H4.54O3.406, which gives the identity and the mole ratios
of atoms present. However, chemical formulas are written with whole numbers. Try to
convert to whole numbers by dividing all the subscripts by the smallest subscript (3.406):
C:
The molecular formula of ascorbic
acid is C6H8O6.
3.407
4.54
3.406
< 1 H:
5 1.33 O:
51
3.406
3.406
3.406
where the < sign means “approximately equal to.” This gives CH1.33O as the formula
for ascorbic acid. Next, we need to convert 1.33, the subscript for H, into an integer.
This can be done by a trial-and-error procedure:
1.33 3 1 5 1.33
1.33 3 2 5 2.66
1.33 3 3 5 3.99 < 4
Because 1.33 3 3 gives us an integer (4), we multiply all the subscripts by 3 and
obtain C3H4O3 as the empirical formula for ascorbic acid.
Check Are the subscripts in C3H4O3 reduced to the smallest whole numbers?
Similar problems: 3.49, 3.50.
Practice Exercise Determine the empirical formula of a compound having the following
percent composition by mass: K: 24.75 percent; Mn: 34.77 percent; O: 40.51 percent.
Chemists often want to know the actual mass of an element in a certain mass of
a compound. For example, in the mining industry, this information will tell the scientists about the quality of the ore. Because the percent composition by mass of the
elements in the substance can be readily calculated, such a problem can be solved in
a rather direct way.
Example 3.10
Chalcopyrite (CuFeS2) is a principal mineral of copper. Calculate the number of
kilograms of Cu in 3.71 3 103 kg of chalcopyrite.
Strategy Chalcopyrite is composed of Cu, Fe, and S. The mass due to Cu is based on its
percentage by mass in the compound. How do we calculate mass percent of an element?
Solution The molar masses of Cu and CuFeS2 are 63.55 g and 183.5 g, respectively.
The mass percent of Cu is therefore
molar mass of Cu
3 100%
molar mass of CuFeS2
63.55 g
3 100% 5 34.63%
5
183.5 g
%Cu 5
To calculate the mass of Cu in a 3.71 3 103 kg sample of CuFeS2, we need to convert the
percentage to a fraction (that is, convert 34.63 percent to 34.63/100, or 0.3463) and write
mass of Cu in CuFeS2 5 0.3463 3 (3.71 3 103 kg) 5 1.28 3 103 kg
(Continued)
Chalcopyrite.
88
Chapter 3 ■ Mass Relationships in Chemical Reactions
Check As a ball-park estimate, note that the mass percent of Cu is roughly 33 percent,
Similar problem: 3.45.
so that a third of the mass should be Cu; that is, 13 3 3.71 3 103 kg < 1.24 3 103 kg.
This quantity is quite close to the answer.
Practice Exercise Calculate the number of grams of Al in 371 g of Al2O3.
Review of Concepts
Without doing detailed calculations, estimate whether the percent composition by
mass of Sr is greater than or smaller than that of O in strontium nitrate [Sr(NO3)2].
3.6 Experimental Determination of Empirical Formulas
The fact that we can determine the empirical formula of a compound if we know the
percent composition enables us to identify compounds experimentally. The procedure
is as follows. First, chemical analysis tells us the number of grams of each element
present in a given amount of a compound. Then, we convert the quantities in grams
to number of moles of each element. Finally, using the method given in Example 3.9,
we find the empirical formula of the compound.
As a specific example, let us consider the compound ethanol. When ethanol is
burned in an apparatus such as that shown in Figure 3.6, carbon dioxide (CO2) and
water (H2O) are given off. Because neither carbon nor hydrogen was in the inlet
gas, we can conclude that both carbon (C) and hydrogen (H) were present in ethanol and that oxygen (O) may also be present. (Molecular oxygen was added in the
combustion process, but some of the oxygen may also have come from the original
ethanol sample.)
The masses of CO2 and of H2O produced can be determined by measuring the
increase in mass of the CO2 and H2O absorbers, respectively. Suppose that in one
experiment the combustion of 11.5 g of ethanol produced 22.0 g of CO2 and 13.5 g
of H2O. We can calculate the mass of carbon and hydrogen in the original 11.5-g
sample of ethanol as follows:
mass of C 5 22.0 g CO2 3
5 6.00 g C
mass of H 5 13.5 g H2O 3
5 1.51 g H
12.01 g C
1 mol CO2
1 mol C
3
3
1 mol CO2
1 mol C
44.01 g CO2
1.008 g H
1 mol H2O
2 mol H
3
3
1 mol H2O
1 mol H
18.02 g H2O
Figure 3.6 Apparatus for
determining the empirical formula
of ethanol. The absorbers are
substances that can retain water
and carbon dioxide, respectively.
CuO is used to ensure complete
combustion of all carbon to CO2.
Sample
CO2
absorber
H2O
absorber
O2
CuO
Furnace
3.6 Experimental Determination of Empirical Formulas
Thus, 11.5 g of ethanol contains 6.00 g of carbon and 1.51 g of hydrogen. The
remainder must be oxygen, whose mass is
mass of O 5 mass of sample 2 (mass of C 1 mass of H)
5 11.5 g 2 (6.00 g 1 1.51 g)
5 4.0 g
The number of moles of each element present in 11.5 g of ethanol is
1 mol C
5 0.500 mol C
12.01 g C
1 mol H
moles of H 5 1.51 g H 3
5 1.50 mol H
1.008 g H
1 mol O
moles of O 5 4.0 g O 3
5 0.25 mol O
16.00 g O
moles of C 5 6.00 g C 3
The formula of ethanol is therefore C0.50H1.5O0.25 (we round off the number of moles
to two significant figures). Because the number of atoms must be an integer, we divide
the subscripts by 0.25, the smallest subscript, and obtain for the empirical formula
C2H6O.
Now we can better understand the word “empirical,” which literally means “based
only on observation and measurement.” The empirical formula of ethanol is determined from analysis of the compound in terms of its component elements. No knowledge of how the atoms are linked together in the compound is required.
Determination of Molecular Formulas
The formula calculated from percent composition by mass is always the empirical
formula because the subscripts in the formula are always reduced to the smallest
whole numbers. To calculate the actual, molecular formula we must know the approximate molar mass of the compound in addition to its empirical formula. Knowing that
the molar mass of a compound must be an integral multiple of the molar mass of its
empirical formula, we can use the molar mass to find the molecular formula, as
Example 3.11 demonstrates.
Example 3.11
A sample of a compound contains 30.46 percent nitrogen and 69.54 percent oxygen by
mass, as determined by a mass spectrometer. In a separate experiment, the molar mass
of the compound is found to be between 90 g and 95 g. Determine the molecular
formula and the accurate molar mass of the compound.
Strategy To determine the molecular formula, we first need to determine the
empirical formula. Comparing the empirical molar mass to the experimentally
determined molar mass will reveal the relationship between the empirical formula
and molecular formula.
Solution We start by assuming that there are 100 g of the compound. Then each
percentage can be converted directly to grams; that is, 30.46 g of N and 69.54 g of O.
Let n represent the number of moles of each element so that
1 mol N
5 2.174 mol N
14.01 g N
1 mol O
5 4.346 mol O
nO 5 69.54 g O 3
16.00 g O
nN 5 30.46 g N 3
(Continued)
It happens that the molecular
formula of ethanol is the same
as its empirical formula.
89
90
Chapter 3 ■ Mass Relationships in Chemical Reactions
Thus, we arrive at the formula N2.174O4.346, which gives the identity and the ratios of
atoms present. However, chemical formulas are written with whole numbers. Try to
convert to whole numbers by dividing the subscripts by the smaller subscript (2.174).
After rounding off, we obtain NO2 as the empirical formula.
The molecular formula might be the same as the empirical formula or some integral
multiple of it (for example, two, three, four, or more times the empirical formula).
Comparing the ratio of the molar mass to the molar mass of the empirical formula will
show the integral relationship between the empirical and molecular formulas. The molar
mass of the empirical formula NO2 is
empirical molar mass 5 14.01 g 1 2(16.00 g) 5 46.01 g
Next, we determine the ratio between the molar mass and the empirical molar mass
90 g
molar mass
5
<2
empirical molar mass
46.01 g
N 2O4
Similar problems: 3.52, 3.53, 3.54.
The molar mass is twice the empirical molar mass. This means that there are two
NO2 units in each molecule of the compound, and the molecular formula is (NO2)2
or N2O4.
The actual molar mass of the compound is two times the empirical molar mass,
that is, 2(46.01 g) or 92.02 g, which is between 90 g and 95 g.
Check Note that in determining the molecular formula from the empirical formula, we
need only know the approximate molar mass of the compound. The reason is that the
true molar mass is an integral multiple (13, 23, 33, . . .) of the empirical molar
mass. Therefore, the ratio (molar mass/empirical molar mass) will always be close to
an integer.
Practice Exercise A sample of a compound containing boron (B) and hydrogen (H)
contains 6.444 g of B and 1.803 g of H. The molar mass of the compound is about 30 g.
What is its molecular formula?
Review of Concepts
What is the molecular formula of a compound containing only carbon and
hydrogen if combustion of 1.05 g of the compound produces 3.30 g CO2 and
1.35 g H2O and its molar mass is about 70 g?
3.7 Chemical Reactions and Chemical Equations
Having discussed the masses of atoms and molecules, we turn next to what happens
to atoms and molecules in a chemical reaction, a process in which a substance (or
substances) is changed into one or more new substances. To communicate with one
another about chemical reactions, chemists have devised a standard way to represent
them using chemical equations. A chemical equation uses chemical symbols to show
what happens during a chemical reaction. In this section, we will learn how to write
chemical equations and balance them.
Writing Chemical Equations
Consider what happens when hydrogen gas (H2) burns in air (which contains
oxygen, O2) to form water (H2O). This reaction can be represented by the chemical equation
H2 1 O2 ¡ H2 O
(3.2)
3.7 Chemical Reactions and Chemical Equations
91
+
Two hydrogen molecules
+
One oxygen molecule
Two water molecules
2H2
+
O2
2H2O
2 moles H2
+
1 mole O2
2 moles H2O
2(2.02 g) = 4.04 g H2
+
32.00 g O2
2(18.02 g) = 36.04 g H2O
36.04 g reactants
36.04 g product
Figure 3.7 Three ways of representing the combustion of hydrogen. In accordance with
the law of conservation of mass, the number of each type of atom must be the same on both sides
of the equation.
where the “plus” sign means “reacts with” and the arrow means “to yield.” Thus,
this symbolic expression can be read: “Molecular hydrogen reacts with molecular
oxygen to yield water.” The reaction is assumed to proceed from left to right as the
arrow indicates.
Equation (3.2) is not complete, however, because there are twice as many oxygen
atoms on the left side of the arrow (two) as on the right side (one). To conform with
the law of conservation of mass, there must be the same number of each type of atom
on both sides of the arrow; that is, we must have as many atoms after the reaction
ends as we did before it started. We can balance Equation (3.2) by placing the appropriate coefficient (2 in this case) in front of H2 and H2O:
2H2 1 O2 ¡ 2H2O
We use the law of conservation of mass
as our guide in balancing chemical
equations.
When the coefficient is 1, as in the case
of O2, it is not shown.
This balanced chemical equation shows that “two hydrogen molecules can combine
or react with one oxygen molecule to form two water molecules” (Figure 3.7). Because
the ratio of the number of molecules is equal to the ratio of the number of moles, the
equation can also be read as “2 moles of hydrogen molecules react with 1 mole of
oxygen molecules to produce 2 moles of water molecules.” We know the mass of a
mole of each of these substances, so we can also interpret the equation as “4.04 g of
H2 react with 32.00 g of O2 to give 36.04 g of H2O.” These three ways of reading
the equation are summarized in Figure 3.7.
We refer to H2 and O2 in Equation (3.2) as reactants, which are the starting
materials in a chemical reaction. Water is the product, which is the substance formed
as a result of a chemical reaction. A chemical equation, then, is just the chemist’s
shorthand description of a reaction. In a chemical equation, the reactants are conventionally written on the left and the products on the right of the arrow:
reactants ¡ products
To provide additional information, chemists often indicate the physical states of
the reactants and products by using the letters g, l, and s to denote gas, liquid, and
solid, respectively. For example,
2CO(g) 1 O2 (g) ¡ 2CO2 (g)
2HgO(s) ¡ 2Hg(l) 1 O2 (g)
To represent what happens when sodium chloride (NaCl) is added to water,
we write
H2O
NaCl(s) ¡ NaCl(aq)
The procedure for balancing chemical
equations is shown on p. 92.
92
Chapter 3 ■ Mass Relationships in Chemical Reactions
where aq denotes the aqueous (that is, water) environment. Writing H2O above the
arrow symbolizes the physical process of dissolving a substance in water, although it
is sometimes left out for simplicity.
Knowing the states of the reactants and products is especially useful in the laboratory. For example, when potassium bromide (KBr) and silver nitrate (AgNO3) react
in an aqueous environment, a solid, silver bromide (AgBr), is formed. This reaction
can be represented by the equation:
KBr(aq) 1 AgNO3 (aq) ¡ KNO3 (aq) 1 AgBr(s)
If the physical states of reactants and products are not given, an uninformed person
might try to bring about the reaction by mixing solid KBr with solid AgNO3. These
solids would react very slowly or not at all. Imagining the process on the microscopic
level, we can understand that for a product like silver bromide to form, the Ag1 and
Br2 ions would have to come in contact with each other. However, these ions are
locked in place in their solid compounds and have little mobility. (Here is an example
of how we explain a phenomenon by thinking about what happens at the molecular
level, as discussed in Section 1.2.)
Balancing Chemical Equations
Suppose we want to write an equation to describe a chemical reaction that we have
just carried out in the laboratory. How should we go about doing it? Because we
know the identities of the reactants, we can write their chemical formulas. The identities of products are more difficult to establish. For simple reactions it is often possible to guess the product(s). For more complicated reactions involving three or more
products, chemists may need to perform further tests to establish the presence of
specific compounds.
Once we have identified all the reactants and products and have written the correct formulas for them, we assemble them in the conventional sequence—reactants
on the left separated by an arrow from products on the right. The equation written at
this point is likely to be unbalanced; that is, the number of each type of atom on one
side of the arrow differs from the number on the other side. In general, we can balance
a chemical equation by the following steps:
1. Identify all reactants and products and write their correct formulas on the left
side and right side of the equation, respectively.
2. Begin balancing the equation by trying different coefficients to make the number
of atoms of each element the same on both sides of the equation. We can change
the coefficients (the numbers preceding the formulas) but not the subscripts (the
numbers within formulas). Changing the subscripts would change the identity of
the substance. For example, 2NO2 means “two molecules of nitrogen dioxide,”
but if we double the subscripts, we have N2O4, which is the formula of dinitrogen
tetroxide, a completely different compound.
3. First, look for elements that appear only once on each side of the equation with
the same number of atoms on each side: The formulas containing these elements
must have the same coefficient. Therefore, there is no need to adjust the coefficients of these elements at this point. Next, look for elements that appear only
once on each side of the equation but in unequal numbers of atoms. Balance these
elements. Finally, balance elements that appear in two or more formulas on the
same side of the equation.
4. Check your balanced equation to be sure that you have the same total number of
each type of atoms on both sides of the equation arrow.
3.7 Chemical Reactions and Chemical Equations
93
Let’s consider a specific example. In the laboratory, small amounts of oxygen gas
can be prepared by heating potassium chlorate (KClO3). The products are oxygen gas
(O2) and potassium chloride (KCl). From this information, we write
KClO3 ¡ KCl 1 O2
(For simplicity, we omit the physical states of reactants and products.) All three elements (K, Cl, and O) appear only once on each side of the equation, but only for K
and Cl do we have equal numbers of atoms on both sides. Thus, KClO3 and KCl must
have the same coefficient. The next step is to make the number of O atoms the same
on both sides of the equation. Because there are three O atoms on the left and two
O atoms on the right of the equation, we can balance the O atoms by placing a 2 in
front of KClO3 and a 3 in front of O2.
Heating potassium chlorate
produces oxygen, which supports
the combustion of a wood splint.
2KClO3 ¡ KCl 1 3O2
Finally, we balance the K and Cl atoms by placing a 2 in front of KCl:
2KClO3 ¡ 2KCl 1 3O2
(3.3)
As a final check, we can draw up a balance sheet for the reactants and products where
the number in parentheses indicates the number of atoms of each element:
Reactants
K (2)
Cl (2)
O (6)
Products
K (2)
Cl (2)
O (6)
Note that this equation could also be balanced with coefficients that are multiples of
2 (for KClO3), 2 (for KCl), and 3 (for O2); for example,
4KClO3 ¡ 4KCl 1 6O2
However, it is common practice to use the simplest possible set of whole-number
coefficients to balance the equation. Equation (3.3) conforms to this convention.
Now let us consider the combustion (that is, burning) of the natural gas component ethane (C2H6) in oxygen or air, which yields carbon dioxide (CO2) and water.
The unbalanced equation is
C2H6 1 O2 ¡ CO2 1 H2O
We see that the number of atoms is not the same on both sides of the equation for
any of the elements (C, H, and O). In addition, C and H appear only once on each
side of the equation; O appears in two compounds on the right side (CO2 and H2O).
To balance the C atoms, we place a 2 in front of CO2:
C2H6 1 O2 ¡ 2CO2 1 H2O
To balance the H atoms, we place a 3 in front of H2O:
C2H6 1 O2 ¡ 2CO2 1 3H2O
At this stage, the C and H atoms are balanced, but the O atoms are not because there
are seven O atoms on the right-hand side and only two O atoms on the left-hand side
C2H6
94
Chapter 3 ■ Mass Relationships in Chemical Reactions
of the equation. This inequality of O atoms can be eliminated by writing
of the O2 on the left-hand side:
7
2
in front
C2H6 1 72 O2 ¡ 2CO2 1 3H2O
The “logic” for using 72 as a coefficient is that there were seven oxygen atoms on the
right-hand side of the equation, but only a pair of oxygen atoms (O2) on the left. To
balance them we ask how many pairs of oxygen atoms are needed to equal seven
oxygen atoms. Just as 3.5 pairs of shoes equal seven shoes, 72 O2 molecules equal seven
O atoms. As the following tally shows, the equation is now balanced:
Reactants
C (2)
H (6)
O (7)
Products
C (2)
H (6)
O (7)
However, we normally prefer to express the coefficients as whole numbers rather than
as fractions. Therefore, we multiply the entire equation by 2 to convert 72 to 7:
2C2H6 1 7O2 ¡ 4CO2 1 6H2O
The final tally is
Reactants
C (4)
H (12)
O (14)
Products
C (4)
H (12)
O (14)
Note that the coefficients used in balancing the last equation are the smallest possible
set of whole numbers.
In Example 3.12 we will continue to practice our equation-balancing skills.
Example 3.12
When aluminum metal is exposed to air, a protective layer of aluminum oxide (Al2O3)
forms on its surface. This layer prevents further reaction between aluminum and oxygen,
and it is the reason that aluminum beverage cans do not corrode. [In the case of iron,
the rust, or iron(III) oxide, that forms is too porous to protect the iron metal underneath,
so rusting continues.] Write a balanced equation for the formation of Al2O3.
Strategy Remember that the formula of an element or compound cannot be changed
when balancing a chemical equation. The equation is balanced by placing the appropriate
coefficients in front of the formulas. Follow the procedure described on p. 92.
Solution The unbalanced equation is
Al 1 O2 ¡ Al2O3
In a balanced equation, the number and types of atoms on each side of the equation
must be the same. We see that there is one Al atom on the reactants side and there are
two Al atoms on the product side. We can balance the Al atoms by placing a coefficient
of 2 in front of Al on the reactants side.
2Al 1 O2 ¡ Al2O3
There are two O atoms on the reactants side, and three O atoms on the product side of
the equation. We can balance the O atoms by placing a coefficient of 32 in front of O2
on the reactants side.
2Al 1 32 O2 ¡ Al2O3
(Continued)
3.8 Amounts of Reactants and Products
This is a balanced equation. However, equations are normally balanced with the smallest
set of whole-number coefficients. Multiplying both sides of the equation by 2 gives
whole-number coefficients.
2(2Al 1 32 O2 ¡ Al2O3 )
4Al 1 3O2 ¡ 2Al2O3
or
Check For an equation to be balanced, the number and types of atoms on each side of
the equation must be the same. The final tally is
Reactants
Al (4)
O (6)
Products
Al (4)
O (6)
The equation is balanced. Also, the coefficients are reduced to the simplest set of
whole numbers.
Similar problems: 3.59, 3.60.
Practice Exercise Balance the equation representing the reaction between iron(III)
oxide, Fe2O3, and carbon monoxide (CO) to yield iron (Fe) and carbon dioxide (CO2).
Review of Concepts
Which parts of the equation shown here are essential for a balanced equation
and which parts are helpful if we want to carry out the reaction in the
laboratory?
BaH2 (s) 1 2H2O(l) ¡ Ba(OH) 2 (aq) 1 2H2 (g)
3.8 Amounts of Reactants and Products
A basic question raised in the chemical laboratory is “How much product will be
formed from specific amounts of starting materials (reactants)?” Or in some cases,
we might ask the reverse question: “How much starting material must be used to
obtain a specific amount of product?” To interpret a reaction quantitatively, we need
to apply our knowledge of molar masses and the mole concept. Stoichiometry is the
quantitative study of reactants and products in a chemical reaction.
Whether the units given for reactants (or products) are moles, grams, liters (for
gases), or some other units, we use moles to calculate the amount of product formed
in a reaction. This approach is called the mole method, which means simply that the
stoichiometric coefficients in a chemical equation can be interpreted as the number
of moles of each substance. For example, industrially ammonia is synthesized from
hydrogen and nitrogen as follows:
N2 (g) 1 3H2 (g) ¡ 2NH3 (g)
The stoichiometric coefficients show that one molecule of N2 reacts with three molecules of H2 to form two molecules of NH3. It follows that the relative numbers of
moles are the same as the relative number of molecules:
N2(g)
1 molecule
6.022 3 1023 molecules
1 mol
1
3H2(g)
3 molecules
3(6.022 3 1023 molecules)
3 mol
¡
2NH3(g)
2 molecules
2(6.022 3 1023 molecules)
2 mol
The synthesis of NH3 from H2
and N2.
95
96
Chapter 3 ■ Mass Relationships in Chemical Reactions
Thus, this equation can also be read as “1 mole of N2 gas combines with 3 moles of
H2 gas to form 2 moles of NH3 gas.” In stoichiometric calculations, we say that three
moles of H2 are equivalent to two moles of NH3, that is,
3 mol H2 ∞ 2 mol NH3
where the symbol ∞ means “stoichiometrically equivalent to” or simply “equivalent
to.” This relationship enables us to write the conversion factors
3 mol H2
2 mol NH3
and
2 mol NH3
3 mol H2
Similarly, we have 1 mol N2 ∞ 2 mol NH3 and 1 mol N2 ∞ 3 mol H2.
Let’s consider a simple example in which 6.0 moles of H2 react completely with
N2 to form NH3. To calculate the amount of NH3 produced in moles, we use the
conversion factor that has H2 in the denominator and write
moles of NH3 produced 5 6.0 mol H2 3
2 mol NH3
3 mol H2
5 4.0 mol NH3
Now suppose 16.0 g of H2 react completely with N2 to form NH3. How many
grams of NH3 will be formed? To do this calculation, we note that the link between
H2 and NH3 is the mole ratio from the balanced equation. So we need to first convert
grams of H2 to moles of H2, then to moles of NH3, and finally to grams of NH3. The
conversion steps are
grams of H2 ¡ moles of H2 ¡ moles of NH3 ¡ grams of NH3
First, we convert 16.0 g of H2 to number of moles of H2, using the molar mass of H2
as the conversion factor:
moles of H2 5 16.0 g H2 3
1 mol H2
2.016 g H2
5 7.94 mol H2
Next, we calculate the number of moles of NH3 produced.
moles of NH3 5 7.94 mol H2 3
2 mol NH3
3 mol H2
5 5.29 mol NH3
Finally, we calculate the mass of NH3 produced in grams using the molar mass of
NH3 as the conversion factor
grams of NH3 5 5.29 mol NH3 3
17.03 g NH3
1 mol NH3
5 90.1 g NH3
These three separate calculations can be combined in a single step as follows:
grams of NH3 5 16.0 g H2 3
5 90.1 g NH3
17.03 g NH3
2 mol NH3
1 mol H2
3
3
2.016 g H2
3 mol H2
1 mol NH3
3.8 Amounts of Reactants and Products
Mass (g)
of compound A
Moles of
compound A
Figure 3.8 The procedure for
calculating the amounts of
reactants or products in a
reaction using the mole method.
Mass (g)
of compound B
Use molar
mass (g/mol)
of compound A
Use molar
mass (g/mol)
of compound B
Use mole ratio
of A and B
from balanced
equation
Moles of
compound B
Similarly, we can calculate the mass in grams of N2 consumed in this reaction.
The conversion steps are
grams of H2 ¡ moles of H2 ¡ moles of N2 ¡ grams of N2
By using the relationship 1 mol N2 ∞ 3 mol H2, we write
grams of N2 5 16.0 g H2 3
5 74.1 g N2
97
28.02 g N2
1 mol H2
1 mol N2
3
3
2.016 g H2
3 mol H2
1 mol N2
The general approach for solving stoichiometry problems is summarized next.
1. Write a balanced equation for the reaction.
2. Convert the given amount of the reactant (in grams or other units) to number of moles.
3. Use the mole ratio from the balanced equation to calculate the number of moles
of product formed.
4. Convert the moles of product to grams (or other units) of product.
Figure 3.8 shows these steps. Sometimes we may be asked to calculate the amount
of a reactant needed to form a specific amount of product. In those cases, we can
reverse the steps shown in Figure 3.8.
Examples 3.13 and 3.14 illustrate the application of this approach.
Example 3.13
The food we eat is degraded, or broken down, in our bodies to provide energy for
growth and function. A general overall equation for this very complex process represents
the degradation of glucose (C6H12O6) to carbon dioxide (CO2) and water (H2O):
C6H12O6 1 6O2 ¡ 6CO2 1 6H2O
If 856 g of C6H12O6 is consumed by a person over a certain period, what is the mass of
CO2 produced?
Strategy Looking at the balanced equation, how do we compare the amounts of
C6H12O6 and CO2? We can compare them based on the mole ratio from the balanced
equation. Starting with grams of C6H12O6, how do we convert to moles of C6H12O6?
Once moles of CO2 are determined using the mole ratio from the balanced equation,
how do we convert to grams of CO2?
Solution We follow the preceding steps and Figure 3.8.
(Continued)
C6H12O6
98
Chapter 3 ■ Mass Relationships in Chemical Reactions
Step 1: The balanced equation is given in the problem.
Step 2: To convert grams of C6H12O6 to moles of C6H12O6, we write
856 g C6H12O6 3
1 mol C6H12O6
5 4.750 mol C6H12O6
180.2 g C6H12O6
Step 3: From the mole ratio, we see that 1 mol C6H12O6 ∞ 6 mol CO2. Therefore, the
number of moles of CO2 formed is
4.750 mol C6H12O6 3
6 mol CO2
5 28.50 mol CO2
1 mol C6H12O6
Step 4: Finally, the number of grams of CO2 formed is given by
28.50 mol CO2 3
44.01 g CO2
5 1.25 3 103 g CO2
1 mol CO2
After some practice, we can combine the conversion steps
grams of C6H12O6 ¡ moles of C6H12O6 ¡ moles of CO2 ¡ grams of CO2
into one equation:
mass of CO2 5 856 g C6H12O6 3
44.01 g CO2
1 mol C6H12O6
6 mol CO2
3
3
180.2 g C6H12 O6
1 mol C6H12O6
1 mol CO2
5 1.25 3 103 g CO2
Check Does the answer seem reasonable? Should the mass of CO2 produced be
Similar problem: 3.72.
larger than the mass of C6H12O6 reacted, even though the molar mass of CO2 is
considerably less than the molar mass of C6H12O6? What is the mole ratio between
CO2 and C6H12O6?
Practice Exercise Methanol (CH3OH) burns in air according to the equation
2CH3OH 1 3O2 ¡ 2CO2 1 4H2O
If 209 g of methanol are used up in a combustion process, what is the mass of H2O
produced?
Example 3.14
All alkali metals react with water to produce hydrogen gas and the corresponding alkali
metal hydroxide. A typical reaction is that between lithium and water:
2Li(s) 1 2H2O(l) ¡ 2LiOH(aq) 1 H2 (g)
Lithium reacting with water to
produce hydrogen gas.
How many grams of Li are needed to produce 9.89 g of H2?
Strategy The question asks for number of grams of reactant (Li) to form a specific
amount of product (H2). Therefore, we need to reverse the steps shown in Figure 3.8.
From the equation we see that 2 mol Li ∞ 1 mol H2.
Solution The conversion steps are
grams of H2 ¡ moles of H2 ¡ moles of Li ¡ grams of Li
(Continued)
99
3.9 Limiting Reagents
Combining these steps into one equation, we write
9.89 g H2 3
6.941 g Li
1 mol H2
2 mol Li
3
3
5 68.1 g Li
2.016 g H2
1 mol H2
1 mol Li
Check There are roughly 5 moles of H2 in 9.89 g H2, so we need 10 moles of Li.
From the approximate molar mass of Li (7 g), does the answer seem reasonable?
Similar problem: 3.66.
Practice Exercise The reaction between nitric oxide (NO) and oxygen to form
nitrogen dioxide (NO2) is a key step in photochemical smog formation:
2NO(g) 1 O2 (g) ¡ 2NO2 (g)
How many grams of O2 are needed to produce 2.21 g of NO2?
Review of Concepts
Animation
Limiting Reagent
Which of the following statements is correct for the equation shown here?
4NH3 (g) 1 5O2 (g) ¡ 4NO(g) 1 6H2O(g)
(a) 6 g of H2O are produced for every 4 g of NH3 reacted.
(b) 1 mole of NO is produced per mole of NH3 reacted.
(c) 2 moles of NO are produced for every 3 moles of O2 reacted.
Before reaction has started
3.9 Limiting Reagents
When a chemist carries out a reaction, the reactants are usually not present in exact
stoichiometric amounts, that is, in the proportions indicated by the balanced equation.
Because the goal of a reaction is to produce the maximum quantity of a useful compound from the starting materials, frequently a large excess of one reactant is supplied
to ensure that the more expensive reactant is completely converted to the desired
product. Consequently, some reactant will be left over at the end of the reaction. The
reactant used up first in a reaction is called the limiting reagent, because the maximum amount of product formed depends on how much of this reactant was originally
present. When this reactant is used up, no more product can be formed. Excess
reagents are the reactants present in quantities greater than necessary to react with
the quantity of the limiting reagent.
The concept of the limiting reagent is analogous to the relationship between men
and women in a dance contest at a club. If there are 14 men and only 9 women, then
only 9 female/male pairs can compete. Five men will be left without partners. The
number of women thus limits the number of men that can dance in the contest, and
there is an excess of men.
Consider the industrial synthesis of methanol (CH3OH) from carbon monoxide
and hydrogen at high temperatures:
CO(g) 1 2H2 (g) ¡ CH3OH(g)
Suppose initially we have 4 moles of CO and 6 moles of H2 (Figure 3.9). One way
to determine which of two reactants is the limiting reagent is to calculate the number
of moles of CH3OH obtained based on the initial quantities of CO and H2. From the
After reaction is complete
H2
CO
CH3OH
Figure 3.9 At the start of the
reaction, there were six H2
molecules and four CO molecules.
At the end, all the H2 molecules
are gone and only one CO
molecule is left. Therefore, H2
molecule is the limiting reagent
and CO is the excess reagent.
Each molecule can also be
treated as one mole of the
substance in this reaction.
100
Chapter 3 ■ Mass Relationships in Chemical Reactions
preceding definition, we see that only the limiting reagent will yield the smaller
amount of the product. Starting with 4 moles of CO, we find the number of moles of
CH3OH produced is
4 mol CO 3
1 mol CH3OH
5 4 mol CH3OH
1 mol CO
and starting with 6 moles of H2, the number of moles of CH3OH formed is
6 mol H2 3
1 mol CH3OH
5 3 mol CH3OH
2 mol H2
Because H2 results in a smaller amount of CH3OH, it must be the limiting reagent.
Therefore, CO is the excess reagent.
In stoichiometric calculations involving limiting reagents, the first step is to decide
which reactant is the limiting reagent. After the limiting reagent has been identified,
the rest of the problem can be solved as outlined in Section 3.8. Example 3.15 illustrates this approach.
Example 3.15
The synthesis of urea, [(NH2)2CO], is considered to be the first recognized example of
preparing a biological compound from nonbiological reactants, challenging the notion
that biological processes involved a “vital force” present only in living systems. Today
urea is produced industrially by reacting ammonia with carbon dioxide:
2NH3 (g) 1 CO2 (g) ¡ (NH2 ) 2CO(aq) 1 H2O(l)
(NH2)2CO
In one process, 637.2 g of NH3 are treated with 1142 g of CO2. (a) Which of the two
reactants is the limiting reagent? (b) Calculate the mass of (NH2)2CO formed. (c) How
much excess reagent (in grams) is left at the end of the reaction?
(a) Strategy The reactant that produces fewer moles of product is the limiting reagent
because it limits the amount of product that can be formed. How do we convert from
the amount of reactant to amount of product? Perform this calculation for each reactant,
then compare the moles of product, (NH2)2CO, formed by the given amounts of NH3
and CO2 to determine which reactant is the limiting reagent.
Solution We carry out two separate calculations. First, starting with 637.2 g of NH3,
we calculate the number of moles of (NH2)2CO that could be produced if all the NH3
reacted according to the following conversions:
grams of NH3 ¡ moles of NH3 ¡ moles of (NH2 ) 2CO
Combining these conversions in one step, we write
1 mol NH3
1 mol (NH2 ) 2CO
3
2 mol NH3
17.03 g NH3
5 18.71 mol (NH2 ) 2CO
moles of (NH2 ) 2CO 5 637.2 g NH3 3
Second, for 1142 g of CO2, the conversions are
grams of CO2 ¡ moles of CO2 ¡ moles of (NH2 ) 2CO
The number of moles of (NH2)2CO that could be produced if all the CO2 reacted is
1 mol CO2
1 mol (NH2 ) 2CO
3
44.01 g CO2
1 mol CO2
5 25.95 mol (NH2 ) 2CO
moles of (NH2 ) 2CO 5 1142 g CO2 3
(Continued)
3.9 Limiting Reagents
It follows, therefore, that NH3 must be the limiting reagent because it produces a
smaller amount of (NH2)2CO.
(b) Strategy We determined the moles of (NH2)2CO produced in part (a), using NH3
as the limiting reagent. How do we convert from moles to grams?
Solution The molar mass of (NH2)2CO is 60.06 g. We use this as a conversion factor
to convert from moles of (NH2)2CO to grams of (NH2)2CO:
mass of (NH2 ) 2CO 5 18.71 mol (NH2 ) 2CO 3
60.06 g (NH2 ) 2CO
1 mol (NH2 ) 2CO
5 1124 g (NH2 ) 2CO
Check Does your answer seem reasonable? 18.71 moles of product are formed. What
is the mass of 1 mole of (NH2)2CO?
(c) Strategy Working backward, we can determine the amount of CO2 that reacted to
produce 18.71 moles of (NH2)2CO. The amount of CO2 left over is the difference
between the initial amount and the amount reacted.
Solution Starting with 18.71 moles of (NH2)2CO, we can determine the mass of CO2
that reacted using the mole ratio from the balanced equation and the molar mass of
CO2. The conversion steps are
moles of (NH2 ) 2CO ¡ moles of CO2 ¡ grams of CO2
so that
mass of CO2 reacted 5 18.71 mol (NH2 ) 2CO 3
44.01 g CO2
1 mol CO2
3
1 mol (NH2 ) 2CO
1 mol CO2
5 823.4 g CO2
The amount of CO2 remaining (in excess) is the difference between the initial amount
(1142 g) and the amount reacted (823.4 g):
mass of CO2 remaining 5 1142 g 2 823.4 g 5 319 g
Practice Exercise The reaction between aluminum and iron(III) oxide can generate
temperatures approaching 30008C and is used in welding metals:
2Al 1 Fe2O3 ¡ Al2O3 1 2Fe
In one process, 124 g of Al are reacted with 601 g of Fe2O3. (a) Calculate the mass
(in grams) of Al2O3 formed. (b) How much of the excess reagent is left at the end of
the reaction?
Example 3.15 brings out an important point. In practice, chemists usually choose
the more expensive chemical as the limiting reagent so that all or most of it will be
converted to products in the reaction. In the synthesis of urea, NH3 is invariably the
limiting reagent because it is more expensive than CO2. At other times, an excess of
one reagent is used to help drive the reaction to completion, or to compensate for a
side reaction that consumes that reagent. Synthetic chemists often have to calculate
the amount of reagents to use based on this need to have one or more components in
excess, as Example 3.16 shows.
Similar problem: 3.86.
101
102
Chapter 3 ■ Mass Relationships in Chemical Reactions
Example 3.16
The reaction between alcohols and halogen compounds to form ethers is important in
organic chemistry, as illustrated here for the reaction between methanol (CH3OH) and
methyl bromide (CH3Br) to form dimethylether (CH3OCH3), which is a useful precursor
to other organic compounds and an aerosol propellant.
CH3OH 1 CH3Br 1 LiC4H9 ¡ CH3OCH3 1 LiBr 1 C4H10
This reaction is carried out in a dry (water-free) organic solvent, and the butyl
lithium (LiC 4H9) serves to remove a hydrogen ion from CH3OH. Butyl lithium
will also react with any residual water in the solvent, so the reaction is typically
carried out with 2.5 molar equivalents of that reagent. How many grams of CH3Br
and LiC4H9 will be needed to carry out the preceding reaction with 10.0 g of
CH 3OH?
Solution We start with the knowledge that CH3OH and CH3Br are present in
stoichiometric amounts and that LiC4H9 is the excess reagent. To calculate the quantities
of CH3Br and LiC4H9 needed, we proceed as shown in Example 3.14.
grams of CH3Br 5 10.0 g CH3OH 3
94.93 g CH3Br
1 mol CH3OH
1 mol CH3Br
3
3
32.04 g CH3OH
1 mol CH3OH
1 mol CH3Br
5 29.6 g CH3Br
grams of LiC4H9 5 10.0 g CH3OH 3
64.05 g LiC4H9
2.5 mol LiC4H9
1 mol CH3OH
3
3
32.04 g CH3OH
1 mol CH3OH
1 mol LiC4H9
5 50.0 g LiC4H9
Similar problems: 3.137, 3.138.
Practice Exercise The reaction between benzoic acid (C6H5COOH) and octanol
(C8H17OH) to yield octyl benzoate (C6H5COOC8H17) and water
C6H5COOH 1 C8H17OH ¡ C6H5COOC8H17 1 H2O
is carried out with an excess of C8H17OH to help drive the reaction to completion and
maximize the yield of product. If an organic chemist wants to use 1.5 molar equivalents
of C8H17OH, how many grams of C8H17OH would be required to carry out the reaction
with 15.7 g of C6H5COOH?
Review of Concepts
Consider the following reaction:
2NO(g) 1 O2 (g) ¡ 2NO2 (g)
Starting with the reactants shown in (a), which of the diagrams shown in (b)–(d) best represents the situation in
which the limiting reagent has completely reacted?
NO
O2
NO2
(a)
(b)
(c)
(d)
3.10 Reaction Yield
103
3.10 Reaction Yield
The amount of limiting reagent present at the start of a reaction determines the
theoretical yield of the reaction, that is, the amount of product that would result
if all the limiting reagent reacted. The theoretical yield, then, is the maximum
obtainable yield, predicted by the balanced equation. In practice, the actual yield,
or the amount of product actually obtained from a reaction, is almost always less
than the theoretical yield. There are many reasons for the difference between
actual and theoretical yields. For instance, many reactions are reversible, and so
they do not proceed 100 percent from left to right. Even when a reaction is 100
percent complete, it may be difficult to recover all of the product from the reaction
medium (say, from an aqueous solution). Some reactions are complex in the sense
that the products formed may react further among themselves or with the reactants
to form still other products. These additional reactions will reduce the yield of the
first reaction.
To determine how efficient a given reaction is, chemists often figure the percent
yield, which describes the proportion of the actual yield to the theoretical yield. It is
calculated as follows:
%yield 5
actual yield
3 100%
theoretical yield
(3.4)
Percent yields may range from a fraction of 1 percent to 100 percent. Chemists strive
to maximize the percent yield in a reaction. Factors that can affect the percent yield
include temperature and pressure. We will study these effects later.
In Example 3.17 we will calculate the yield of an industrial process.
Example 3.17
Titanium is a strong, lightweight, corrosion-resistant metal that is used in rockets,
aircraft, jet engines, and bicycle frames. It is prepared by the reaction of titanium(IV)
chloride with molten magnesium between 9508C and 11508C:
TiCl4 (g) 1 2Mg(l) ¡ Ti(s) 1 2MgCl2 (l)
In a certain industrial operation 3.54 3 107 g of TiCl4 are reacted with 1.13 3 107 g of
Mg. (a) Calculate the theoretical yield of Ti in grams. (b) Calculate the percent yield if
7.91 3 106 g of Ti are actually obtained.
(a) Strategy Because there are two reactants, this is likely to be a limiting
reagent problem. The reactant that produces fewer moles of product is the limiting
reagent. How do we convert from amount of reactant to amount of product?
Perform this calculation for each reactant, then compare the moles of product,
Ti, formed.
Solution Carry out two separate calculations to see which of the two reactants is the
limiting reagent. First, starting with 3.54 3 107 g of TiCl4, calculate the number of
moles of Ti that could be produced if all the TiCl4 reacted. The conversions are
grams of TiCl4 ¡ moles of TiCl4 ¡ moles of Ti
(Continued)
Keep in mind that the theoretical yield
is the yield that you calculate using the
balanced equation. The actual yield is
the yield obtained by carrying out the
reaction.
104
Chapter 3 ■ Mass Relationships in Chemical Reactions
so that
moles of Ti 5 3.54 3 107 g TiCl4 3
5 1.87 3 105 mol Ti
1 mol TiCl4
1 mol Ti
3
189.7 g TiCl4
1 mol TiCl4
Next, we calculate the number of moles of Ti formed from 1.13 3 107 g of Mg.
The conversion steps are
grams of Mg ¡ moles of Mg ¡ moles of Ti
and we write
moles of Ti 5 1.13 3 107 g Mg 3
1 mol Mg
1 mol Ti
3
24.31 g Mg
2 mol Mg
5 2.32 3 105 mol Ti
Therefore, TiCl4 is the limiting reagent because it produces a smaller amount of Ti.
The mass of Ti formed is
1.87 3 105 mol Ti 3
47.88 g Ti
5 8.95 3 106 g Ti
1 mol Ti
(b) Strategy The mass of Ti determined in part (a) is the theoretical yield. The amount
given in part (b) is the actual yield of the reaction.
Solution The percent yield is given by
An artificial hip joint made of
titanium and the structure of
solid titanium.
Similar problems: 3.89, 3.90.
actual yield
3 100%
theoretical yield
6
7.91 3 10 g
5
3 100%
8.95 3 106 g
5 88.4%
%yield 5
Check Should the percent yield be less than 100 percent?
Practice Exercise Industrially, vanadium metal, which is used in steel alloys, can be
obtained by reacting vanadium(V) oxide with calcium at high temperatures:
5Ca 1 V2O5 ¡ 5CaO 1 2V
In one process, 1.54 3 103 g of V2O5 react with 1.96 3 103 g of Ca. (a) Calculate the
theoretical yield of V. (b) Calculate the percent yield if 803 g of V are obtained.
Industrial processes usually involve huge quantities (thousands to millions of
tons) of products. Thus, even a slight improvement in the yield can significantly
reduce the cost of production. A case in point is the manufacture of chemical fertilizers, discussed in the Chemistry in Action essay on p. 105.
Review of Concepts
Can the percent yield ever exceed the theoretical yield of a reaction?
CHEMISTRY in Action
Chemical Fertilizers
F
eeding the world’s rapidly increasing population requires
that farmers produce ever-larger and healthier crops. Every
year they add hundreds of millions of tons of chemical fertilizers to the soil to increase crop quality and yield. In addition to
carbon dioxide and water, plants need at least six elements for
satisfactory growth. They are N, P, K, Ca, S, and Mg. The
preparation and properties of several nitrogen- and phosphoruscontaining fertilizers illustrate some of the principles introduced
in this chapter.
Nitrogen fertilizers contain nitrate (NO23) salts, ammonium
1
(NH4 ) salts, and other compounds. Plants can absorb nitrogen
in the form of nitrate directly, but ammonium salts and ammonia
(NH3) must first be converted to nitrates by the action of soil
bacteria. The principal raw material of nitrogen fertilizers is
ammonia, prepared by the reaction between hydrogen and
nitrogen:
3H2 (g) 1 N2 (g) ¡ 2NH3 (g)
Another method of preparing ammonium sulfate requires two
steps:
2NH3 (aq) 1 CO2 (aq) 1 H2O(l) ¡ (NH4 ) 2CO3 (aq) (1)
(NH4 ) 2CO3 (aq) 1 CaSO4 (aq) ¡
(NH4 ) 2SO4 (aq) 1 CaCO3 (s) (2)
This approach is desirable because the starting materials—
carbon dioxide and calcium sulfate—are less costly than sulfuric acid. To increase the yield, ammonia is made the limiting
reagent in Reaction (1) and ammonium carbonate is made the
limiting reagent in Reaction (2).
The table lists the percent composition by mass of nitrogen
in some common fertilizers. The preparation of urea was discussed in Example 3.15.
Percent Composition by Mass of Nitrogen
in Five Common Fertilizers
Fertilizer
(This reaction will be discussed in detail in Chapters 13 and 14.)
In its liquid form, ammonia can be injected directly into the soil.
Alternatively, ammonia can be converted to ammonium
nitrate, NH4NO3, ammonium sulfate, (NH4)2SO4, or ammonium
hydrogen phosphate, (NH4)2HPO4, in the following acid-base
reactions:
NH3 (aq) 1 HNO3 (aq) ¡ NH4NO3 (aq)
2NH3 (aq) 1 H2SO4 (aq) ¡ (NH4 ) 2SO4 (aq)
2NH3 (aq) 1 H3PO4 (aq) ¡ (NH4 ) 2HPO4 (aq)
NH3
NH4NO3
(NH4)2SO4
(NH4)2HPO4
(NH2)2CO
% N by Mass
82.4
35.0
21.2
21.2
46.7
Several factors influence the choice of one fertilizer over another: (1) cost of the raw materials needed to prepare the fertilizer;
(2) ease of storage, transportation, and utilization; (3) percent
composition by mass of the desired element; and (4) suitability of
the compound, that is, whether the compound is soluble in water
and whether it can be readily taken up by plants. Considering all
these factors together, we find that NH4NO3 is the most important
nitrogen-containing fertilizer in the world, even though ammonia
has the highest percentage by mass of nitrogen.
Phosphorus fertilizers are derived from phosphate rock,
called fluorapatite, Ca5(PO4)3F. Fluorapatite is insoluble in
water, so it must first be converted to water-soluble calcium
dihydrogen phosphate [Ca(H2PO4)2]:
2Ca5 (PO4 ) 3F(s) 1 7H2SO4 (aq) ¡
3Ca(H2PO4 ) 2 (aq) 1 7CaSO4 (aq) 1 2HF(g)
Liquid ammonia being applied to the soil before planting.
For maximum yield, fluorapatite is made the limiting reagent in
this reaction.
The reactions we have discussed for the preparation of fertilizers all appear relatively simple, yet much effort has been
expended to improve the yields by changing conditions such as
temperature, pressure, and so on. Industrial chemists usually
run promising reactions first in the laboratory and then test them
in a pilot facility before putting them into mass production.
105
106
Chapter 3 ■ Mass Relationships in Chemical Reactions
Key Equations
percent composition of an element in a compound 5
n 3 molar mass of element
3 100% (3.1)
molar mass of compound
actual yield
%yield 5
3 100% (3.4)
theoretical yield
Summary of Facts & Concepts
1. Atomic masses are measured in atomic mass units (amu),
a relative unit based on a value of exactly 12 for the C-12
isotope. The atomic mass given for the atoms of a particular element is the average of the naturally occurring isotope distribution of that element. The molecular mass of a
molecule is the sum of the atomic masses of the atoms in
the molecule. Both atomic mass and molecular mass can
be accurately determined with a mass spectrometer.
2. A mole is Avogadro’s number (6.022 3 1023) of atoms,
molecules, or other particles. The molar mass (in grams)
of an element or a compound is numerically equal to its
mass in atomic mass units (amu) and contains Avogadro’s
number of atoms (in the case of elements), molecules
(in the case of molecular substances), or simplest formula units (in the case of ionic compounds).
3. The percent composition by mass of a compound is the
percent by mass of each element present. If we know
the percent composition by mass of a compound, we
can deduce the empirical formula of the compound and
also the molecular formula of the compound if the
approximate molar mass is known.
4. Chemical changes, called chemical reactions, are represented by chemical equations. Substances that undergo
change—the reactants—are written on the left and the
substances formed—the products—appear to the right
of the arrow. Chemical equations must be balanced, in
accordance with the law of conservation of mass. The
number of atoms of each element in the reactants must
equal the number in the products.
5. Stoichiometry is the quantitative study of products and
reactants in chemical reactions. Stoichiometric calculations are best done by expressing both the known
and unknown quantities in terms of moles and then
converting to other units if necessary. A limiting
reagent is the reactant that is present in the smallest
stoichiometric amount. It limits the amount of product
that can be formed. The amount of product obtained in
a reaction (the actual yield) may be less than the maximum possible amount (the theoretical yield). The
ratio of the two multiplied by 100 percent is expressed
as the percent yield.
Key Words
Actual yield, p. 103
Atomic mass, p. 76
Atomic mass unit (amu), p. 76
Avogadro’s number (NA), p. 77
Chemical equation, p. 90
Chemical reaction, p. 90
Excess reagent, p. 99
Limiting reagent, p. 99
Molar mass (m), p. 78
Mole (mol), p. 77
Mole method, p. 95
Molecular mass, p. 81
Percent composition by
mass, p. 85
Percent yield, p. 103
Product, p. 91
Reactant, p. 91
Stoichiometric amount, p. 99
Stoichiometry, p. 95
Theoretical yield, p. 103
Questions & Problems
• Problems available in Connect Plus
3.2
Red numbered problems solved in Student Solutions Manual
Atomic Mass
Review Questions
3.1
What is an atomic mass unit? Why is it necessary to
introduce such a unit?
3.3
What is the mass (in amu) of a carbon-12 atom?
Why is the atomic mass of carbon listed as
12.01 amu in the table on the inside front cover of
this book?
Explain clearly what is meant by the statement “The
atomic mass of gold is 197.0 amu.”
Questions & Problems
3.4
What information would you need to calculate the
average atomic mass of an element?
Problems
• 3.5
3.6
• 3.7
• 3.8
37
The atomic masses of 35
17Cl (75.53 percent) and 17Cl
(24.47 percent) are 34.968 amu and 36.956 amu,
respectively. Calculate the average atomic mass of
chlorine. The percentages in parentheses denote the
relative abundances.
The atomic masses of 63Li and 73Li are 6.0151 amu
and 7.0160 amu, respectively. Calculate the natural
abundances of these two isotopes. The average
atomic mass of Li is 6.941 amu.
What is the mass in grams of 13.2 amu?
How many amu are there in 8.4 g?
Avogadro’s Number and Molar Mass
Review Questions
3.9
3.10
Define the term “mole.” What is the unit for mole in
calculations? What does the mole have in common
with the pair, the dozen, and the gross? What does
Avogadro’s number represent?
What is the molar mass of an atom? What are the
commonly used units for molar mass?
• 3.21
• 3.22
3.12
• 3.13
• 3.14
• 3.15
• 3.16
• 3.17
•
3.18
• 3.19
• 3.20
Earth’s population is about 7.2 billion. Suppose that
every person on Earth participates in a process of
counting identical particles at the rate of two particles per second. How many years would it take to
count 6.0 3 1023 particles? Assume that there are
365 days in a year.
The thickness of a piece of paper is 0.0036 in. Suppose
a certain book has an Avogadro’s number of pages;
calculate the thickness of the book in light-years. (Hint:
See Problem 1.49 for the definition of light-year.)
How many atoms are there in 5.10 moles of
sulfur (S)?
How many moles of cobalt (Co) atoms are there in
6.00 3 109 (6 billion) Co atoms?
How many moles of calcium (Ca) atoms are in 77.4 g
of Ca?
How many grams of gold (Au) are there in 15.3 moles
of Au?
What is the mass in grams of a single atom of each
of the following elements? (a) Hg, (b) Ne.
What is the mass in grams of a single atom of each
of the following elements? (a) As, (b) Ni.
What is the mass in grams of 1.00 3 1012 lead (Pb)
atoms?
A modern penny weighs 2.5 g but contains only
0.063 g of copper (Cu). How many copper atoms are
present in a modern penny?
Which of the following has more atoms: 1.10 g of
hydrogen atoms or 14.7 g of chromium atoms?
Which of the following has a greater mass: 2 atoms
of lead or 5.1 3 10223 mole of helium.
Molecular Mass
Problems
• 3.23
3.24
• 3.25
• 3.26
• 3.27
• 3.28
Problems
• 3.11
107
• 3.29
• 3.30
Calculate the molecular mass or formula mass (in
amu) of each of the following substances: (a) CH4,
(b) NO2, (c) SO3, (d) C6H6, (e) NaI, (f) K2SO4,
(g) Ca3(PO4)2.
Calculate the molar mass of the following substances:
(a) Li2CO3, (b) CS2, (c) CHCl3 (chloroform),
(d) C6H8O6 (ascorbic acid, or vitamin C), (e) KNO3,
(f) Mg3N2.
Calculate the molar mass of a compound if 0.372 mole
of it has a mass of 152 g.
How many molecules of ethane (C2H6) are present
in 0.334 g of C2H6?
Calculate the number of C, H, and O atoms in 1.50 g
of glucose (C6H12O6), a sugar.
Dimethyl sulfoxide [(CH 3) 2SO], also called
DMSO, is an important solvent that penetrates
the skin, enabling it to be used as a topical
drug-delivery agent. Calculate the number of C,
S, H, and O atoms in 7.14 3 103 g of dimethyl
sulfoxide.
Pheromones are a special type of compound
secreted by the females of many insect species to
attract the males for mating. One pheromone has
the molecular formula C19H38O. Normally, the
amount of this pheromone secreted by a female insect is about 1.0 3 10212 g. How many molecules
are there in this quantity?
The density of water is 1.00 g/mL at 48C. How many
water molecules are present in 2.56 mL of water at
this temperature?
Mass Spectrometry
Review Questions
3.31
3.32
Describe the operation of a mass spectrometer.
Describe how you would determine the isotopic
abundance of an element from its mass spectrum.
Problems
• 3.33
3.34
Carbon has two stable isotopes, 126C and 136C, and
fluorine has only one stable isotope, 199F. How many
peaks would you observe in the mass spectrum of
the positive ion of CF14? Assume that the ion does
not break up into smaller fragments.
Hydrogen has two stable isotopes, 11H and 21H, and
33
34
36
sulfur has four stable isotopes, 32
16S, 16S, 16S, and 16S.
108
Chapter 3 ■ Mass Relationships in Chemical Reactions
How many peaks would you observe in the mass
spectrum of the positive ion of hydrogen sulfide,
H2S1? Assume no decomposition of the ion into
smaller fragments.
Percent Composition and Chemical Formulas
• 3.45
• 3.46
Review Questions
3.35
3.36
3.37
3.38
Use ammonia (NH3) to explain what is meant by the
percent composition by mass of a compound.
Describe how the knowledge of the percent composition by mass of an unknown compound can help us
identify the compound.
What does the word “empirical” in empirical formula mean?
If we know the empirical formula of a compound,
what additional information do we need to determine its molecular formula?
• 3.47
• 3.48
• 3.49
• 3.50
Problems
• 3.39
• 3.40
• 3.41
3.42
• 3.43
3.44
Tin (Sn) exists in Earth’s crust as SnO2. Calculate
the percent composition by mass of Sn and O in
SnO2.
For many years chloroform (CHCl3) was used as an
inhalation anesthetic in spite of the fact that it is also
a toxic substance that may cause severe liver, kidney,
and heart damage. Calculate the percent composition by mass of this compound.
Cinnamic alcohol is used mainly in perfumery,
particularly in soaps and cosmetics. Its molecular
formula is C9H10O. (a) Calculate the percent composition by mass of C, H, and O in cinnamic alcohol.
(b) How many molecules of cinnamic alcohol are
contained in a sample of mass 0.469 g?
All of the substances listed here are fertilizers that
contribute nitrogen to the soil. Which of these is
the richest source of nitrogen on a mass percentage
basis?
(a) Urea, (NH2)2CO
(b) Ammonium nitrate, NH4NO3
(c) Guanidine, HNC(NH2)2
(d) Ammonia, NH3
Allicin is the compound responsible for the characteristic smell of garlic. An analysis of the compound
gives the following percent composition by mass:
C: 44.4 percent; H: 6.21 percent; S: 39.5 percent;
O: 9.86 percent. Calculate its empirical formula.
What is its molecular formula given that its molar
mass is about 162 g?
Peroxyacylnitrate (PAN) is one of the components of smog. It is a compound of C, H, N, and
O. Determine the percent composition of oxygen
and the empirical formula from the following
percent composition by mass: 19.8 percent C,
2.50 percent H, 11.6 percent N. What is its
• 3.51
• 3.52
• 3.53
• 3.54
molecular formula given that its molar mass is
about 120 g?
The formula for rust can be represented by Fe2O3.
How many moles of Fe are present in 24.6 g of the
compound?
How many grams of sulfur (S) are needed to react
completely with 246 g of mercury (Hg) to form
HgS?
Calculate the mass in grams of iodine (I2) that will
react completely with 20.4 g of aluminum (Al) to
form aluminum iodide (AlI3).
Tin(II) fluoride (SnF2) is often added to toothpaste
as an ingredient to prevent tooth decay. What is the
mass of F in grams in 24.6 g of the compound?
What are the empirical formulas of the compounds
with the following compositions? (a) 2.1 percent H,
65.3 percent O, 32.6 percent S, (b) 20.2 percent Al,
79.8 percent Cl.
What are the empirical formulas of the compounds
with the following compositions? (a) 40.1 percent
C, 6.6 percent H, 53.3 percent O, (b) 18.4 percent C,
21.5 percent N, 60.1 percent K.
The anticaking agent added to Morton salt is calcium silicate, CaSiO3. This compound can absorb
up to 2.5 times its mass of water and still remains a
free-flowing powder. Calculate the percent composition of CaSiO3.
The empirical formula of a compound is CH. If the
molar mass of this compound is about 78 g, what is
its molecular formula?
The molar mass of caffeine is 194.19 g. Is the
molecular formula of caffeine C 4H 5N 2O or
C8H10N4O2?
Monosodium glutamate (MSG), a food-flavor
enhancer, has been blamed for “Chinese restaurant
syndrome,” the symptoms of which are headaches
and chest pains. MSG has the following composition by mass: 35.51 percent C, 4.77 percent H,
37.85 percent O, 8.29 percent N, and 13.60 percent
Na. What is its molecular formula if its molar mass
is about 169 g?
Chemical Reactions and Chemical Equations
Review Questions
3.55
3.56
3.57
3.58
Use the formation of water from hydrogen and oxygen to explain the following terms: chemical reaction, reactant, product.
What is the difference between a chemical reaction
and a chemical equation?
Why must a chemical equation be balanced? What
law is obeyed by a balanced chemical equation?
Write the symbols used to represent gas, liquid,
solid, and the aqueous phase in chemical
equations.
Questions & Problems
109
Problems
• 3.59
3.60
Balance the following equations using the method
outlined in Section 3.7:
(a) C 1 O2 ¡ CO
(b) CO 1 O2 ¡ CO2
(c) H2 1 Br2 ¡ HBr
(d) K 1 H2O ¡ KOH 1 H2
(e) Mg 1 O2 ¡ MgO
(f) O3 ¡ O2
(g) H2O2 ¡ H2O 1 O2
(h) N2 1 H2 ¡ NH3
(i) Zn 1 AgCl ¡ ZnCl2 1 Ag
(j) S8 1 O2 ¡ SO2
(k) NaOH 1 H2SO4 ¡ Na2SO4 1 H2O
(l) Cl2 1 NaI ¡ NaCl 1 I2
(m) KOH 1 H3PO4 ¡ K3PO4 1 H2O
(n) CH4 1 Br2 ¡ CBr4 1 HBr
Balance the following equations using the method
outlined in Section 3.7:
(a) N2O5 ¡ N2O4 1 O2
(b) KNO3 ¡ KNO2 1 O2
(c) NH4NO3 ¡ N2O 1 H2O
(d) NH4NO2 ¡ N2 1 H2O
(e) NaHCO3 ¡ Na2CO3 1 H2O 1 CO2
(f) P4O10 1 H2O ¡ H3PO4
(g) HCl 1 CaCO3 ¡ CaCl2 1 H2O 1 CO2
(h) Al 1 H2SO4 ¡ Al2 (SO4 ) 3 1 H2
(i) CO2 1 KOH ¡ K2CO3 1 H2O
(j) CH4 1 O2 ¡ CO2 1 H2O
(k) Be2C 1 H2O ¡ Be(OH) 2 1 CH4
(l) Cu 1 HNO3 ¡ Cu(NO3 ) 2 1 NO 1 H2O
(m) S 1 HNO3 ¡ H2SO4 1 NO2 1 H2O
(n) NH3 1 CuO ¡ Cu 1 N2 1 H2O
A
8n
C
D
3.64
3.61
3.62
8n
D
• 3.65
Consider the combustion of carbon monoxide (CO)
in oxygen gas:
2CO(g) 1 O2 (g) ¡ 2CO2 (g)
• 3.66
Starting with 3.60 moles of CO, calculate the number of moles of CO2 produced if there is enough
oxygen gas to react with all of the CO.
Silicon tetrachloride (SiCl4) can be prepared by
heating Si in chlorine gas:
Si(s) 1 2Cl2 (g) ¡ SiCl4 (l)
• 3.67
On what law is stoichiometry based? Why is it
essential to use balanced equations in solving stoichiometric problems?
Describe the steps involved in the mole method.
Which of the following equations best represents the
reaction shown in the diagram?
(a) 8A 1 4B ¡ C 1 D
(b) 4A 1 8B ¡ 4C 1 4D
(c) 2A 1 B ¡ C 1 D
(d) 4A 1 2B ¡ 4C 1 4D
(e) 2A 1 4B ¡ C 1 D
B
C
In one reaction, 0.507 mole of SiCl4 is produced.
How many moles of molecular chlorine were used
in the reaction?
Ammonia is a principal nitrogen fertilizer. It is
prepared by the reaction between hydrogen and
nitrogen.
3H2 (g) 1 N2 (g) ¡ 2NH3 (g)
Problems
• 3.63
Which of the following equations best represents the
reaction shown in the diagram?
(a) A 1 B ¡ C 1 D
(b) 6A 1 4B ¡ C 1 D
(c) A 1 2B ¡ 2C 1 D
(d) 3A 1 2B ¡ 2C 1 D
(e) 3A 1 2B ¡ 4C 1 2D
A
Amounts of Reactants and Products
Review Questions
B
• 3.68
In a particular reaction, 6.0 moles of NH3 were produced. How many moles of H2 and how many moles
of N2 were reacted to produce this amount of NH3?
Certain race cars use methanol (CH3OH, also called
wood alcohol) as a fuel. The combustion of methanol occurs according to the following equation:
2CH3OH(l) 1 3O2 (g) ¡ 2CO2 (g) 1 4H2O(l)
In a particular reaction, 9.8 moles of CH3OH are
reacted with an excess of O2. Calculate the number
of moles of H2O formed.
110
• 3.69
Chapter 3 ■ Mass Relationships in Chemical Reactions
The annual production of sulfur dioxide from burning coal and fossil fuels, auto exhaust, and other
sources is about 26 million tons. The equation for
the reaction is
• 3.76
S(s) 1 O2 (g) ¡ SO2 (g)
• 3.70
• 3.71
How much sulfur (in tons), present in the original
materials, would result in that quantity of SO2?
When baking soda (sodium bicarbonate or sodium
hydrogen carbonate, NaHCO3) is heated, it releases
carbon dioxide gas, which is responsible for the rising of cookies, donuts, and bread. (a) Write a balanced equation for the decomposition of the
compound (one of the products is Na2CO3). (b) Calculate the mass of NaHCO3 required to produce
20.5 g of CO2.
If chlorine bleach is mixed with other cleaning products containing ammonia, the toxic gas NCl3(g) can
form according to the equation:
• 3.77
Nitrous oxide (N2O) is also called “laughing gas.” It
can be prepared by the thermal decomposition of
ammonium nitrate (NH4NO3). The other product is
H2O. (a) Write a balanced equation for this reaction.
(b) How many grams of N2O are formed if 0.46
mole of NH4NO3 is used in the reaction?
The fertilizer ammonium sulfate [(NH4)2SO4] is prepared by the reaction between ammonia (NH3) and
sulfuric acid:
2NH3 (g) 1 H2SO4 (aq) ¡ (NH4 ) 2SO4 (aq)
• 3.78
How many kilograms of NH3 are needed to produce
1.00 3 105 kg of (NH4)2SO4?
A common laboratory preparation of oxygen gas is
the thermal decomposition of potassium chlorate
(KClO3). Assuming complete decomposition, calculate the number of grams of O2 gas that can be
obtained from 46.0 g of KClO3. (The products are
KCl and O2.)
3NaClO(aq) 1 NH3 (aq) ¡ 3NaOH(aq) 1 NCl3 (g)
• 3.72
When 2.94 g of NH3 reacts with an excess of NaClO
according to the preceding reaction, how many
grams of NCl3 are formed?
Fermentation is a complex chemical process of wine
making in which glucose is converted into ethanol
and carbon dioxide:
C6H12O6 ¡ 2C2H5OH 1 2CO2
glucose
• 3.73
ethanol
Starting with 500.4 g of glucose, what is the maximum amount of ethanol in grams and in liters that can
be obtained by this process? (Density of ethanol 5
0.789 g/mL.)
Each copper(II) sulfate unit is associated with five
water molecules in crystalline copper(II) sulfate
pentahydrate (CuSO4 ? 5H2O). When this compound
is heated in air above 1008C, it loses the water molecules and also its blue color:
Limiting Reagents
Review Questions
3.79
3.80
Problems
• 3.81
Consider the reaction
2A 1 B ¡ C
(a) In the diagram here that represents the reaction,
which reactant, A or B, is the limiting reagent?
(b) Assuming complete reaction, draw a molecularmodel representation of the amounts of reactants
and products left after the reaction. The atomic
arrangement in C is ABA.
CuSO4 ? 5H2O ¡ CuSO4 1 5H2O
• 3.74
Define limiting reagent and excess reagent. What is
the significance of the limiting reagent in predicting
the amount of the product obtained in a reaction?
Can there be a limiting reagent if only one reactant
is present?
Give an everyday example that illustrates the limiting reagent concept.
If 9.60 g of CuSO4 are left after heating 15.01 g of
the blue compound, calculate the number of moles
of H2O originally present in the compound.
For many years the recovery of gold—that is, the
separation of gold from other materials—involved
the use of potassium cyanide:
A
B
4Au 1 8KCN 1 O2 1 2H2O ¡
4KAu(CN) 2 1 4KOH
• 3.75
What is the minimum amount of KCN in moles
needed to extract 29.0 g (about an ounce) of gold?
Limestone (CaCO3) is decomposed by heating to
quicklime (CaO) and carbon dioxide. Calculate how
many grams of quicklime can be produced from 1.0 kg
of limestone.
3.82
Consider the reaction
N2 1 3H2 ¡ 2NH3
Questions & Problems
Assuming each model represents 1 mole of the
substance, show the number of moles of the product and the excess reagent left after the complete
reaction.
111
Problems
• 3.89
Hydrogen fluoride is used in the manufacture of
Freons (which destroy ozone in the stratosphere)
and in the production of aluminum metal. It is prepared by the reaction
CaF2 1 H2SO4 ¡ CaSO4 1 2HF
H2
N2
• 3.90
NH3
In one process, 6.00 kg of CaF2 are treated with an
excess of H2SO4 and yield 2.86 kg of HF. Calculate
the percent yield of HF.
Nitroglycerin (C3H5N3O9) is a powerful explosive.
Its decomposition may be represented by
4C3H5N3O9 ¡ 6N2 1 12CO2 1 10H2O 1 O2
• 3.83
Nitric oxide (NO) reacts with oxygen gas to form
nitrogen dioxide (NO2), a dark-brown gas:
2NO(g) 1 O2 (g) ¡ 2NO2 (g)
• 3.84
• 3.85
In one experiment 0.886 mole of NO is mixed with
0.503 mole of O2. Calculate which of the two reactants is the limiting reagent. Calculate also the number of moles of NO2 produced.
Ammonia and sulfuric acid react to form ammonium sulfate. (a) Write an equation for the reaction. (b) Determine the starting mass (in g) of
each reactant if 20.3 g of ammonium sulfate is
produced and 5.89 g of sulfuric acid remains
unreacted.
Propane (C3H8) is a component of natural gas and is
used in domestic cooking and heating. (a) Balance
the following equation representing the combustion
of propane in air:
• 3.91
FeTiO3 1 H2SO4 ¡ TiO2 1 FeSO4 1 H2O
3.92
• 3.86
• 3.93
Reaction Yield
Review Questions
3.87
3.88
Why is the theoretical yield of a reaction determined
only by the amount of the limiting reagent?
Why is the actual yield of a reaction almost always
smaller than the theoretical yield?
If the yield of ethylene production is 42.5 percent, what
mass of hexane must be reacted to produce 481 g of
ethylene?
When heated, lithium reacts with nitrogen to form
lithium nitride:
6Li(s) 1 N2 (g) ¡ 2Li3N(s)
MnO2 1 4HCl ¡ MnCl2 1 Cl2 1 2H2O
If 0.86 mole of MnO2 and 48.2 g of HCl react, which
reagent will be used up first? How many grams of
Cl2 will be produced?
Its opaque and nontoxic properties make it suitable
as a pigment in plastics and paints. In one process,
8.00 3 103 kg of FeTiO3 yielded 3.67 3 103 kg of
TiO2. What is the percent yield of the reaction?
Ethylene (C2H4), an important industrial organic
chemical, can be prepared by heating hexane (C6H14)
at 8008C:
C6H14 ¡ C2H4 1 other products
C3H8 1 O2 ¡ CO2 1 H2O
(b) How many grams of carbon dioxide can be
produced by burning 3.65 moles of propane?
Assume that oxygen is the excess reagent in this
reaction.
Consider the reaction
This reaction generates a large amount of heat and
many gaseous products. It is the sudden formation
of these gases, together with their rapid expansion,
that produces the explosion. (a) What is the maximum amount of O2 in grams that can be obtained
from 2.00 3 102 g of nitroglycerin? (b) Calculate
the percent yield in this reaction if the amount of O2
generated is found to be 6.55 g.
Titanium(IV) oxide (TiO2) is a white substance produced by the action of sulfuric acid on the mineral
ilmenite (FeTiO3):
3.94
What is the theoretical yield of Li3N in grams when
12.3 g of Li are heated with 33.6 g of N2? If the
actual yield of Li3N is 5.89 g, what is the percent
yield of the reaction?
Disulfide dichloride (S2Cl2) is used in the vulcanization
of rubber, a process that prevents the slippage of rubber
molecules past one another when stretched. It is prepared by heating sulfur in an atmosphere of chlorine:
S8 (l) 1 4Cl2 (g) ¡ 4S2Cl2 (l)
What is the theoretical yield of S2Cl2 in grams when
4.06 g of S8 are heated with 6.24 g of Cl2? If the actual
yield of S2Cl2 is 6.55 g, what is the percent yield?
112
Chapter 3 ■ Mass Relationships in Chemical Reactions
Additional Problems
3.95
3.96
• 3.97
3.99
Gallium is an important element in the production of
semiconductors. The average atomic mass of 69
31 Ga
(68.9256 amu) and 71
31 Ga (70.9247 amu) is 69.72 amu.
Calculate the natural abundances of the gallium
isotopes.
Rubidium is used in “atomic clocks” and other precise electronic equipment. The average atomic mass
87
of 85
37 Rb (84.912 amu) and 37 Rb (86.909 amu) is
85.47 amu. Calculate the natural abundances of the
rubidium isotopes.
The following diagram represents the products
(CO2 and H2O) formed after the combustion of a
hydrocarbon (a compound containing only C and H
atoms). Write an equation for the reaction. (Hint:
The molar mass of the hydrocarbon is about 30 g.)
C2H4 (g) 1 HCl(g) ¡ C2H5Cl(g)
3.100
CO2
H2O
Ethylene reacts with hydrogen chloride to form
ethyl chloride:
• 3.101
Calculate the mass of ethyl chloride formed if 4.66 g
of ethylene reacts with an 89.4 percent yield.
Write balanced equations for the following reactions
described in words.
(a) Pentane burns in oxygen to form carbon
dioxide and water.
(b) Sodium bicarbonate reacts with hydrochloric
acid to form carbon dioxide, sodium chloride,
and water.
(c) When heated in an atmosphere of nitrogen,
lithium forms lithium nitride.
(d) Phosphorus trichloride reacts with water to
form phosphorus acid and hydrogen chloride.
(e) Copper(II) oxide heated with ammonia will
form copper, nitrogen gas, and water.
Industrially, nitric acid is produced by the Ostwald
process represented by the following equations:
4NH3 (g) 1 5O2 (g) ¡ 4NO(g) 1 6H2O(l)
2NO(g) 1 O2 (g) ¡ 2NO2 (g)
2NO2 (g) 1 H2O(l) ¡ HNO3 (aq) 1 HNO2 (aq)
• 3.98
Consider the reaction of hydrogen gas with oxygen gas:
2H2 (g) 1 O2 (g) ¡ 2H2O(g)
H2
O2
H2O
•
Assuming complete reaction, which of the diagrams
shown next represents the amounts of reactants and
products left after the reaction?
(a)
(b)
(c)
(d)
•
What mass of NH3 (in g) must be used to produce
1.00 ton of HNO3 by the above procedure, assuming
an 80 percent yield in each step? (1 ton 5 2000 lb;
1 lb 5 453.6 g.)
3.102 A sample of a compound of Cl and O reacts with an
excess of H2 to give 0.233 g of HCl and 0.403 g of H2O.
Determine the empirical formula of the compound.
3.103 How many grams of H2O will be produced from the
complete combustion of 26.7 g of butane (C4H10)?
3.104 A 26.2-g sample of oxalic acid hydrate (H2C2O4 ?
2H2O) is heated in an oven until all the water is
driven off. How much of the anhydrous acid is left?
3.105 The atomic mass of element X is 33.42 amu. A
27.22-g sample of X combines with 84.10 g of
another element Y to form a compound XY. Calculate the atomic mass of Y.
3.106 How many moles of O are needed to combine with
0.212 mole of C to form (a) CO and (b) CO2?
3.107 A research chemist used a mass spectrometer to study
the two isotopes of an element. Over time, she recorded
a number of mass spectra of these isotopes. On analysis, she noticed that the ratio of the taller peak (the
more abundant isotope) to the shorter peak (the less
abundant isotope) gradually increased with time. Assuming that the mass spectrometer was functioning
normally, what do you think was causing this change?
3.108 The aluminum sulfate hydrate [Al2(SO4)3 ? xH2O]
contains 8.10 percent Al by mass. Calculate x, that
is, the number of water molecules associated with
each Al2(SO4)3 unit.
113
Questions & Problems
3.109
3.110
• 3.111
• 3.112
• 3.113
• 3.114
The explosive nitroglycerin (C3H5N3O9) has also been
used as a drug to treat heart patients to relieve pain
(angina pectoris). We now know that nitroglycerin produces nitric oxide (NO), which causes muscles to relax
and allows the arteries to dilate. If each nitroglycerin
molecule releases one NO per atom of N, calculate the
mass percent of NO available from nitroglycerin.
The carat is the unit of mass used by jewelers. One
carat is exactly 200 mg. How many carbon atoms
are present in a 24-carat diamond?
An iron bar weighed 664 g. After the bar had been
standing in moist air for a month, exactly one-eighth
of the iron turned to rust (Fe2O3). Calculate the final
mass of the iron bar and rust.
A certain metal oxide has the formula MO where M
denotes the metal. A 39.46-g sample of the compound is strongly heated in an atmosphere of hydrogen to remove oxygen as water molecules. At the
end, 31.70 g of the metal is left over. If O has an
atomic mass of 16.00 amu, calculate the atomic
mass of M and identify the element.
An impure sample of zinc (Zn) is treated with an
excess of sulfuric acid (H2SO4) to form zinc sulfate
(ZnSO4) and molecular hydrogen (H2). (a) Write a
balanced equation for the reaction. (b) If 0.0764 g of
H2 is obtained from 3.86 g of the sample, calculate
the percent purity of the sample. (c) What assumptions must you make in (b)?
One of the reactions that occurs in a blast furnace,
where iron ore is converted to cast iron, is
3.119
•
Hemoglobin (C2952H4664N812O832S8Fe4) is the oxygen
carrier in blood. (a) Calculate its molar mass. (b) An
average adult has about 5.0 L of blood. Every milliliter
of blood has approximately 5.0 3 109 erythrocytes, or
red blood cells, and every red blood cell has about
2.8 3 108 hemoglobin molecules. Calculate the mass
of hemoglobin molecules in grams in an average adult.
3.120 Myoglobin stores oxygen for metabolic processes in
muscle. Chemical analysis shows that it contains
0.34 percent Fe by mass. What is the molar mass of
myoglobin? (There is one Fe atom per molecule.)
3.121 Calculate the number of cations and anions in each
of the following compounds: (a) 0.764 g of CsI,
(b) 72.8 g of K2Cr2O7, (c) 6.54 g of Hg2(NO3)2.
3.122 A mixture of NaBr and Na2SO4 contains 29.96 percent Na by mass. Calculate the percent by mass of
each compound in the mixture.
3.123 Consider the reaction 3A 1 2B S 3C. A student
mixed 4.0 moles of A with 4.0 moles of B and
obtained 2.8 moles of C. What is the percent yield of
the reaction?
3.124 Balance the following equation shown in molecular
models.
1
• 3.125
Fe2O3 1 3CO ¡ 2Fe 1 3CO2
• 3.115
3.116
3.117
• 3.118
Suppose that 1.64 3 103 kg of Fe are obtained from
a 2.62 3 103-kg sample of Fe2O3. Assuming that the
reaction goes to completion, what is the percent purity of Fe2O3 in the original sample?
Carbon dioxide (CO2) is the gas that is mainly
responsible for global warming (the greenhouse
effect). The burning of fossil fuels is a major cause of
the increased concentration of CO2 in the atmosphere.
Carbon dioxide is also the end product of metabolism
(see Example 3.13). Using glucose as an example
of food, calculate the annual human production of
CO2 in grams, assuming that each person consumes
5.0 3 102 g of glucose per day. The world’s population is 7.2 billion, and there are 365 days in a year.
Carbohydrates are compounds containing carbon,
hydrogen, and oxygen in which the hydrogen to
oxygen ratio is 2:1. A certain carbohydrate contains
40.0 percent carbon by mass. Calculate the empirical and molecular formulas of the compound if the
approximate molar mass is 178 g.
Which of the following has the greater mass: 0.72 g of
O2 or 0.0011 mole of chlorophyll (C55H72MgN4O5)?
Analysis of a metal chloride XCl3 shows that it contains 67.2 percent Cl by mass. Calculate the molar
mass of X and identify the element.
8n
Aspirin or acetyl salicylic acid is synthesized by
reacting salicylic acid with acetic anhydride:
C7H6O3 1 C4H6O3
salicylic acid
• 3.126
3.127
3.128
1
acetic anhydride
¡ C9H8O4 1 C2H4O2
aspirin
acetic acid
(a) How much salicylic acid is required to produce
0.400 g of aspirin (about the content in a tablet), assuming acetic anhydride is present in excess?
(b) Calculate the amount of salicylic acid needed if
only 74.9 percent of salicylic acid is converted to
aspirin. (c) In one experiment, 9.26 g of salicylic
acid is reacted with 8.54 g of acetic anhydride. Calculate the theoretical yield of aspirin and the percent
yield if only 10.9 g of aspirin is produced.
Calculate the percent composition by mass of all the
elements in calcium phosphate [Ca3(PO4)2], a major
component of bone.
Lysine, an essential amino acid in the human body,
contains C, H, O, and N. In one experiment, the
complete combustion of 2.175 g of lysine gave 3.94 g
CO2 and 1.89 g H2O. In a separate experiment,
1.873 g of lysine gave 0.436 g NH3. (a) Calculate the
empirical formula of lysine. (b) The approximate
molar mass of lysine is 150 g. What is the molecular
formula of the compound?
Does 1 g of hydrogen molecules contain as many
H atoms as 1 g of hydrogen atoms?
114
Chapter 3 ■ Mass Relationships in Chemical Reactions
3.129
•
Avogadro’s number has sometimes been described
as a conversion factor between amu and grams. Use
the fluorine atom (19.00 amu) as an example to
show the relation between the atomic mass unit and
the gram.
3.130 The natural abundances of the two stable isotopes of
hydrogen (hydrogen and deuterium) are 11H: 99.985
percent and 21H: 0.015 percent. Assume that water
exists as either H2O or D2O. Calculate the number
of D2O molecules in exactly 400 mL of water.
(Density 5 1.00 g/mL.)
3.131 A compound containing only C, H, and Cl was examined in a mass spectrometer. The highest mass
peak seen corresponds to an ion mass of 52 amu.
The most abundant mass peak seen corresponds to
an ion mass of 50 amu and is about three times as
intense as the peak at 52 amu. Deduce a reasonable
molecular formula for the compound and explain
the positions and intensities of the mass peaks
mentioned. (Hint: Chlorine is the only element that
has isotopes in comparable abundances: 35
17Cl: 75.5
1
percent; 35
17Cl: 24.5 percent. For H, use 1H; for C,
use 121C.)
3.132 In the formation of carbon monoxide, CO, it is
found that 2.445 g of carbon combine with 3.257 g
of oxygen. What is the atomic mass of oxygen if the
atomic mass of carbon is 12.01 amu?
3.133 What mole ratio of molecular chlorine (Cl2) to
molecular oxygen (O2) would result from the
breakup of the compound Cl2O7 into its constituent
elements?
3.134 Which of the following substances contains the
greatest mass of chlorine? (a) 5.0 g Cl2, (b) 60.0 g
NaClO3, (c) 0.10 mol KCl, (d) 30.0 g MgCl2,
(e) 0.50 mol Cl2.
3.135 A compound made up of C, H, and Cl contains 55.0
percent Cl by mass. If 9.00 g of the compound contain 4.19 3 1023 H atoms, what is the empirical formula of the compound?
3.136 Platinum forms two different compounds with
chlorine. One contains 26.7 percent Cl by mass,
and the other contains 42.1 percent Cl by mass.
Determine the empirical formulas of the two
compounds.
3.137 The following reaction is stoichiometric as written
•
•
because of their role in systems that convert solar
energy to electricity. The compound [Ru(C10H8N2)3]
Cl2 is synthesized by reacting RuCl3 ? 3H2O(s)
with three molar equivalents of C10H8N2(s), along
with an excess of triethylamine, N(C2H5)3(l), to
convert ruthenium(III) to ruthenium(II). The density of triethylamine is 0.73 g/mL, and typically
eight molar equivalents are used in the synthesis.
(a) Assuming that you start with 6.5 g of RuCl3 ?
3H2O, how many grams of C10H8N2 and what volume of N(C2H5)3 should be used in the reaction?
(b) Given that the yield of this reaction is 91 percent, how many grams of [Ru(C10H8N2)3]Cl2 will
be obtained?
3.139 Heating 2.40 g of the oxide of metal X (molar
mass of X 5 55.9 g/mol) in carbon monoxide
(CO) yields the pure metal and carbon dioxide.
The mass of the metal product is 1.68 g. From the
data given, show that the simplest formula of the
oxide is X2O3 and write a balanced equation for
the reaction.
3.140 A compound X contains 63.3 percent manganese
(Mn) and 36.7 percent O by mass. When X is heated,
oxygen gas is evolved and a new compound Y containing 72.0 percent Mn and 28.0 percent O is
formed. (a) Determine the empirical formulas of X
and Y. (b) Write a balanced equation for the conversion of X to Y.
3.141 The formula of a hydrate of barium chloride is
BaCl2 ? xH2O. If 1.936 g of the compound gives
1.864 g of anhydrous BaSO4 upon treatment with
sulfuric acid, calculate the value of x.
3.142 It is estimated that the day Mt. St. Helens erupted
(May 18, 1980), about 4.0 3 105 tons of SO2 were
released into the atmosphere. If all the SO2 were
eventually converted to sulfuric acid, how many tons
of H2SO4 were produced?
3.143 Cysteine, shown here, is one of the 20 amino acids
found in proteins in humans. Write the molecular
formula and calculate its percent composition
by mass.
H
C4H9Cl 1 NaOC2H5 ¡ C4H8 1 C2H5OH 1 NaCl
but it is often carried out with an excess of
NaOC2H5 to react with any water present in the
reaction mixture that might reduce the yield. If the
reaction shown was carried out with 6.83 g of
C4H9Cl, how many grams of NaOC2H5 would be
needed to have a 50 percent molar excess of that
reactant?
3.138 Compounds containing ruthenium(II) and bipyridine, C10H8N2, have received considerable interest
S
O
C
Questions & Problems
• 3.144
Isoflurane, shown here, is a common inhalation
anesthetic. Write its molecular formula and calculate its percent composition by mass.
• 3.151
Cl
O
C
• 3.152
H
F
• 3.145
•
•
•
•
A mixture of CuSO4 ? 5H2O and MgSO4 ? 7H2O is
heated until all the water is lost. If 5.020 g of the
mixture gives 2.988 g of the anhydrous salts, what
is the percent by mass of CuSO4 ? 5H2O in the
mixture?
3.146 When 0.273 g of Mg is heated strongly in a nitrogen
(N2) atmosphere, a chemical reaction occurs. The
product of the reaction weighs 0.378 g. Calculate
the empirical formula of the compound containing
Mg and N. Name the compound.
3.147 A mixture of methane (CH4) and ethane (C2H6) of
mass 13.43 g is completely burned in oxygen. If the
total mass of CO2 and H2O produced is 64.84 g,
calculate the fraction of CH4 in the mixture.
3.148 Leaded gasoline contains an additive to prevent
engine “knocking.” On analysis, the additive compound is found to contain carbon, hydrogen, and
lead (Pb) (hence, “leaded gasoline”). When 51.36 g
of this compound are burned in an apparatus such as
that shown in Figure 3.6, 55.90 g of CO2 and 28.61 g
of H2O are produced. Determine the empirical formula of the gasoline additive.
3.149 Because of its detrimental effect on the environment,
the lead compound described in Problem 3.148 has
been replaced by methyl tert-butyl ether (a compound
of C, H, and O) to enhance the performance of gasoline. (This compound is also being phased out because
of its contamination of drinking water.) When 12.1 g
of the compound are burned in an apparatus like the
one shown in Figure 3.6, 30.2 g of CO2 and 14.8 g of
H2O are formed. What is the empirical formula of the
compound?
3.150 Suppose you are given a cube made of magnesium
(Mg) metal of edge length 1.0 cm. (a) Calculate
the number of Mg atoms in the cube. (b) Atoms
are spherical in shape. Therefore, the Mg atoms in
the cube cannot fill all of the available space. If
only 74 percent of the space inside the cube is
taken up by Mg atoms, calculate the radius in picometers of a Mg atom. (The density of Mg is
3.153
3.154
3.155
• 3.156
• 3.157
115
1.74 g/cm3 and the volume of a sphere of radius r
is 43πr3.)
A certain sample of coal contains 1.6 percent sulfur by mass. When the coal is burned, the sulfur is
converted to sulfur dioxide. To prevent air pollution, this sulfur dioxide is treated with calcium
oxide (CaO) to form calcium sulfite (CaSO3).
Calculate the daily mass (in kilograms) of CaO
needed by a power plant that uses 6.60 3 106 kg
of coal per day.
Air is a mixture of many gases. However, in calculating its “molar mass” we need consider only the
three major components: nitrogen, oxygen, and
argon. Given that one mole of air at sea level is
made up of 78.08 percent nitrogen, 20.95 percent
oxygen, and 0.97 percent argon, what is the molar
mass of air?
(a) Determine the mass of calcium metal that contains the same number of moles as 89.6 g of zinc
metal. (b) Calculate the number of moles of molecular fluorine that has the same mass as 36.9 moles of
argon. (c) What is the mass of sulfuric acid that contains 0.56 mole of oxygen atoms? (d) Determine the
number of moles of phosphoric acid that contains
2.12 g of hydrogen atoms.
A major industrial use of hydrochloric acid is in
metal pickling. This process involves the removal
of metal oxide layers from metal surfaces to prepare them for coating. (a) Write an equation
between iron(III) oxide, which represents the rust
layer over iron, and HCl to form iron(III) chloride
and water. (b) If 1.22 moles of Fe2O3 and 289.2 g
of HCl react, how many grams of FeCl3 will be
produced?
Octane (C8H18) is a component of gasoline. Complete combustion of octane yields H2O and CO2. Incomplete combustion produces H2O and CO, which
not only reduces the efficiency of the engine using
the fuel but is also toxic. In a certain test run, 1.000 gal
of octane is burned in an engine. The total mass of
CO, CO2, and H2O produced is 11.53 kg. Calculate
the efficiency of the process; that is, calculate the
fraction of octane converted to CO2. The density of
octane is 2.650 kg/gal.
Industrially, hydrogen gas can be prepared by reacting propane gas (C3H8) with steam at about
4008C. The products are carbon monoxide (CO)
and hydrogen gas (H2). (a) Write a balanced equation for the reaction. (b) How many kilograms
of H2 can be obtained from 2.84 3 103 kg of
propane?
In a natural product synthesis, a chemist prepares a
complex biological molecule entirely from nonbiological starting materials. The target molecules are
often known to have some promise as therapeutic
agents, and the organic reactions that are developed
116
3.158
Chapter 3 ■ Mass Relationships in Chemical Reactions
along the way benefit all chemists. The overall synthesis, however, requires many steps, so it is important to have the best possible percent yields at each
step. What is the overall percent yield for such a
synthesis that has 24 steps with an 80 percent yield
at each step?
What is wrong or ambiguous with each of the statements here?
(a) NH4NO2 is the limiting reagent in the reaction
NH4NO2 (s) ¡ N2 (g) 1 2H2O(l)
(b) The limiting reagents for the reaction shown
here are NH3 and NaCl.
• 3.161
NH3 (aq) 1 NaCl(aq) 1 H2CO3 (aq) ¡
NaHCO3 (aq) 1 NH4Cl(aq)
•
3.159 (a) For molecules having small molecular masses,
mass spectrometry can be used to identify their formulas. To illustrate this point, identify the molecule that most likely accounts for the observation
of a peak in a mass spectrum at: 16 amu, 17 amu,
18 amu, and 64 amu. (b) Note that there are (among
others) two likely molecules that would give rise to
a peak at 44 amu, namely, C3H8 and CO2. In such
cases, a chemist might try to look for other peaks
generated when some of the molecules break apart
in the spectrometer. For example, if a chemist sees
a peak at 44 amu and also one at 15 amu, which
molecule is producing the 44-amu peak? Why?
(c) Using the following precise atomic masses—
1
H (1.00797 amu), 12C (12.00000 amu), and 16O
(15.99491 amu)—how precisely must the masses
of C3H8 and CO2 be measured to distinguish
between them?
3.160 Potash is any potassium mineral that is used for its
potassium content. Most of the potash produced in
• 3.162
3.163
• 3.164
3.165
the United States goes into fertilizer. The major
sources of potash are potassium chloride (KCl) and
potassium sulfate (K2SO4). Potash production is
often reported as the potassium oxide (K2O) equivalent or the amount of K2O that could be made
from a given mineral. (a) If KCl costs $0.55 per kg,
for what price (dollar per kg) must K2SO4 be sold
to supply the same amount of potassium on a per
dollar basis? (b) What mass (in kg) of K2O contains
the same number of moles of K atoms as 1.00 kg
of KCl?
A 21.496-g sample of magnesium is burned in air to
form magnesium oxide and magnesium nitride.
When the products are treated with water, 2.813 g of
gaseous ammonia are generated. Calculate the
amounts of magnesium nitride and magnesium
oxide formed.
A certain metal M forms a bromide containing 53.79
percent Br by mass. What is the chemical formula of
the compound?
A sample of iron weighing 15.0 g was heated with
potassium chlorate (KClO3) in an evacuated container. The oxygen generated from the decomposition of KClO3 converted some of the Fe to Fe2O3. If
the combined mass of Fe and Fe2O3 was 17.9 g, calculate the mass of Fe2O3 formed and the mass of
KClO3 decomposed.
A sample containing NaCl, Na2SO4, and NaNO3
gives the following elemental analysis: Na: 32.08 percent; O: 36.01 percent; Cl: 19.51 percent. Calculate
the mass percent of each compound in the sample.
A sample of 10.00 g of sodium reacts with oxygen
to form 13.83 g of sodium oxide (Na2O) and sodium
peroxide (Na2O2). Calculate the percent composition of the mixture.
Interpreting, Modeling & Estimating
3.166
While most isotopes of light elements such as oxygen and phosphorus contain relatively equal numbers of protons and neutrons, recent results indicate
that a new class of isotopes called neutron-rich isotopes can be prepared. These neutron-rich isotopes
push the limits of nuclear stability as the large number of neutrons approach the “neutron drip line.”
They may play a critical role in the nuclear reactions
of stars. An unusually heavy isotope of aluminum
(43
13Al) has been reported. How many more neutrons
does this atom contain compared to an average
aluminum atom?
3.167
3.168
Without doing any detailed calculations, arrange the
following substances in the increasing order of number of moles: 20.0 g Cl, 35.0 g Br, and 94.0 g I.
Without doing any detailed calculations, estimate
which element has the highest percent composition
by mass in each of the following compounds:
(a) Hg(NO3)2
(b) NF3
(c) K2Cr2O7
(d) C2952H4664N812O832S8Fe4
117
Answers to Practice Exercises
3.169
Consider the reaction
number using stearic acid (C18H36O2) shown here.
When stearic acid is added to water, its molecules
collect at the surface and form a monolayer; that is,
the layer is only one molecule thick. The crosssectional area of each stearic acid molecule has been
measured to be 0.21 nm2. In one experiment it is
found that 1.4 3 1024 g of stearic acid is needed to
form a monolayer over water in a dish of diameter
20 cm. Based on these measurements, what is
Avogadro’s number?
6Li(s) 1 N2 (g) ¡ 2Li3N(s)
3.170
3.171
Without doing any detailed calculations, choose one
of the following combinations in which nitrogen is
the limiting reagent:
(a) 44 g Li and 38 g N2
(b) 1380 g Li and 842 g N2
(c) 1.1 g Li and 0.81 g N2
Estimate how high in miles you can stack up an
Avogadro’s number of oranges covering the entire
Earth.
The following is a crude but effective method for
estimating the order of magnitude of Avogadro’s
H3C
CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 OH
CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 C
O
Answers to Practice Exercises
3.1 63.55 amu. 3.2 3.59 moles. 3.3 2.57 3 103 g.
3.4 1.0 3 10220 g. 3.5 32.04 amu. 3.6 1.66 moles.
3.7 5.81 3 1024 H atoms. 3.8 H: 2.055%; S: 32.69%;
O: 65.25%. 3.9 KMnO4 (potassium permanganate).
3.10 196 g. 3.11 B2H6. 3.12 Fe2O3 1 3CO S 2Fe 1 3CO2.
3.13 235 g. 3.14 0.769 g. 3.15 (a) 234 g, (b) 234 g.
3.16 25.1 g. 3.17 (a) 863 g, (b) 93.0%.
CHAPTER
4
Reactions in
Aqueous Solutions
Black smokers form when superheated water, rich in
minerals, flows out onto the ocean floor through the lava
from an ocean volcano. The hydrogen sulfide present
converts the metal ions to insoluble metal sulfides.
CHAPTER OUTLINE
A LOOK AHEAD
4.1
General Properties of
Aqueous Solutions
4.2
4.3
4.4
Precipitation Reactions
We begin by studying the properties of solutions prepared by dissolving
substances in water, called aqueous solutions. Aqueous solutions can be
classified as nonelectrolyte or electrolyte, depending on their ability to conduct electricity. (4.1)
We will see that precipitation reactions are those in which the product is an
insoluble compound. We learn to represent these reactions using ionic equations and net ionic equations. (4.2)
4.5
4.6
4.7
4.8
Concentration of Solutions
Next, we learn acid-base reactions, which involve the transfer of proton
(H1) from an acid to a base. (4.3)
We then learn oxidation-reduction (redox) reactions in which electrons are
transferred between reactants. We will see that there are several types of
redox reactions. (4.4)
To carry out quantitative studies of solutions, we learn how to express the
concentration of a solution in molarity. (4.5)
Finally, we will apply our knowledge of the mole method from Chapter 3 to
the three types of reactions studied here. We will see how gravimetric analysis is used to study precipitation reactions, and the titration technique is used
to study acid-base and redox reactions. (4.6, 4.7, and 4.8)
118
Acid-Base Reactions
Oxidation-Reduction
Reactions
Gravimetric Analysis
Acid-Base Titrations
Redox Titrations
4.1 General Properties of Aqueous Solutions
119
M
any chemical reactions and virtually all biological processes take place in water. In this
chapter, we will discuss three major categories of reactions that occur in aqueous solutions: precipitation reactions, acid-base reactions, and redox reactions. In later chapters, we will
study the structural characteristics and properties of water—the so-called universal solvent—
and its solutions.
4.1 General Properties of Aqueous Solutions
A solution is a homogeneous mixture of two or more substances. The solute is the
substance present in a smaller amount, and the solvent is the substance present in a
larger amount. A solution may be gaseous (such as air), solid (such as an alloy), or
liquid (seawater, for example). In this section we will discuss only aqueous solutions,
in which the solute initially is a liquid or a solid and the solvent is water.
Electrolytic Properties
All solutes that dissolve in water fit into one of two categories: electrolytes and
nonelectrolytes. An electrolyte is a substance that, when dissolved in water, results
in a solution that can conduct electricity. A nonelectrolyte does not conduct electricity when dissolved in water. Figure 4.1 shows an easy and straightforward method
of distinguishing between electrolytes and nonelectrolytes. A pair of inert electrodes
(copper or platinum) is immersed in a beaker of water. To light the bulb, electric
current must flow from one electrode to the other, thus completing the circuit. Pure
water is a very poor conductor of electricity. However, if we add a small amount of
sodium chloride (NaCl), the bulb will glow as soon as the salt dissolves in the water.
Solid NaCl, an ionic compound, breaks up into Na1 and Cl2 ions when it dissolves
in water. The Na1 ions are attracted to the negative electrode, and the Cl2 ions to
the positive electrode. This movement sets up an electric current that is equivalent
to the flow of electrons along a metal wire. Because the NaCl solution conducts
electricity, we say that NaCl is an electrolyte. Pure water contains very few ions, so
it cannot conduct electricity.
Comparing the lightbulb’s brightness for the same molar amounts of dissolved
substances helps us distinguish between strong and weak electrolytes. A characteristic of strong electrolytes is that the solute is assumed to be 100 percent dissociated into ions in solution. (By dissociation we mean the breaking up of the
Tap water does conduct electricity
because it contains many dissolved ions.
Animation
Strong Electrolytes, Weak Electrolytes,
and Nonelectrolytes
Figure 4.1 An arrangement for
distinguishing between electrolytes
and nonelectrolytes. A solution's
ability to conduct electricity
depends on the number of ions it
contains. (a) A nonelectrolyte
solution does not contain ions, and
the lightbulb is not lit. (b) A weak
electrolyte solution contains a
small number of ions, and the
lightbulb is dimly lit. (c) A strong
electrolyte solution contains a
large number of ions, and the
lightbulb is brightly lit. The molar
amounts of the dissolved solutes
are equal in all three cases.
(a)
(b)
(c)
120
Chapter 4
■
Reactions in Aqueous Solutions
Table 4.1
Classification of Solutes in Aqueous Solution
Strong Electrolyte
Weak Electrolyte
Nonelectrolyte
HCl
HNO3
HClO4
H2SO4*
NaOH
Ba(OH)2
Ionic compounds
CH3COOH
HF
HNO2
NH3
H2O†
(NH2)2CO (urea)
CH3OH (methanol)
C2H5OH (ethanol)
C6H12O6 (glucose)
C12H22O11 (sucrose)
*H2SO4 has two ionizable H1 ions, but only one of the H1 ions is totally ionized.
†
Pure water is an extremely weak electrolyte.
compound into cations and anions.) Thus, we can represent sodium chloride dissolving in water as
H2O
NaCl(s) ¡ Na1 (aq) 1 Cl2 (aq)
Animation
Hydration
This equation says that all sodium chloride that enters the solution ends up as Na1
and Cl2 ions; there are no undissociated NaCl units in solution.
Table 4.1 lists examples of strong electrolytes, weak electrolytes, and nonelectrolytes. Ionic compounds, such as sodium chloride, potassium iodide (KI), and calcium
nitrate [Ca(NO3)2], are strong electrolytes. It is interesting to note that human body
fluids contain many strong and weak electrolytes.
Water is a very effective solvent for ionic compounds. Although water is an
electrically neutral molecule, it has a positive region (the H atoms) and a negative
region (the O atom), or positive and negative “poles”; for this reason it is a polar
solvent. When an ionic compound such as sodium chloride dissolves in water, the
three-dimensional network of ions in the solid is destroyed. The Na1 and Cl2 ions
are separated from each other and undergo hydration, the process in which an ion is
surrounded by water molecules arranged in a specific manner. Each Na1 ion is surrounded by a number of water molecules orienting their negative poles toward the
cation. Similarly, each Cl2 ion is surrounded by water molecules with their positive
poles oriented toward the anion (Figure 4.2). Hydration helps to stabilize ions in solution and prevents cations from combining with anions.
Acids and bases are also electrolytes. Some acids, including hydrochloric acid
(HCl) and nitric acid (HNO3), are strong electrolytes. These acids are assumed to
ionize completely in water; for example, when hydrogen chloride gas dissolves in
water, it forms hydrated H1 and Cl2 ions:
H2O
HCl(g) ¡ H1 (aq) 1 Cl2 (aq)
In other words, all the dissolved HCl molecules separate into hydrated H1 and Cl2
ions. Thus, when we write HCl(aq), it is understood that it is a solution of only H1(aq)
Figure 4.2
Hydration of Na1
and Cl2 ions.
1
2
4.2 Precipitation Reactions
121
and Cl2(aq) ions and that there are no hydrated HCl molecules present. On the other
hand, certain acids, such as acetic acid (CH3COOH), which gives vinegar its tart
flavor, do not ionize completely and are weak electrolytes. We represent the ionization
of acetic acid as
CH3COOH(aq) Δ CH3COO2 (aq) 1 H1 (aq)
where CH3COO2 is called the acetate ion. We use the term ionization to describe the
separation of acids and bases into ions. By writing the formula of acetic acid as
CH3COOH, we indicate that the ionizable proton is in the COOH group.
The ionization of acetic acid is written with a double arrow to show that it is
a reversible reaction; that is, the reaction can occur in both directions. Initially,
a number of CH3COOH molecules break up into CH3COO2 and H1 ions. As time
goes on, some of the CH3COO2 and H1 ions recombine into CH3COOH molecules. Eventually, a state is reached in which the acid molecules ionize as fast as
the ions recombine. Such a chemical state, in which no net change can be observed
(although activity is continuous on the molecular level), is called chemical equilibrium. Acetic acid, then, is a weak electrolyte because its ionization in water is
incomplete. By contrast, in a hydrochloric acid solution the H1 and Cl2 ions have
no tendency to recombine and form molecular HCl. We use a single arrow to
represent complete ionizations.
CH3COOH
There are different types of chemical
equilibrium. We will return to this very
important topic in Chapter 14.
Review of Concepts
The diagrams here show three compounds AB2 (a), AC2 (b), and AD2 (c) dissolved
in water. Which is the strongest electrolyte and which is the weakest? (For
simplicity, water molecules are not shown.)
(a)
(b)
(c)
4.2 Precipitation Reactions
One common type of reaction that occurs in aqueous solution is the precipitation
reaction, which results in the formation of an insoluble product, or precipitate. A
precipitate is an insoluble solid that separates from the solution. Precipitation reactions usually involve ionic compounds. For example, when an aqueous solution of
lead(II) nitrate [Pb(NO3)2] is added to an aqueous solution of potassium iodide (KI),
a yellow precipitate of lead(II) iodide (PbI2) is formed:
Pb(NO3 ) 2 (aq) 1 2KI(aq) ¡ PbI2 (s) 1 2KNO3 (aq)
Potassium nitrate remains in solution. Figure 4.3 shows this reaction in progress.
The preceding reaction is an example of a metathesis reaction (also called a
double-displacement reaction), a reaction that involves the exchange of parts
between the two compounds. (In this case, the cations in the two compounds
exchange anions, so Pb21 ends up with I2 as PbI2 and K1 ends up with NO2
3 as
Animation
Precipitation Reactions
122
Chapter 4
■
Reactions in Aqueous Solutions
Pb21
K1
NO2
3
K1
88n
I2
Pb21
NO2
3
Figure 4.3
I2
Formation of yellow PbI2 precipitate as a solution of Pb(NO3)2 is added to a solution of KI.
KNO3.) As we will see, the precipitation reactions discussed in this chapter are
examples of metathesis reactions.
Solubility
How can we predict whether a precipitate will form when a compound is added to a
solution or when two solutions are mixed? It depends on the solubility of the solute,
which is defined as the maximum amount of solute that will dissolve in a given quantity of solvent at a specific temperature. Chemists refer to substances as soluble,
slightly soluble, or insoluble in a qualitative sense. A substance is said to be soluble
if a fair amount of it visibly dissolves when added to water. If not, the substance is
described as slightly soluble or insoluble. All ionic compounds are strong electrolytes,
but they are not equally soluble.
Table 4.2 classifies a number of common ionic compounds as soluble or insoluble. Keep in mind, however, that even insoluble compounds dissolve to a certain
extent. Figure 4.4 shows several precipitates.
Table 4.2 Solubility Rules for Common Ionic Compounds in Water at 258C
Soluble Compounds
Insoluble Exceptions
Compounds containing alkali metal
ions (Li1, Na1, K1, Rb1, Cs1) and
the ammonium ion (NH14 )
Nitrates (NO23 ), acetates (CH3COO2),
bicarbonates (HCO23 ), chlorates
(ClO23 ), and perchlorates (ClO24 )
Halides (Cl2, Br2, I2)
Halides of Ag1, Hg221, and Pb21
22
Sulfates (SO4 )
Sulfates of Ag1, Ca21, Sr21, Ba21, Hg221, and Pb21
Insoluble Compounds
(CO322),
Carbonates
phosphates
(PO432), chromates (CrO422),
sulfides (S22)
Hydroxides (OH2)
Soluble Exceptions
Compounds containing alkali metal ions
and the ammonium ion
Compounds containing alkali metal ions
and the Ba21 ion
4.2 Precipitation Reactions
123
Figure 4.4 Appearance of
several precipitates. From left to
right: CdS, PbS, Ni(OH)2, and
Al(OH)3.
Example 4.1 applies the solubility rules in Table 4.2.
Example 4.1
Classify the following ionic compounds as soluble or insoluble: (a) silver sulfate
(Ag2SO4), (b) calcium carbonate (CaCO3), (c) sodium phosphate (Na3PO4).
Strategy Although it is not necessary to memorize the solubilities of compounds, you
should keep in mind the following useful rules: All ionic compounds containing alkali
metal cations; the ammonium ion; and the nitrate, bicarbonate, and chlorate ions are
soluble. For other compounds, we need to refer to Table 4.2.
Solution
(a) According to Table 4.2, Ag2SO4 is insoluble.
(b) This is a carbonate and Ca is a Group 2A metal. Therefore, CaCO3 is insoluble.
(c) Sodium is an alkali metal (Group 1A) so Na3PO4 is soluble.
Practice Exercise Classify the following ionic compounds as soluble or insoluble:
(a) CuS, (b) Ca(OH)2, (c) Zn(NO3)2.
Molecular Equations, Ionic Equations, and Net Ionic Equations
The equation describing the precipitation of lead(II) iodide on page 121 is called a
molecular equation because the formulas of the compounds are written as though all
species existed as molecules or whole units. A molecular equation is useful because
it identifies the reagents [that is, lead(II) nitrate and potassium iodide]. If we wanted
to bring about this reaction in the laboratory, we would use the molecular equation.
However, a molecular equation does not describe in detail what actually is happening
in solution.
As pointed out earlier, when ionic compounds dissolve in water, they break apart
into their component cations and anions. To be more realistic, the equations should
show the dissociation of dissolved ionic compounds into ions. Therefore, returning to
the reaction between potassium iodide and lead(II) nitrate, we would write
Pb21 (aq) 1 2NO23 (aq) 1 2K1 (aq) 1 2I2 (aq) ¡
PbI2 (s) 1 2K 1 (aq) 1 2NO23 (aq)
Similar problems: 4.19, 4.20.
124
Chapter 4
■
Reactions in Aqueous Solutions
The preceding equation is an example of an ionic equation, which shows dissolved species as free ions. To see whether a precipitate might form from this solution,
we first combine the cation and anion from different compounds; that is, PbI2 and
KNO3. Referring to Table 4.2, we see that PbI2 is an insoluble compound and KNO3
is soluble. Therefore, the dissolved KNO3 remains in solution as separate K1 and
NO32 ions, which are called spectator ions, or ions that are not involved in the overall reaction. Because spectator ions appear on both sides of an equation, they can be
eliminated from the ionic equation
Pb21 (aq) 1 2NO23 (aq) 1 2K1 (aq) 1 2I2 (aq) ¡
PbI2 (s) 1 2K1 (aq) 1 2NO23 (aq)
Finally, we end up with the net ionic equation, which shows only the species that
actually take part in the reaction:
Figure 4.5 Formation of BaSO4
Pb21 (aq) 1 2I2 (aq) ¡ PbI2 (s)
precipitate.
Looking at another example, we find that when an aqueous solution of barium
chloride (BaCl2) is added to an aqueous solution of sodium sulfate (Na2SO4), a white
precipitate is formed (Figure 4.5). Treating this as a metathesis reaction, the products
are BaSO4 and NaCl. From Table 4.2 we see that only BaSO4 is insoluble. Therefore,
we write the molecular equation as
BaCl2 (aq) 1 Na2SO4 (aq) ¡ BaSO4 (s) 1 2NaCl(aq)
The ionic equation for the reaction is
Ba21 (aq) 1 2Cl2 (aq) 1 2Na1 (aq) 1 SO22
4 (aq) ¡
BaSO4 (s) 1 2Na1 (aq) 1 2Cl2 (aq)
Canceling the spectator ions (Na1 and Cl2) on both sides of the equation gives us the
net ionic equation
Ba21 (aq) 1 SO22
4 (aq) ¡ BaSO4 (s)
The following four steps summarize the procedure for writing ionic and net ionic
equations:
1. Write a balanced molecular equation for the reaction, using the correct formulas
for the reactant and product ionic compounds. Refer to Table 4.2 to decide which
of the products is insoluble and therefore will appear as a precipitate.
2. Write the ionic equation for the reaction. The compound that does not appear as
the precipitate should be shown as free ions.
3. Identify and cancel the spectator ions on both sides of the equation. Write the net
ionic equation for the reaction.
4. Check that the charges and number of atoms balance in the net ionic equation.
These steps are applied in Example 4.2.
Example 4.2
Precipitate formed by the reaction
between K3PO4 (aq) and
Ca(NO3)2(aq).
Predict what happens when a potassium phosphate (K3PO4) solution is mixed with a
calcium nitrate [Ca(NO3)2] solution. Write a net ionic equation for the reaction.
(Continued)
4.2 Precipitation Reactions
Strategy From the given information, it is useful to first write the unbalanced equation
K3PO4 (aq) 1 Ca(NO3 ) 2 (aq) ¡ ?
What happens when ionic compounds dissolve in water? What ions are formed from the
dissociation of K3PO4 and Ca(NO3)2? What happens when the cations encounter the
anions in solution?
Solution In solution, K3PO4 dissociates into K1 and PO432 ions and Ca(NO3)2
dissociates into Ca21 and NO23 ions. According to Table 4.2, calcium ions (Ca21) and
phosphate ions (PO432) will form an insoluble compound, calcium phosphate [Ca3(PO4)2],
while the other product, KNO3, is soluble and remains in solution. Therefore, this is a
precipitation reaction. We follow the stepwise procedure just outlined.
Step 1: The balanced molecular equation for this reaction is
2K3PO4 (aq) 1 3Ca(NO3 ) 2 (aq) ¡ Ca3 (PO4 ) 2 (s) 1 6KNO3 (aq)
Step 2: To write the ionic equation, the soluble compounds are shown as dissociated ions:
21
2
6K1 (aq) 1 2PO32
4 (aq) 1 3Ca (aq) 1 6NO3 (aq) ¡
1
6K (aq) 1 6NO23 (aq) 1 Ca3 (PO4 ) 2 (s)
Step 3: Canceling the spectator ions (K1 and NO32) on each side of the equation, we
obtain the net ionic equation:
3Ca21 (aq) 1 2PO32
4 (aq) ¡ Ca3 (PO4 ) 2 (s)
Step 4: Note that because we balanced the molecular equation first, the net ionic
equation is balanced as to the number of atoms on each side and the number of
positive (16) and negative (26) charges on the left-hand side is the same.
Practice Exercise Predict the precipitate produced by mixing an Al(NO3)3 solution
with a NaOH solution. Write the net ionic equation for the reaction.
Review of Concepts
Which of the diagrams here accurately describes the reaction between
Ca(NO3)2(aq) and Na2CO3(aq)? For simplicity, only the Ca21 (yellow) and
CO322 (blue) ions are shown.
(a)
(b)
(c)
The Chemistry in Action essay on p. 126 discusses some practical problems
associated with precipitation reactions.
Similar problems: 4.21, 4.22.
125
CHEMISTRY in Action
An Undesirable Precipitation Reaction
L
imestone (CaCO3) and dolomite (CaCO3 ? MgCO3), which
are widespread on Earth’s surface, often enter the water supply. According to Table 4.2, calcium carbonate is insoluble in
water. However, in the presence of dissolved carbon dioxide
(from the atmosphere), calcium carbonate is converted to soluble calcium bicarbonate [Ca(HCO3)2]:
FPO
CaCO3 (s) 1 CO2 (aq) 1 H2O(l) ¡
Ca21 (aq) 1 2HCO23 (aq)
where HCO2
3 is the bicarbonate ion.
Water containing Ca21 and/or Mg21 ions is called hard water, and water that is mostly free of these ions is called soft water.
Hard water is unsuitable for some household and industrial uses.
When water containing Ca21 and HCO2
3 ions is heated or
boiled, the solution reaction is reversed to produce the CaCO3
precipitate
Ca21 (aq) 1 2HCO23 (aq) ¡
CaCO3 (s) 1 CO2 (aq) 1 H2O(l)
and gaseous carbon dioxide is driven off:
CO2 (aq) ¡ CO2 (g)
Solid calcium carbonate formed in this way is the main component of the scale that accumulates in boilers, water heaters,
pipes, and teakettles. A thick layer of scale reduces heat transfer and decreases the efficiency and durability of boilers, pipes,
and appliances. In household hot-water pipes it can restrict or
Boiler scale almost fills this hot-water pipe. The deposits consist mostly of
CaCO3 with some MgCO3.
totally block the flow of water. A simple method used by
plumbers to remove scale deposits is to introduce a small
amount of hydrochloric acid, which reacts with (and therefore
dissolves) CaCO3:
CaCO3 (s) 1 2HCl(aq) ¡
CaCl2 (aq) 1 H2O(l) 1 CO2 (g)
In this way, CaCO3 is converted to soluble CaCl2.
4.3 Acid-Base Reactions
Acids and bases are as familiar as aspirin and milk of magnesia although many
people do not know their chemical names—acetylsalicylic acid (aspirin) and magnesium hydroxide (milk of magnesia). In addition to being the basis of many medicinal
and household products, acid-base chemistry is important in industrial processes and
essential in sustaining biological systems. Before we can discuss acid-base reactions,
we need to know more about acids and bases themselves.
General Properties of Acids and Bases
In Section 2.7 we defined acids as substances that ionize in water to produce H1
ions and bases as substances that ionize in water to produce OH2 ions. These
definitions were formulated in the late nineteenth century by the Swedish chemist
126
4.3 Acid-Base Reactions
127
Svante Arrhenius† to classify substances whose properties in aqueous solutions were
well known.
Acids
• Acids have a sour taste; for example, vinegar owes its sourness to acetic acid,
and lemons and other citrus fruits contain citric acid.
• Acids cause color changes in plant dyes; for example, they change the color of
litmus from blue to red.
• Acids react with certain metals, such as zinc, magnesium, and iron, to produce
hydrogen gas. A typical reaction is that between hydrochloric acid and magnesium:
2HCl(aq) 1 Mg(s) ¡ MgCl2 (aq) 1 H2 (g)
•
Acids react with carbonates and bicarbonates, such as Na2CO3, CaCO3, and
NaHCO3, to produce carbon dioxide gas (Figure 4.6). For example,
2HCl(aq) 1 CaCO3 (s) ¡ CaCl2 (aq) 1 H2O(l) 1 CO2 (g)
HCl(aq) 1 NaHCO3 (s) ¡ NaCl(aq) 1 H2O(l) 1 CO2 (g)
•
Aqueous acid solutions conduct electricity.
Bases
• Bases have a bitter taste.
• Bases feel slippery; for example, soaps, which contain bases, exhibit this property.
• Bases cause color changes in plant dyes; for example, they change the color of
litmus from red to blue.
• Aqueous base solutions conduct electricity.
Brønsted Acids and Bases
Arrhenius’ definitions of acids and bases are limited in that they apply only to aqueous solutions. Broader definitions were proposed by the Danish chemist Johannes
Brønsted‡ in 1932; a Brønsted acid is a proton donor, and a Brønsted base is a
proton acceptor. Note that Brønsted’s definitions do not require acids and bases to be
in aqueous solution.
Hydrochloric acid is a Brønsted acid because it donates a proton in water:
HCl(aq) ¡ H1 (aq) 1 Cl2 (aq)
Note that the H1 ion is a hydrogen atom that has lost its electron; that is, it is just
a bare proton. The size of a proton is about 10215 m, compared to a diameter of
10210 m for an average atom or ion. Such an exceedingly small charged particle
cannot exist as a separate entity in aqueous solution owing to its strong attraction
†
Svante August Arrhenius (1859–1927). Swedish chemist. Arrhenius made important contributions in
the study of chemical kinetics and electrolyte solutions. He also speculated that life had come to Earth
from other planets, a theory now known as panspermia. Arrhenius was awarded the Nobel Prize in
Chemistry in 1903.
‡
Johannes Nicolaus Brønsted (1879–1947). Danish chemist. In addition to his theory of acids and
bases, Brønsted worked on thermodynamics and the separation of mercury isotopes. In some texts, Brønsted
acids and bases are called Brønsted-Lowry acids and bases. Thomas Martin Lowry (1874–1936).
English chemist. Brønsted and Lowry developed essentially the same acid-base theory independently
in 1923.
Figure 4.6 A piece of blackboard
chalk, which is mostly CaCO3,
reacts with hydrochloric acid.
128
Chapter 4
■
Reactions in Aqueous Solutions
Figure 4.7 Ionization of HCl in
water to form the hydronium ion
and the chloride ion.
8n
1
HCl
1
H2O
8n
1
H3O1
1
Cl2
for the negative pole (the O atom) in H2O. Consequently, the proton exists in the
hydrated form, as shown in Figure 4.7. Therefore, the ionization of hydrochloric acid
should be written as
HCl(aq) 1 H2O(l) ¡ H3O1 (aq) 1 Cl2 (aq)
Electrostatic potential map of the
H3O1 ion. In the rainbow color
spectrum representation, the most
electron-rich region is red and the
most electron-poor region is blue.
In most cases, acids start with H in the
formula or have a COOH group.
Table 4.3
Some Common Strong
and Weak Acids
Strong Acids
Hydrochloric
acid
Hydrobromic
acid
Hydroiodic
acid
Nitric acid
Sulfuric acid
Perchloric acid
HCl
HCl(aq) ¡ H1 (aq) 1 Cl2 (aq)
HNO3 (aq) ¡ H1 (aq) 1 NO23 (aq)
CH3COOH(aq) Δ CH3COO2 (aq) 1 H1 (aq)
As mentioned earlier, because the ionization of acetic acid is incomplete (note the
double arrows), it is a weak electrolyte. For this reason it is called a weak acid (see
Table 4.1). On the other hand, HCl and HNO3 are strong acids because they are strong
electrolytes, so they are completely ionized in solution (note the use of single arrows).
Sulfuric acid (H2SO4) is a diprotic acid because each unit of the acid gives up
two H1ions, in two separate steps:
H2SO4 (aq) ¡ H1 (aq) 1 HSO24 (aq)
HSO24 (aq) Δ H1 (aq) 1 SO22
4 (aq)
HBr
HI
HNO3
H2SO4
HClO4
Weak Acids
Hydrofluoric
acid
Nitrous acid
Phosphoric acid
Acetic acid
The hydrated proton, H3O1, is called the hydronium ion. This equation shows a reaction in which a Brønsted acid (HCl) donates a proton to a Brønsted base (H2O).
Experiments show that the hydronium ion is further hydrated so that the proton
may have several water molecules associated with it. Because the acidic properties of
the proton are unaffected by the degree of hydration, in this text we will generally
use H1(aq) to represent the hydrated proton. This notation is for convenience, but
H3O1 is closer to reality. Keep in mind that both notations represent the same species
in aqueous solution.
Acids commonly used in the laboratory include hydrochloric acid (HCl), nitric
acid (HNO3), acetic acid (CH3COOH), sulfuric acid (H2SO4), and phosphoric acid
(H3PO4). The first three are monoprotic acids; that is, each unit of the acid yields one
hydrogen ion upon ionization:
HF
HNO2
H3PO4
CH3COOH
H2SO4 is a strong electrolyte or strong acid (the first step of ionization is complete),
but HSO2
4 is a weak acid or weak electrolyte, and we need a double arrow to represent its incomplete ionization.
Triprotic acids, which yield three H1ions, are relatively few in number. The best
known triprotic acid is phosphoric acid, whose ionizations are
H3PO4 (aq) Δ H1 (aq) 1 H2PO24 (aq)
H2PO24 (aq) Δ H1 (aq) 1 HPO22
4 (aq)
1
32
HPO22
(aq)
Δ
H
(aq)
1
PO
4
4 (aq)
22
All three species (H3PO4, H2PO2
4 , and HPO4 ) in this case are weak acids, and we
use the double arrows to represent each ionization step. Anions such as H2PO2
4 and
HPO22
4 are found in aqueous solutions of phosphates such as NaH2PO4 and Na2HPO4.
Table 4.3 lists several common strong and weak acids.
4.3 Acid-Base Reactions
34
1
NH3
34
H2O
1
129
1
NH14
1
OH2
Figure 4.8 Ionization of ammonia in water to form the ammonium ion and the hydroxide ion.
Review of Concepts
Which of the following diagrams best represents a weak acid? Which represents
a very weak acid? Which represents a strong acid? The proton exists in water as
the hydronium ion. All acids are monoprotic. (For simplicity, water molecules
are not shown.)
(a)
(b)
(c)
Table 4.1 shows that sodium hydroxide (NaOH) and barium hydroxide [Ba(OH)2]
are strong electrolytes. This means that they are completely ionized in solution:
H2O
NaOH(s) ¡ Na1(aq) 1 OH2 (aq)
H2O
Ba(OH) 2 (s) ¡ Ba21(aq) 1 2OH2 (aq)
The OH2 ion can accept a proton as follows:
H1 (aq) 1 OH2 (aq) ¡ H2O(l)
Thus, OH2 is a Brønsted base.
Ammonia (NH3) is classified as a Brønsted base because it can accept a H1 ion
(Figure 4.8):
NH3 (aq) 1 H2O(l) Δ NH 14 (aq) 1 OH2 (aq)
Ammonia is a weak electrolyte (and therefore a weak base) because only a small
fraction of dissolved NH3 molecules react with water to form NH14 and OH2 ions.
The most commonly used strong base in the laboratory is sodium hydroxide.
It is cheap and soluble. (In fact, all of the alkali metal hydroxides are soluble.)
The most commonly used weak base is aqueous ammonia solution, which is
sometimes erroneously called ammonium hydroxide. There is no evidence that
the species NH4OH actually exists other than the NH14 and OH2 ions in solution.
All of the Group 2A elements form hydroxides of the type M(OH)2, where M
denotes an alkaline earth metal. Of these hydroxides, only Ba(OH)2 is soluble.
Note that this bottle of aqueous
ammonia is erroneously labeled.
130
Chapter 4
■
Reactions in Aqueous Solutions
Magnesium and calcium hydroxides are used in medicine and industry. Hydroxides of other metals, such as Al(OH)3 and Zn(OH)2 are insoluble and are not used
as bases.
Example 4.3 classifies substances as Brønsted acids or Brønsted bases.
Example 4.3
Classify each of the following species in aqueous solution as a Brønsted acid or base:
(a) HBr, (b) NO22 , (c) HCO23 .
Strategy What are the characteristics of a Brønsted acid? Does it contain at least an
H atom? With the exception of ammonia, most Brønsted bases that you will encounter
at this stage are anions.
Solution
(a) We know that HCl is an acid. Because Br and Cl are both halogens (Group 7A), we
expect HBr, like HCl, to ionize in water as follows:
HBr(aq) ¡ H1 (aq) 1 Br2 (aq)
Therefore HBr is a Brønsted acid.
(b) In solution the nitrite ion can accept a proton from water to form nitrous acid:
NO22 (aq) 1 H1 (aq) ¡ HNO2 (aq)
This property makes NO22 a Brønsted base.
(c) The bicarbonate ion is a Brønsted acid because it ionizes in solution as follows:
HCO23 (aq) Δ H1 (aq) 1 CO22
3 (aq)
It is also a Brønsted base because it can accept a proton to form carbonic acid:
HCO23 (aq) 1 H1 (aq) Δ H2CO3 (aq)
Comment The HCO23 species is said to be amphoteric because it possesses both acidic
Similar problems: 4.31, 4.32.
and basic properties. The double arrows show that this is a reversible reaction.
Practice Exercise Classify each of the following species as a Brønsted acid or base:
(a) SO22
4 , (b) HI.
Acid-Base Neutralization
Animation
Neutralization Reactions
A neutralization reaction is a reaction between an acid and a base. Generally, aqueous acid-base reactions produce water and a salt, which is an ionic compound made
up of a cation other than H1 and an anion other than OH2 or O22:
acid 1 base ¡ salt 1 water
The substance we know as table salt, NaCl, is a product of the acid-base reaction
Acid-base reactions generally go to
completion.
HCl(aq) 1 NaOH(aq) ¡ NaCl(aq) 1 H2O(l)
However, because both the acid and the base are strong electrolytes, they are completely ionized in solution. The ionic equation is
H1 (aq) 1 Cl2 (aq) 1 Na1 (aq) 1 OH2 (aq) ¡ Na1 (aq) 1 Cl1 (aq) 1 H2O(l)
4.3 Acid-Base Reactions
Therefore, the reaction can be represented by the net ionic equation
H1 (aq) 1 OH2 (aq) ¡ H2O(l)
Both Na1 and Cl2 are spectator ions.
If we had started the preceding reaction with equal molar amounts of the acid
and the base, at the end of the reaction we would have only a salt and no leftover
acid or base. This is a characteristic of acid-base neutralization reactions.
A reaction between a weak acid such as hydrocyanic acid (HCN) and a strong base is
HCN(aq) 1 NaOH(aq) ¡ NaCN(aq) 1 H2O(l)
Because HCN is a weak acid, it does not ionize appreciably in solution. Thus, the
ionic equation is written as
HCN(aq) 1 Na1 (aq) 1 OH2 (aq) ¡ Na1 (aq) 1 CN2 (aq) 1 H2O(l)
and the net ionic equation is
HCN(aq) 1 OH2 (aq) ¡ CN2 (aq) 1 H2O(l)
Note that only Na1 is a spectator ion; OH2 and CN2 are not.
The following are also examples of acid-base neutralization reactions, represented
by molecular equations:
HF(aq) 1 KOH(aq) ¡ KF(aq) 1 H2O(l)
H2SO4 (aq) 1 2NaOH(aq) ¡ Na2SO4 (aq) 1 2H2O(l)
HNO3 (aq) 1 NH3 (aq) ¡ NH4NO3 (aq)
The last equation looks different because it does not show water as a product. However, if we express NH3(aq) as NH14 (aq) and OH2(aq), as discussed earlier, then the
equation becomes
2
HNO3 (aq) 1 NH1
4 (aq) 1 OH (aq) ¡ NH4NO3 (aq) 1 H2O(l)
Example 4.4
Write molecular, ionic, and net ionic equations for each of the following acid-base reactions:
(a) hydrobromic acid(aq) 1 barium hydroxide(aq) ¡
(b) sulfuric acid(aq) 1 potassium hydroxide(aq) ¡
Strategy The first step is to identify the acids and bases as strong or weak. We see
that HBr is a strong acid and H2SO4 is a strong acid for the first step ionization and a
weak acid for the second step ionization. Both Ba(OH)2 and KOH are strong bases.
Solution
(a) Molecular equation:
2HBr(aq) 1 Ba(OH) 2 (aq) ¡ BaBr2 (aq) 1 2H2O(l)
Ionic equation:
2H1 (aq) 1 2Br2 (aq) 1 Ba21 (aq) 1 2OH2 (aq) ¡
Ba21 (aq) 1 2Br2 (aq) 1 2H2O(l)
(Continued)
131
132
Chapter 4
■
Reactions in Aqueous Solutions
Net ionic equation:
2H 1 (aq) 1 2OH 2 (aq) ¡ 2H2O(l)
H 1 (aq) 1 OH 2 (aq) ¡ H2O(l)
or
Both Ba21 and Br2 are spectator ions.
(b) Molecular equation:
H2SO4 (aq) 1 2KOH(aq) ¡ K2SO4 (aq) 1 2H2O(l)
Ionic equation:
H 1 (aq) 1 HSO24 (aq) 1 2K 1 (aq) 1 2OH 2 (aq) ¡
2K 1 (aq) 1 SO22
4 (aq) 1 2H2O(l)
Net ionic equation:
H 1 (aq) 1 HSO24 (aq) 1 2OH 2 (aq) ¡ SO22
4 (aq) 1 2H2O(l)
Similar problem: 4.33(b).
Note that because HSO24 is a weak acid and does not ionize appreciably in water,
the only spectator ion is K1.
Practice Exercise Write a molecular equation, an ionic equation, and a net ionic
equation for the reaction between aqueous solutions of phosphoric acid and sodium
hydroxide.
Acid-Base Reactions Leading to Gas Formation
Certain salts like carbonates (containing the CO322 ion), bicarbonates (containing
the HCO23 ion), sulfites (containing the SO322 ion), and sulfides (containing the
S22 ion) react with acids to form gaseous products. For example, the molecular
equation for the reaction between sodium carbonate (Na2CO3) and HCl(aq) is
(see Figure 4.6)
Na2CO3 (aq) 1 2HCl(aq) ¡ 2NaCl(aq) 1 H2CO3 (aq)
Carbonic acid is unstable and if present in solution in sufficient concentrations decomposes as follows:
H2CO3 (aq) ¡ H2O(l) 1 CO2 (g)
Similar reactions involving other mentioned salts are
NaHCO3 (aq) 1 HCl(aq) ¡ NaCl(aq) 1 H2O(l) 1 CO2 (g)
Na2SO3 (aq) 1 2HCl(aq) ¡ 2NaCl(aq) 1 H2O(l) 1 SO2 (g)
K2S(aq) 1 2HCl(aq) ¡ 2KCl(aq) 1 H2S(g)
4.4 Oxidation-Reduction Reactions
Animation
Oxidation-Reduction Reactions
Whereas acid-base reactions can be characterized as proton-transfer processes, the
class of reactions called oxidation-reduction, or redox, reactions are considered electrontransfer reactions. Oxidation-reduction reactions are very much a part of the world
around us. They range from the burning of fossil fuels to the action of household
4.4 Oxidation-Reduction Reactions
133
Mg
1
88n
Mg21
O22
O2
Figure 4.9 Magnesium burns in oxygen to form magnesium oxide.
bleach. Additionally, most metallic and nonmetallic elements are obtained from their
ores by the process of oxidation or reduction.
Many important redox reactions take place in water, but not all redox reactions
occur in aqueous solution. We begin our discussion with a reaction in which two
elements combine to form a compound. Consider the formation of magnesium oxide
(MgO) from magnesium and oxygen (Figure 4.9):
2Mg(s) 1 O2 (g) ¡ 2MgO(s)
Magnesium oxide (MgO) is an ionic compound made up of Mg21 and O22 ions. In
this reaction, two Mg atoms give up or transfer four electrons to two O atoms (in O2).
For convenience, we can think of this process as two separate steps, one involving
the loss of four electrons by the two Mg atoms and the other being the gain of four
electrons by an O2 molecule:
2Mg ¡ 2Mg21 1 4e 2
O2 1 4e 2 ¡ 2O22
Animation
Reaction of Magnesium and Oxygen
Animation
Formation of Ag2S by OxidationReduction
Note that in an oxidation half-reaction,
electrons appear as the product; in a
reduction half-reaction, electrons appear
as the reactant.
Each of these steps is called a half-reaction, which explicitly shows the electrons
involved in a redox reaction. The sum of the half-reactions gives the overall reaction:
2Mg 1 O2 1 4e 2 ¡ 2Mg21 1 2O22 1 4e 2
or, if we cancel the electrons that appear on both sides of the equation,
2Mg 1 O2 ¡ 2Mg21 1 2O22
Finally, the Mg21 and O22 ions combine to form MgO:
2Mg21 1 2O22 ¡ 2MgO
The term oxidation reaction refers to the half-reaction that involves loss of electrons. Chemists originally used “oxidation” to denote the combination of elements
with oxygen. However, it now has a broader meaning that includes reactions not
involving oxygen. A reduction reaction is a half-reaction that involves gain of electrons. In the formation of magnesium oxide, magnesium is oxidized. It is said to act
A useful mnemonic for redox is OILRIG:
Oxidation Is Loss (of electrons) and
Reduction Is Gain (of electrons).
134
Chapter 4
■
Reactions in Aqueous Solutions
Oxidizing agents are always reduced and
reducing agents are always oxidized. This
statement may be somewhat confusing,
but it is simply a consequence of the
definitions of the two processes.
as a reducing agent because it donates electrons to oxygen and causes oxygen to be
reduced. Oxygen is reduced and acts as an oxidizing agent because it accepts electrons from magnesium, causing magnesium to be oxidized. Note that the extent of
oxidation in a redox reaction must be equal to the extent of reduction; that is, the
number of electrons lost by a reducing agent must be equal to the number of electrons
gained by an oxidizing agent.
The occurrence of electron transfer is more apparent in some redox reactions than
others. When metallic zinc is added to a solution containing copper(II) sulfate (CuSO4),
zinc reduces Cu21 by donating two electrons to it:
Zn(s) 1 CuSO4 (aq) ¡ ZnSO4 (aq) 1 Cu(s)
In the process, the solution loses the blue color that characterizes the presence of
hydrated Cu21 ions (Figure 4.10):
Zn(s) 1 Cu21 (aq) ¡ Zn21 (aq) 1 Cu(s)
The Zn
bar is in
aqueous
solution
of CuSO4
Zn
Cu2+
2e–
Zn
Zn2+
Cu
2e–
Ag+
Zn2+
Cu
Cu2+
Cu2+
Ag
Cu
Cu2+ ions are
converted to Cu atoms.
Zn atoms enter the
solution as Zn2+ ions.
(a)
When a piece of copper wire is
placed in an aqueous AgNO3 solution
Cu atoms enter the solution as Cu2+ ions,
and Ag+ ions are converted to solid Ag.
Ag
(b)
Figure 4.10 Metal displacement reactions in solution. (a) First beaker: A zinc strip is placed in a blue CuSO4 solution. Immediately Cu21
ions are reduced to metallic Cu in the form of a dark layer. Second beaker: In time, most of the Cu21 ions are reduced and the solution
becomes colorless. (b) First beaker: A piece of Cu wire is placed in a colorless AgNO3 solution. Ag1 ions are reduced to metallic Ag.
Second beaker: As time progresses, most of the Ag1 ions are reduced and the solution acquires the characteristic blue color due to the
presence of hydrated Cu21 ions.
4.4 Oxidation-Reduction Reactions
The oxidation and reduction half-reactions are
21
Cu
Zn ¡ Zn21 1 2e 2
1 2e 2 ¡ Cu
Similarly, metallic copper reduces silver ions in a solution of silver nitrate (AgNO3):
Animation
Reaction of Cu with AgNO3
Cu(s) 1 2AgNO3 (aq) ¡ Cu(NO3 ) 2 (aq) 1 2Ag(s)
or
Cu(s) 1 2Ag 1 (aq) ¡ Cu21 (aq) 1 2Ag(s)
Oxidation Number
The definitions of oxidation and reduction in terms of loss and gain of electrons apply
to the formation of ionic compounds such as MgO and the reduction of Cu21 ions
by Zn. However, these definitions do not accurately characterize the formation of
hydrogen chloride (HCl) and sulfur dioxide (SO2):
H2 (g) 1 Cl2 (g) ¡ 2HCl(g)
S(s) 1 O2 (g) ¡ SO2 (g)
Because HCl and SO2 are not ionic but molecular compounds, no electrons are actually transferred in the formation of these compounds, as they are in the case of MgO.
Nevertheless, chemists find it convenient to treat these reactions as redox reactions
because experimental measurements show that there is a partial transfer of electrons
(from H to Cl in HCl and from S to O in SO2).
To keep track of electrons in redox reactions, it is useful to assign oxidation
numbers to the reactants and products. An atom’s oxidation number, also called
oxidation state, signifies the number of charges the atom would have in a
molecule (or an ionic compound) if electrons were transferred completely. For
example, we can rewrite the previous equations for the formation of HCl and SO2
as follows:
0
0
1121
H2(g) 1 Cl2(g) 88n 2HCl(g)
0
0
14 22
S(s) 1 O2(g) 88n SO2(g)
The numbers above the element symbols are the oxidation numbers. In both of
the reactions shown, there is no charge on the atoms in the reactant molecules.
Thus, their oxidation number is zero. For the product molecules, however, it is
assumed that complete electron transfer has taken place and that atoms have
gained or lost electrons. The oxidation numbers reflect the number of electrons
“transferred.”
Oxidation numbers enable us to identify elements that are oxidized and reduced
at a glance. The elements that show an increase in oxidation number—hydrogen and
sulfur in the preceding examples—are oxidized. Chlorine and oxygen are reduced,
so their oxidation numbers show a decrease from their initial values. Note that the
sum of the oxidation numbers of H and Cl in HCl (11 and 21) is zero. Likewise,
if we add the oxidation numbers of S (14) and two atoms of O [2 3 (22)], the
135
136
Chapter 4
■
Reactions in Aqueous Solutions
total is zero. The reason is that the HCl and SO2 molecules are neutral, so the charges
must cancel.
We use the following rules to assign oxidation numbers:
1. In free elements (that is, in the uncombined state), each atom has an oxidation
number of zero. Thus, each atom in H2, Br2, Na, Be, K, O2, and P4 has the same
oxidation number: zero.
2. For ions composed of only one atom (that is, monatomic ions), the oxidation
number is equal to the charge on the ion. Thus, Li1 ion has an oxidation number
of 11; Ba21 ion, 12; Fe31 ion, 13; I2 ion, 21; O22 ion, 22; and so on. All
alkali metals have an oxidation number of 11 and all alkaline earth metals have
an oxidation number of 12 in their compounds. Aluminum has an oxidation
number of 13 in all its compounds.
3. The oxidation number of oxygen in most compounds (for example, MgO and
H2O) is 22, but in hydrogen peroxide (H2O2) and peroxide ion (O22
2 ), it is 21.
4. The oxidation number of hydrogen is 11, except when it is bonded to metals in
binary compounds. In these cases (for example, LiH, NaH, CaH2), its oxidation
number is 21.
5. Fluorine has an oxidation number of 21 in all its compounds. Other halogens
(Cl, Br, and I) have negative oxidation numbers when they occur as halide ions
in their compounds. When combined with oxygen—for example in oxoacids and
oxoanions (see Section 2.7)—they have positive oxidation numbers.
6. In a neutral molecule, the sum of the oxidation numbers of all the atoms must
be zero. In a polyatomic ion, the sum of oxidation numbers of all the elements
in the ion must be equal to the net charge of the ion. For example, in the ammonium ion, NH1
4 , the oxidation number of N is 23 and that of H is 11. Thus, the
sum of the oxidation numbers is 23 1 4(11) 5 11, which is equal to the net
charge of the ion.
7. Oxidation numbers do not have to be integers. For example, the oxidation number
of O in the superoxide ion, O22 , is 212.
We apply the preceding rules to assign oxidation numbers in Example 4.5.
Example 4.5
Assign oxidation numbers to all the elements in the following compounds and ion:
(a) Li2O, (b) HNO3, (c) Cr2O722.
Strategy In general, we follow the rules just listed for assigning oxidation numbers.
Remember that all alkali metals have an oxidation number of 11, and in most cases
hydrogen has an oxidation number of 11 and oxygen has an oxidation number of 22
in their compounds.
Solution
(a) By rule 2 we see that lithium has an oxidation number of 11 (Li1) and oxygen’s
oxidation number is 22 (O22).
(b) This is the formula for nitric acid, which yields a H1 ion and a NO23 ion in
solution. From rule 4 we see that H has an oxidation number of 11. Thus the other
group (the nitrate ion) must have a net oxidation number of 21. Oxygen has an
(Continued)
4.4 Oxidation-Reduction Reactions
137
oxidation number of 22, and if we use x to represent the oxidation number of
nitrogen, then the nitrate ion can be written as
3N(x)O3(22) 4 2
x 1 3(22) 5 21
so that
x 5 15
or
(c) From rule 6 we see that the sum of the oxidation numbers in the dichromate ion
Cr2O722 must be 22. We know that the oxidation number of O is 22, so all that
remains is to determine the oxidation number of Cr, which we call y. The
dichromate ion can be written as
3Cr2( y)O7(22) 4 22
so that
2(y) 1 7(22) 5 22
y 5 16
or
Check In each case, does the sum of the oxidation numbers of all the atoms equal the
net charge on the species?
Similar problems: 4.47, 4.49.
Practice Exercise Assign oxidation numbers to all the elements in the following
compound and ion: (a) PF3, (b) MnO24 .
Figure 4.11 shows the known oxidation numbers of the familiar elements, arranged
according to their positions in the periodic table. We can summarize the content of
this figure as follows:
• Metallic elements have only positive oxidation numbers, whereas nonmetallic
elements may have either positive or negative oxidation numbers.
• The highest oxidation number an element in Groups 1A–7A can have is its group
number. For example, the halogens are in Group 7A, so their highest possible
oxidation number is 17.
• The transition metals (Groups 1B, 3B–8B) usually have several possible oxidation
numbers.
Types of Redox Reactions
Among the most common oxidation-reduction reactions are combination, decomposition, combustion, and displacement reactions. A more involved type is called disproportionation reactions, which will also be discussed in this section.
Combination Reactions
A combination reaction is a reaction in which two or more substances combine to
form a single product. Figure 4.12 shows some combination reactions. For example,
0
0
14 22
S(s) 1 O2(g) 88n SO2(g)
0
0
13 21
2Al(s) 1 3Br2(l) 88n 2AlBr3(s)
Not all combination reactions are redox in
nature. The same holds for decomposition
reactions.
138
Chapter 4
■
Reactions in Aqueous Solutions
1
1A
18
8A
1
2
H
He
+1
–1
2
2A
13
3A
14
4A
15
5A
16
6A
17
7A
3
4
5
6
7
8
9
10
Li
Be
B
C
N
O
F
Ne
+1
+2
+3
+4
+2
–4
+5
+4
+3
+2
+1
–3
+2
– 12
–1
–2
–1
11
12
13
14
15
16
17
18
Na
Mg
Al
Si
P
S
Cl
Ar
+1
+3
+2
3
3B
4
4B
5
5B
6
6B
7
7B
8
9
8B
10
11
1B
+4
–4
+5
+3
–3
12
2B
+6
+4
+2
–2
+7
+6
+5
+4
+3
+1
–1
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
K
Ca
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
Ga
Ge
As
Se
Br
Kr
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
Rb
Sr
Y
Zr
Nb
Mo
Tc
Ru
Rh
Pd
Ag
Cd
In
Sn
Sb
Te
I
Xe
+1
+1
+3
+2
+3
+2
+4
+3
+2
+4
+5
+4
+3
+2
+5
+4
+6
+5
+4
+3
+2
+6
+4
+3
+7
+6
+4
+3
+2
+7
+6
+4
+3
+2
+8
+6
+4
+3
+2
+3
+2
+4
+3
+2
+4
+2
+2
+1
+1
+2
+2
+3
+3
+4
–4
+4
+2
+5
+3
–3
+5
+3
–3
+6
+4
–2
+5
+3
+1
–1
+6
+4
–2
+7
+5
+1
–1
+4
+2
+6
+4
+2
55
56
57
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
Cs
Ba
La
Hf
Ta
W
Re
Os
Ir
Pt
Au
Hg
Tl
Pb
Bi
Po
At
Rn
+1
+3
+2
+4
+5
+6
+4
+7
+6
+4
+8
+4
+4
+3
+4
+2
+3
+1
+2
+1
+3
+1
+4
+2
+5
+3
+2
–1
Figure 4.11 The oxidation numbers of elements in their compounds. The more common oxidation numbers are in color.
(a)
(b)
(c)
Figure 4.12 Some simple combination redox reactions. (a) Sulfur burning in air to form sulfur dioxide. (b) Sodium burning in chlorine to
form sodium chloride. (c) Aluminum reacting with bromine to form aluminum bromide.
4.4 Oxidation-Reduction Reactions
139
Figure 4.13 (a) On heating,
mercury(II) oxide (HgO)
decomposes to form mercury and
oxygen. (b) Heating potassium
chlorate (KClO3) produces oxygen,
which supports the combustion of
the wood splint.
(a)
(b)
Decomposition Reactions
Decomposition reactions are the opposite of combination reactions. Specifically, a
decomposition reaction is the breakdown of a compound into two or more components
(Figure 4.13). For example,
12 22
0
We show oxidation numbers only for
elements that are oxidized or reduced.
All combustion reactions are redox
processes.
0
2HgO(s) 88n 2Hg(l) 1 O2(g)
15 22
21
0
2KClO3(s) 88n 2KCl(s) 1 3O2(g)
1121
0
0
2NaH(s) 88n 2Na(s) 1 H2(g)
Combustion Reactions
A combustion reaction is a reaction in which a substance reacts with oxygen, usually
with the release of heat and light to produce a flame. The reactions between magnesium and sulfur with oxygen described earlier are combustion reactions. Another
example is the burning of propane (C3H8), a component of natural gas that is used
for domestic heating and cooking:
(a)
C3H8 (g) 1 5O2 (g) ¡ 3CO2 (g) 1 4H2O(l)
Assigning an oxidation number to C atoms in organic compounds is more involved. Here,
we focus only on the oxidation number of O atoms, which changes from 0 to 22.
Displacement Reactions
In a displacement reaction, an ion (or atom) in a compound is replaced by an ion
(or atom) of another element: Most displacement reactions fit into one of three
subcategories: hydrogen displacement, metal displacement, or halogen displacement.
1. Hydrogen Displacement. All alkali metals and some alkaline earth metals (Ca,
Sr, and Ba), which are the most reactive of the metallic elements, will displace hydrogen from cold water (Figure 4.14):
0
11
11
11
0
2Na(s) 1 2H2O(l) 88n 2NaOH(aq) 1 H2(g)
0
11
12
11
0
Ca(s) 1 2H2O(l) 88n Ca(OH)2(s) 1 H2(g)
(b)
Figure 4.14 Reactions of
(a) sodium (Na) and (b) calcium
(Ca) with cold water. Note that the
reaction is more vigorous with Na
than with Ca.
140
Chapter 4
■
Reactions in Aqueous Solutions
(a)
(b)
(c)
Figure 4.15 Reactions of (a) iron (Fe), (b) zinc (Zn), and (c) magnesium (Mg) with hydrochloric acid to form hydrogen gas and the metal
chlorides (FeCl2, ZnCl2, MgCl2). The reactivity of these metals is reflected in the rate of hydrogen gas evolution, which is slowest for the
least reactive metal, Fe, and fastest for the most reactive metal, Mg.
Many metals, including those that do not react with water, are capable of displacing hydrogen from acids. For example, zinc (Zn) and magnesium (Mg) do not react
with cold water but do react with hydrochloric acid, as follows:
0
11
12
0
Zn(s) 1 2HCl(aq) 88n ZnCl2(aq) 1 H2(g)
0
11
12
0
Mg(s) 1 2HCl(aq) 88n MgCl2(aq) 1 H2(g)
Figure 4.15 shows the reactions between hydrochloric acid (HCl) and iron (Fe), zinc
(Zn), and magnesium (Mg). These reactions are used to prepare hydrogen gas in the
laboratory.
2. Metal Displacement. A metal in a compound can be displaced by another metal
in the elemental state. We have already seen examples of zinc replacing copper ions
and copper replacing silver ions (see p. 134). Reversing the roles of the metals would
result in no reaction. Thus, copper metal will not displace zinc ions from zinc sulfate,
and silver metal will not displace copper ions from copper nitrate.
An easy way to predict whether a metal or hydrogen displacement reaction will
actually occur is to refer to an activity series (sometimes called the electrochemical
series), shown in Figure 4.16. Basically, an activity series is a convenient summary
of the results of many possible displacement reactions similar to the ones already
discussed. According to this series, any metal above hydrogen will displace it from
water or from an acid, but metals below hydrogen will not react with either water or
an acid. In fact, any metal listed in the series will react with any metal (in a compound) below it. For example, Zn is above Cu, so zinc metal will displace copper
ions from copper sulfate.
4.4 Oxidation-Reduction Reactions
Reducing strength increases
Li n Li1 1 e2
K n K1 1 e2
Ba n Ba21 1 2e2
Ca n Ca21 1 2e2
Na n Na1 1 e2
Mg n Mg21 1 2e2
Al n Al31 1 3e2
Zn n Zn21 1 2e2
Cr n Cr31 1 3e2
Fe n Fe21 1 2e2
Cd n Cd21 1 2e2
Co n Co21 1 2e2
Ni n Ni21 1 2e2
Sn n Sn21 1 2e2
Pb n Pb21 1 2e2
H2 n 2H1 1 2e2
Cu n Cu21 1 2e2
Ag n Ag1 1 e2
Hg n Hg21 1 2e2
Pt n Pt21 1 2e2
Au n Au31 1 3e2
141
React with cold
water to produce H2
React with steam
to produce H2
React with acids
to produce H2
Do not react with water
or acids to produce H2
Figure 4.16 The activity series for metals. The metals are arranged according to their ability to
displace hydrogen from an acid or water. Li (lithium) is the most reactive metal, and Au (gold) is the
least reactive.
Metal displacement reactions find many applications in metallurgical processes,
the goal of which is to separate pure metals from their ores. For example, vanadium
is obtained by treating vanadium(V) oxide with metallic calcium:
V2O5 (s) 1 5Ca(l) ¡ 2V(l) 1 5CaO(s)
Similarly, titanium is obtained from titanium(IV) chloride according to the reaction
TiCl4 (g) 1 2Mg(l) ¡ Ti(s) 1 2MgCl2 (l)
In each case, the metal that acts as the reducing agent lies above the metal that is
reduced (that is, Ca is above V and Mg is above Ti) in the activity series. We will
see more examples of this type of reaction in Chapter 18.
3. Halogen Displacement. Another activity series summarizes the halogens’ behavior
in halogen displacement reactions:
1A
8A
2A
F2 . Cl2 . Br2 . I2
The power of these elements as oxidizing agents decreases as we move down Group
7A from fluorine to iodine, so molecular fluorine can replace chloride, bromide, and
iodide ions in solution. In fact, molecular fluorine is so reactive that it also attacks
water; thus these reactions cannot be carried out in aqueous solutions. On the other
hand, molecular chlorine can displace bromide and iodide ions in aqueous solution.
The displacement equations are
0
21
21
0
Cl2(g) 1 2KBr(aq) 88n 2KCl(aq) 1 Br2(l)
0
21
21
0
Cl2(g) 1 2NaI(aq) 88n 2NaCl(aq) 1 I2(s)
The halogens.
3A 4A 5A 6A 7A
F
Cl
Br
I
142
Chapter 4
■
Reactions in Aqueous Solutions
The ionic equations are
0
21
21
0
Cl2(g) 1 2Br(aq) 88n 2Cl2(aq) 1 Br2(l)
0
21
21
0
Cl2(g) 1 2I2(aq) 88n 2Cl2(aq) 1 I2(s)
Molecular bromine, in turn, can displace iodide ion in solution:
0
21
21
0
Br2(l) 1 2I2(aq) 88n 2Br2(aq) 1 I2(s)
Reversing the roles of the halogens produces no reaction. Thus, bromine cannot displace chloride ions, and iodine cannot displace bromide and chloride ions.
The halogen displacement reactions have a direct industrial application. The halogens as a group are the most reactive of the nonmetallic elements. They are all strong
oxidizing agents. As a result, they are found in nature in the combined state (with
metals) as halides and never as free elements. Of these four elements, chlorine is by
far the most important industrial chemical. In 2010 the amount of chlorine produced
in the United States was about 25 billion pounds, making chlorine the tenth-ranking
industrial chemical. The annual production of bromine is only one-hundredth that of
chlorine, while the amounts of fluorine and iodine produced are even less.
Recovering the halogens from their halides requires an oxidation process, which
is represented by
2X 2 ¡ X2 1 2e 2
where X denotes a halogen element. Seawater and natural brine (for example, underground water in contact with salt deposits) are rich sources of Cl2, Br2, and I2 ions.
Minerals such as fluorite (CaF2) and cryolite (Na3AlF6) are used to prepare fluorine.
Because fluorine is the strongest oxidizing agent known, there is no way to convert
F2 ions to F2 by chemical means. The only way to carry out the oxidation is by
electrolytic means, the details of which will be discussed in Chapter 18. Industrially,
chlorine, like fluorine, is produced electrolytically.
Bromine is prepared industrially by oxidizing Br2 ions with chlorine, which is a
strong enough oxidizing agent to oxidize Br2 ions but not water:
Bromine is a fuming red liquid.
2Br 2 (aq) ¡ Br2 (l) 1 2e 2
Figure 4.17 The industrial
manufacture of liquid bromine by
oxidizing an aqueous solution
containing Br2 ions with
chlorine gas.
7B
Mn
1B 2B
Cu
5A 6A 7A
N O
P S Cl
Br
I
Au Hg
Elements that are most likely
to undergo disproportionation
reactions.
One of the richest sources of Br2 ions is the Dead Sea—about 4000 parts per
million (ppm) by mass of all dissolved substances in the Dead Sea is Br. Following the oxidation of Br2 ions, bromine is removed from the solution by blowing
air over the solution, and the air-bromine mixture is then cooled to condense the
bromine (Figure 4.17).
Iodine is also prepared from seawater and natural brine by the oxidation of I2
ions with chlorine. Because Br2 and I2 ions are invariably present in the same source,
they are both oxidized by chlorine. However, it is relatively easy to separate Br2 from
I2 because iodine is a solid that is sparingly soluble in water. The air-blowing procedure will remove most of the bromine formed but will not affect the iodine present.
Disproportionation Reaction
A special type of redox reaction is the disproportionation reaction. In a disproportionation reaction, an element in one oxidation state is simultaneously oxidized and
reduced. One reactant in a disproportionation reaction always contains an element that
can have at least three oxidation states. The element itself is in an intermediate
oxidation state; that is, both higher and lower oxidation states exist for that element
4.4 Oxidation-Reduction Reactions
in the products. The decomposition of hydrogen peroxide is an example of a disproportionation reaction:
21
22
0
2H2O2(aq) 88n 2H2O(l) 1 O2(g)
Note that the oxidation number of H
remains unchanged at 11.
Here the oxidation number of oxygen in the reactant (21) both increases to zero in
O2 and decreases to 22 in H2O. Another example is the reaction between molecular
chlorine and NaOH solution:
0
11
21
Cl2(g) 1 2OH2(aq) 88n ClO2(aq) 1 Cl2(aq) 1 H2O(l)
This reaction describes the formation of household bleaching agents, for it is the
hypochlorite ion (ClO2) that oxidizes the color-bearing substances in stains, converting them to colorless compounds.
Finally, it is interesting to compare redox reactions and acid-base reactions. They
are analogous in that acid-base reactions involve the transfer of protons while redox
reactions involve the transfer of electrons. However, while acid-base reactions are
quite easy to recognize (because they always involve an acid and a base), there is no
simple procedure for identifying a redox process. The only sure way is to compare
the oxidation numbers of all the elements in the reactants and products. Any change
in oxidation number guarantees that the reaction is redox in nature.
The classification of different types of redox reactions is illustrated in Example 4.6.
Example 4.6
Classify the following redox reactions and indicate changes in the oxidation numbers of
the elements:
(a) 2N2O(g) ¡ 2N2(g) 1 O2(g)
(b) 6Li(s) 1 N2(g) ¡ 2Li3N(s)
(c) Ni(s) 1 Pb(NO3)2(aq) ¡ Pb(s) 1 Ni(NO3)2(aq)
(d) 2NO2(g) 1 H2O(l) ¡ HNO2(aq) 1 HNO3(aq)
Strategy Review the definitions of combination reactions, decomposition reactions,
displacement reactions, and disproportionation reactions.
Solution
(a) This is a decomposition reaction because one reactant is converted to two different
products. The oxidation number of N changes from 11 to 0, while that of O
changes from 22 to 0.
(b) This is a combination reaction (two reactants form a single product). The oxidation
number of Li changes from 0 to 11 while that of N changes from 0 to 23.
(c) This is a metal displacement reaction. The Ni metal replaces (reduces) the Pb21 ion. The
oxidation number of Ni increases from 0 to 12 while that of Pb decreases from 12 to 0.
(d) The oxidation number of N is 14 in NO2 and it is 13 in HNO2 and 15 in HNO3.
Because the oxidation number of the same element both increases and decreases,
this is a disproportionation reaction.
Practice Exercise Identify the following redox reactions by type:
(a)
(b)
(c)
(d)
Fe 1 H2SO4 ¡ FeSO4 1 H2
S 1 3F2 ¡ SF6
2CuCl ¡ Cu 1 CuCl2
2Ag 1 PtCl2 ¡ 2AgCl 1 Pt
Similar problems: 4.55, 4.56.
143
CHEMISTRY in Action
Breathalyzer
E
very year in the United States about 25,000 people are killed
and 500,000 more are injured as a result of drunk driving. In
spite of efforts to educate the public about the dangers of driving
while intoxicated and stiffer penalties for drunk driving offenses,
law enforcement agencies still have to devote a great deal of
work to removing drunk drivers from America’s roads.
The police often use a device called a breathalyzer to
test drivers suspected of being drunk. The chemical basis of
this device is a redox reaction. A sample of the driver’s
breath is drawn into the breathalyzer, where it is treated with
an acidic solution of potassium dichromate. The alcohol
(ethanol) in the breath is converted to acetic acid as shown in
the following equation:
3CH3CH2OH
ethanol
1
2K2Cr2O7
1
8H2SO4 ¡
potassium
dichromate
(orange yellow)
sulfuric
acid
A driver being tested for blood alcohol content with a handheld breathalyzer.
3CH3COOH 1 2Cr2(SO4)3 1 2K2SO4 1 11H2O
acetic acid
chromium(III)
sulfate (green)
to the green chromium(III) ion (see Figure 4.22). The driver’s
blood alcohol level can be determined readily by measuring the
degree of this color change (read from a calibrated meter on the
instrument). The current legal limit of blood alcohol content is
0.08 percent by mass. Anything higher constitutes intoxication.
potassium
sulfate
In this reaction, the ethanol is oxidized to acetic acid and the
chromium(VI) in the orange-yellow dichromate ion is reduced
Breath
Meter
Filter
Light
source
Photocell
detector
Schematic diagram of a breathalyzer.
The alcohol in the driver’s breath is
reacted with a potassium dichromate
solution. The change in the absorption
of light due to the formation of
chromium(III) sulfate is registered by
the detector and shown on a meter,
which directly displays the alcohol
content in blood. The filter selects only
one wavelength of light for
measurement.
K2Cr2O7
solution
Review of Concepts
Which of the following combination reactions is not a redox reaction?
(a) 2Mg(s) 1 O2(g) ¡ 2MgO(s)
(b) H2(g) 1 F2(g) ¡ 2HF(g)
(c) NH3(g) 1 HCl(g) ¡ NH4Cl(s)
(d) 2Na(s) 1 S(s) ¡ Na2S(s)
The above Chemistry in Action essay describes how law enforcement makes use
of a redox reaction to apprehend drunk drivers.
144
4.5 Concentration of Solutions
145
4.5 Concentration of Solutions
To study solution stoichiometry, we must know how much of the reactants are present
in a solution and also how to control the amounts of reactants used to bring about a
reaction in aqueous solution.
The concentration of a solution is the amount of solute present in a given amount
of solvent, or a given amount of solution. (For this discussion, we will assume the
solute is a liquid or a solid and the solvent is a liquid.) The concentration of a solution can be expressed in many different ways, as we will see in Chapter 12. Here we
will consider one of the most commonly used units in chemistry, molarity (M), or
molar concentration, which is the number of moles of solute per liter of solution.
Molarity is defined as
molarity 5
moles of solute
liters of solution
(4.1)
Equation (4.1) can also be expressed algebraically as
M5
n
V
(4.2)
Keep in mind that volume (V) is liters of
solution, not liters of solvent. Also, the
molarity of a solution depends on
temperature.
where n denotes the number of moles of solute and V is the volume of the solution
in liters.
A 1.46 molar glucose (C6H12O6) solution, written as 1.46 M C6H12O6, contains
1.46 moles of the solute (C6H12O6) in 1 L of the solution. Of course, we do not always
work with solution volumes of 1 L. Thus, a 500-mL solution containing 0.730 mole
of C6H12O6 also has a concentration of 1.46 M:
molarity 5
0.730 mol C6H12O6
1000 mL soln
3
5 1.46 M C6H12O6
500 mL soln
1 L soln
Note that concentration, like density, is an intensive property, so its value does not
depend on how much of the solution is present.
It is important to keep in mind that molarity refers only to the amount of solute
originally dissolved in water and does not take into account any subsequent processes,
such as the dissociation of a salt or the ionization of an acid. Consider what happens
when a sample of potassium chloride (KCl) is dissolved in enough water to make a
1 M solution:
H2O
KCl(s) ¡ K1(aq) 1 Cl2(aq)
Because KCl is a strong electrolyte, it undergoes complete dissociation in solution. Thus, a 1 M KCl solution contains 1 mole of K1 ions and 1 mole of Cl2
ions, and no KCl units are present. The concentrations of the ions can be expressed
as [K1] 5 1 M and [Cl2] 5 1 M, where the square brackets [ ] indicate that the
concentration is expressed in molarity. Similarly, in a 1 M barium nitrate
[Ba(NO3)2] solution
H2O
Ba(NO3 ) 2 (s) ¡ Ba21 (aq) 1 2NO23 (aq)
we have [Ba21] 5 1 M and [NO23 ] 5 2 M and no Ba(NO3)2 units at all.
The procedure for preparing a solution of known molarity is as follows. First, the
solute is accurately weighed and transferred to a volumetric flask through a funnel
Animation
Making a Solution
146
Chapter 4
■
Reactions in Aqueous Solutions
Figure 4.18 Preparing a solution
of known molarity. (a) A known
amount of a solid solute is
transferred into the volumetric
flask; then water is added through
a funnel. (b) The solid is slowly
dissolved by gently swirling the
flask. (c) After the solid has
completely dissolved, more water
is added to bring the level of
solution to the mark. Knowing the
volume of the solution and the
amount of solute dissolved in it,
we can calculate the molarity of
the prepared solution.
Meniscus
Marker showing
known volume
of solution
(a)
(b)
(c)
(Figure 4.18). Next, water is added to the flask, which is carefully swirled to dissolve
the solid. After all the solid has dissolved, more water is added slowly to bring the
level of solution exactly to the volume mark. Knowing the volume of the solution in
the flask and the quantity of compound (the number of moles) dissolved, we can
calculate the molarity of the solution using Equation (4.1). Note that this procedure
does not require knowing the amount of water added, as long as the volume of the
final solution is known.
Examples 4.7 and 4.8 illustrate the applications of Equations (4.1) and (4.2).
Example 4.7
How many grams of potassium dichromate (K2Cr2O7) are required to prepare a 250-mL
solution whose concentration is 2.16 M?
Strategy How many moles of K2Cr2O7 does a 1-L (or 1000 mL) 2.16 M K2Cr2O7
solution contain? A 250-mL solution? How would you convert moles to grams?
Solution The first step is to determine the number of moles of K2Cr2O7 in 250 mL or
0.250 L of a 2.16 M solution. Rearranging Equation (4.1) gives
moles of solute 5 molarity 3 L soln
A K2Cr2O7 solution.
Thus,
2.16 mol K2Cr2O7
3 0.250 L soln
1 L soln
5 0.540 mol K2Cr2O7
moles of K2Cr2O7 5
The molar mass of K2Cr2O7 is 294.2 g, so we write
grams of K2Cr2O7 needed 5 0.540 mol K2Cr2O7 3
294.2 g K2Cr2O7
1 mol K2Cr2O7
5 159 g K2Cr2O7
(Continued)
4.5 Concentration of Solutions
147
Check As a ball-park estimate, the mass should be given by [molarity (mol/L) 3
volume (L) 3 molar mass (g/mol)] or [2 mol/L 3 0.25 L 3 300 g/mol] 5 150 g. So
the answer is reasonable.
Similar problems: 4.65, 4.68.
Practice Exercise What is the molarity of an 85.0-mL ethanol (C2H5OH) solution
containing 1.77 g of ethanol?
Example 4.8
A chemist needs to add 3.81 g of glucose to a reaction mixture. Calculate the volume in
milliliters of a 2.53 M glucose solution she should use for the addition.
Strategy We must first determine the number of moles contained in 3.81 g of glucose
and then use Equation (4.2) to calculate the volume.
Solution From the molar mass of glucose, we write
3.81 g C6H12O6 3
1 mol C6H12O6
5 2.114 3 1022 mol C6H12O6
180.2 g C6H12O6
Note that we have carried an additional
digit past the number of significant figures
for the intermediate step.
Next, we calculate the volume of the solution that contains 2.114 3 1022 mole of the
solute. Rearranging Equation (4.2) gives
n
M
2.114 3 1022 mol C6H12O6
1000 mL soln
5
3
2.53 mol C6H12O6/L soln
1 L soln
V5
5 8.36 mL soln
Check One liter of the solution contains 2.53 moles of C6H12O6. Therefore, the number
of moles in 8.36 mL or 8.36 3 1023 L is (2.53 mol 3 8.36 3 1023) or 2.12 3 1022 mol.
The small difference is due to the different ways of rounding off.
Similar problem: 4.67.
Practice Exercise What volume (in milliliters) of a 0.315 M NaOH solution contains
6.22 g of NaOH?
Dilution of Solutions
Concentrated solutions are often stored in the laboratory stockroom for use as needed.
Frequently we dilute these “stock” solutions before working with them. Dilution is the
procedure for preparing a less concentrated solution from a more concentrated one.
Suppose that we want to prepare 1 L of a 0.400 M KMnO4 solution from a solution of 1.00 M KMnO4. For this purpose we need 0.400 mole of KMnO4. Because
there is 1.00 mole of KMnO4 in 1 L of a 1.00 M KMnO4 solution, there is 0.400
mole of KMnO4 in 0.400 L of the same solution:
Animation
Preparing a Solution by Dilution
1.00 mol
0.400 mol
5
1 L soln
0.400 L soln
Therefore, we must withdraw 400 mL from the 1.00 M KMnO4 solution and dilute
it to 1000 mL by adding water (in a 1-L volumetric flask). This method gives us 1 L
of the desired solution of 0.400 M KMnO4.
In carrying out a dilution process, it is useful to remember that adding more
solvent to a given amount of the stock solution changes (decreases) the concentration
Two KMnO4 solutions of different
concentrations.
148
Chapter 4
■
Reactions in Aqueous Solutions
Figure 4.19 The dilution of a
more concentrated solution (a) to
a less concentrated one (b) does
not change the total number of
solute particles (18).
(a)
(b)
of the solution without changing the number of moles of solute present in the solution
(Figure 4.19). In other words,
moles of solute before dilution 5 moles of solute after dilution
Molarity is defined as moles of solute in 1 liter of solution, so the number of moles
of solute is given by [see Equation (4.2)]
moles of solute
3 volume of soln (in liters) 5 moles of solute
liters of soln
M
V
n
or
MV 5 n
Because all the solute comes from the original stock solution, we can conclude that
n remains the same; that is,
MiVi
moles of solute
before dilution
5
MfVf
(4.3)
moles of solute
after dilution
where Mi and Mf are the initial and final concentrations of the solution in molarity
and Vi and Vf are the initial and final volumes of the solution, respectively. Of course,
the units of Vi and Vf must be the same (mL or L) for the calculation to work. To
check the reasonableness of your results, be sure that Mi . Mf and Vf . Vi.
We apply Equation (4.3) in Example 4.9.
Example 4.9
Describe how you would prepare 5.00 3 102 mL of a 1.75 M H2SO4 solution, starting
with an 8.61 M stock solution of H2SO4.
Strategy Because the concentration of the final solution is less than that of the original
one, this is a dilution process. Keep in mind that in dilution, the concentration of the
solution decreases but the number of moles of the solute remains the same.
Solution We prepare for the calculation by tabulating our data:
Mi 5 8.61 M
Vi 5 ?
Mf 5 1.75 M
Vf 5 5.00 3 102 mL
(Continued)
4.6 Gravimetric Analysis
149
Substituting in Equation (4.3),
(8.61 M) (Vi ) 5 (1.75 M) (5.00 3 102 mL)
(1.75 M) (5.00 3 102 mL)
Vi 5
8.61 M
5 102 mL
Thus, we must dilute 102 mL of the 8.61 M H2SO4 solution with sufficient water to
give a final volume of 5.00 3 102 mL in a 500-mL volumetric flask to obtain the
desired concentration.
Check The initial volume is less than the final volume, so the answer is reasonable.
Similar problems: 4.75, 4.76.
Practice Exercise How would you prepare 2.00 3 102 mL of a 0.866 M NaOH
solution, starting with a 5.07 M stock solution?
Review of Concepts
What is the final concentration of a 0.6 M NaCl solution if its volume is doubled
and the number of moles of solute is tripled?
Now that we have discussed the concentration and dilution of solutions, we can
examine the quantitative aspects of reactions in aqueous solution, or solution stoichiometry. Sections 4.6–4.8 focus on two techniques for studying solution stoichiometry:
gravimetric analysis and titration. These techniques are important tools of quantitative
analysis, which is the determination of the amount or concentration of a substance
in a sample.
4.6 Gravimetric Analysis
Gravimetric analysis is an analytical technique based on the measurement of mass.
One type of gravimetric analysis experiment involves the formation, isolation, and
mass determination of a precipitate. Generally, this procedure is applied to ionic compounds. First, a sample substance of unknown composition is dissolved in water and
allowed to react with another substance to form a precipitate. Then the precipitate is
filtered off, dried, and weighed. Knowing the mass and chemical formula of the precipitate formed, we can calculate the mass of a particular chemical component (that
is, the anion or cation) of the original sample. Finally, from the mass of the component
and the mass of the original sample, we can determine the percent composition by
mass of the component in the original compound.
A reaction that is often studied in gravimetric analysis, because the reactants can
be obtained in pure form, is
AgNO3 (aq) 1 NaCl(aq) ¡ NaNO3 (aq) 1 AgCl(s)
The net ionic equation is
Ag 1 (aq) 1 Cl 2 (aq) ¡ AgCl(s)
The precipitate is silver chloride (see Table 4.2). As an example, let us say that we
wanted to determine experimentally the percent by mass of Cl in NaCl. First, we
would accurately weigh out a sample of NaCl and dissolve it in water. Next, we would
add enough AgNO3 solution to the NaCl solution to cause the precipitation of all
This procedure would enable us to
determine the purity of the NaCl sample.
150
Chapter 4
■
Reactions in Aqueous Solutions
(a)
(b)
(c)
Figure 4.20 Basic steps for gravimetric analysis. (a) A solution containing a known amount of NaCl in a beaker. (b) The precipitation of
AgCl upon the addition of AgNO3 solution from a measuring cylinder. In this reaction, AgNO3 is the excess reagent and NaCl is the limiting
reagent. (c) The solution containing the AgCl precipitate is filtered through a preweighed sintered-disk crucible, which allows the liquid (but
not the precipitate) to pass through. The crucible is then removed from the apparatus, dried in an oven, and weighed again. The difference
between this mass and that of the empty crucible gives the mass of the AgCl precipitate.
th
Pablo Simba Ramírez