Periodic Table of the Elements
Hydrogen
1
H
1
2
3
4
5
6
7
MAIN GROUP METALS
1.008
1A
(1)
2A
(2)
Lithium
3
Beryllium
4
Li
TRANSITION METALS
Uranium
92
U
METALLOIDS
Be
6.94
9.0122
Sodium Magnesium
12
11
Na
Mg
22.990
24.305
Potassium
19
Calcium
20
39.098
3B
(3)
4B
(4)
5B
(5)
6B
(6)
Symbol
238.03
NONMETALS
7B
(7)
Atomic number
Atomic weight
8B
(8)
(9)
(10)
Scandium Titanium Vanadium Chromium Manganese
25
22
23
24
21
Iron
26
Cobalt
27
Nickel
28
40.078
44.956
47.867
55.845
58.933
58.693
Rubidium Strontium
38
37
Yttrium
39
Zirconium Niobium Molybdenum Technetium Ruthenium Rhodium Palladium
45
42
43
40
44
46
41
85.468
Cesium
55
87.62
Barium
56
88.906
91.224
92.906
Lanthanum Hafnium Tantalum
57
72
73
132.91
Francium
87
137.33
Radium
88
138.91
178.49
180.95
183.84
186.21
Actinium Rutherfordium Dubnium Seaborgium Bohrium
105
107
104
106
89
(223)
(226)
K
Rb
Cs
Fr
Ca
Sr
Ba
Ra
Sc
Y
La
Ac
(227)
Note: Atomic weights are
IUPAC values. For elements
for which IUPAC recommends
ranges of atomic weights,
conventional values are shown.
Numbers in parentheses are
mass numbers of the most
stable isotope of an element.
Ti
V
50.942
Zr
Nb
Hf
Ta
Rf
(267)
Lanthanides
Db
(268)
Cerium
58
Ce
140.12
Actinides
Cr
51.996
Mn
54.938
Mo
Tc
W
Re
95.95
(98)
Tungsten Rhenium
75
74
Sg
(269)
Bh
(270)
Fe
Ru
101.07
Osmium
76
Os
Co
Rh
102.91
Iridium
77
Ir
Ni
Pd
106.42
Platinum
78
Pt
190.23
192.22
195.08
Hassium Meitnerium Darmstadtium
109
110
108
Hs
(277)
Mt
(276)
Ds
(281)
Praseodymium Neodymium Promethium Samarium Europium
59
60
61
63
62
Pr
140.91
Nd
144.24
Pm
(145)
Sm
150.36
Eu
151.96
Thorium Protactinium Uranium Neptunium Plutonium Americium
92
94
91
90
93
95
Th
232.04
Pa
231.04
U
238.03
Np
(237)
Pu
(244)
Am
(243)
For the latest information see: https://iupac.org/what-we-do/periodic-table-of-elements/
and https://www.nist.gov/pml/periodic-table-elements
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8A
(18)
Helium
2
4A
(14)
5A
(15)
6A
(16)
7A
(17)
4.0026
He
hydrogen atoms
Boron
5
Carbon
6
Nitrogen
7
Oxygen
8
Fluorine
9
Neon
10
oxygen atoms
10.81
Aluminum
13
12.011
Silicon
14
14.007
Phosphorus
15
15.999
Sulfur
16
18.998
Chlorine
17
20.180
Argon
18
Al
C
Si
N
P
O
S
F
Cl
Ne
Ar
1B
(11)
2B
(12)
26.982
28.085
30.974
32.06
35.45
39.95
Copper
29
Zinc
30
Gallium
31
Germanium
32
Arsenic
33
Selenium
34
Bromine
35
Krypton
36
63.546
65.38
69.723
72.630
74.922
78.971
79.904
83.798
Silver
47
Cadmium
48
Indium
49
Tin
50
Antimony Tellurium
51
52
Iodine
53
Xenon
54
107.87
Gold
79
112.41
Mercury
80
114.82
Thallium
81
127.60
126.90
Polonium Astatine
84
85
131.29
Radon
86
Ag
Au
Zn
Cd
Hg
carbon atoms
3A
(13)
B
Cu
Standard Colors for Atoms
in Molecular Models
Ga
In
Tl
Ge
Sn
118.71
Lead
82
Pb
As
Sb
121.76
Bismuth
83
Bi
Se
Te
Po
Br
I
At
nitrogen atoms
chlorine atoms
Kr
Xe
Rn
200.59
204.38
207.2
208.98
(209)
(210)
(222)
196.97
Roentgenium Copernicium Nihonium Flerovium Moscovium Livermorium Tennessine Oganesson
114
111
112
113
115
116
117
118
Rg
(282)
Cn
Nh
Mc
Lv
Ts
(293)
Gadolinium Terbium Dysprosium Holmium
66
67
65
64
Erbium
68
Thulium
69
Ytterbium Lutetium
71
70
167.26
168.93
173.05
Dy
Ho
Er
Tm
Yb
Lu
158.93
Curium
96
Berkelium Californium Einsteinium Fermium Mendelevium Nobelium Lawrencium
97
100
98
99
101
102
103
Cm
Bk
(247)
Cf
(251)
164.93
(294)
157.25
(247)
162.50
(294)
Og
(290)
Tb
(286)
Fl
(289)
Gd
(285)
Es
(252)
Fm
(257)
Md
(258)
No
(259)
174.97
Lr
(262)
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Chemistry
11th Edition
& Chemical Reactivity
Kotz
Treichel
Townsend
Treichel
Australia • Brazil • Canada • Mexico • Singapore • United Kingdom • United States
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Chemistry and Chemical Reactivity,
­Eleventh Edition
© 2023, 2019, 2015 Cengage Learning, Inc. ALL RIGHTS RESERVED.
John C. Kotz, Paul M. Treichel,
John R. Townsend, and David A. Treichel
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Brief Contents
Part One The Basic Tools of Chemistry
Part Five The Chemistry of the Elements
1 Basic Concepts of Chemistry 2
1R Let’s Review: The Tools of Quantitative
Chemistry 30
2 Atoms, Molecules, and Ions 62
3 Chemical Reactions 138
4 Stoichiometry: Quantitative Information about
Chemical Reactions 190
5 Principles of Chemical Reactivity: Energy and
Chemical Reactions 254
20 Nuclear Chemistry 992
21 The Chemistry of the Main Group Elements 1038
22 The Chemistry of the Transition Elements 1104
23 Carbon: Not Just Another Element 1150
24 Biochemistry 1206
25 Environmental Chemistry—Earth’s Environment,
Energy, and Sustainability 1244
Part Two Atoms and Molecules
6 The Structure of Atoms 304
7 The Structure of Atoms and Periodic Trends 344
8 Bonding and Molecular Structure 386
9 Bonding and Molecular Structure: Orbital
Hybridization and Molecular Orbitals 458
Part Three States of Matter
10 Gases and Their Properties 498
11 Intermolecular Forces and Liquids 540
12 The Solid State 580
13 Solutions and Their Behavior 626
Part Four The Control of Chemical Reactions
14 Chemical Kinetics: The Rates of Chemical
Reactions 672
15 Principles of Chemical Reactivity: Equilibria 736
16 Principles of Chemical Reactivity: The Chemistry of
Acids and Bases 776
17 Principles of Chemical Reactivity: Other Aspects of
Aqueous Equilibria 830
18 Principles of Chemical Reactivity: Entropy and Free
Energy 886
19 Principles of Chemical Reactivity: Electron Transfer
Reactions 932
List of Appendices
A Using Logarithms and Solving Quadratic
Equations A-2
B Some Important Physical Concepts A-6
C Abbreviations and Useful Conversion Factors A-9
D Physical Constants A-13
E A Brief Guide to Naming Organic
Compounds A-15
F Values for the Ionization Energies and Electron
Attachment Enthalpies of the Elements A-18
G Vapor Pressure of Water at Various
Temperatures A-19
H Ionization Constants for Aqueous Weak Acids at
25 °C A-20
I Ionization Constants for Aqueous Weak Bases at
25 °C A-22
J Solubility Product Constants for Some Inorganic
Compounds at 25 °C A-23
K Formation Constants for Some Complex Ions in
Aqueous Solution at 25 °C A-24
L Selected Thermodynamic Values A-25
M Standard Reduction Potentials in Aqueous Solution
at 25 °C A-32
N Answers to Study Questions, Check Your
Understanding, and Applying Chemical Principles
Questions A-36
Index of Names I-1
Index and Glossary I-4
iii
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Contents
Preface xii
Part One The Basic Tools of Chemistry
1
1.1
1.2
1.3
1.4
1.5
1.6
1.7
Basic Concepts of Chemistry 2
Chemistry and Its Methods 3
Sustainability and Green Chemistry 7
Classifying Matter 8
Elements 12
Compounds 13
Properties and Changes 15
Energy: Some Basic Principles 20
Applying Chemical Principles 1.1:
CO2 in the Oceans 21
Review: The Tools of
1R Let’s
Quantitative Chemistry 30
1R.1 Units of Measurement 31
A Closer Look: The SI Base Units
34
A Closer Look: Energy and Food
37
2.3
A Closer Look: Mendeleev and the Periodic
Table 78
2.4
2.5
2.6
1R.3
1R.4
1R.5
1R.6
2
2.1
2.2
Atoms, Molecules, and Ions 62
Atomic Structure, Atomic Number, and Atomic
Mass 63
Atomic Weight 67
A Closer Look: Marie Curie (1867–1934) 82
Molecules: Formulas, Models, and Names 84
Ions 89
Ionic Compounds: Formulas, Names, and
Properties 93
A Closer Look: Hydrated Ionic Compounds
2.7
Atoms, Molecules, and the Mole
98
99
A Closer Look: Amedeo Avogadro and His
Number 100
2.8
2.9
1R.2 Making Measurements: Precision, Accuracy,
Experimental Error, and Standard Deviation 37
Mathematics of Chemistry 41
Problem Solving by Dimensional Analysis 47
Graphs and Graphing 48
Problem Solving and Chemical Arithmetic 49
Applying Chemical Principles 1R.1:
Out of Gas! 51
Applying Chemical Principles 1R.2:
Ties in Swimming and Significant
Figures 52
A Closer Look: Isotopic Abundances and Atomic
Weights 70
Key Experiments: The Nature of the Atom and Its
Components 72
The Periodic Table 74
3
A Closer Look: The Mole, a Counting Unit 103
Chemical Analysis: Determining Compound
Formulas 106
Instrumental Analysis:
Determining Compound Formulas 114
Applying Chemical Principles 2.1:
Using Isotopes: Ötzi, the Iceman of the Alps 117
Applying Chemical Principles 2.2:
Arsenic, Medicine, and the Formula of
Compound 606 118
Applying Chemical Principles 2.3:
Argon—An Amazing Discovery 118
Chemical Reactions 138
3.1
Introduction to Chemical Equations
3.2
3.3
3.4
3.5
3.6
A Closer Look: Antoine Laurent Lavoisier
(1743–1794) 141
Balancing Chemical Equations 142
Introduction to Chemical Equilibrium 145
Aqueous Solutions 147
Precipitation Reactions 151
Acids and Bases 156
139
A Closer Look: Sulfuric Acid 162
iv
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3.7
3.8
3.9
Acid–Base Reactions 163
Oxidation–Reduction Reactions 167
A Closer Look: Are Oxidation Numbers
Real? 171
Classifying Reactions in Aqueous
Solution 173
A Closer Look: Alternative Organizations of
Reaction Types 174
Applying Chemical Principles 3.1:
Superconductors 177
Applying Chemical Principles 3.2:
Sequestering Carbon Dioxide 177
Applying Chemical Principles 3.3:
Black Smokers and Volcanoes 178
4
Stoichiometry: Quantitative
Information about Chemical
Reactions 190
4.1
Mass Relationships in Chemical Reactions:
Stoichiometry 191
4.2
Reactions in Which One Reactant
Is Present in Limited Supply 195
Percent Yield 200
Chemical Equations and Chemical
Analysis 202
4.3
4.4
4.5
A Closer Look: Nuclear Magnetic Resonance
(NMR) Spectroscopy 208
Measuring Concentrations of Compounds in
Solution 209
A Closer Look: Serial Dilutions 215
pH, a Concentration Scale for Acids and
Bases 215
4.7 Stoichiometry of Reactions
in Aqueous Solution—Fundamentals 217
4.8 Stoichiometry of Reactions in
Aqueous Solution—Titrations 220
4.9 Spectrophotometry 227
Applying Chemical Principles 4.1:
Atom Economy 232
Applying Chemical Principles 4.2:
Bleach 232
Applying Chemical Principles 4.3:
How Much Salt is There in
Seawater? 233
Applying Chemical Principles 4.4:
The Martian 234
4.6
5
Principles of Chemical
Reactivity: Energy and Chemical
Reactions 254
5.1
Energy: Some Basic Principles
5.2
5.3
5.4
A Closer Look: What Is Heat? 257
Specific Heat Capacity: Heating and Cooling 258
Energy and Changes of State 262
The First Law of Thermodynamics 266
A Closer Look: P–V Work
5.5
5.6
5.7
5.8
255
268
A Closer Look: Enthalpy, Internal Energy, and
Non-Expansion Work 270
Enthalpy Changes for Chemical Reactions 271
Calorimetry 274
Enthalpy Calculations 278
A Closer Look: Hess’s Law and Equation
5.7 284
Product- or Reactant-Favored
Reactions and Thermodynamics 285
Applying Chemical Principles 5.1:
Gunpowder 286
Applying Chemical Principles 5.2:
The Fuel Controversy—Alcohol and
Gasoline 287
Part Two Atoms and Molecules
6
6.1
6.2
6.3
6.4
6.5
6.6
6.7
The Structure of Atoms 304
Electromagnetic Radiation 305
Quantization: Planck, Einstein, Energy, and
Photons 308
Atomic Line Spectra and Niels Bohr 312
A Closer Look: Niels Bohr (1885–1962) 314
Wave–Particle Duality: Prelude to Quantum
Mechanics 319
The Modern View of Electronic Structure:
Wave or Quantum Mechanics 321
The Shapes of Atomic Orbitals 325
A Closer Look: More about H Atom Orbital
Shapes and Wavefunctions 329
One More Electron Property: Electron Spin
Applying Chemical Principles 6.1:
Sunburn, Sunscreens, and Ultraviolet
Radiation 330
Contents
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330
v
Applying Chemical Principles 6.2:
What Makes the Colors in Fireworks?
Applying Chemical Principles 6.3:
Chemistry of the Sun 332
7
7.1
7.2
331
8.9
A Closer Look: Measuring Molecular Polarity and
Debye Units 426
The Structure of Atoms
and Periodic Trends 344
7.3
The Pauli Exclusion Principle 345
Atomic Subshell Energies and Electron
Assignments 347
Electron Configurations of Atoms 350
7.4
A Closer Look: Orbital Energies, Z*, and Electron
Configurations 359
Electron Configurations of Ions 360
8.10
8.11
A Closer Look: Questions about Transition Element
Electron Configurations 361
7.5
7.6
8
8.1
A Closer Look: Paramagnetism and
Ferromagnetism 364
Atomic Properties and Periodic Trends 364
A Closer Look: Photoelectron Spectroscopy 371
Periodic Trends and Chemical Properties 374
Applying Chemical Principles 7.1:
The Not-So-Rare Earths 375
Applying Chemical Principles 7.2:
Metals in Biochemistry 376
Bonding and Molecular
Structure 386
8.5
Lewis Electron Dot Symbols and Chemical Bond
Formation 388
Electronegativity and Bond Polarity 390
Lewis Structures of Molecules and Polyatomic
Ions 393
Common Patterns of Bonding in Lewis
Structures 399
Resonance 404
8.6
A Closer Look: Resonance 405
Exceptions to the Octet Rule 406
8.2
8.3
8.4
A Closer Look: A Scientific Controversy—Resonance,
Formal Charges, and the ­Question of Double
Bonds in Sulfate and Phosphate Ions 417
Molecular Polarity 424
9
9.1
9.2
9.3
A Closer Look: Visualizing Charge Distributions
and Molecular Polarity—Electrostatic Potential
Surfaces and Partial Charge 428
Bond Properties: Order, Length,
and Dissociation Enthalpy 432
DNA 437
Applying Chemical Principles 8.1:
Ibuprofen, A Study in Green Chemistry 441
Applying Chemical Principles 8.2:
van Arkel Triangles and Bonding 442
Applying Chemical Principles 8.3:
Linus Pauling and the Origin of the Concept
of Electronegativity 443
Bonding and Molecular Structure:
Orbital Hybridization and
Molecular Orbitals 458
Valence Bond Theory 459
Molecular Orbital Theory 473
A Closer Look: Molecular Orbitals for Molecules
Formed from p-Block Elements 481
Theories of Chemical Bonding:
A Summary 483
A Closer Look: Three-Center Bonds in HF22, B2H6,
and SF6 484
Applying Chemical Principles 9.1:
Probing Molecules with Photoelectron
Spectroscopy 485
Applying Chemical Principles 9.2:
Green Chemistry, Safe Dyes, and Molecular
Orbitals 486
Part Three States of Matter
10 Gases and Their Properties 498
A Closer Look: Structure and Bonding for
Hypervalent Molecules 409
Formal Charges in Covalent Molecules and
Ions 410
10.1 Modeling a State of Matter: Gases and Gas
8.8
A Closer Look: Comparing Oxidation Number
and Formal Charge 411
Molecular Shapes 416
10.2 Gas Laws: The Experimental Basis 502
vi
Contents
8.7
Pressure
499
A Closer Look: Measuring Gas Pressure
500
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10.3
10.4
10.5
10.6
10.7
10.8
A Closer Look Studies on Gases—Robert Boyle
and Jacques Charles 508
The Ideal Gas Law 508
Gas Laws and Chemical Reactions 513
Gas Mixtures and Partial Pressures 514
The Kinetic-Molecular Theory of Gases 517
Diffusion and Effusion 521
A Closer Look: Surface Science and the Need for
Ultrahigh Vacuum Systems 523
Nonideal Behavior of Gases 524
Applying Chemical Principles 10.1:
The Atmosphere and Altitude Sickness 526
Applying Chemical Principles 10.2:
The Chemistry of Airbags 527
Forces and
11 Intermolecular
Liquids 540
11.1 States of Matter and Intermolecular Forces 541
11.2 Interactions between Ions and
Molecules with a Permanent Dipole
11.3
11.4
11.5
11.6
543
A Closer Look: Hydrated Salts: A Result of Ion–
Dipole Bonding 545
Interactions between Molecules with a Permanent
Dipole 546
A Closer Look: Hydrogen Bonding in
Biochemistry 551
Intermolecular Forces Involving Nonpolar
Molecules 552
A Summary of van der Waals Intermolecular
Forces 555
A Closer Look: Geckos Can Climb Up der
Waals 556
Properties of Liquids 557
Applying Chemical Principles 11.1:
Chromatography 567
Applying Chemical Principles 11.2:
A Pet Food Catastrophe 568
12
The Solid State 580
12.1 Crystal Lattices and Unit Cells 581
12.2
A Closer Look: Packing Oranges, Marbles, and
Atoms 588
Structures and Formulas of Ionic Solids 589
A Closer Look: Using X-Rays to Determine Crystal
Structures 593
12.3Bonding in Ionic Compounds: Lattice
Energy
594
12.4 Bonding in Metals and Semiconductors 596
12.5 Other Types of Solid Materials 602
12.6
A Closer Look: Glass 605
Phase Changes 609
Applying Chemical Principles 12.1:
Lithium and Electric Vehicles 613
Applying Chemical Principles 12.2:
Nanotubes and Graphene: Network
Solids 614
Applying Chemical Principles 12.3:
Tin Disease 615
13 Solutions and Their Behavior 626
13.1 Units of Concentration 627
13.2 The Solution Process 630
A Closer Look: Supersaturated Solutions
631
13.3 Factors Affecting Solubility: Pressure and
13.4
Temperature 637
Colligative Properties
640
A Closer Look: Growing Crystals 641
A Closer Look: Hardening of Trees
13.5
646
A Closer Look: Reverse Osmosis for Pure
Water 649
A Closer Look: Osmosis and Medicine 651
Colloids 655
Applying Chemical Principles 13.1:
Distillation 659
Applying Chemical Principles 13.2:
Henry’s Law and Exploding Lakes 660
Applying Chemical Principles 13.3:
Narcosis and the Bends 661
Part Four The Control of Chemical Reactions
Kinetics: The Rates
14 Chemical
of ­Chemical Reactions 672
14.1
14.2
14.3
14.4
14.5
Rates of Chemical Reactions 673
Reaction Conditions and Rate 678
Effect of Concentration on Reaction Rate 680
Concentration–Time Relationships: Integrated
Rate Laws 686
A Microscopic View of Reaction Rates 694
Contents
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vii
14.6
14.7
A Closer Look: Rate Laws, Rate Constants, and
Reaction Stoichiometry 695
A Closer Look: More about Molecular Orientation
and Reaction Coordinate Diagrams 697
Catalysts 702
A Closer Look: Thinking about Kinetics, Catalysis,
and Bond Energies 702
Reaction Mechanisms 706
A Closer Look: Organic Bimolecular Substitution
Reactions 709
Applying Chemical Principles 14.1:
Enzymes–Nature’s Catalysts 716
Applying Chemical Principles 14.2:
Kinetics and Mechanisms: A 70-Year-Old
Mystery Solved 717
of Chemical Reactivity:
15 Principles
Equilibria 736
15.1 Chemical Equilibrium: A Review 737
15.2The Equilibrium Constant and Reaction
Quotient 738
A Closer Look: Activities and Units of K
16.7Calculations with Equilibrium Constants 796
16.8 Polyprotic Acids and Bases 805
16.9 Molecular Structure, Bonding, and Acid–Base
Behavior
A Closer Look: Acid Strengths and Molecular
Structure 811
16.10The Lewis Concept of Acids and Bases 812
Applying Chemical Principles 16.1:
Would You Like Some Belladonna Juice in
Your Drink? 816
Applying Chemical Principles 16.2:
The Leveling Effect, Nonaqueous Solvents,
and Superacids 817
of Chemical Reactivity:
17 Principles
Other Aspects of Aqueous
Equilibria 830
17.1 Buffers 831
17.2 Acid–Base Titrations 843
17.3 Solubility of Salts 853
A Closer Look: Minerals and Gems—The
Importance of Solubility 859
740
A Closer Look: Equilibrium Constant Expressions
for Gases—Kc and Kp 742
15.3Determining an Equilibrium Constant 746
15.4 Using Equilibrium Constants in Calculations 748
15.5More about Balanced Equations and Equilibrium
Constants 754
15.6 Disturbing a Chemical Equilibrium 757
Applying Chemical Principles 15.1:
Applying Equilibrium Concepts—
The Haber–Bosch Ammonia Process 762
Applying Chemical Principles 15.2:
Trivalent Carbon 763
of Chemical Reactivity:
16 Principles
The Chemistry of Acids and
17.4
17.5
Bases 777
16.2 Water and the pH Scale 780
16.3Equilibrium Constants for Acids and Bases 783
16.4 Acid–Base Properties of Salts 789
16.5Predicting the Direction of Acid–Base
Reactions 791
16.6 Types of Acid–Base Reactions 794
viii
A Closer Look: Solubility Calculations 860
Precipitation Reactions 863
Equilibria Involving Complex Ions 867
Applying Chemical Principles 17.1:
Everything that Glitters . . . 871
Applying Chemical Principles 17.2:
Take a Deep Breath 872
of Chemical Reactivity:
18 Principles
Entropy and Free Energy 886
18.1Spontaneity and Dispersal of Energy:
18.2
Entropy 887
Entropy: A Microscopic Understanding
18.3
18.4
A Closer Look: Reversible and Irreversible
Processes 890
Entropy Measurement and Values 894
Entropy Changes and Spontaneity 898
Bases 776
16.1 The Brønsted-Lowry Concept of Acids and
807
18.5
18.6
18.7
889
A Closer Look: Entropy and Spontaneity? 899
Gibbs Free Energy 902
Calculating and Using Standard Free Energies,
DrG° 906
The Interplay of Kinetics and
Thermodynamics 914
Contents
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Applying Chemical Principles 18.1:
Thermodynamics and Living Things
Applying Chemical Principles 18.2:
Are Diamonds Forever? 917
19
19.1
19.2
19.3
19.4
19.5
19.6
19.7
19.8
19.9
Principles of Chemical Reactivity:
Electron Transfer Reactions 932
Oxidation–Reduction Reactions 933
Voltaic Cells 940
Commercial Voltaic Cells 946
Standard Electrochemical Potentials 951
A Closer Look: EMF, Cell Potential, and
Voltage 952
Electrochemical Cells Under Nonstandard
Conditions 960
Electrochemistry and Thermodynamics 964
Electrolysis: Chemical Change Using Electrical
Energy 968
A Closer Look: Electrochemistry and Michael
Faraday 970
Counting Electrons 973
Corrosion: Redox Reactions in
the Environment 975
Applying Chemical Principles 19.1:
Electric Batteries versus Gasoline 978
Applying Chemical Principles 19.2:
Sacrifice! 978
Part Five The Chemistry of the Elements
20 Nuclear Chemistry 992
20.1 Natural Radioactivity 994
20.2 Nuclear Reactions and Radioactive Decay 995
20.3
20.4
20.5
20.6
20.7
20.8
20.9
Applying Chemical Principles 20.2:
Technetium-99m and Medical Imaging
Applying Chemical Principles 20.3:
The Age of Meteorites 1029
916
A Closer Look: Radioactive Decay Series 997
Stability of Atomic Nuclei 1000
Origin of the Elements: Nucleosynthesis 1006
Rates of Nuclear Decay 1008
Artificial Nuclear Reactions 1014
Nuclear Fission and Nuclear Fusion 1018
A Closer Look: Lise Meitner (1878–1968) 1020
Radiation Health and Safety 1021
Applications of Nuclear Chemistry 1023
Applying Chemical Principles 20.1:
A Primordial Nuclear Reactor 1027
1028
Chemistry of the Main
21 The
Group Elements 1038
21.1 Abundance of the Elements 1039
21.2 The Periodic Table: A Guide to the
Elements
1040
21.3 Hydrogen 1046
21.4
21.5
A Closer Look: Hydrogen in
Transportation 1049
The Alkali Metals, Group 1A (1)
1049
A Closer Look: The Reactivity of the Alkali
Metals 1054
The Alkaline Earth Elements,
Group 2A (2) 1054
A Closer Look: Alkaline Earth Metals and
Biology 1057
A Closer Look: Cement—The Second Most Used
Substance 1058
21.6Boron, Aluminum, and the Group 3A (13)
Elements 1059
A Closer Look: Complexity in Boron
Chemistry 1064
21.7Silicon and the Group 4A (14) Elements 1065
21.8Nitrogen, Phosphorus, and the Group 5A (15)
Elements 1070
A Closer Look: Alchemists Making
Phosphorus 1072
A Closer Look: Ammonium Nitrate—A Mixed
Blessing 1075
21.9Oxygen, Sulfur, and the Group 6A (16)
Elements 1079
21.10 The Halogens, Group 7A (17) 1082
A Closer Look: Iodine and Your Thyroid
Gland 1084
A Closer Look: The Many Uses of FluorineContaining Compounds 1085
21.11 The Noble Gases, Group 8A (18) 1087
A Closer Look: The Noble Gases—Not So
Inert 1088
Applying Chemical Principles 21.1:
Lead in the Environment 1089
Applying Chemical Principles 21.2:
Hydrogen Storage 1090
Contents
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ix
Chemistry of the
22 The
Transition Elements 1104
24.3 Nucleic Acids 1219
22.1 Overview of the Transition Elements 1105
22.2Periodic Properties of the Transition
24.4
22.3
22.4
Elements 1107
Metallurgy 1111
Coordination Compounds 1114
A Closer Look: Hemoglobin: A Molecule with a
Tetradentate Ligand 1118
22.5Structures of Coordination Compounds 1122
22.6 Bonding in Coordination Compounds 1128
22.7 Colors of Coordination Compounds 1133
Applying Chemical Principles 22.1:
Blue! 1137
Applying Chemical Principles 22.2:
Cisplatin: Accidental Discovery of a
Chemotherapy Agent 1138
Applying Chemical Principles 22.3:
The Rare Earth Elements 1139
23
Carbon: Not Just Another
Element 1150
23.1 Why Carbon? 1151
23.3
23.4
A Closer Look: Writing Formulas and Drawing
Structures 1153
Hydrocarbons 1155
A Closer Look: Flexible Molecules 1161
Alcohols, Ethers, and Amines 1170
Compounds with a Carbonyl Group 1176
23.5
A Closer Look: Omega-3 Fatty Acids 1180
Polymers 1184
23.2
A Closer Look: Green Chemistry: Recycling
PET 1190
Applying Chemical Principles 23.1:
An Awakening with l-DOPA 1192
Applying Chemical Principles 23.2:
Green Adhesives 1193
Applying Chemical Principles 23.3:
Bisphenol A (BPA) 1193
24
Biochemistry 1206
24.1 Proteins 1208
24.2 Carbohydrates 1216
x
24.5
A Closer Look: Genetic Engineering with
CRISPR-Cas9 1222
Lipids and Cell Membranes 1225
A Closer Look: mRNA Vaccines 1228
Metabolism 1230
Applying Chemical Principles 24.1: Polymerase
Chain Reaction 1236
Chemistry—Earth’s
25 Environmental
­Environment, Energy, and
Sustainability 1244
25.1 The Atmosphere 1246
A Closer Look: The Earth’s Atmosphere
A Closer Look: Methane Hydrates
1247
1256
25.2 The Aqua Sphere (Water) 1256
25.3
A Closer Look: Perfluoroalkyl Substances
(PFAS) 1263
Energy 1263
A Closer Look: Fracking 1266
25.4 Fossil Fuels 1267
A Closer Look: Petroleum Chemistry
1271
25.5 Environmental Impact of Fossil Fuels 1271
25.6 Alternative Sources of Energy 1277
25.7 Green Chemistry and Sustainability 1281
Applying Chemical Principles 25.1:
Chlorination of Water Supplies 1282
Applying Chemical Principles 25.2: Hard
Water 1283
List of Appendices A-1
A
Using Logarithms and Solving Quadratic
Equations A-2
B
C
D
E
F
Some Important Physical Concepts
G
Vapor Pressure of Water at Various
Temperatures A-19
H
Ionization Constants for Aqueous Weak Acids at
25 °C A-20
A-6
Abbreviations and Useful Conversion Factors
Physical Constants
A-9
A-13
A Brief Guide to Naming Organic Compounds
A-15
Values for the Ionization Energies and Electron
Attachment Enthalpies of the Elements A-18
Contents
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
I
Ionization Constants for Aqueous Weak Bases at
25 °C A-22
J
Solubility Product Constants for Some Inorganic
Compounds at 25 °C A-23
K
Formation Constants for Some Complex Ions in
Aqueous Solution at 25 °C A-24
L
Selected Thermodynamic Values
A-25
M
Standard Reduction Potentials in Aqueous Solution at
25 °C A-32
N
Answers to Study Questions, Check Your
Understanding, and Applying Chemical Principles
Questions A-36
Index of Names I-1
Index and Glossary I-4
Contents
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xi
Preface
David Kotz
The first edition of this book
problems as a scientist. We have
was conceived over 40 years ago.
tried to provide the tools to help
Since that time, there have been
you become a chemically and
ten editions, and over one milscientifically literate citizen.
lion students worldwide have
used the book to begin their
Audience for
study of chemistry. Although the
Chemistry &
details of the book and its orgaChemical Reactivity
nization have changed over the
years, our fundamental goal has
This textbook is designed for
remained the same: to provide a
students interested in furbroad overview of the principles
ther study in science, whether
of chemistry, the reactivity of the
that science is chemistry, biolchemical elements and their comogy, medicine, e­ ngineering,
pounds, and the applications of
geology, physics, or related
chemistry. To reach this goal, we
subjects. Our assumption is
have tried to show the close relathat students in a course using
tionship between the observathis book have had some preptions of chemical and physical
aration in ­algebra and general
changes made by chemists in the
science. Although u
­ ndeniably
laboratory and in nature and the
helpful, a ­previous ­exposure to
way these changes are viewed at
chemistry is neither assumed
the atomic and molecular levels. Fireworks. See Chapter 6 for the chemistry of
nor required.
We have also tried to convey the ­fireworks.
sense that chemistry not only has
a lively history but is also interesting and dynamic, with Philosophy and Approach of
important new developments occurring every year. Further- Chemistry & Chemical Reactivity
more, we wanted to provide some insight into the chemical
aspects of the world around us.
We have had several major but not independent objectives
The authors of this text have collectively taught chem- since the first edition of the book. Our first goal has been
istry for over 100 years, and we have engaged in years of to write a book that students will find useful and interestfundamental research. Like countless other scientists, our ing and that presents chemistry and chemical principles in
goals in our research and in writing this textbook have a format and organization typical of college and university
been to satisfy our curiosity about areas of chemistry, to courses today. Second, we want to convey the utility and
document what we found, and to convey that to students importance of chemistry by introducing the properties of
and other scientists. Our results, and many others, may the elements, their compounds, and their reactions.
be used, perhaps only years later, to make a better mateThe American Chemical Society has been urging edurial or better pharmaceutical. Every person eventually cators to put chemistry back into introductory chemistry
benefits from the work of the worldwide community of courses. We agree wholeheartedly. Therefore, we have
scientists.
tried to describe the elements, their compounds, and
In recent years, when people around the world have their reactions as early and as often as possible by:
experienced various epidemics and increasing evidence of
climate change has been published, science has come under • Bringing material on the properties of elements and
compounds into the Examples and Study Questions.
attack. Some distrust the scientific community and dismiss
the results of carefully done research. Therefore, a key objec- • Using numerous photographs of the elements and comtive of this book, and of a course in general chemistry, is to
mon compounds, of chemical reactions, and of comdescribe basic chemical “facts”—chemical processes and
mon laboratory operations and industrial processes.
principles; how chemists came to understand those prin• Each chapter incorporates Applying Chemical Principles and new ideas; how they can be applied in industry,
ciples questions that delve into the applications of
medicine, and the environment; and how to think about
chemistry.
xii
Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
General Organization
units are introduced in Chapter 1, and thermochemistry
is introduced in Chapter 5.
Through its many editions, Chemistry & Chemical
­Reactivity has had two broad themes: Chemical ­Reactivity
and ­B onding and Molecular Structure. The chapters on
­P rinciples of Reactivity introduce the factors that lead
chemical ­r eactions to be successful in converting
­reactants to products: common types of reactions, the
energy involved in reactions, and the factors that affect
the speed of a reaction. One reason for the enormous
advances in chemistry and molecular biology in the last
several decades has been an understanding of molecular structure. The sections of the book on Principles of
­Bonding and Molecular Structure lay the groundwork for
understanding these developments. Particular attention
is paid to an understanding of the structural aspects of
such biologically important molecules as hemoglobin
and DNA.
Sections of the Book — Organization
and Purpose
Part One: The Basic Tools of Chemistry
The basic ideas and methods of chemistry are i­ ntroduced
in Part One. Chapter 1 defines important terms, and
the accompanying Chapter 1R reviews measurement
units and some fundamental mathematical methods
used throughout the text. Chapter 2 introduces atoms,
molecules, and ions as well as the most important
­organizational device in chemistry, the periodic table. In
Chapter 3, we begin to discuss the principles of chemical reactivity. Writing chemical equations is covered
here, and there is a short introduction to the concept of
­chemical equilibrium. In addition, some major types of
chemical reactions in aqueous solution are introduced.
Then, in Chapter 4, we describe the numerical methods
used by chemists to extract quantitative information
from chemical reactions. Chapter 5 is an introduction to
the energy changes involved in chemical processes.
As we look at the introductory chemistry texts currently
available and talk with colleagues at other universities,
it is evident there is a generally accepted order of topics
in the course. With minor variations, we followed that
order. That is not to say that the chapters in our book
cannot be used in some other order. We have w
­ ritten
this book to be as flexible as possible. An example is the
flexibility in covering the behavior of gases (­Chapter 10).
It has been placed with the chapters on liquids, solids,
and solutions (Chapters 11–13) because it logically
fits with those topics. However, it can also be read and
­understood after covering only the first four chapters of
the book.
Similarly, the chapters on atomic and molecular
structure (Chapters 6–9) can be used in an atoms-first
approach before the chapters on stoichiometry and common reactions (Chapters 3 and 4). To facilitate this, there
is an introduction to energy and its units in C
­ hapter
1. Also, the chapters on chemical equilibria (Chapters 15–17) can be covered before those on solutions
and ­kinetics ­(Chapters 13 and 14). Although organic
­chemistry (Chapter 23) is one of the final chapters in
the textbook, the topics of this chapter can also be presented to ­students following the chapters on structure
and bonding (Chapters 9 and 10).
The order of topics in the text was also devised
to introduce as early as possible the background
required for the laboratory experiments that are usually
­performed in introductory chemistry courses. For this
reason, ­chapters on chemical and physical properties,
­common reaction types, and stoichiometry begin the
book. In addition, because an understanding of energy is
very important for the study of chemistry, energy and its
© Charles D. Winters/Cengage
Flexibility of Chapter Organization
Elemental sulfur.
Part Two: Atoms and Molecules
The current theories that explain the arrangement of
electrons in atoms and monatomic ions are presented in
Chapters 6 and 7. This discussion is tied closely to the
arrangement of elements in the periodic table and to
their periodic properties. In Chapter 8, we discuss the
details of chemical bonding and the properties of these
bonds. In addition, we show how to derive the threedimensional structure and charge distribution of simple
molecules. Finally, Chapter 9 considers the major theories of chemical bonding in more detail.
Preface
Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
xiii
Part Four: The Control of Chemical
Reactions
© Charles D. Winters/Cengage
This section is wholly concerned with the Principles
of Reactivity. Chapter 14 examines the rates of chemical processes and the factors controlling these rates.
Next, Chapters 15–17 describe chemical equilibrium.
After an introduction to equilibrium in Chapter 15,
we highlight reactions involving acids and bases in
water (Chapters 16 and 17) and reactions leading to
slightly soluble salts (Chapter 17). To tie together the
discussion of chemical equilibria and thermodynamics, we explore entropy and free energy in C
­ hapter 18.
As a final topic in this section, we describe in
­C hapter 19 chemical reactions that involve the transfer of e­ lectrons and the use of these reactions in electrochemical cells.
Liquid oxygen is attracted to a strong magnet. See ­Chapter 9 for
an explanation of the magnetic properties of oxygen.
Part Three: States of Matter
© Charles D. Winters/Cengage
The behavior of the three states of matter—gases, liquids, and solids—is described in Chapters 10–12. The
discussion of liquids and solids is tied to gases through
the description of intermolecular forces in Chapter 11,
with particular attention given to liquid and solid water.
In Chapter 13, we describe the properties of solutions—
intimate mixtures of gases, liquids, and solids.
The explosive reaction of hydrogen and oxygen.
John C. Kotz
Part Five: The Chemistry of the Elements
Hot air balloon takes off. See Chapter 5 for an introduction to
transfers of energy as heat and work. See Chapter 10 for a
discussion of the gas laws.
xiv
Although the chemistry of many elements and compounds is described throughout the book, Part Five considers this topic in a more systematic way. ­Chapter 20
is an overview of nuclear chemistry. C
­ hapter 21 is
devoted to the chemistry of the main group elements,
and Chapter 22 is a discussion of the transition elements and their compounds. Chapter 23 is a brief
introduction to organic chemistry with an emphasis
on molecular structure, basic reaction types, and polymers, and Chapter 24 is an introduction to biochemistry. Finally, Chapter 25 brings together many of the
Preface
Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
What’s New to this Edition?
concepts in earlier chapters into a discussion of “Environmental Chemistry—Earth’s Environment, Energy,
and Sustainability.”
The entire book was thoroughly reviewed. Many parts are
rewritten, and new Study Questions were added. In addition, there are several new features that occur in each chapter.
Think–Pair–Share Questions
values per mole and per gram. Which provides more energy
For the purposes of this analysis, octane (C8H18) is used as a
substitute for the complex mixture of hydrocarbons in gasoline. Data needed for this question (in addition to the data in
Appendix L) are:
∆fH° [C8H18()] = −250.1 kJ/mol
Density of ethanol = 0.785 g/mL
Density of octane = 0.699 g/mL
1. Calculate ∆rH° for the combustion of ethanol and octane producing carbon dioxide gas and liquid water, and compare the
Elements: yellow phosphorus in water (left) and shiny
­potassium under oil (right).
Photos: © Charles D. Winters/Cengage
Questions
Many
instructors
have
with moving
per mole?
Which provides
more experimented
energy per gram?
beyond
lecturing
in liter
general
One
2.
Comparesimply
the energy
produced per
of the chemistry.
two fuels.
Which produces
more
energy for
a given volume
approach
that has
worked
especially
well(something
has been to take
useful to know when filling your gas tank)?
class
time for students to work together on ­questions that
3. What mass of CO2, a greenhouse gas, is produced per liter of
help
learn
course
­topics. This is the purpose of the
fuelthem
(assuming
complete
combustion)?
Think–Pair–Share
­q uestions.
Studentsbasis.
work
indepen4.
Now compare the fuels
on an energy-equivalent
What
volumeon
of the
ethanol
would havefirst,
to bethen
burnedform
to getsmall
the same
dently
questions
groups to
energy as 1.00 L of octane? When you burn enough ethanol
discuss
their answers, and finally, present their results to
to have the same energy as a liter of octane, which fuel protheduces
class.
These
more
CO2? questions do not generally require many
­calculations. Instead, they focus on helping ­students to
think more deeply about the ­concepts of the chapter.
They are placed after the A
­ pplying Chemical ­P rinciples
questions and before the Chapter Goals Revisited.
Think–Pair–Share
1. You are tasked with determining the specific heat capacity
(in J/g ? K) for an unknown metal. The following equipment
and supplies are available: a 25.0-g piece of the metal, a
200-mL insulated container, a graduated cylinder, water,
ice, and a thermometer that can measure temperature
changes accurately to ±0.02 °C.
(a) Outline the steps for an experimental procedure to determine the specific heat capacity of the metal using the
available equipment and supplies.
(b) Identify possible sources of error in your experimental procedure that might cause an inaccurate result. (Assume the
graduated cylinder and thermometer are both accurate
and precise, and that human error is not a source of error.)
(c) Suggest some ideas for how to correct for some of these
error sources.
2. For introductory laboratories, resealable plastic bags are a
convenient way to conduct experiments involving gas-forming
reactions. Energy as heat or work can transfer in or out of
the plastic bag, but reactants and products remain trapped.
In an experiment, nitric acid reacts with a small amount of
copper in a sealed plastic bag according to the following balanced chemical equation:
Cu(s) + 4 HNO3(aq) n Cu(NO3)2(aq) + 2 NO2(g) + 2 H2O()
As the reaction proceeds, the contents of the bag become
warm, and the bag inflates with a brown gas (NO2).
(a) Define the system and the surroundings for this
experiment.
(b) Does energy as heat (q) flow into the system or out of the
system? What is the sign of q?
(c) Is work (w) done in this experiment? If so, is work done by
the system on the surroundings, or by the surroundings on
the system? What is the sign of w?
3. The following table shows the enthalpy of combustion for
some fuels in units of kJ/mol, kJ/L, and kJ/g. The enthalpy of
combustion per volume assumes a temperature of 25 °C and
atmospheric pressure (1 atm). Note that although gasoline
is a mixture of many hydrocarbons, it is often represented
as octane.
288
Fuel
∆H°
(kJ/mol)
∆H°
(kJ/L)
∆H°
(kJ/g)
Hydrogen,
H2(g)
−285.8
−11.7
−141.8
Methane,
CH4(g)
−890.2
−36.4
−55.5
Ethane,
C2H6(g)
−1560.6
−63.8
−51.9
Ethanol,
C2H6O()
−1376.5
−23,590
−29.9
Octane,
C8H18()
−5490
−33,814
−48.1
(a) When determining which fuel provides the most energy
in a car engine, is it best to compare the enthalpy change
in a combustion reaction by moles, volume, or mass?
Explain your reasoning? Based on your decision, which fuel
provides the most energy (at the given temperature and
pressure).
(b) Gasoline sold in the United States is often a blend of 10%,
15%, or even up to 85% ethanol by volume. Identify at least
one advantage and one disadvantage of using gasoline
blended with ethanol versus ethanol-free gasoline?
(c) Why are the enthalpies of combustion per liter of hydrogen, methane, and ethane much lower than those of ethanol and octane?
(d) The enthalpy of combustion of hydrogen per gram is
nearly three times that of a hydrocarbon. And unlike a
hydrocarbon, which produces the greenhouse gas CO 2
upon combustion, the combustion of hydrogen produces only water. Unfortunately, there are multiple issues
in replacing gasoline with hydrogen as a fuel. What are
some of the problems that must be overcome if hydrogen
is to compete with, or possibly replace, gasoline as a fuel
in vehicles?
Chapter 5 / Principles of Chemical Reactivity: Energy and Chemical Reactions
Preface
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
xv
Dr. Jessica N. Isaac
Dr. Jessica N. Isaac
Jessica N. Isaac is a Clinical Assistant Professor at at the chemical interactions between medications
Binghamton University’s School of Pharmacy and and the chemicals in the human body; creating
and refining
drug development;
and
Pharmaceutical Sciences, where her work
is “a comstudents
that they
have amolecules
place infor
chemistry
and STEM
Chemistry in Your Career
bination of administration, teaching,courses.
research, clin- combining, mixing, or altering chemicals to create
All of us have students who have
gone on to wonderful
ical pharmacy, and mentorship.” As an instructor, a medication tailored to the needs of a patient.
careers, some in traditional chemistry
careers,
Dr. Isaac describes how her identity as a an
one of Dr. Isaac’s
goalsbut
is tooften
help “student pharmaRedesigned Strategy
Maps
in some other field where acists
background
in
chemistry
develop the skills necessary to safely and African American/Jamaican American woman has
For
many
students,
a
visual
Strategy and
Mapprofessionally.
can be a useful
is useful. Each chapter features
a
short
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a
shaped her both personally
“Being
accurately prepare medications.”
tool
in problem
(as onand
page
194). These have
been
person who studied chemistry and
perhaps
Black woman
a first-generation
student
in
“Thenow
studyworks,
of chemistry
is integral
to the
study a solving
pharmacy
many obstacles;
of pharmacy,”
explains Dr.
Isaac, including
“stoi-to be
redesigned
more school
fully presented
incorporated
into thehowever,
Soluas a chemist, but more generally
in a profession
where
experiences
also helped
me toare
develop
certain
dimensional
analysis, acid–base
prop- of these
tion section
the Example
problems.
There
44 Strategy
they use their background chiometry,
in chemistry.
This feature
. .including
social intelligence,
perseverance/
erties,
and physiochemical
properties.”
The
study strengths.
accompanying
Example
problems
in the book.
highlights people from diverse
backgrounds
to show
all Maps
of pharmacy has three core components: looking resilience, creativity, gratitude, and critical thinking.”
Exam pl e 4 .1
Mass Relations in Chemical Reactions
Strategy Map
Problem
Calculate mass of O2 required
for combustion of 25.0 g of
glucose.
Data/Information
Formulas for reactants and
products and the mass of one
reactant (glucose)
Problem Glucose, C6H12O6, reacts with oxygen to give CO2 and H2O. What mass of
oxygen (in grams) is required to completely react with 25.0 g of glucose? What masses
of carbon dioxide and water (in grams) are formed?
What Do You Know? You are given the mass of one of the reactants (glucose)
and are asked to determine the masses of the other substances in the reaction. You know
formulas for the reactants and products and need to calculate their molar masses.
Strategy Write the balanced chemical equation for this reaction. Then, follow the
scheme outlined in Problem Solving Tip 4.1 and in the Strategy Map for this example.
Solution
Step 1
Write the balanced equation.
C6H12O6(s) + 6 O2(g) n 6 CO2(g) + 6 H2O(ℓ)
Step 2
Convert the given mass (grams) to amount (moles).
Use the molar mass of glucose to convert its mass (25.0 g) to the amount of glucose.
25.0 g glucose Step 3
1 mol glucose
0.1387 mol glucose
180.2 g glucose
Use the stoichiometric factor to convert the amount (moles) of glucose to the
amount (moles) of O2.
Use the coefficients of the balanced equation to obtain the stoichiometric factor of 6 mol
O2 per 1 mol glucose, and then convert the amount of glucose to the amount of O2.
0.1387 mol glucose Step 4
6 mol O2
0.8324 mol O2
1 mol glucose
Convert the amount (moles) of the requested substance to its mass (grams).
Use the molar mass of O2 to convert from the amount of O2 to the mass of O2.
0.8324 mol O2 32.00 g O2
26.6 g O2
1 mol O2
194
Chapter 4 / Stoichiometry: Quantitative Information about Chemical Reactions
xvi
Preface
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Some of the Other Changes for This
Edition
•
Recent changes to the definitions of the SI Base Units
are explained in a new Closer Look box (page 34).
•
The atomic weights of the elements on the periodic
tables and other tables have been updated based
on values from the International Union of Pure
and Applied Chemistry (IUPAC) and the National
­Institute of Standards and Technology (NIST).
environmental chemistry can serve as an overarching
conclusion to the book.
•
A new Closer Look box about radioactive decay series
(page 997) was added to Chapter 20.
•
The section on the origin of the elements (in Chapter
20, “Nuclear Chemistry”) was expanded and revised.
•
In Chapter 24, “Biochemistry,” new material has
been added on mRNA vaccines, electron carriers in
biochemical oxidation–reduction reactions, and the
metabolism of glucose in respiration.
•
Chapter 25, “Environmental Chemistry—Earth’s
Environment, Energy, and Sustainability” has been
updated and reorganized.
•
All Examples have been reviewed, some have been
revised to make the steps of the strategy clearer, and
nine new Examples were written (Example R.3 “Precision and Standard Deviation;” Example 2.5 “Binary
Molecular Compounds;” Example 2.8 “Naming Ionic
Compounds;” Example 3.5 “Separating a Mixture
by Selective Precipitation;” Example 3.9 “Recognizing Oxidation–Reduction Reactions;” Example 4.3
“Calculating Percent Yield;” Example 4.8 “Relating
Amount and Molarity;” Example 4.12 “Acid–Base
Titration;” and Example 6.6 “Orbitals and Quantum
Numbers”).
•
The issue of ranges being recommended by IUPAC
for the atomic weights of some elements is discussed
in a new Closer Look box (page 70).
•
References to element groups in the periodic table
are now given in both the traditional system used in
the United States (Group 5A, for example) and the
1–18 system recommended by IUPAC.
•
Biographies of some important scientists and their
discoveries have been prepared (Marie Curie [page
82]; Antoine and Marie-Anne Lavoisier [page 141];
Niels Bohr [page 314]; James Watson, Francis Crick,
and Rosalind Franklin [page 439]; and Lise Meitner
[page 1020]).
•
A new subsection “Naming Common Acids” (page
157) was added.
•
The classification scheme for acid–base and gas-­
forming acid–base reactions (Sections 3.6–3.7)
has been revised as well as the overall classification scheme of reactions in aqueous solution
(Section 3.9).
•
There are now over 2650 Study Questions in the book.
Of these, over 360 are either new or revised in this
edition.
•
•
All appendices have been updated to ensure they
contain the latest information.
A Closer Look box (page 208) on nuclear magnetic
resonance spectroscopy has been added.
•
•
A greater distinction between heat capacity and
­specific heat capacity was made in Chapter 5.
•
A new Closer Look box “Enthalpy, Internal Energy and
Non-Expansion Work” (page 270) was added.
Appendix N, “Answers to Study Questions, Check
Your Understanding, and Applying Chemical Principles Questions,” has been accuracy checked by the
book authors and the author of the Student Solutions
Manual, Professor Charles Atwood.
•
•
A new Problem-Solving Tip (page 355) was included
about the different methods used for writing the
electron configuration of an atom.
An Index of Names has been added so readers can
find the contributions of generations of chemists.
Features of the Book
•
Chapter 8, “Bonding and Molecular Structure,” was
restructured.
•
The story of the unraveling of the structure of DNA
was expanded in Chapter 8.
•
Two Closer Look boxes were added in Chapter 12, one
on using X-rays to determine crystal structure and the
other about glass.
•
The positions of the chapters about nuclear chemistry (Chapter 20 in the current edition) and environmental chemistry (Chapter 25) were swapped so that
information about nuclear chemistry can inform the
content of later chapters and so that the chapter on
Some years ago, a former student of one of the authors,
now an accountant, shared his perspective on his experience in general chemistry. He said that, while chemistry was one of his hardest subjects, it was also the most
useful course he had taken because it taught him how to
solve problems in addition to having learned to appreciate a bit of chemistry. We were certainly pleased because
we have always thought that an important goal in general
chemistry is not only to teach students chemistry but also
to help them learn critical thinking and problem-solving
skills. Many of the features of the book are meant to support those goals.
Preface
Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
xvii
Problem-Solving Approach: Organization
and Strategy Maps
A Closer Look Essays and
Problem-Solving Tips
Worked-out Examples are an essential part of each ­chapter.
To better help students follow the logic of a solution, all
Examples are organized around the following outline:
As in the tenth edition, there are boxed essays titled
A Closer Look that take a more in-depth look at relevant chemistry. While retaining and updating many
from previous editions, we wrote several new ones and
­h eavily revised some others including: “The SI Base
Units” (Chapter 1R), “Isotopic Abundances and Atomic
Weights” (Chapter 2), “Marie Curie (1867–1934)”
(Chapter 2), “Amedeo Avogadro and His Number”
(Chapter 2), “Nuclear Magnetic Resonance (NMR)
Spectroscopy” (Chapter 4), “Enthalpy, Internal Energy,
and Non-­E xpansion Work” (Chapter 5), “Niels Bohr
(1885–1962)” (Chapter 6), “Lise Meitner (1878–1968)”
­(Chapter 20), “Hydrogen in Transportation” (Chapter
21), “mRNA Vaccines” (Chapter 24), “Methane Hydrates”
(Chapter 25), and “Perfluoroalkyl Substances (PFAS)”
(Chapter 25).
From our teaching experience, we have learned some
“tricks of the trade” and try to pass on some of those in
Problem-Solving Tips.
Problem: A statement of the problem.
What Do You Know?: The information given is
outlined.
Strategy: The information available is combined
with the objective, and we begin to devise a pathway to a solution.
Solution: We work through the steps, both logical
and mathematical, to the answer.
Think About Your Answer: We ask if the answer is
reasonable or what it means.
Check Your Understanding: This is a similar problem for the student to try. A solution to the problem is in Appendix N.
Chapter Goals Revisited
The learning goals for each chapter section are listed
at the beginning of the section. The goals are revisited on
the last pages of the chapter, and specific end-of-chapter
Study Questions are listed that can help students determine if they have met those goals.
End-of-Chapter Study Questions
There are between 48 and 178 Study Questions for each
chapter, and answers to the odd-numbered questions are
given in Appendix N. Questions are grouped as follows:
Practicing Skills: These questions are grouped by the
topic covered by the questions.
General Questions: There is no indication regarding
the pertinent section of the chapter. They generally cover several chapter sections.
In the Laboratory: These are problems that may be
encountered in a laboratory experiment on the
chapter material.
Summary and Conceptual Questions: These questions use concepts from the current chapter as
well as preceding chapters.
Finally, some questions are marked with a small red
triangle (▲). These are more challenging than other
questions.
xviii
Applying Chemical Principles
At the end of each chapter are longer ­questions that use
the principles learned in the chapter to study examples of
forensic chemistry, environmental chemistry, ­medicinal
chemistry, or other areas. Examples are “Atom Economy”
(Chapter 4), “What Makes the ­C olors in Fireworks”
(Chapter 6), “A Pet Food ­C atastrophe” (Chapter 11),
“Lithium and Electric ­Vehicles” (Chapter 12), “The Age
of Meteorites” ­(Chapter 20), and “Blue!” (Chapter 22).
Online Learning
Created by teaching chemists, OWLv2 is a powerful
online learning solution for chemistry with a unique
Mastery Learning approach. It enables students to practice at their own pace, receive meaningful feedback, and
access a variety of learning resources to help them master
chemistry and achieve better grades.
The textbook’s Study Questions are available in the
OWLv2 online learning system. OWLv2 now has over
1800 of the roughly 2650 Study Questions in the book.
The OWLv2 course and MindTap eReader both contain nearly 300 videos on specific topics narrated by the
authors to help students visualize concepts and master
difficult problems by watching them be solved on screen.
Preface
Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Acknowledgments
Preparing this new edition of Chemistry & Chemical
­Reactivity took about two years of continuous effort. As
was true for our work on the first ten editions, we have
had the support and encouragement of our colleagues at
Cengage and our families, friends, faculty colleagues, and
students.
Cengage
The ten previous editions of this book have been published by Cengage and its predecessor companies, and
once again we had an excellent production team in place
for this, the eleventh edition. Maureen McLaughlin and
Helene Alfaro led the team with Mona Zeftel overseeing
many aspects of book design. Various people helped with
content organization: James Nash, Breanna Holmes, and
Kelly Aull. They were invaluable.
The first half of the book in this edition was thoroughly reviewed and edited by Margy Kuntz. This is the
eleventh edition of a book that has been used successfully in its previous editions by over a million students.
Nonetheless, Margy found ways to better organize and
clarify sections in these chapters.
The Chemistry in Your Career boxes are a new feature
of the book, and we want to acknowledge the many people who told us their stories. We hope these will help the
many students who take a chemistry course see how it
can be important in their careers. Rebecca Heider at Cengage was masterful in putting their stories into small,
readable vignettes.
Chemistry & Chemistry Reactivity has been supported
by OWL for many editions. The relationship of the book
and OWL has continued to be very well managed by
­Theresa Dearborn.
Art, Design, and Photography
Many of the color photographs in this book have been
beautifully created by Charles D. Winters over many
years and ten editions.
The book still profits from the design and illustration
skills of Patrick Harman. Pat designed the first edition of
our Interactive General Chemistry CD-ROM (published in
the 1990s). For the fifth through the tenth editions of
the book, Pat revised many of the figures in the book to
bring a fresh perspective to ways to communicate chemistry. All these illustrations remain in use in this edition.
Other Collaborators
We have been fortunate to have had several colleagues
play valuable roles in this project over its many editions.
One who has been especially important to this edition
is Professor Charles (Butch) Atwood. He has been very
helpful in ensuring the accuracy of the Study Question
answers in the book and producing the Student Solutions
Manual.
Eleventh Edition Reviewers
We encourage users, both faculty and students, to contact
us about book content and with suggestions for improvement. There have been many instances of this over the
years and they have improved the book. In particular, we
would like to thank Roger Barth (West Chester University
of Pennsylvania) for many useful comments that assisted
us as we planned changes for this edition. The following
reviewed the book for this edition:
•
•
•
John M. Farrar, Northern Kentucky University
•
•
•
•
•
Jessica A. Parr, University of Southern California
Bernard Majdi, South Georgia State College
Danica A. Nowosielski, Hudson Valley Community
College
Dr. Jeff Seyler, University of Southern Indiana
Jeffrey Stephens, North Iowa Area Community College
Tarek Trad, Sam Houston State University
Saul R. Trevino, Houston Christian University
xix
Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
About the Authors
John (Jack) Kotz graduated from Washington and
Lee University in 1959 and earned a Ph.D. in c­ hemistry
at Cornell University in 1963. He was a National
­Institutes of Health postdoctoral fellow at the ­University
of ­M anchester in England and at Indiana University.
He was an Assistant Professor of Chemistry at Kansas
State University before moving to the State University of New York at Oneonta in 1970. He taught general chemistry and inorganic chemistry, and in 1986
was appointed a State University of New York Distinguished Teaching Professor of Chemistry. He retired
from active teaching in 2005. He is the author or coauthor of sixteen chemistry textbooks, among them
two in advanced inorganic chemistry, two introductory
general chemistry books in numerous editions, and various manuals and study guides. The general chemistry
book has been published as an interactive CD-ROM,
as an interactive ebook, and has been translated into
­ ublished research papers
five languages. He has also p
in organometallic chemistry, and among his awards
are the SUNY Award for Research and Scholarship and the
Catalyst Award in ­Education from the Chemical Manufacturers Association. He was a ­Fulbright Senior Lecturer
in Portugal and a mentor for the U.S. National Chemistry O
­ lympiad team. He has served on the boards of
trustees for the ­C ollege at Oneonta Foundation, the
Kiawah Island Nature C
­ onservancy, and Camp Dudley.
He is also an avid ­photographer, primarily of wildlife
(www.greensward.smugmug.com). His email address is
johnkotz@mac.com.
Emeritus Professor of ­Chemistry. During his faculty career
he taught courses in general ­chemistry, ­inorganic chemistry, ­organometallic c­ hemistry, and s­ cientific ethics. Professor Treichel’s research in ­organometallic and metal-cluster
­chemistry and in mass spectrometry, aided by 75 graduate and undergraduate students, has led to more than
170 papers in scientific journals. He may be contacted by
email at treichelpaul@me.com.
Paul M. Treichel received his B.S. degree from the
­ niversity of Wisconsin in 1958 and a Ph.D. from H
U
­ arvard
University in 1962. After a year of postdoctoral study in
London, he assumed a faculty position at the University
of Wisconsin–Madison. He served as department chair
from 1986 through 1995 and was awarded a Helfaer Professorship in 1996. He has held visiting faculty positions
in South Africa (1975) and in Japan (1995). Retiring
after 44 years as a faculty member in 2007, he is currently
David A. Treichel, Professor of Chemistry at Nebraska
Wesleyan University, received a B.A. degree from ­Carleton
College. He earned a M.S. and a Ph.D. in analytical
chemistry at Northwestern University. After postdoctoral
research at the University of Texas in Austin, he joined
the faculty at Nebraska Wesleyan University. His research
interests are in the fields of electrochemistry and surface laser spectroscopy. He may be contacted by email at
dat@nebrwesleyan.edu.
John R. Townsend completed his B.A. in C
­ hemistry as
well as the Approved Program for Teacher ­Certification
in Chemistry at the University of Delaware. After a career
teaching high school science and mathem­atics, he earned
his M.S. and Ph.D. in biophysical c­ hemistry at Cornell
University, where he also received the DuPont Teaching
Award for his work as a teaching assistant. After teaching
at Bloomsburg University of P
­ ennsylvania, he joined
the faculty at West Chester University of ­Pennsylvania
where he coordinated the chemistry e­ ducation program
for prospective high school teachers and the general
­c hemistry lecture programs, taught undergraduate
courses in general chemistry and biochemistry, and was
the university supervisor for 78 prospective high school
chemistry teachers during their student teaching semester.
In 2021, he was the recipient of the Award for Excellence
in Undergraduate Teaching in Chemical Science from
the Philadelphia (Pennsylvania) S­ ection of the ­American
Chemical Society. Retiring in 2021, he is an Emeritus
Professor of Chemistry at West Chester ­University. He
may be contacted by email at jtownsend@wcupa.edu.
xx
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
About the Cover
Kotz • Treichel • Townsend • Treichel
Chemistry
11th Edition
& Chemical Reactivity
Chemistry & Chemical Reactivity
11th Edition
Ralph Lee Hopkins
Kotz
Treichel
Townsend
Treichel
SE/Kotz • Treichel • Townsend • Treichel, Chemistry & Chemical Reactivity, 11th Edition ISBN 9780357851401
Printer: Binding: Casebound Trim: 8.5” x 10.875” CMYK
©2024
Designer: Chris Doughman
Cover photo by Ralph Lee Hopkins,
Fagradalsfjall volcano, Iceland, 2021.
Fagradalsfjall volcano, on the Reykjanes Peninsula not far from Reykjavik
in Iceland, started erupting on March 19, 2021, and continued to erupt for
six months. The volcano erupted again August 3, 2022, but went quiet after
only 10 days.
For many observers, the most spectacular part of a volcanic e­ ruption
is the lava or molten rock that comes from the Earth’s mantle during an
­e ruption. For others, particularly scientists, volcanic activity provides
­opportunities to explore the chemical makeup of the Earth’s mantle and to
study the effects of volcanic activity on the environment.
Volcanoes are part of the story of the history and chemistry of the Earth.
Volcanic events have always occurred and continue to occur around the globe
as the result of tectonic plate activity. One massive volcanic event occurred
in 1883 on the island of Krakatoa in Indonesia. Thousands of people in
the volcano’s vicinity perished quickly, but the eruption also had global
consequences. For example, temperatures across the northern h
­ emisphere
dropped by an average of 0.4 °C in the year following the eruption.
Environmental scientists study the effects of volcanic activity. ­Scientists
know that volcanic activity injects large amounts of water vapor, carbon
dioxide, and sulfur dioxide into the atmosphere. Other gases released
include hydrogen chloride and stinky “sewer gas” or hydrogen sulfide.
There is a significant cooling effect in the atmosphere, partly from the conversion of sulfur dioxide to sulfuric acid and sulfate aerosols. These reflect
sunlight back into space and cool the Earth. You might think that the
­carbon ­dioxide from volcanoes would increase global warming, a topic of
current concern. However, scientists have shown that volcanoes emit less
than 1% of the massive amount of carbon dioxide put into the atmosphere
by human activities today.
The elements and compounds that arise from volcanic activity are
­fundamentally important. You will encounter these substances and many
others in this general introduction to the field of chemistry.
Personal statement from Ralph Lee Hopkins, the photographer: It was a dream
come true for a geologist-photographer to trek to an active volcano in
­Iceland. Words can’t describe the sights, sounds, and smells of new earth
being created. I spent two weeks and made five treks in between bad
weather and poison gas warnings. I was very lucky to witness flowing lava
up close and a sinuous river of lava spilling from the crater at sundown.
xxi
Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Dedication
We wish to dedicate this edition of Chemistry & Chemical Reactivity to our
colleagues who have contributed to our knowledge of chemistry and teaching and to our many students, some of whom became good friends and
who helped us understand better how to communicate our science. We
also acknowledge and thank Professor Paul Treichel who helped shape
this book with his work on many of the previous editions. His expertise,
good humor, and friendship over the years are appreciated. And, finally, we
thank our families who supported the years of work needed to produce this
book and for their support throughout our careers.
xxii
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
1 Basic Concepts of Chemistry
Dnsphotography/iStock/Getty Images
Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
C hapt e r O ut li n e
1.1
Chemistry and Its Methods
1.2
Sustainability and Green Chemistry
1.3
Classifying Matter
1.4
Elements
1.5
Compounds
1.6
Properties and Changes
1.7
Energy: Some Basic Principles
Chemistry is the scientific study of the composition, structure, and properties of
matter and the changes in both composition and energy that matter undergoes during reactions. Although chemistry is endlessly fascinating—at least to ­chemists—
why should you study chemistry? Each person probably has a different answer, but
many students take a chemistry course because those who are professional scientists, who teach and are involved daily in scientific work, realize how important
chemistry is in any curriculum leading to a career in a science-related discipline.
You will come to appreciate that chemistry is central to understanding disciplines
as diverse as biology, geology, materials science, medicine, physics, and some
branches of engineering. This is why chemistry is sometimes referred to as the
central science. In addition, chemistry plays a major role in national economies, and
chemistry and chemicals affect our daily lives in a wide variety of ways.
A course in chemistry can also help you see how a scientist thinks about the world
and solves problems. The knowledge and skills developed in chemistry courses will
benefit you in many career paths and help you become a better-informed citizen in a
world that is becoming technologically more complex—and more interesting.
Matter Anything that occupies
space and has mass — all
substances and mixtures in
the universe are composed of
matter.
1.1 Chemistry and Its Methods
Goal for Section 1.1
• Recognize the difference between a hypothesis and a theory, and understand how
laws are established.
This book provides a foundation for learning chemistry, a discipline that
has ­d eveloped over many centuries through the work of people around the
planet. H
­ owever, chemistry is about far more than historical knowledge. New
◀ Methane Bubbles Trapped in Ice. Bodies of water are often filled with and surrounded by
vegetation. Over time, the vegetation will decay, slowly being digested by bacteria that
release methane, a greenhouse gas, as a product of the digestion. Some of the methane
bubbles rise to the surface, and in the winter the bubbles can be trapped in ice. The white
patches in the photo are trapped methane bubbles in a lake in Alberta, Canada.
3
Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
discoveries occur frequently, and many recent discoveries are highlighted in this
book. As you read, please do not overlook the special features that explore some
of these d
­ iscoveries, in particular “A Closer Look” boxes and “Applying Chemical
Principles” sections.
Are you interested in medicine or medical advances? Do not miss the story on
the development of mRNA vaccines (“A Closer Look: mRNA Vaccines,” page 1228)
that have been valuable in the fight against the COVID-19 viruses or the story on
how gene editing holds the promise for correcting genetic mutations that lead to
diseases (“A Closer Look: Genetic Engineering with CRISPR-Cas9,” page 1222).
Chemists, physicists, and material scientists work together to develop electrical devices using atomically thin films of pure carbon (“Applying Chemical
Principles 12.2: Nanotubes and Graphene: Network Solids,” page 614), create high-­
temperature superconductors that may one day replace inefficient power transmission lines (“Applying Chemical Principles 3.1: S­ uperconductors,” page 177), and
find ways to create naturally occurring, but rare materials, such as diamonds in laboratories (“Applying Chemical Principles 18.2: Are Diamonds Forever?,” page 917).
Perhaps most importantly, scientists across multiple disciplines are s­ tudying
ways to slow climate change. Increasing levels of carbon dioxide, methane,
and other greenhouse gases are changing the conditions on earth, both on land
and in the oceans (“Applying Chemical Principles 1.1: CO 2 in the Oceans,”
page 21). It is accepted by the scientific community that significant changes to
the ­e nvironment will continue to occur if greenhouse gas emissions are not
reduced. Some hope comes from sequestering greenhouse gases (“Applying
­ ioxide,” page 177) and reducChemical Principles 3.2: Sequestering Carbon D
ing our reliance on fossil fuels (“Applying Chemical Principles 5.2: The Fuel
Controversy—Alcohol and Gasoline,” page 287). The environment is so important
that an entire chapter (Chapter 25) is devoted to the subject.
As you use this book in your study of chemistry and chemical principles, be
sure to understand that it is just the beginning. It provides an introduction to the
most important topics of chemistry, but we hope that it will also help you appreciate those topics and their interconnections as well as their uses and importance in
your lives.
Cinnabar
Mercury
droplets
Figure 1.1 Cinnabar and
mercury. Heating cinnabar
(mercury(II) sulfide) in air changes
it into orange mercury(II) oxide,
which, upon further heating,
decomposes into the elements
mercury and oxygen gas.
4
© Charles D. Winters/Cengage
Chemistry and Change
Chemistry is about change. It was once only about changing one natural substance into
another—wood and oil burn, grape juice turns into wine, and cinnabar (Figure 1.1),
a red mineral, ultimately changes into shiny quicksilver (mercury) when heated. The
emphasis was largely on finding a recipe to complete a desired change with little
understanding of the underlying structure of the materials or explanations for why
particular changes occurred. Chemistry is still about change, but now chemists focus
on the change of one pure substance, whether natural or synthetic, into another
and on understanding that change (Figure 1.2). Chemists now picture an exciting
world of submicroscopic atoms and molecules interacting with each other, and they
have developed ways to predict whether or not a particular reaction may occur.
Methods of Science
Neil deGrasse Tyson, noted physicist, author, and TV personality once said “Science
is a method of inquiry. Science is a way of expressing doubt and knowing when it’s
time to embrace what’s discovered and move on to something else to doubt.”
While there is no one scientific method by which all scientists conduct their
studies, there are certain common practices. You almost always start the process by
asking questions. These can be questions of your own choosing or ones that someone else poses. Having posed a reasonable question, the next step is often to look at
the experimental work done in the field so that you have some notion of the possible answers. Based on this work, you may form a hypothesis, a tentative explanation or prediction of experimental observations.
Chapter 1 / Basic Concepts of Chemistry
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Martyn F. Chillmaid/Science Source
© Charles D. Winters/Cengage
Sodium solid, Na
Sodium chloride solid, NaCl
Figure 1.2 Forming a chemical
compound. Combining sodium
Chlorine gas, Cl 2
metal (Na) and yellow chlorine
gas (Cl2) gives sodium chloride.
After formulating a hypothesis, systematic investigations are conducted, which
may include formal experiments designed to give results that will confirm or invalidate the hypothesis. Systematic investigations require the collection of information
or data, which may be either quantitative or qualitative. Quantitative information
is numerical data, such as the mass of a substance (Figure 1.3) or the temperature
at which it melts. Qualitative information, in contrast, consists of nonnumerical
observations, such as the color of a substance or its physical appearance.
The data from your investigations must be analyzed and interpreted in order
to derive meaning. Based on the analysis of your investigations, and perhaps studies from other researchers, you may have evidence supporting your hypothesis.
However, it is also possible, and quite common, that you will need to revise your
hypothesis and continue to test it with more experiments, or that the investigation
will end up raising more questions for you to answer. After you have checked to
ensure that your results are truly reproducible, a pattern of behavior or results might
begin to emerge. At this point, you may be able to summarize your observations in
the form of a law, a concise verbal or mathematical statement of a relation that is
always the same under any condition.
Figure 1.3 Qualitative and
quantitative observations. ​
Weighing a compound on a
laboratory balance.
Quantitative:
mass is 28.331 grams
© Charles D. Winters/Cengage
Qualitative:
blue, granular solid
1.1 Chemistry and Its Methods
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
5
© Charles D. Winters/Cengage
Figure 1.4 The metallic
element sodium reacts with
water.
Much of the work in science is based on laws because they help predict what
may occur under a new set of circumstances. For example, chemists know from
­experience that if the element sodium comes in contact with water, a violent r­ eaction
occurs and new substances are formed (Figure 1.4). They also know the mass of the
substances produced in the reaction is the same as the mass of the sodium and
water used in the reaction. That is, mass is always conserved in c­ hemical ­reactions, a
statement of the law of conservation of matter.
Once enough reproducible studies are conducted and experimental results
generalized as a law or general rule, it may be possible to conceive a theory to explain
the observations. A theory is a well-tested, unifying principle that explains a body
of facts and the laws based on them. Theories can suggest new hypotheses that can
be tested experimentally.
Sometimes nonscientists use the word theory to imply that someone has made
a guess and that an idea is not yet substantiated. To scientists, however, a theory
is based on carefully determined and reproducible evidence that is being continuously tested. Theories are the cornerstone of our understanding of the natural world
at any given time. Remember, though, that theories are inventions of the human
mind. Theories can and do change as new facts are uncovered.
Goals of Science
Scientists, including chemists, have several goals. Two of these are prediction and
control. Scientists do experiments and look for generalities because they want to
predict what may occur under other circumstances. They also want to learn how to
control the outcome of a chemical reaction or process.
Understanding and explaining are two other important goals. For example, certain elements such as sodium react vigorously with water. But why is this true? To
explain and understand this, you need a background in chemical concepts.
Dilemmas and Integrity in Science
You may think research in science is straightforward: Do experiments, collect
information, and draw a conclusion. But, research is seldom that easy. Frustrations and disappointments are common, and results can be inconclusive. Experiments always have some level of uncertainty, and sometimes the data collected
are contradictory. For example, suppose you do an experiment expecting to find
a direct relation between two experimental quantities. You collect six data sets.
When plotted on a graph, four of the sets lie on a straight line, but two others lie
far away from the line. Should you ignore the last two sets of data? Or should you
do more experiments when you know that others could publish their results first
and thus get the credit? Or should you consider that the two points not on the
line might indicate that your original hypothesis is wrong and abandon a favorite
idea you have worked on for many months? Scientists have a responsibility to
remain objective in these situations, but sometimes it is hard to do.
It is important to remember that a scientist is subject to the same moral pressures and dilemmas as any other person. To help ensure integrity in science, some
simple principles have emerged over time that guide scientific practice:
6
•
Experimental results should be reproducible. Furthermore, these results should
be reported in the scientific literature in enough detail to be used or reproduced
by others.
•
Research reports should be reviewed before publication by experts in the
field to ensure that the experiments were conducted properly and that the
conclusions are logical. (Scientists call this peer review.)
•
Conclusions should be reasonable and unbiased.
•
Credit should be given where it is due.
Chapter 1 / Basic Concepts of Chemistry
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Chemistry in Your Career
Darius Z. Brown
Darius Z. Brown
adequate industry experience in various fields as a
c­ hemist, has allowed me to focus on the relationships needed to teach high school students,” says
Brown, who believes that his own disadvantaged
background helps him connect with students who
face similar challenges. “I believe that once you
can see the world in terms of atoms and chemical reactions, your perspective . . . changes, and
you become more conscious and aware of the
little things in life, which ultimately helps . . . with
problem-solving and working together.”
Darius Z. Brown (he/him/his) began his journey
in the world of chemistry by obtaining a B.S.
(University of Buffalo) and M.S. (University of
­
­Illinois), with a focus on materials chemistry. He
first put his degrees to use in a variety of industrial
and research settings, ­including with a food manufacturer, university, and a paint producer.
Missing something in his industry work,
Brown returned to school to become a high
school chemistry teacher. “Having a solid educational background in chemistry, along with
1.2 Sustainability and Green Chemistry
Goal for Section 1.2
• Understand the principles of green chemistry.
•
“It is better to prevent waste than to treat or clean up waste after it is formed.”
•
“Synthetic methods should be designed to maximize the incorporation of all
materials used in the final product.”
•
Synthetic methods “should be designed to use and generate substances that
possess little or no toxicity to human health or the environment.”
•
“Chemical products should be designed to [function effectively] while still
reducing toxicity.”
© Charles D. Winters/Cengage
The world’s population is over 8.0 billion people, with about 99 million added per year.
Each new person needs shelter, food, and medical care, and each uses increasingly scarce
resources like fresh water and energy. And each produces by-products in the act of living
and working that can affect our environment. With such a large population, these individual effects can have large consequences for our planet. The focus of scientists, planners, and politicians is increasingly turning to the concept of sustainable development.
James Cusumano, a chemist and former president of a chemical company, said
that “On one hand, society, governments, and industry seek economic growth to create
greater value, new jobs, and a more enjoyable and fulfilling lifestyle. Yet, on the other,
regulators, environmentalists, and citizens of the globe demand that we do so with sustainable development—meeting today’s global economic and environmental needs while
preserving the options of future generations to meet theirs. How do nations resolve
these potentially conflicting goals?” This conflict is even more evident now than it was
in 1995 when Dr. Cusumano made this statement in the Journal of Chemical Education.
Much of the increase in life expectancy and quality of life, at least in the developed
world, is derived from advances in science. But it comes at a cost to the environment,
with increases in polluting gases such as nitrogen oxides and sulfur oxides in the atmosphere, acid rain falling in many parts of the world, and waste pharmaceuticals entering
the water supply. Among many others, chemists are seeking answers to these problems,
and one response has been to practice green chemistry.
The concept of green chemistry began to take root more than 30 years ago and
now leads to new chemical methods and lower pollutant levels. Paul Anastas and
John Warner stated 12 principles of green chemistry in their book Green Chemistry:
Theory and Practice (Oxford, 1998) that have become hallmarks for chemists
attempting to devise processes and products that are more environmentally
sustainable. Among these are
GREEN
C H E M I S T RY
1.2 Sustainability and Green Chemistry
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
7
•
“Energy requirements should be recognized for their environmental and economic impacts and should be minimized. Synthetic methods should be conducted at ambient temperature and pressure.”
•
Raw materials “should be renewable whenever technically and economically
practical.”
•
“Chemical products should be designed so that at the end of their function, they
do not persist in the environment or break down into dangerous products.”
•
“Substances used in a chemical process should be chosen to minimize the
potential for chemical accidents, including releases, explosions, and fires.”
You will be reminded about these principles at various points in Chemistry &
Chemical Reactivity as they are applied to modern applications in chemistry. Stating
these fundamental ideas is good, but the real challenge is to put them into practice.
1.3 Classifying Matter
Goals for Section 1.3
• Understand the basic ideas of kinetic-molecular theory.
• Recognize the importance of representing matter at the macroscopic, submicroscopic,
and symbolic levels.
• Recognize the different states of matter (solids, liquids, and gases) and know their
characteristics.
• Recognize the difference between pure substances and mixtures as well as the
difference between homogeneous and heterogeneous mixtures.
This section is an introduction to how chemists think about science in general and
about matter in particular. Terms such as atom, element, molecule, and compound
may appear to describe similar things, but each term has a unique definition. It
is important that you know these definitions as they will be used throughout the
book.
States of Matter and Kinetic-Molecular Theory
Solid
Bromine
solid and liquid
Bromine
gas and liquid
Liquid
Figure 1.5 States of matter—
solid, liquid, and gas. Elemental
bromine exists in all three states
near room temperature.
8
© Charles D. Winters/Cengage
Gas
One key property of matter is its state—that is, whether a substance is a solid,
liquid, or gas (Figure 1.5). A solid has a rigid shape and fixed volume that changes
little as temperature and pressure change. Like solids, liquids have a fixed volume,
but a liquid is fluid—it takes on the shape of its container and has no definite shape
of its own. Gases are fluid as well, but the volume of a gas is determined by the
size of its container. The volume of a gas varies with changes in temperature and
pressure.
At low enough temperatures, virtually all matter is in the solid state. As the temperature is raised, solids usually melt to form liquids. Eventually, if the temperature
is high enough, liquids evaporate to form gases. Volume changes typically accompany changes in state. For a given mass of material, there is usually a small increase
in volume upon melting—water being a significant exception—and then a large
increase in volume occurs upon evaporation.
The kinetic-molecular theory of matter helps you interpret the properties of
solids, liquids, and gases. According to this theory, all matter consists of extremely
tiny particles (atoms, molecules, or ions) in constant motion.
Solids: In solids, particles are packed closely together, usually in a regular pattern.
The particles vibrate back and forth about their average positions, but seldom do
particles in a solid squeeze past their immediate neighbors to come into contact
with a new set of particles.
Chapter 1 / Basic Concepts of Chemistry
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Liquids: The particles in liquids are arranged randomly rather than in the regular
patterns found in solids. Liquids and gases are fluid because the particles are not
confined to specific locations and can move past one another.
Gases: Under normal conditions, the particles in a gas are far apart. Gas molecules
move extremely rapidly and are not constrained by their neighbors. The ­molecules
of a gas fly about, colliding with one another and with the container walls.
This ­random motion allows gas molecules to fill their container, so the volume of
the gas sample is the volume of the container.
There are net forces of attraction between particles in all states—they are
­g enerally small in gases and large in liquids and solids. These forces have a
­significant role in determining the properties of matter. An important aspect of the
kinetic-­molecular theory is that the higher the temperature, the faster the particles
move. The energy of motion of the particles (their kinetic energy, Section 1.7) acts
to ­overcome the forces of attraction between particles. A solid melts to form a liquid
when the ­temperature of the solid is raised to the point at which the particles vibrate
fast enough and far enough to push one another out of the way and move out of
their regularly spaced positions. As the temperature increases even more, the particles
move faster still until finally they escape the clutches of their neighbors and enter the
gaseous state.
Matter at the Macroscopic and Particulate Levels
The characteristic properties of gases, liquids, and solids can be observed by the
unaided human senses. They are determined using samples of matter large enough
to be seen, measured, and handled. You can determine, for example, the color
of a substance, whether it dissolves in water, whether it conducts electricity, and
if it reacts with oxygen. Observations such as these generally take place in the
­macroscopic world of chemistry (Figure 1.6). This is the world of experiments and
observations.
Now imagine taking a macroscopic sample of material and dividing it again
and again, past the point that the sample can be seen by the naked eye, and past the
point where it can be seen using an optical microscope. Eventually, you reach the
level of individual particles that make up all matter, a level that chemists refer to as
the submicroscopic or particulate world of atoms and molecules (Figure 1.6).
Figure 1.6 Levels of
matter. Chemical and physical
A beaker of boiling water can be
modeled at the particulate level as
rapidly moving H2O molecules.
PI
C
PA
© Charles D. Winters/Cengage
M ACR
L E V E L S
O F
M A T T E R
S Y
M B O L I
ATE
The process is
symbolized by a
chemical equation.
T
UL
Observe
R
IC
O
S
C
O
processes are observed at the
macroscopic level. To understand
or illustrate these processes,
scientists often imagine what
has occurred at the particulate
atomic and molecular levels and
write symbols to represent these
observations.
Imagine
C
H2O (liquid) 88n H2O (gas)
Represent
1.3 Classifying Matter
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
9
Chemists are interested in the structure of matter at this particulate level.
Atoms, molecules, and ions cannot be seen in the same way that one views the macroscopic world, but they are no less real. Chemists imagine what atoms must look
like and how they might fit together to form molecules. They create models to represent atoms and molecules (Figure 1.6)—where tiny spheres are used to represent
atoms—and then use these models to think about chemistry and to explain their
observations about the macroscopic world. For example, the macroscopic properties of gases are explained using the kinetic-molecular theory that assumes the existence of submicroscopic particles in motion.
Chemists carry out experiments at the macroscopic level, but they think about
chemistry at the particulate level. They then record their observations as symbols,
formulas (such as H2O for water or NH3 for ammonia molecules), and drawings
that represent the elements and compounds involved. As you study chemistry, try
to make the connections in your own mind among the symbolic, particulate, and
macroscopic worlds of chemistry.
Pure Substances
A chemist looks at a glass of drinking water and sees a liquid. This liquid could be
the pure chemical compound water. However, it is also possible the liquid is a mixture of water and dissolved substances. Specifically, matter can be classified as either
a pure substance or a mixture (Figure 1.7).
A pure substance has a set of unique properties by which it can be recognized.
Pure water, for example, is colorless and odorless. If you want to identify a s­ ubstance
conclusively as water, however, you would have to examine its properties more c­ arefully
and compare them against the known properties of pure water. For example, pure
water melts at 0 °C and boils at 100 °C under normal atmospheric pressure. If you can
show that the substance melts and boils at these temperatures, you can be certain it is
water. No other known substance melts and boils at precisely those temperatures.
A second feature of a pure substance is that it cannot be separated into two or
more different species by any physical technique at ordinary temperatures. If it can
be separated, the sample is classified as a mixture.
Mixtures: Heterogeneous and Homogeneous
A mixture consists of two or more pure substances that can be separated by physical techniques. In a heterogeneous mixture, the composition of the mixture is not
uniform. The uneven distribution of the substances in a heterogeneous mixture
can often be detected by the naked eye (Figure 1.8). However, keep in mind there
are heterogeneous mixtures that may appear completely uniform but upon closer
examination are not. Milk, for example, appears smooth in texture to the unaided
eye, but magnification reveals fat and protein globules within the liquid.
MATTER
(may be solid, liquid, or gas)
Anything that occupies
space and has mass
HETEROGENEOUS MIXTURE
Nonuniform composition
MIXTURES
More than one pure
substance present.
Composition can be varied.
HOMOGENEOUS MIXTURE
Uniform composition
throughout
Physically
separable
into...
COMPOUNDS
Elements united
in fixed ratios
PURE SUBSTANCES
Fixed composition;
cannot be
further purified
Chemically
separable
into...
Combine
chemically
to form...
ELEMENTS
Cannot be subdivided
by chemical
or physical processes
Figure 1.7 Classifying matter.
10
Chapter 1 / Basic Concepts of Chemistry
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
A solution of salt in water.
The model shows that salt in
water consists of separate,
electrically charged particles
(ions), but the particles
cannot be seen with an
optical microscope.
−
+
John C. Kotz
−
A heterogeneous mixture.
+
−
+
A homogeneous mixture.
© Charles D. Winters/Cengage
Granite, an
igneous rock,
has grains large
enough to see with
the naked eye.
Figure 1.8 Heterogeneous and homogeneous mixtures.
A homogeneous mixture consists of two or more substances that are uniformly distributed down to the molecular level (Figure 1.8). No amount of optical
­magnification will reveal different properties in different regions of a homogeneous
mixture. Homogeneous mixtures are often called solutions. Common examples
include air (mostly a mixture of nitrogen and oxygen gases), gasoline (a mixture of
carbon- and hydrogen-containing compounds called hydrocarbons), and a soft drink
in an unopened container.
When a mixture is separated into its pure components, the components are
said to be purified. From the earliest days to modern times, the development of
methods to separate mixtures is an important goal for chemists. In pharmaceutical
­laboratories, synthesizing a new drug requires pure starting materials. And, the synthesis usually results in a mixture of compounds from which the desired drug must
be separated. Outside of a laboratory, purifying compounds may be even more
important. For example, your health relies on the removal of harmful bacteria and
toxic substances from water before it reaches your kitchen sink.
One means of separation, filtration, is used to remove an undissolved solid
from a solution. In this process, the mixture is applied to one side of a filter, a
porous barrier whose openings are small enough that most of the undissolved
solid particles cannot pass through, but the solution can. The solution emerges on
the other side of the barrier, leaving the solid particles behind. Efforts at separation are often not completed in a single step, however, and repetition almost always
increases purity. For example, soil particles can be separated from water by filtration (Figure 1.9). When the mixture is passed through a filter, many of the particles
© Charles D. Winters/Cengage
A heterogeneous mixture of
soil and water
When the mixture is poured
through the filter paper, the
larger soil particles are trapped
and the water passes through.
The water passing through
the filter is more pure than
in the mixture.
Figure 1.9 Purifying a hetero­geneous mixture by filtration.
1.3 Classifying Matter
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11
are removed. Repeated filtrations will give water with a higher and higher state of
purity. This purification process uses a property of the mixture, its clarity, to measure the extent of purification. When a perfectly clear sample of water is obtained,
all of the soil particles are assumed to have been removed.
1.4 Elements
Goals for Section 1.4
Hydrogen—gas
© Charles D. Winters/Cengage
Oxygen—gas
Water—liquid
Figure 1.10 Decomposing water
to yield hydrogen and oxygen
gases.
• Identify the name or symbol for an element, given its symbol or name, respectively.
• Use the terms atom, element, and molecule correctly.
Passing an electric current through water can decompose it into gaseous ­hydrogen
and oxygen (Figure 1.10). Substances like hydrogen and oxygen that are composed of only one type of atom are classified as elements. Several elements are
shown in ­Figure 1.11. Though a matter of some debate, 94–98 of the currently
known 118 ­elements have been confirmed to exist in nature, though some occur
only in trace quantities. The remainder have been created by scientists and have not
yet been found in any naturally occurring samples. The names of the elements and
their chemical symbols, one- or two-letter abbreviations used in place of the names,
are listed in the tables inside the front and back covers of this book. Carbon (C),
sulfur (S), iron (Fe), copper (Cu), silver (Ag), tin (Sn), gold (Au), mercury (Hg), and
lead (Pb) were known to the early Greeks and Romans as well as the alchemists of
ancient China, the medieval Islamic world, and medieval Europe. However, many
other elements—such as aluminum (Al), silicon (Si), iodine (I), and helium (He)—
were not discovered until the eighteenth and nineteenth centuries. Finally, scientists
in the twentieth and twenty-first centuries have made elements that do not exist in
nature, such as technetium (Tc) and plutonium (Pu).
The stories behind some of the names of the elements are fascinating. Many
elements have names and symbols with Latin or Greek origins. Examples include
helium (He), from the Greek word helios meaning “sun,” and lead, whose symbol,
Pb, comes from the Latin word for “heavy,” plumbum. More recently discovered
­elements have been named for their place of discovery or place of significance, such
as americium (Am), californium (Cf), scandium (Sc), europium (Eu), francium
(Fr), and polonium (Po). Some elements are named for their discoverers or famous
scientists. For example, curium (Cm), einsteinium (Es), fermium (Fm), mendelevium (Md), nobelium (No), and meitnerium (Mt) were named after Marie and
Pierre Curie, Albert Einstein, Enrico Fermi, Dmitri Mendeleev, Alfred Nobel, and
Lise Meitner, respectively.
Figure 1.11 Elements. ​
© Charles D. Winters/Cengage
Chemical elements can often be
distinguished by their color and
their state at room temperature.
12
Mercury—
liquid
Powdered
sulfur—solid
Copper wire—
solid
Iron chips—
solid
Aluminum—
solid
Chapter 1 / Basic Concepts of Chemistry
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
© Charles D. Winters/Cengage
When writing the symbol for an element, notice that the first letter (but not the
second) of an element’s symbol is capitalized. For example, cobalt is Co, not co or
CO. The notation co has no chemical meaning, whereas CO represents the chemical
compound carbon monoxide. Also note that the element name is not capitalized,
except at the beginning of a sentence.
The table inside the front cover of this book, in which the symbol and other
information for the elements are enclosed in a box, is called the periodic table. This
important tool of chemistry is described in more detail beginning in Chapter 2.
An atom is the smallest particle of an element that retains the characteristic
chemical properties of that element. Some elements, such as neon and argon, are
found in nature as isolated atoms. Others are found as molecules, particles consisting of more than one atom in which the atoms are held together by chemical bonds.
Examples of molecular elements are the colorless gases of the air, nitrogen (N2) and
oxygen (O2) as well as deep purple iodine (I2) and orange liquid bromine (Br2). For
each of these cases, the subscript “2” following the element symbol indicates that two
atoms of the element exist in a single molecule. Yet other elements consist of infinite
networks of atoms; an example of this is diamond, one form of the element carbon.
Diamond.
Diamond. A diamond consists of
a network of carbon atoms linked
by chemical bonds.
1.5 Compounds
Goals for Section 1.5
•
Sodium is a shiny metal that reacts violently with water (Figure 1.4). Its solidstate structure has sodium atoms tightly packed together.
•
Chlorine is a light yellow-green gas that has a distinctive, suffocating odor and
is a powerful irritant to lungs and other tissues. The element is composed of Cl2
molecules in which two chlorine atoms are tightly bound together.
•
Sodium chloride, or common salt (NaCl), is a colorless, crystalline solid composed of sodium and chlorine ions bound tightly together. Its properties are
completely unlike those of the two elements from which it is made.
It is important to distinguish between a mixture of elements and a chemical
c­ ompound of two or more elements. Pure metallic iron and yellow, powdered sulfur
can be mixed in varying proportions. In the chemical compound iron pyrite,
­however, there is no variation in composition. Not only does iron pyrite exhibit
properties unique to itself and different from those of iron, sulfur, or a mixture of
these two elements, it also has a definite percentage composition by mass (46.55%
Fe and 53.45% S). That a compound has a definite composition (by mass) of its
combining elements is a basic principle of chemistry and is often referred to as the
law of definite proportions or the law of constant composition. Thus, two major
differences exist between a mixture and a pure compound: A compound has
­distinctly different characteristics from its parent elements, and it has a definite
­percentage composition (by mass) of its combining elements.
Mixture The material in the dish is a
mixture of elements, iron and sulfur.
The iron can be separated easily
from the sulfur by using a magnet.
© Charles D. Winters/Cengage
A pure substance like sugar, salt, or water, which is composed of two or more
­different elements held together by chemical bonds, is referred to as a chemical
compound. Even though only 118 elements are known, there appears to be no limit
to the number of compounds that can be made from those elements. As of 2021,
over 193 million different compounds were identified in CAS, a ­database created by
the American Chemical Society.
The properties of a compound, such as its color, hardness, and melting point, are
different than those of its constituent elements. Consider common table salt (sodium
chloride), which is composed of two elements, sodium and chlorine (­Figure 1.2):
© Charles D. Winters/Cengage
• Use the term compound correctly.
• Understand the law of definite proportions (law of constant composition).
Chemical compound Iron pyrite is
a chemical compound composed
of iron and sulfur. It is often found
in nature as perfect, golden cubes.
1.5 Compounds
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13
Name
Water
Methane
Ammonia
Carbon dioxide
Formula
H2O
CH4
NH3
CO2
Model
Figure 1.12 Names, formulas, and models of some common molecular compounds. A common color scheme
used in molecular models is C atoms are gray or black, H atoms are white, N atoms are blue, and O atoms are red.
Compounds with Indefinite
Proportions Though the vast
majority of compounds have
definite proportions, a few break
this law. For example, a class of
high temperature superconductors
have the formula La2-xBaxCuO4,
in which x varies from 0 to
0.5. (See “Applying Chemical
Principles 3.1: Superconductors,”
page 177).
Some compounds—such as table salt, NaCl—are composed of ions, which are
electrically charged atoms or groups of atoms (Section 2.5). Other compounds—
such as water and sugar—consist of molecules.
The composition of any compound is represented by its chemical formula. In
the formula for water, H2O, for example, the symbol for hydrogen, H, is followed
by a subscript “2” that indicates that two atoms of hydrogen occur in a single water
molecule. The symbol for oxygen appears without a subscript, indicating that one
oxygen atom occurs in the molecule.
As you shall see throughout this text, molecules can be represented with
models that depict their composition and structure. Figure 1.12 illustrates the
names, formulas, and models of the structures of four common molecular
compounds.
E xamp le 1.1
Elements and Compounds
Problem Identify whether each of the following is an element or compound: bromine,
Br2, and hydrogen peroxide, H2O2.
What Do You Know? Elements and compounds are both pure substances. An
element is composed of only one type of atom. A compound is composed of more than
one type of atom, where the atoms are connected by chemical bonds and occur in a definite proportion by mass.
Strategy Look at each formula given and use the guidelines in the “What Do You
Know?” section to determine whether the formula is that for an element or a compound.
Solution Bromine, Br2, is an element because both of the atoms present in the molecule are the same type, bromine atoms. H2O2 is a compound. In H2O2, there are two
different types of atoms present, hydrogen atoms and oxygen atoms. They are bonded
together in hydrogen peroxide molecules that have a definite composition by mass; each
molecule of H2O2 has two atoms of hydrogen and two atoms of oxygen.
Think about Your Answer If all of the atoms in a molecule are the same type,
such as in Br2, then it is a molecule of an element.
Check Your Understanding
Identify whether white phosphorus (P 4) and carbon monoxide (CO) are elements or
compounds.
14
Chapter 1 / Basic Concepts of Chemistry
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1.6 Properties and Changes
Goals for Section 1.6
• Identify several physical properties of common substances.
• Relate density to the volume and mass of a substance.
• Understand the difference between extensive and intensive properties and give
examples.
• Explain the difference between chemical and physical changes.
• Identify several chemical properties of common substances.
You recognize your friends by their physical appearance: their height and weight and
the color of their eyes and hair. The same is true of chemical substances. You can
tell the difference between an ice cube and a cube of lead of the same size by their
appearance (one is clear and colorless, the other is a lustrous metal), and because
one is heavier (lead) than the other (ice). Properties such as these, which can be
observed and measured without changing the composition of a substance, are called
­physical properties. The chemical elements in Figure 1.11, for example, differ in
terms of their color, appearance, and state (solid, liquid, or gas). Substances can
often be classified and identified by their physical properties. Table 1.1 lists several
physical properties of matter that chemists commonly use.
Density, the ratio of the mass of an object to its volume, is a physical property
useful for identifying substances.
Density 5
mass
volume
(1.1)
For example, you can tell the difference between an ice cube and a cube of lead of
identical size because lead has a high density, 11.35 g/cm3 (11.35 grams per cubic
centimeter), whereas ice has a density slightly less than 0.917 g/cm3. An ice cube
with a volume of 16.0 cm3 has a mass of 14.7 g, whereas a cube of lead with the
same volume has a mass of 182 g.
The temperature of a sample of matter often affects the numerical values of its
­properties. Density is a particularly important example. Although the change in water
density with temperature seems small (Table 1.2), it affects the environment profoundly.
Table 1.1
Ice
© Charles D. Winters/Cengage
Lead
Units of Density The decimal
system of units in the sciences is
called Le Système International
d’Unités, often referred to as
SI units. The SI unit of density is
kg/m3. In chemistry, the more
commonly used unit is g/cm3.
To convert from kg/m3 to g/cm3,
divide by 1000.
Some Physical Properties
Property
Using the Property to Distinguish Substances
Color
Is the substance colored or colorless? What is the color, and what is its
intensity?
State of matter
Is it a solid, liquid, or gas? If it is a solid, what is the shape of the particles?
Melting point
At what temperature does a solid melt?
Boiling point
At what temperature does a liquid boil (at 1 atmosphere pressure)?
Density
What is the substance’s density (mass per unit volume)?
Solubility
What mass of substance can dissolve in a given volume of water or other
solvent?
Electric
conductivity
Does the substance conduct electricity?
Malleability
Table 1.2
Temperature Dependence
of Water Density
Temperature
(° C)
Density
of Water
(g/cm3)
0 (ice)
0.917
0 (liq water)
0.99984
2
0.99994
4
0.99997
How easily can a solid be deformed?
10
0.99970
Ductility
How easily can a solid be drawn into a wire?
25
0.99707
Viscosity
How easily will a liquid flow?
100
0.95836
1.6 Properties and Changes
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15
For example, as the water in a lake cools, the density of the water increases and the denser
water sinks as shown in Figure 1.13. This continues until the water temperature reaches
3.98 °C, the point at which water has its maximum density (0.999973 g/cm3). If the
water temperature drops farther, the density decreases slightly and the colder water floats
on top of water at 3.98 °C. If water is cooled below about 0 °C, solid ice forms. Water
has a rare property: Its solid form is less dense than its liquid form, so ice floats on water.
The volume of a given mass of liquid changes with temperature, so its density does also. This is why laboratory glassware used to measure precise volumes of
solutions always specifies the temperature at which it was calibrated (Figure 1.14).
E xamp le 1.2
Density
Problem A piece of a polypropylene rope (used for water skiing) floats on water,
whereas a terephthalate polymer from a soda bottle sinks in water. Place polypropylene,
the terephthalate polymer, and water in order of increasing density.
What Do You Know? Density is given by Equation 1.1. An object with a higher
density sinks in a liquid of lower density, whereas an object with a lower density floats in a
liquid of higher density.
Strategy Use the observations as to whether a material sinks or floats in water to
determine the order of the densities.
Solution The polypropylene rope floats in water, therefore polypropylene is less dense
than water. The soda bottle plastic sinks in water; therefore the soda bottle plastic is more
dense than the water. This gives the overall order of densities:
polypropylene < water < soda bottle plastic.
Think about Your Answer In a material with a greater density, the matter is
more tightly packed for a given mass than in materials of lower density.
Check Your Understanding
© Charles D. Winters/Cengage
Blue dye was added to the left side of
the water-filled tank, and ice cubes
were placed in the right side.
The water beneath the ice is cooler
and denser than the surrounding
water, so it sinks. The convection
current created by this movement of
water is traced by the dye movement
as the denser, cooler water sinks.
Figure 1.13 Temperature dependence of physical properties. The density of
water and other substances changes with temperature.
16
© Charles D. Winters/Cengage
Some salad dressings are made from a mixture of olive oil and vinegar. These two liquids do not
dissolve significantly in each other. If this mixture is allowed to sit, the two liquids separate from
each other with the olive oil floating on top of the vinegar. Which liquid has the greater density?
Figure 1.14 Dependence of density on
temperature. Water and other substances
change in density with temperature so laboratory
glassware is calibrated for a particular
temperature.
Chapter 1 / Basic Concepts of Chemistry
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Extensive and Intensive Properties
The physical properties of matter can be classified as either extensive properties or
intensive properties. Extensive properties depend on the amount of a substance present. The mass and volume of samples of elements or compounds or the amount of
energy transferred as heat from burning gasoline are extensive properties, for example.
In contrast, intensive properties do not depend on the amount of substance.
A sample of ice will melt at 0 °C, no matter whether it is an ice cube or an iceberg.
Although mass and volume are extensive properties, it is interesting that density
(the quotient of these two quantities) is an intensive property. The density of gold,
for example, is the same (19.3 g/cm3 at 20 °C) whether it is a flake of pure gold or
a solid gold ring.
Intensive properties are often useful in identifying a material. For example, the
temperature at which a material melts (its melting point) is often so characteristic
that it can be used to identify the solid (Figure 1.15).
Exam p le 1.3
Extensive and Intensive Properties
Problem A sample of liquid mercury has a shiny surface, melts at 234 K, has a mass of
27.2 g, has a volume of 2.00 cm3, and has a density of 13.6 g/cm3. Which of these properties are extensive properties and which are intensive properties?
What Do You Know? Extensive properties depend on the amount of a substance
present. Intensive properties do not depend on the amount of substance present.
Strategy Determine which of the properties listed depend on the amount of material
present and which do not.
Solution The mass and volume of the sample each depend on the amount of material
present; the greater the amount of material present, the greater will be the mass and the
volume. Mass and volume are extensive properties. The shininess of the surface, the melting point, and the density are properties that are the same regardless of the amount of
material present, so they are intensive properties.
Think about Your Answer Mass and volume are both extensive properties, but
their quotient, density, is an intensive property.
Check Your Understanding
Identify whether each of the following properties is extensive or intensive: boiling point,
hardness, volume of a solution, number of atoms, number of atoms dissolved per volume
of solution.
© Charles D. Winters/Cengage
Figure 1.15 An intensive
property used to distinguish
compounds.
Naphthalene, a white solid at
room temperature, melts at
80.2 °C and so is molten at the
temperature of boiling water.
Aspirin, a white solid at room
temperature, melts at 135 °C
and so remains a solid at the
temperature of boiling water.
1.6 Properties and Changes
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17
Physical and Chemical Changes
Changes in physical properties are called physical changes. In a physical change the
identity of a substance is preserved even though it may have changed its physical
state or the gross size and shape of its pieces. A physical change does not result in a
new chemical substance being produced. The particles (atoms, molecules, or ions)
present before and after the change are the same. An example of a physical change is
the melting of a solid (Figure 1.15) or the evaporation of a liquid (Figure 1.16). In
either case, the same molecules are present both before and after the change. Their
chemical identities have not changed.
Physical change • The same molecules are present both before and after the change.
O2 molecules in
the gas phase
© Charles D. Winters/Cengage
Liquid oxygen (boiling
point, –183 °C) is a pale
blue liquid.
O2 molecules in
the liquid phase
Chemical change • One or more substances (reactants) are transformed into one or more different substances (products)
A balloon filled with
molecules of hydrogen
gas and surrounded by
molecules of oxygen in
the air. (The balloon floats
in air because gaseous
hydrogen is less dense
than air.)
© Charles D. Winters/Cengage
When ignited with a
burning candle, H2 and O2
react to form water, H2O.
O2 (gas)
2 H2 (gas)
2 H2O(gas)
A symbolic and particulate view • The reaction of O2 and H2
Symbolic view
2 H2(gas)
O2(gas)
2 H2O(gas)
Particulate view
Reactants
Products
Figure 1.16 Physical and chemical change.
18
Chapter 1 / Basic Concepts of Chemistry
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A physical property of hydrogen gas (H2) is its low density, so a balloon filled
with H2 floats in air. Suppose, however, that a lighted candle is brought up to the
balloon. When the heat causes the skin of the balloon to rupture, the hydrogen
mixes with the oxygen (O2) in the air, and the heat of the burning candle sets off a
chemical reaction that produces water, H2O (Figure 1.16). This reaction is an example of a chemical change, in which one or more substances (the reactants) are
transformed into one or more different substances (the products).
The chemical change in Figure 1.16 is also shown at the particulate level:
hydrogen molecules and oxygen molecules react to form water molecules. The representation of the change using chemical formulas is called a chemical equation.
It shows that the substances on the left (the reactants) produce the substances on
the right (the products). This equation illustrates an important principle of chemical
reactions: matter is conserved. The number and identity of the atoms found in the
reactants are the same as in the products. Here, there are four atoms of H and two
atoms of O before and after the reaction, but the molecules before the reaction are
different from those after the reaction.
The term chemical property refers to chemical reactions that a substance might
undergo. For example, a chemical property of hydrogen gas is that it reacts with
oxygen gas, and quite vigorously so.
Exam p le 1.4
Physical and Chemical Changes
Problem Identify each of the following as being either a physical or a chemical change:
boiling water and rusting of an iron nail (which converts iron (Fe) to Fe2O3).
What Do You Know? In a physical change, the chemical identities of the materials do not change, whereas in a chemical change, they do.
Strategy Examine each of the changes to determine if the chemical identities of the
materials change.
Solution In liquid water, the chemical species present is H2O molecules. When water
boils, molecules move to the gaseous state. The chemical species is still H2O molecules;
the molecules have merely changed state. Boiling water is thus a physical change. Rusting
of an iron nail is a chemical change because you begin with iron and oxygen and end up
with rust, which is predominantly the chemical compound iron(III) oxide, Fe 2O3. The substance at the end of the process is a different chemical species than the ones with which
you started.
Think about Your Answer Students sometimes confuse changes of state with
chemical changes; changes of state are physical changes.
Check Your Understanding
Identify whether each of the following is a physical change or a chemical change: (a)
melting butter, (b) burning wood, (c) dissolving sugar in water.
1.6 Properties and Changes
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19
1.7 Energy: Some Basic Principles
Goals for Section 1.7
• Identify types of potential and kinetic energy.
• Recognize and apply the law of conservation of energy.
Units of Energy Energy in
chemistry is measured in units
of joules. (See Chapter 1R and
Chapter 5 for calculations
involving energy units.)
Energy, a crucial part of many chemical and physical changes, is defined as the capacity to do work. You do work against the force of gravity when carrying yourself and
hiking equipment up a mountain. The energy to do this is provided by the food you
have eaten. Food is a source of chemical energy—energy stored in chemical compounds and released when the compounds undergo the chemical reactions of metabolism in your body. Chemical reactions almost always either release or absorb energy.
Energy can be classified as kinetic or potential. Kinetic energy is energy associated with motion, such as:
•
The motion of atoms, molecules, or ions at the submicroscopic (particulate)
level (thermal energy). All matter has thermal energy.
•
The motion of macroscopic objects such as a moving tennis ball or automobile
(mechanical energy).
•
The movement of electrons in a conductor (electrical energy).
•
The compression and expansion of the spaces between molecules in the transmission of sound (acoustic energy).
Potential energy results from an object’s position or state and includes:
•
Energy possessed by a ball held above the floor and by water at the top of a
water wheel (gravitational energy) (Figure 1.17a).
•
Energy stored in an extended spring.
•
Energy stored in fuels (chemical energy) (Figure 1.17b).
•
Energy associated with the separation of electrical charges (electrostatic energy)
(Figure 1.17c).
(a) Potential energy is converted into
mechanical energy.
(b) Chemical potential energy of a fuel
and oxygen is converted into thermal and
mechanical energy in the launch of a rocket.
Aleksandr Gogolin/Shutterstock.com
Salajean/Shutterstock.com
Handout/Getty Images News/Getty Images
Potential energy and kinetic energy can be interconverted. For example, as water
falls over a waterfall, its potential energy is converted into kinetic energy. S­ imilarly,
kinetic energy can be converted into potential energy: The kinetic energy of f­ alling
water can turn a turbine to produce electricity, which can then be used to c­ onvert
(c) Electrostatic energy is converted to
radiant and thermal energy.
Figure 1.17 Energy and its conversion.
20
Chapter 1 / Basic Concepts of Chemistry
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Kinetic energy
(energy of motion)
Diego Barbieri
/Shutterstock.com
The diver has potential energy
when standing a distance
above the water surface.
Heat and work
(thermal and
mechanical energy)
AkosHorvath
/Shutterstock.com
Paolo Bona/Shutterstock.com
Potential energy
(energy of position)
Figure 1.18 The law of energy conservation.
The diver’s potential energy is first
converted to kinetic energy, which
is then transferred to the water.
water into H2 and O2 by electrolysis. Hydrogen gas contains stored chemical ­potential
energy because it can be burned to produce heat and light or electricity.
Conservation of Energy
Standing on a diving board, you have considerable potential energy because of your position above the water. Once you dive off the board, some of that potential energy is converted into kinetic energy (Figure 1.18). During the dive, the force of gravity accelerates
your body so that it moves faster and faster. Your kinetic energy increases and your potential
energy decreases. At the moment you hit the water, your velocity is abruptly reduced and
much of your kinetic energy is transferred to the water as your body moves it aside. Eventually you float to the surface, and the water becomes still again. If you could see the water
molecules, however, you would find that they are moving a little faster in the vicinity of
your entry into the water; that is, the kinetic energy of the water molecules is slightly higher.
This series of energy conversions illustrates the law of conservation of energy, which
states that energy can neither be created nor destroyed. Or, to state this law differently, the
total energy of the universe is constant. The law of conservation of energy summarizes
the results of many experiments in which the amounts of energy transferred were measured and the total energy content was found to be the same before and after an event.
This law can also be examined in the case of a chemical reaction, the reaction
of hydrogen and oxygen to form water (Figure 1.16). In this reaction, the reactants
(­hydrogen and oxygen) have a certain amount of energy associated with them. When
they react, some of this energy is released to their surroundings. If you add up all the
energy present before the reaction and all the energy present after the reaction, you will
find that the energy was only redistributed; the total amount of energy in the universe
has remained constant. Energy has been conserved.
Applying Chemical Principles
1.1 CO2 in the Oceans
A 2019 report on the absorption of carbon in the world’s
ocean stated that “The global ocean absorbed 34 billion metric tons of carbon from the burning of fossil fuels from 1994
to 2007—a fourfold increase of 2.6 billion metric tons per
year when compared to the period starting from the Industrial
Revolution in 1800 to 1994.
The amount of carbon in the form of CO2 dissolved in the
oceans is of great concern and interest because it affects the pH
of the water, that is, its level of acidity. This in turn can affect
the growth of marine organisms such as corals, sea urchins,
and microscopic coccolithophores (single-cell phytoplankton).
Studies have indicated that, in water with a high CO2 content, the spines of sea urchins are greatly impaired, the larvae
of orange clown fish lose their homing ability, and the concentrations of calcium, copper, manganese, and iron in sea water
are affected, sometimes drastically.
Applying Chemical Principles
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21
and magnitude of ocean geochemical changes potentially
unparalleled in at least the last 300 million years of Earth
history, raising the possibility that we are entering an unknown
territory of marine ecosystem change.”
Cbpix/Shutterstock.com
Questions
Clown fish. The larvae of the clown fish are affected by
increased levels of CO2 in the ocean.
An investigation of the history of ocean acidification ended
with the statement that “the current rate of (mainly fossil fuel)
CO2 release stands out as capable of driving a combination
1. Much has been written about CO2. What is its name?
2. Give the symbols for the four metals mentioned in this article.
3. Of the four metals mentioned here, which is the most
dense? The least dense? (Use an Internet tool such as
www.ptable.com to find this information.)
4. The spines of sea urchins, corals, and coccolithophores all are
built of the compound CaCO3. Which elements are involved in
this compound? Do you know this compound’s name? If not,
you might be able to find it using an Internet search.
References
1. https://www.ncei.noaa.gov/news/global-ocean-absorbing-more-carbon
2. “The Geological Record of Ocean Acidification,” B. Hönisch, et al.,
Science, 2012, 335, 1058–1063.
Think–Pair–Share
1. Hypotheses, Laws, and Theories
(a) What are the differences between a hypothesis and a
theory?
(b) Why is the law of constant composition a law and not a
theory?
2. Carbon dioxide consists of molecules containing one carbon atom and two oxygen atoms. In each molecule the three
atoms are arranged in a line with the carbon in between the
two oxygen atoms. Make a drawing, based on the kineticmolecular theory, of the arrangement of carbon dioxide molecules in (a) the solid state, (b) the liquid state, and (c) the
gas state. In your drawing, represent the carbon atoms with
a solid circle and the oxygen atoms with an open circle. For
each case, draw 10 molecules. It is acceptable for your diagram to be two dimensional.
3. Some students come to a chemistry class thinking that all
samples of elements consist of isolated atoms and all samples of compounds consist of molecules.
(a) Are all elements found in nature as isolated atoms? If not,
how else can the atoms be arranged?
(b) Are all compounds composed of molecules? If not, what
other types of particles are found in some compounds?
(c) Is N2 an element or a compound? Explain your answer.
4. Answer the following.
(a) Zinc is a shiny silver-colored solid that reacts vigorously
with hydrochloric acid. Identify each of these properties as
being either a physical property or a chemical property.
Explain your reasoning.
(b) The mass of a piece of aluminum is 10.0 g, its volume is 3.70
cm3, and its density is 2.70 g/cm3. Identify each of these
properties as being either an intensive property or an
­extensive property. Explain your reasoning.
(c) Identify each of the following as being either a physical
change or a chemical change. Explain your reasoning.
(i) Bubbles of carbon dioxide gas form when baking soda
is added to vinegar.
(ii) Paper burns.
(iii) Sugar dissolves in water.
(iv) Water boils.
5. Give one or more examples of each of the following types of
energy conversions.
(a) Chemical potential energy into electrical energy.
(b) Chemical potential energy into thermal energy.
(c) Thermal energy into mechanical energy.
(d) Chemical potential energy into sound and/or light energy.
Chapter Goals Revisited
The goals for this chapter are keyed to specific Study Questions to help you organize
your review.
1.1 Chemistry and Its Methods
•
22
Recognize the difference between a hypothesis and a theory and
understand how laws are established. 1, 2.
Chapter 1 / Basic Concepts of Chemistry
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1.2 Sustainability and Green Chemistry
•
Understand the principles of green chemistry. 3–6.
1.3 Classifying Matter
•
Understand the basic ideas of kinetic-molecular theory. 43, 44.
•
Recognize the importance of representing matter at the macroscopic,
submicroscopic, and symbolic levels. 37, 38.
•
Recognize the different states of matter (solids, liquids, and gases) and
know their characteristics. 31, 43, 53.
•
Recognize the difference between pure substances and mixtures and the
difference between homogeneous and heterogeneous mixtures. 33, 34, 44.
1.4 Elements
•
Identify the name or symbol for an element, given its symbol or name,
respectively. 7–10, 31, 32.
•
Use the terms atom, element, and molecule correctly. 11, 12, 41, 42.
1.5 Compounds
•
Use the term compound correctly. 11, 12, 41, 42.
•
Understand the law of definite proportions (law of constant composition). 13, 14.
1.6 Properties and Changes
•
Identify several physical properties of common substances. 15, 16, 19, 20,
32, 46, 48.
•
Relate density to the volume and mass of a substance. 27, 28, 39, 40, 45, 47,
49, 50, 51, 54, 55, 58.
•
Understand the difference between extensive and intensive properties and
give examples. 27, 28.
•
Explain the difference between chemical and physical changes. 17, 18, 35,
36, 53, 57.
•
Identify several chemical properties of common substances. 15, 16, 19, 20,
29, 30.
1.7 Energy: Some Basic Principles
•
Identify types of potential and kinetic energy. 21–24.
•
Recognize and apply the law of conservation of energy. 25, 26.
Key Equation
Equation 1.1 (page 15) Density: In chemistry the common unit of density is g/cm3,
whereas kg/m3 is commonly used in geology and oceanography.
Density 5
mass
volume
Key Equation
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
23
Study Questions
▲ denotes challenging questions. Blue-numbered questions have answers in Appendix N and fully
worked solutions in the ­Student Solutions Manual.
Practicing Skills
Nature of Science
(See Section 1.1.)
1. In the following scenario, identify which of the
statements represents a theory, law, or hypothesis.
(a) A student exploring the properties of gases proposes that if she decreases the volume of a sample
of gas then the pressure exerted by the sample will
increase. (b) Many scientists over time have conducted similar experiments and have concluded
that pressure and volume are inversely proportional. (c) She proposes that the reason this occurs
is that if the volume is decreased, more molecules
will collide with a given area of the container walls,
causing the pressure to be greater.
2. State whether the following is a hypothesis,
theory, or law of science. Global climate change
is occurring because of human-generated carbon
dioxide. Explain.
Green Chemistry
(See Section 1.2.)
3. What is meant by the phrase “sustainable
development”?
4. What is meant by the phrase “green chemistry”?
5. Chronic cough is a condition estimated to affect as
many as 5% to 10% of adults. Merck and Co., Inc.
earned a Presidential Green Chemistry Challenge
Award in 2021 for developing a new method of synthesizing gefapixant citrate, a drug to treat chronic
cough. The process allows for a six-fold reduction
in raw material costs, requires fewer synthetic steps,
and achieves a higher yield of the final product
than the previous synthetic method. A synthetic
step that involved highly hazardous chemicals was
replaced, resulting in a safer production process.
Finally, the synthesis produces less carbon dioxide
and carbon monoxide emissions. Which principles
of green chemistry are being followed by this new
process compared to the old process?
6. Surfactants are substances used in detergents to
reduce the surface tension of liquids in which they
are dissolved. Most surfactants are petroleumbased, and their production requires considerable
energy and produces toxic waste products. Colonial
Chemical was a winner of a 2021 Presidential
Green Chemistry Award for the development
24
of Suga®Boost surfactants that use more
environmentally-friendly chemicals than traditional
surfactants. These surfactants are produced from
renewable plant-based materials, require less
energy to create, require only water as a solvent,
and biodegrade into nontoxic substances. Which
principles of green chemistry are being followed by
this new process compared to the old process?
Matter: Elements and Atoms, Compounds,
and Molecules
(See Example 1.1.)
7. Give the name of each of the following elements:
(a) N
(d) I
(b) Ca
(e) Li
(c) Br
(f) Fe
8. Give the name of each of the following elements:
(a) Cr
(d) Cl
(b) Ni
(e) Ar
(c) Mg
(f) Ti
9. Give the symbol for each of the following elements:
(a) strontium
(d) mercury
(b) cadmium
(e) selenium
(c) cobalt
(f) bismuth
10. Give the symbol for each of the following elements:
(a) platinum
(d) thallium
(b) gallium
(e) tungsten
(c) uranium
(f) xenon
11. In each of the following pairs, decide which is an
element and which is a compound.
(a) Na or NaCl
(b) sugar or carbon
(c) gold or gold(III) chloride
12. In each of the following pairs, decide which is an
element and which is a compound.
(a) Pt(NH3)2Cl2 or Pt
(b) copper or copper(II) oxide
(c) silicon or SiO2
13. An 18 g sample of water decomposes into 2 g of
hydrogen gas and 16 g of oxygen gas. What masses
of hydrogen and oxygen gases would be prepared
from 27 g of water? What law of chemistry is used
in solving this problem?
Chapter 1 / Basic Concepts of Chemistry
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(See Section 1.6)
15. In each case, decide if the underlined property is a
physical or chemical property.
(a) The color of elemental bromine is orange-red.
(b) Iron turns to rust in the presence of air and water.
(c) Hydrogen can explode when ignited in air
(Figure 1.16).
(d) The density of titanium metal is 4.5 g/cm3.
(e) Tin metal melts at 505 K.
(f) Chlorophyll, a plant pigment, is green.
16. In each case, decide if the underlined property is a
physical or chemical property.
(a) Copper wire is used for the transmission of electricity because it is a good electrical conductor.
(b) Olive oil pours more slowly from a container
than water due to its greater viscosity.
(c) The density of liquid water is 1.00 g/mL at 4°C.
(d) In wine making, grapes ferment to produce
wine, an alcoholic beverage.
(e) A pigment used in white paint is titanium(IV)
oxide.
(f) Gasoline burns when ignited in air.
17. In each case, decide if the change is a chemical or
physical change.
(a) A puddle of water evaporates after a rain
shower ends.
(b) Milk turns sour after it is left in a warm place
for too long.
(c) When baking soda and vinegar are combined, the
mixture produces bubbles of carbon dioxide gas.
(d) Table sugar, or sucrose, dissolves in water.
(e) A tomato turns red as it ripens.
(f) A firework explodes filling the sky with bright
flashes of light.
18. In each case, decide if the change is a chemical or
physical change.
(a) A cup of household bleach changes the color
of your favorite T-shirt from purple to pink.
(b) Water vapor in your exhaled breath condenses
in the air on a cold day.
19. Which part of the description of a compound or
element refers to its physical properties and which
to its chemical properties?
(a) Chlorine, a yellow-green gas, reacts with
sodium metal to produce sodium chloride.
(b) Sodium bicarbonate is a white powdery solid that
reacts with an acid to produce carbon dioxide.
20. Which part of the description of a compound or
element refers to its physical properties and which
to its chemical properties?
(a) Copper(II) sulfide is a black solid with a
density of 2.71 g/cm3. It reacts readily with an
acid to produce gaseous hydrogen sulfide.
(b) Magnesium, a silvery metal, reacts with oxygen
gas to produce a white compound.
Energy
(See Section 1.7.)
21. The flashlight in the photo does not use batteries.
Instead, you move a lever, which turns a geared
mechanism that results in light from the bulb. What
type of energy is used to move the lever? What type
or types of energy are produced?
A hand-operated flashlight
22. A solar panel is pictured in the photo. When light
shines on the panel, it generates an electric current
that can be used to recharge the batteries in an electric
car. What types of energy are involved in this setup?
John C. Kotz
Physical and Chemical Properties
(c) Plants use carbon dioxide from the air to make
sugar.
(d) Butter melts when placed in the sun.
© Charles D. Winters/Cengage
14. A sample of the compound magnesium oxide is
synthesized as follows: 60. g of magnesium is burned
and produces 100. g of magnesium oxide, indicating
that the magnesium combined with 40. g of oxygen
in the air. If 30. g of magnesium were used, what
mass of oxygen would combine with it? What law of
chemistry is used in solving this problem?
A solar panel
Study Questions
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25
23. Determine which of the following represent potential energy and which represent kinetic energy.
(a) acoustic energy
(b) thermal energy
(c) gravitational energy
(d) chemical energy
(e) electrostatic energy
24. Determine whether kinetic energy is being
converted to potential energy, or vice versa, in the
following processes.
(a) Water cascades downward in a waterfall.
(b) A furnace burns natural gas to heat a house.
(c) An electric current is used to recharge a battery.
(d) A piston moves when a gas is produced in a
chemical reaction.
25. A hot metal block is plunged into water in a
well-insulated container. The temperature of the
metal block goes down, and the temperature of the
water goes up until their temperatures are the same.
A total of 1500 J of energy is lost by the metal object.
How much energy was absorbed by the water? What
law of science is illustrated by this problem?
29. Which observations below describe chemical
properties?
(a) Mixing vinegar and sodium bicarbonate
produces bubbles of carbon dioxide gas.
(b) Sugar is soluble in water.
(c) Water boils at 100 °C.
(d) Ultraviolet light converts O3 (ozone) to O2
(oxygen).
(e) Ice is less dense than liquid water.
30. Which observations below describe chemical
properties?
(a) A reddish-brown coating of rust appears on the
surface of a piece of iron.
(b) Sodium metal reacts violently with water.
(c) The combustion of octane (a compound in
gasoline) gives CO2 and H2O.
(d) Chlorine is a yellow-green gas.
(e) Heat is required to melt ice.
31. The mineral fluorite contains the elements calcium
and fluorine and can have various colors, including
blue, violet, green, and yellow.
These questions are not designated as to type or location in
the chapter. They may combine several concepts.
27. A piece of turquoise is a blue-green solid; it has a
density of 2.65 g/cm3 and a mass of 2.5 g.
(a) Which of these observations are qualitative and
which are quantitative?
(b) Which of the observations are extensive and
which are intensive?
(c) What is the volume of the piece of turquoise?
28. Iron pyrite (fool’s gold, page 13) has a shiny golden
metallic appearance. Crystals are often in the form
of perfect cubes. A cube 0.40 cm on each side has a
mass of 0.32 g.
(a) Which of these observations are qualitative and
which are quantitative?
(b) Which of the observations are extensive and
which are intensive?
(c) What is the density of the sample of iron pyrite?
26
The mineral fluorite, calcium fluoride
(a) What are the symbols of these elements?
(b) How would you describe the shape of the
fluorite crystals in the photo? What can this
tell you about the arrangement of the particles
(ions) inside the crystal?
32. Azurite, a blue, crystalline mineral, is composed of
copper, carbon, and oxygen.
© Charles D. Winters/Cengage
General Questions
© Charles D. Winters/Cengage
26. A book is held at a height above the floor. It has a
certain amount of potential energy. When the book
is released, its potential energy is converted into
kinetic energy as it falls to the floor. The book hits
the floor and comes to rest. According to the law of
conservation of energy, the amount of energy in the
universe is constant. Where has the energy gone?
Azurite is a deep blue crystalline mineral. It is surrounded
by copper pellets and powdered carbon (in the dish).
Chapter 1 / Basic Concepts of Chemistry
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(a) What are the symbols of the three elements
that combine to make the mineral azurite?
(b) Based on the photo, describe some of the
physical properties of the elements and the
mineral. Are any the same? Are any properties
different?
33. You have a solution of NaCl dissolved in water.
Describe a method by which these two compounds
can be separated.
39. Carbon tetrachloride, CCl4, a liquid compound,
has a density of 1.58 g/cm3. If you place a piece of a
plastic soda bottle (d 5 1.37 g/cm3) and a piece of
aluminum (d 5 2.70 g/cm3) in liquid CCl4, will the
plastic and aluminum float or sink?
40. The following photo shows copper balls, immersed
in water, floating on top of mercury. What are the
liquids and solids in this photo? Which substance is
most dense? Which is least dense?
Chips of iron mixed with sand
35. Identify the following as either physical changes or
chemical changes.
(a) Dry ice (solid CO2) sublimes (converts directly
from solid to gaseous CO2).
(b) The density of mercury metal decreases as the
temperature increases.
(c) Energy is given off as heat when natural gas
(mostly methane, CH4) burns.
(d) NaCl dissolves in water.
36. Identify the following as either physical changes or
chemical changes.
(a) The desalination of sea water (separation of
pure water from dissolved salts).
(b) The formation of SO2 (an air pollutant) when
coal containing sulfur is burned.
(c) Silver tarnishes.
(d) Iron is heated to red heat.
37. In Figure 1.2 you see a piece of salt and a representation of its internal structure. Which is the macroscopic view and which is the particulate view? How
are the macroscopic and particulate views related?
38. In Figure 1.5 you see macroscopic and particulate
views of the element bromine. Which are the
macroscopic views and which are the particulate
views? Describe how the particulate views explain
properties of this element related to the state of
matter.
© Charles D. Winters/Cengage
© Charles D. Winters/Cengage
34. Small chips of iron are mixed with sand (see photo).
Is this a homogeneous or heterogeneous mixture?
Suggest a way to separate the iron from the sand.
Water, copper, and mercury
41. Categorize each of the following as an element, a
compound, or a mixture.
(a) iron pyrite (also known as fool’s gold)
(b) carbonated mineral water
(c) molybdenum
(d) sucrose (also known as table sugar)
42. Categorize each of the following as an element, a
compound, or a mixture.
(a) sea water
(c) bronze
(b) sodium chloride
(d) 24-carat gold
43. ▲ Make a drawing of the arrangement of particles
in each of the following cases based on the kineticmolecular theory and the ideas about atoms and
molecules presented in this chapter. For each case,
draw 10 particles of each substance. It is acceptable
for your diagram to be two dimensional. Represent
each atom as a circle, and distinguish each different
kind of atom by shading.
(a) A sample of solid iron (which consists of iron
atoms).
(b) A sample of liquid water (which consists of
H2O molecules).
(c) A sample of water vapor.
44. ▲ Make a drawing of the arrangement of particles
in each of the following cases based on the kineticmolecular theory and the ideas about atoms and
molecules presented in this chapter. For each case,
draw 10 particles of each substance. It is acceptable
for your diagram to be two dimensional. Represent
each atom as a circle, and distinguish each different
kind of atom by shading.
Study Questions
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27
45. Hexane (C6H14, density 5 0.766 g/cm3), perfluorohexane (C6F14, density 5 1.669 g/cm3), and water
are immiscible liquids; that is, they do not dissolve
in one another. You place 10 mL of each in a graduated cylinder, along with pieces of high-density
polyethylene (HDPE, density 5 0.97 g/cm3), polyvinyl chloride (PVC, density 5 1.36 g/cm3), and
Teflon™ (density 5 2.3 g/cm3). None of these common plastics dissolves in these liquids. Describe
what you expect to see.
46. ▲ You have a sample of a white crystalline
substance from your kitchen. You know that it is
either salt or sugar. Although you could decide
by taste, suggest another property that you could
use to decide. (Hint: You may use the Internet or a
handbook of chemistry in the library to find some
information.)
47. You can figure out whether a solid floats or sinks
if you know its density and the density of the
liquid. In which of the liquids listed below will
high-density polyethylene (HDPE) float? (HDPE,
a common plastic, has a density of 0.97 g/cm3.
It does not dissolve in any of these liquids.)
Density
(g/cm3)
50. Describe an experimental method that can be used
to determine the density of an irregularly shaped
piece of metal.
51. Diabetes can alter the density of urine, so urine density
can be used as a diagnostic tool. Diabetics can excrete
too much sugar or excrete too much water. What do
you predict will happen to the density of urine under
each of these conditions? (Hint: Water containing dissolved sugar is more dense than pure water.)
52. Suggest a way to determine if the colorless liquid in
a beaker is water. How could you discover if there is
salt dissolved in the water?
53. The following photo shows the element potassium
reacting with water to form the element hydrogen, a gas,
and a solution of the compound potassium hydroxide.
Properties, Uses
Ethylene
glycol
1.1088
Toxic; major component
of automobile antifreeze
Water
0.9997
Ethanol
0.7893
Alcohol in alcoholic
beverages
Methanol
0.7914
Toxic; gasoline additive to prevent gas line
freezing
Acetic acid
1.0492
Component of vinegar
Glycerol
1.2613
Solvent used in home
care products
48. You are given a sample of a silvery metal. What
information could you use to prove the metal is silver?
49. Milk in a glass bottle was placed in the ­freezing
compartment of a refrigerator overnight.
28
Frozen milk in a glass bottle
© Charles D. Winters/Cengage
Substance
By morning, a column of frozen milk emerged from
the bottle. Explain this observation.
© Charles D. Winters/Cengage
(a) A homogeneous mixture of water vapor and
helium gas (which consists of helium atoms).
(b) A heterogeneous mixture consisting of liquid
water and solid aluminum; show a region of
the sample that includes both substance.
(c) A sample of brass (which is a homogeneous
solid mixture of copper and zinc).
Potassium reacting with water to produce
hydrogen gas and potassium hydroxide.
(a) What states of matter are involved in the
reaction?
(b) Is the observed change chemical or physical?
(c) What are the reactants in this reaction, and
what are the products?
(d) What qualitative observations can be made
concerning this reaction?
Chapter 1 / Basic Concepts of Chemistry
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54. Three liquids of different densities are mixed.
Because they are not miscible (do not form a
homogeneous solution with one another), they
form discrete layers, one on top of the other. Sketch
the result of mixing carbon tetrachloride (CCl4,
d 5 1.58 g/cm3), mercury (d 5 13.546 g/cm3), and
water (d 5 1.00 g/cm3).
55. Four balloons are each filled with a different gas,
each having a different density:
chlorine, d 5 2.897 g/L
propane, d 5 1.802 g/L
methane, d 5 0.656 g/L
hydrogen, d 5 0.082 g/L
If the density of dry air is 1.12 g/L, which
balloon(s) will float in air?
56. A copper-colored metal is found to conduct an
electric current. Can you say with certainty that
it is copper? Why or why not? Suggest additional
information that could provide additional evidence
that it is copper.
© Charles D. Winters/Cengage
57. The photo below shows elemental iodine dissolving
in ethanol to give a solution. Is this a physical or
chemical change?
Elemental iodine dissolving in ethanol.
58. ▲ You want to determine the density of a
compound but have only a tiny crystal, and it
would be difficult to measure mass and volume
accurately. There is another way to determine
density, however, called the flotation method. If
you place the crystal in a liquid whose density is
precisely that of the substance, it will be suspended
in the liquid, neither sinking to the bottom of the
beaker nor floating to the surface. However, for such
an experiment you need a liquid with the precise
density of the crystal. You can accomplish this by
mixing two liquids of different densities to obtain
a liquid of the desired density.
(a) Consider the following: you mix 10.0 mL of
CHCl3 (d 5 1.492 g/mL) and 5.0 mL of CHBr3
(d 5 2.890 g/mL), giving 15.0 mL of solution.
What is the density of this mixture?
(b) Suppose now that you want to determine the
density of a small yellow crystal to confirm that
it is sulfur. From the literature, you know that
sulfur has a density of 2.07 g/cm3. How would
you prepare 20.0 mL of the liquid mixture with
that density from pure samples of CHCl3 and
CHBr3? (Note: 1 mL 5 1 cm3.)
59. A young chemist in Vienna, Austria, wanted to see
just how permanent the gold was in his wedding
band. The ring was 18-carat gold. (18-carat gold is
75% gold with the remainder copper and silver.)
One week after his wedding day he took off the
ring, cleaned it carefully, and weighed it. It had a
mass of 5.58387 g. He weighed it weekly from then
on, and after 1 year it had lost 6.15 mg just from
normal wear and tear. He found that the activities
that took the greatest toll on the gold were vacationing on a sandy beach and gardening.
(a) What are the symbols of the elements that
make up 18-carat gold?
(b) The density of gold is 19.3 g/cm3. Use one of
the periodic tables on the Internet (such as
www.ptable.com) to find out if gold is the
most dense of all of the known elements. If it
is not gold, then what element is the most
dense [considering only the elements from
hydrogen (H) through uranium (U)]?
(c) If a wedding band is 18-carat gold and has a
mass of 5.58 g, what mass of gold is contained
within the ring?
(d) Assume there are 56 million married couples in
the United States, and each person has an
18-carat gold ring. What mass of gold is lost by
all the wedding rings in the United States in
1 year (in units of grams) if each ring loses 6.15
mg of mass per year? Assuming gold is $1815
per troy ounce (where 1 troy ounce 5 31.1 g),
what is the lost gold worth?
Study Questions
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
29
© Charles D. Winters/Cengage
1R
Let’s Review: The Tools of Quantitative
Chemistry
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
C hapt e r O ut li n e
1R.1
Units of Measurement
1R.2
Making Measurements: Precision, Accuracy, Experimental Error,
and Standard Deviation
1R.3
Mathematics of Chemistry
1R.4
Problem Solving by Dimensional Analysis
1R.5
Graphs and Graphing
1R.6
Problem Solving and Chemical Arithmetic
Doing chemistry requires observing chemical reactions and physical changes. You
will make qualitative observations—such as changes in color or the evolution of
heat—and quantitative measurements of temperature, time, volume, mass, and
length or size. To be successful in chemistry, you will need to be familiar with the
units used by scientists to make measurements and be able to use quantitative data
properly to reach conclusions. These issues are very important to scientists and engineers, and a failure to pay attention to them can have disastrous results. For example, on September 23, 1999, nine months after its launch, the Mars Climate Orbiter,
which cost 125 million dollars, reached Mars and began its maneuvers to enter orbit
around the planet. The Orbiter failed to reestablish contact with Earth after passing
behind Mars. It was determined that the Orbiter failed to enter orbit and fell into
the atmosphere of Mars, where it disintegrated. What went wrong? The problem was
eventually traced back to a problem with units. Navigational signals were sent from
Earth to the Orbiter using English units of pound-seconds, but the Orbiter was
designed to use metric units of newton-seconds. As you take measurements in the
lab and perform calculations, it is very important that you keep track of all the units
used so that you can interpret the results properly.
1R.1 Units of Measurement
Goal for Section 1R.1
•
Use the common units for measurements in chemistry and make unit conversions
(such as from liters to milliliters).
To record and report measurements, the scientific community has chosen a ­modified
version of the metric system. This decimal system is called the Système ­International
d’Unités (International System of Units), abbreviated SI.
◀ Scientific Instruments and Glassware. Chemistry is a quantitative science. Many different
instruments and pieces of glassware have been invented to measure the properties of matter.
31
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Table 1R.1
The Seven SI Base Units
Measured Property
0.82 nm
Structure of the aspirin molecule
Name of Unit
Abbreviation
Mass
kilogram
kg
Length
meter
m
Time
second
s
Temperature
kelvin
K
Amount of substance
mole
mol
Electric current
ampere
A
Luminous intensity
candela
cd
All SI units are derived from base units, listed in Table 1R.1. Larger and smaller
quantities are expressed by using appropriate prefixes with the base unit (Table 1R.2).
The nanometer (nm), for example, is 1 billionth of a meter, that is, 1 3 1029 m
(meter). Dimensions on the nanometer scale are common in chemistry and biology.
For example, a typical molecule (such as aspirin) is about 1 nm in length and a
bacterium is about 1000 nm in length. Indeed, the prefix nano- is also used in the
name for a whole area of science, nanotechnology, which involves the synthesis and
study of materials having this tiny scale.
Temperature Scales
Two temperature scales are commonly used in scientific work: Celsius and Kelvin
(Figure 1R.1). The Celsius scale is generally used worldwide for measurements in
the laboratory. When calculations incorporate temperature data, however, the
­Kelvin scale is almost always used.
Table 1R.2
Common Conversion Factors
1000 g 5 1 kg
1 3 109 nm 5 1 m
10 mm 5 1 cm
100 cm 5 10 dm 5 1 m
1000 m 5 1 km
Conversion factors for SI units
are given in Appendix C and
inside the back cover of this
book.
32
Selected Prefixes Used in the Metric System
Prefix
Abbreviation
Meaning
Example
Giga-
G
109 (billion)
1 gigahertz 5 1 3 109 Hz
Mega-
M
106 (million)
1 megaton 5 1 3 106 tons
Kilo-
k
103 (thousand)
1 kilogram (kg) 5 1 3 103 g
Deci-
d
1021 (tenth)
1 decimeter (dm) 5 1 3 1021 m
Centi-
c
1022 (one hundredth)
1 centimeter (cm) 5 1 3 1022 m
Milli-
m
1023 (one thousandth)
1 millimeter (mm) 5 1 3 1023 m
Micro-
μ
1026 (one millionth)
1 micrometer (μm) 5 1 3 1026 m
Nano-
n
1029 (one billionth)
1 nanometer (nm) 5 1 3 1029 m
Pico-
p
10212
1 picometer (pm) 5 1 3 10212 m
Femto-
f
10215
1 femtometer (fm) 5 1 3 10215 m
Chapter 1R / Let’s Review: The Tools of Quantitative Chemistry
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Boiling point
212 ° F
of water
100 ° C
373.15 K
180 ° F
100 ° C
100 K
Freezing point
32 ° F
of water
0 °C
273.15 K
iStock.com/Magnascan
Kelvin
(or absolute)
Celsius
Figure 1R.1 Comparison of
Fahrenheit, Celsius, and Kelvin
scales. Note that the degree sign
(°) is not used with the Kelvin
scale.
iStock.com/ValentynVolkov
Fahrenheit
The Celsius Temperature Scale
The size of the Celsius degree is defined by assigning zero as the freezing point of
pure water (0 °C) and 100 as its boiling point (100 °C). A comfortable room temperature is around 20 °C, and your normal body temperature is 37 °C. The warmest
water you can stand to immerse a finger in is probably about 60 °C.
The Kelvin Temperature Scale
William Thomson, known as Lord Kelvin (1824–1907), first suggested the temperature scale that now bears his name. The Kelvin scale assigns zero as the lowest
temperature that can be achieved, a point called absolute zero. Many experiments
have found that this limiting temperature is 2273.15 °C. Kelvin units and Celsius
degrees are the same size. Thus, the freezing point of water is reached at 273.15 K;
that is, 0 °C 5 273.15 K. The normal boiling point of pure water is 373.15 K. Temperatures in Celsius degrees are converted to kelvins, and vice versa, using the
relation
T (K) 5
1K
(T °C 1 273.15 °C)
1 °C
(1R.1)
Lord Kelvin William Thomson
(1824–1907), known as
Lord Kelvin, was a professor
of natural philosophy at the
University in Glasgow, Scotland,
from 1846 to 1899. He was
best known for his study of heat
and work, from which came
the concept of the absolute
temperature scale.
Using this equation, you can show that a common room temperature of 23.5 °C is
equivalent to 296.7 K.
T (K) 5
1K
(23.5  C 1 273.15  C ) 5 296.7 K
1 C
Three things to notice about the Kelvin scale are:
•
The degree symbol (°) is not used with Kelvin temperatures.
•
The name of the unit is the kelvin (not capitalized).
•
Temperatures are designated with a capital K.
1R.1 Units of Measurement
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33
In 1790, during the French Revolution, the National Assembly of France
asked the French Academy of Sciences to come up with a new system
of weights and measures. This was
the beginning of the metric system.
The meter was initially intended to be
one ten millionth the distance between the equator and the North Pole
along a meridian that crossed Dunkirk,
France and Barcelona, Spain. Based on
measurements of this value (which were
later shown to be in error), a platinum bar
was constructed with the distance between its ends at a specified temperature
defined as a meter. Since then, the definition of a meter has been revised at various
times to allow its value to be determined
more accurately and precisely. In 1875,
the Treaty of the Meter was signed by 17
countries, and the Bureau Intérnational
des Poids et Mésures (BIPM, International
Bureau of Weights and Measures) was established. In 1889, a new International
Prototype Meter was constructed of a platinum-iridium alloy and adopted by the
BIPM. In 1960, a revised system of units,
called the Système International d’Unités
(SI, International System of Units) was adopted by the BIPM and with it a new definition of the meter. This definition was
based on something that could be universally examined in the laboratory. It related
the meter to a fraction of the wavelength
of light emitted by a type of atom of the
element krypton when it undergoes a particular type of electronic transition. Yet
another change in the definition occurred
in 1983. At this point, the BIPM based
the length of a meter on one of the fundamental constants of nature: the speed of
light. They defined the speed of light in a
vacuum to be exactly 299,792,458 meters per second. This means that 1 meter
is now defined as the distance that light
travels in one 299,792,458th of a second.
You might note that this definition also
­depends on the definition of the second,
another base unit in the SI.
On May 20, 2019, revisions to the definitions of four SI base units went into effect.
All seven of the base units are now tied to
values for seven physical constants of the
universe. You will learn about a number of
these constants as you proceed through
this course. Two of the base units whose
definitions were changed are the ­kilogram
(the standard unit of mass) and the kelvin
(the standard unit of temperature).
The kilogram was previously defined as the
mass of a cylinder of a particular platinumiridium alloy nicknamed Le Grand K, that
was housed in Paris, France. Unfortunately,
this block was shown to be mysteriously
­losing mass, clearly not desirable for a standard for mass. Following the revision, the
value of a kilogram is now tied to the value
of Planck’s constant, which you will learn
about in Chapter 6. Planck’s constant is assigned a value of 6.62607015 3 10234 kg
m2/s2 (or J· s). The kilogram is thus the mass
that makes this value true, given the definitions of the second and the meter.
To define the kelvin, the value of
Boltzmann’s constant, which you will learn
about in Chapter 18, is assigned a value of
1.380649 3 10223 kg m2/s2 · K (or J/K).
The kelvin can thus be determined using
this value and the previously defined values for the second, meter, and kilogram.
What consequences do these changes
have for you in your study of general
­chemistry? Not many. The values of the
universal constants were selected so that
the new definitions and the previous definitions give as close to the same answer
as possible. Thus, using a laboratory balance, 10.00 g using the old definition will
still be 10.00 g using the new definition.
The new definitions, however, relate the
base units to fundamental constants of
nature, freeing them from physical standards such as a block of a particular alloy
for the kilogram, and will increase the
­accuracy and precision of measurements
in the future.
SOTK2011/Alamy Stock Photo
A Closer Look
The SI Base Units
Figure Le Grand K. The International
Prototype of the Kilogram was a cylinder
of a platinum-iridium alloy that was the
standard for a kilogram from 1889 to
2019. Since 2019, the kilogram and all
the other SI base units are defined based
on their relationship to fundamental
constants of nature, whose values are
now defined.
Length, Volume, and Mass
The meter is the standard unit of length, but objects observed in chemistry are frequently smaller than 1 meter. Measurements are often reported in units of centi­
meters (cm), millimeters (mm), or micrometers (μm), and objects on the atomic
and molecular scale have dimensions of nanometers (nm; 1 nm 5 1 3 1029 m) or
picometers (pm; 1 pm 5 1 3 10212 m) (Figure 1R.2).
34
Chapter 1R / Let’s Review: The Tools of Quantitative Chemistry
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Ångstrom Units An older but often-
used non-SI unit for molecular
distances is the Ångstrom unit (Å),
where 1 Å 5 1.0 3 10210 m.
The distance between two carbon
atoms in diamond (Figure 1R.2)
is 1.54 Å.
© Charles D. Winters/Cengage
0.154 nm
3.0 mm
Figure 1R.2 Dimensions in the molecular world. Dimensions
on the molecular scale are often given in terms of nanometers
(1 nm 5 1 3 1029 m) or picometers (1 pm 5 1 3 10212 m).
Here, the distance between C atoms in diamond is 0.154 nm.
Exam p le 1R.1
Distances on the Molecular Scale
Problem The distance between an O atom and an H atom in a water molecule is
95.8 pm. What is this distance in nanometers (nm)?
What Do You Know? You are given the interatomic O–H distance. You will need
to know (or look up) the relationships of the metric units.
Strategy You can solve this problem by knowing the conversion
95.8 pm
factor between the units in the information you are given (picometers)
and the desired units (meters or nanometers). (For more about
conversion factors and their use in problem solving, see page 47.)
There is no conversion factor given in Table 1R.2 to change nanometers to picometers directly, but relationships are listed between
meters and picometers and between meters and nanometers. Therefore, you will first convert picometers to meters, and then convert ­meters to nanometers.
picometers
x m pm
y nm m
→ meters
→ nanometers
Solution First, use the conversion factor 1 pm 5 1 3 10212 m to convert 95.8 ­picometers
to meters.
95.8 pm 3
1 3 10212 m
5 9.58 3 10211 m
1 pm
Then use the conversion factor 1 nm 5 1 3 1029 m to convert from meters to nanometers.
9.58 3 10211 m 3
1 nm
5 9.58 3 1022 nm or 0.0958 nm
1 3 1029 m
Think about Your Answer A nanometer is a larger unit than a picometer, so the
same distance expressed in nanometers will have a smaller numerical value. The answer
agrees with this. Notice how the units cancel in the calculation to leave an answer whose
unit is that of the numerator of the conversion factor. The process of using units to guide
a calculation is called dimensional analysis. It is explored further on pages 47–48.
Check Your Understanding
The distance between two carbon atoms in diamond (Figure 1R.2) is 0.154 nm. What is this
distance in picometers (pm)? In centimeters (cm)?
1R.1 Units of Measurement
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35
Figure 1R.3 Dimensions in chemistry and
biology. These photos are from the research
1 cm
of Professor Joanna Aizenberg of Harvard
University.
(c) Scanning electron microscope
(SEM) image of a single strand showing
its ceramic-composite structure.
Scale bar = 20 μm.
5 mm
20 μm
© Charles D. Winters/Cengage
(b) Fragment of the structure
showing the square grid of the
lattice with diagonal supports.
Scale bar = 5 mm.
Figure 1R.4 Some common
laboratory glassware. Volumes
are marked in units of milliliters
(mL). Remember that 1 mL is
equivalent to 1 cm3.
36
Photos courtesy of Joanna Aizenberg, Bell
Laboratories. Reference: J. Aizenberg, et al.,
Science, Vol. 309, pages 275-278, 2005
(a) Photograph of the glassy sea sponge Euplectella.
Scale bar = 1 cm.
The glassy skeleton of a sea sponge (Figure 1R.3) can give you an idea of the range
of dimensions used in science. The sea sponge is about 20 cm long and a few centimeters in diameter. A closer look shows more detail of the lattice-like structure. Scientists
at Bell Laboratories found that each strand of the lattice is a ceramic-fiber composite
of silica (SiO2) and protein less than 100 μm in diameter. Each strand is composed of
spicules, which consist of silica nanoparticles 50 to 200 nanometers in diameter.
Chemists often use glassware such as beakers, flasks, pipets, graduated cylinders, and burets, which are marked in volume units (Figure 1R.4). Because the SI
unit of volume [the cubic meter (m3)] is too large for everyday laboratory use,
chemists usually use the liter (L) or the milliliter (mL) for volume measurements.
One liter is equivalent to the volume of a cube with sides equal to 10 cm
[V 5 (0.1 m)3 5 0.001 m3].
1 liter (L) 5 1000 cm3 5 1000 mL 5 0.001 m3
Because there are exactly 1000 mL (5 1000 cm3) in a liter, this means that
1 mL 5 0.001 L 5 1 cm3
The units milliliter and cubic centimeter (sometimes abbreviated cc by medical
­professionals) are interchangeable. Therefore, a flask that contains exactly 125 mL
has a volume of 125 cm3.
Although not widely used in the United States, the cubic decimeter (dm3) is a
common unit of volume in the rest of the world. A length of 10 cm is called a
­decimeter (dm), a tenth of a meter. Because a cube with a 10-cm side defines a volume of 1 liter, a liter is equivalent to a cubic decimeter: 1 L 5 1 dm3. Products in
Europe, Africa, and other parts of the world are often sold by the cubic decimeter.
The deciliter, dL, which is exactly equivalent to 1/10 of a liter (0.100 L) or
100 mL, is widely used in medicine. Standards for concentrations of environmental
contaminants are often set as a certain mass per deciliter. For example, the U.S.
­Centers for Disease Control and Prevention recommends that children with more
than 3.5 micrograms (3.5 3 1026 g) of lead per deciliter of blood undergo further
testing for lead poisoning.
Finally, when chemists prepare chemicals for reactions, they often measure the
mass of each material. Mass is the fundamental measure of the quantity of matter,
Chapter 1R / Let’s Review: The Tools of Quantitative Chemistry
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A Closer Look
Energy and Food
The U.S. Food and Drug Administration
(FDA) mandates that nutritional data, including energy content in Calories (where
1 Cal 5 1 kilocalorie), be included on
almost all packaged food. The Nutrition
Labeling and Education Act of 1990
requires that the total ­energy from protein, carbohydrates, fat, and alcohol be
specified. How is this determined? Initially, the method used was calorimetry.
In this method (described in Chapter 5), a
food product is burned, and the energy
transferred as heat in the combustion is
measured. Now, however, energy contents
are estimated using the Atwater system.
This specifies the following average values
for energy sources in foods:
1 g protein 5 4 kcal (17 kJ)
1 g carbohydrate 5 4 kcal (17 kJ)
1 g fat 5 9 kcal (38 kJ)
1 g alcohol 5 7 kcal (29 kJ)
Because carbohydrates may include some
indigestible fiber, the mass of fiber is subtracted from the mass of carbohydrate when
calculating the energy from carbohydrates.
As an example, one serving of cashew
nuts (about 28 g) has
14 g fat 5 126 kcal
6 g protein 5 24 kcal
7 g carbohydrates 2 1 g fiber 5 24 kcal
Total 5 174 kcal (728 kJ)
A value of 170 Calories (5 170 kcal) is
reported on the package. The calculated
and reported values agree to two s­ ignificant
figures (see Section 1R.3).
and the SI unit of mass is the kilogram (kg). Smaller masses are expressed in grams
(g) or milligrams (mg).
Energy Units
When expressing energy quantities, most chemists (and much of the world outside
the United States) use the joule (J), the SI unit. The joule is related directly to the
units used for mechanical energy: 1 J equals 1 kg · m2/s2. Because the joule is inconveniently small for most uses in chemistry, the kilojoule (kJ), equivalent to 1000
joules, is often the unit of choice.
To give you some feeling for joules, suppose you drop a six-pack of soft-drink
cans, each full of liquid, on your foot. Although you probably will not take time to
calculate it, the kinetic energy at the moment of impact is 10 or more joules.
The calorie (cal) is an older energy unit. It is defined as the energy transferred
as heat that is required to raise the temperature of 1.00 g of pure liquid water from
14.5 °C to 15.5 °C. A kilocalorie (kcal) is equivalent to 1000 calories. The conversion factor relating joules and calories is
1 calorie (cal) 5 4.184 joules (J)
The dietary Calorie (with a capital C) is often used in the United States to represent the energy content of foods. The dietary Calorie (Cal) is equivalent to a kilocalorie or 1000 calories. Thus, a breakfast cereal that gives you 100.0 Calories of
nutritional energy per serving provides 100.0 kcal or 418.4 kJ.
Oesper Collection in the
History of Chemistry/
University of Cincinnati
1 kg 5 1000 g and 1 g 5 1000 mg
James Joule The joule is named
for James P. Joule (1818–1889),
the son of a wealthy brewer
in Manchester, England. The
family wealth and a workshop
in the brewery gave Joule the
opportunity to pursue scientific
studies. Among the topics that
Joule studied was whether heat
was a massless fluid. Scientists
at that time referred to this idea
as the caloric hypothesis. Joule’s
careful experiments showed that
heat and mechanical work are
related, providing evidence that
heat is not a fluid.
1R.2 Making Measurements: Precision,
Accuracy, Experimental Error,
and Standard Deviation
Goal for Section 1R.2
•
Recognize and express uncertainties in measurements.
The precision of a measurement indicates how well several determinations of the same
quantity agree. This is illustrated by the results of throwing darts at a target. In Figures
1R.5a and 1R.5b, the dart thrower was not consistent, and the precision of the darts’
placement on the target is low. In Figures 1R.5c and 1R.5d, the darts are clustered together,
indicating much better consistency on the part of the thrower—that is, greater precision.
1R.2 Making Measurements: Precision, Accuracy, Experimental Error, and Standard Deviation
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37
Chemistry in Your Career
Thalia I. Navedo
Thalia I. Navedo
Accuracy and NIST The National
Institute of Standards and
Technology (NIST) is an important
resource for standards and data
used in science. Comparison
with NIST data is a test of the
accuracy of the measurement
(see www.nist.gov).
At first glance, a knowledge of chemistry may not
seem essential to Thalia I. Navedo’s job as a project
accountant in construction design. ­Navedo (she/
her/hers) spends her workday on finance and
­accounting tasks such as project set-up, budgeting,
subcontracts, billing, safety and compliance, and
much more. Yet Navedo, who identifies as ­Hispanic/
Latinx/Black, says that her study of chemistry during
college provided skills that she uses daily in her
work. “Working in finance involves a good amount of
math and problem solving, which were prevalent
all throughout my study of chemistry. Be it dimensional analysis or problem solving, learning chemistry helped me a lot on my analytical skills. . . .
Chemistry taught me how to apply the mathematical knowledge into solving real problems and training my brain muscles.”
In addition to analytical skills, Navedo’s study
of chemistry informs the safety and compliance
aspects of her job as she understands and manages
the impact of chemicals used in her workplace.
Accuracy is the agreement of a measurement with the accepted value of the
quantity. In Figures 1R.5a and 1R.5c, the thrower was not accurate because the average location of the darts is not the bull’s eye. In Figure 1R.5b, the average location
of the darts is the bull’s eye, so the thrower was accurate, even though the results
were not precise. Figure 1R.5d shows the optimal case where the thrower was accurate as well as precise—the average of all shots is close to the targeted position, the
bull’s eye, and the darts are all clustered around this value.
Notice that, as shown by Figure 1R.5c, it is possible to be precise without being
accurate—the thrower has consistently missed the bull’s eye, although all the darts
are clustered precisely around one point on the target. This is analogous to an
experiment with some flaw (either in design or in a measuring device) that causes
all results to differ from the correct value by the same amount.
The accuracy of a result in the laboratory is often expressed in terms of percent
error relative to a standard or accepted value, whereas the precision is expressed as
a standard deviation.
Experimental Error
If you measure a quantity in the laboratory, you may be required to report the error
in the result, the difference between your result and the accepted value,
Error in measurement 5 experimentally determined value − accepted value (1R.2)
or the percent error.
(a) Poor precision and poor
accuracy
(b) Poor precision and good
accuracy
(c) Good precision and poor
accuracy
(d) Good precision and good
accuracy
Figure 1R.5 ​Precision and accuracy.
38
Chapter 1R / Let’s Review: The Tools of Quantitative Chemistry
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error in measurement
3 100%
accepted value
experimentally determined value 2 accepted value
5
accepted value
Percent error 5
3 100% (1R.3)
Exam p le 1R.2
Accuracy and Error
Problem Suppose a coin has a diameter of 28.054 mm. In an experiment, two students
measure this diameter. Student A makes four measurements of the diameter of the coin
using a precision tool called a micrometer. Student B measures the same coin using a
simple plastic ruler. The two students report the following results:
Student A
Student B
28.246 mm
27.9 mm
28.244
28.0
28.246
27.8
28.248
28.1
What is the average diameter and percent error obtained in each case? Which student’s
data are more accurate?
What Do You Know? You know the data collected by the two students and want
to compare them with the actual value by calculating the percent error.
Strategy
Step 1. For each set of values, calculate the average of the four measurements.
Step 2. Calculate the percent error.
Solution
Step 1. Obtain the average for each set of data by summing the four values and dividing
by four, the number of measurements
Average value for Student A 5
28.246 mm 1 28.244 mm 1 28.246 mm 1 28.248 mm
4
5 28.246 mm
Average value for Student B 5
27.9 mm 1 28.0 mm 1 27.8 mm 1 28.1 mm
4
5 27.95 mm 5 28.0 mm
Step 2. Use Equation 1R.3 to calculate the percent error for each student.
Percent error for Student A 5
28.246 mm 2 28.054 mm
3 100% 5 0.684%
28.054 mm
Percent error for Student B 5
27.95 mm 2 28.054 mm
3 100% 5 20.4%
28.054 mm
Student B’s average is more accurate because it is closer to the accepted value, and thus
it has a smaller percent error.
Percent Error Percent error
can be positive or negative,
indicating whether the
experimental value is too high
or too low compared to the
accepted value. In Example 1R.2,
Student B’s percent error is
20.4%, indicating it is 0.4%
lower than the accepted value.
1R.2 Making Measurements: Precision, Accuracy, Experimental Error, and Standard Deviation
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39
Think about Your Answer Student A’s results were less accurate, as shown in
this example. But Student A’s results were more closely grouped around their average and
so were more precise. Possible reasons for the error in Student A’s result are incorrect use
of the micrometer or a flaw in the instrument. The rules for determining how many digits
should be retained in the calculations and answers will be provided later in the chapter.
Check Your Understanding
A student checked the accuracy of two standard top-loading balances by testing them
with a standard 5.000-g mass. The results were as follows:
Balance 1: 4.99 g, 5.04 g, 5.03 g, 5.01 g
Balance 2: 4.97 g, 4.99 g, 4.95 g, 4.96 g
Calculate the average values for balances 1 and 2 and calculate the percent error for each.
Which balance is more accurate?
Standard Deviation
Laboratory measurements can be in error for two basic reasons. First, determinate
errors are errors that can be identified and theoretically avoided or corrected like those
caused by faulty instruments, human errors such as incorrect record keeping, and
errors inherent in the methods used in the experiment such as other reaction products
being formed in a chemical reaction in addition to the desired product. Second, indeterminate (or random) errors arise from uncertainties in a m
­ easurement. One way to
judge the indeterminate error in a result is to calculate the standard deviation.
The standard deviation of a series of measurements is equal to the square root
of the sum of the squares of the deviations for each measurement from the average,
divided by one less than the number of measurements.
∑ ( xi 2 x )
N 21
2
Standard deviation 5
(1R.4)
In this equation, the Greek capital letter sigma, , means to add up, xi is each
individual measurement, x is the average value, and N is the number of measurements. The use of this equation is shown in Example 1R.3.
The standard deviation has a precise statistical significance: Assuming a large
number of measurements is used to calculate the average, slightly more than 68%
of the values collected are expected to be within one standard deviation of the value
determined, and 95% are within two standard deviations.
E xamp le 1R.3
Precision and Standard Deviation
Problem Suppose you carefully measure the mass of water delivered by a 10-mL pipet.
(A pipet containing a green solution is shown on the right in Figure 1R.4.) Your results for
five measurements are 9.990 g, 9.993 g, 9.975 g, 9.980 g, and 9.982 g. Calculate the
­standard deviation in the mass for this series of measurements.
What Do You Know? You know the masses obtained for the five samples and the
equation for determining the standard deviation.
Strategy
Step 1. Calculate the average of the measurements.
Step 2. Determine the deviation of each individual measurement from the average by
subtracting the average from each of the measurements.
40
Chapter 1R / Let’s Review: The Tools of Quantitative Chemistry
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Step 3. Square the individual deviations, and then add these squared values together.
Step 4. Finally, calculate the standard deviation by dividing the sum of the squares by
one less than the number of measurements and taking the square root.
Solution
Step 1. Calculate the average of the measurements.
The average of the measurements is calculated by adding together the measured masses
and dividing by five, the number of measurements.
Average ( x ) 5
9.990 g 1 9.993 g 1 9.975 g 1 9.980 g 1 9.982 g
5 9.984 g
5
Steps 2-3. Determine the deviation of each individual measurement (xi) from the aver_
age (x ), square the individual deviations, and then add together these squared values.
The deviations (difference between each measurement and the average) and squares of
the deviations for each measurement are summarized in the following table.
Determination
Measured
Mass (g)
Deviation 5 xi2 x (g)
Square of
Deviation (g2)
1
9.990
0.006
0.00004
2
9.993
0.009
0.00008
3
9.975
20.009
0.00008
4
9.980
20.004
0.00002
5
9.982
20.002
0.000004
Add together the squares of the deviations.
Sum of the squares of the deviations 5 (0.00004 g2 ) 1 (0.00008 g2 ) 1 (0.00008 g2)
1 (0.00002 g2 ) 1 (0.000004 g2) 5 0.00022 g2
Step 4. Use Equation 1R.4 to calculate the standard deviation.
Standard deviation 5
0.00022 g2
5
521
0.007 g
Think about Your Answer The standard deviation tells you that if this e­ xperiment
were repeated, most of the values would fall in the range from 9.977 g to 9.991 g
(±0.007 g from the average value). Even though standard deviations are often reported,
it is rare for people today to carry out the detailed calculations shown in this example
because many calculators and spreadsheet programs (such as Microsoft Excel or Apple’s
Numbers) have methods to perform the calculation using just a few keystrokes.
Check Your Understanding
A student obtained the following masses for the mass of an object: 4.99 g, 5.04 g, 5.03 g,
and 5.01 g. Determine the standard deviation for these data.
1R.3 Mathematics of Chemistry
Goals for Section 1R.3
•
•
Express and use numbers in exponential or scientific notation.
Report the answer of a calculation to the correct number of significant figures.
1R.3 Mathematics of Chemistry
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41
Exponential or Scientific Notation
The Eiffel Tower, built in 1889, is the tallest building in Paris. It was designed by the
French architect Gustave Eiffel to mark the centennial of the French Revolution. The
Tower, constructed of very pure iron, is as tall as an 81-story building. It was supposed to be dismantled in 1909, but the building still stands as a symbol of Paris.
Some quantitative information on the structure is given in the following table:
RomanSlavik.com/Shutterstock.com
Eiffel Tower Characteristics
Eiffel Tower (Paris, France)
Quantitative Information
Height
324 meters (m)
Mass of iron
7.3 3 106 kilograms (kg)
Volume of iron
930 cubic meters (m3)
Number of iron pieces
1.8 3 104 pieces
Approximate number of visitors annually
7 3 106 people
Some of the data on the Tower are expressed in exponential notation, or scientific
notation, (for example, 7.3 3 106 kilograms). Scientific notation is a way of presenting
very large or very small numbers in a compact and consistent form that simplifies calculations. Because of its convenience, scientific notation is widely used in the sciences.
In scientific notation a number is expressed as the product of two numbers:
N 3 10n. N is the digit term and is a number between 1 and 9.9999. . . . The second
number, 10n, the exponential term, is some integer power of 10. For example, 1234 is
written in scientific notation as 1.234 3 103, or 1.234 multiplied by 10 three times:
1234 5 1.234 3 101 3 101 3 101 5 1.234 3 103
Conversely, a number less than 1, such as 0.01234, is written as 1.234 3 1022. This
notation tells you that 1.234 should be divided twice by 10 to obtain 0.01234:
0.01234 5
1.234
5 1.234 3 1021 3 1021 5 1.234 3 1022
101 3 101
When converting a positive number to scientific notation, notice that the exponent n is positive if the number is greater than 1 and negative if the number is less
than 1. The value of n is the number of places by which the decimal is shifted to
obtain the number in scientific notation; if the decimal is shifted to the left, n is
positive, and if the decimal is shifted to the right, n is negative:
1 2 3 4 5. = 1.2345 × 104
(a) Decimal shifted four places to the left. Therefore, n is positive 4.
Comparing Earth and a Plant Cell—
Powers of 10
Earth
5 12,760,000 meters wide
5 1.276 3 107 meters
Plant cell
5 0.00001276 meters wide
5 1.276 3 1025 meters
0.0 0 1 2 = 1.2 × 10–3
(b) Decimal shifted three places to the right. Therefore, n is negative 3.
If you wish to convert a number in the opposite direction, from scientific notation to one that does not use scientific notation, the procedure is reversed:
6 . 2 7 3 × 102 = 627.3
(a) Decimal point moved two places to the right because n is positive 2.
0 0 6.273 × 10–3 = 0.006273
(b) Decimal point shifted three places to the left because n is negative 3.
42
Chapter 1R / Let’s Review: The Tools of Quantitative Chemistry
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Significant Figures
In most experiments, several kinds of measurements must be made, and some can
be made more precisely than others. A result calculated from experimental data
­cannot be more precise than the least precise piece of information used in the calculation. This is where the rules for significant figures come in. Significant figures
(also called significant digits) are the digits in a measured quantity that are known
exactly plus one inexact digit.
Suppose you place a new U.S. dime on the pan of an analytical laboratory balance
such as the one pictured in Figure 1R.6 and observe a mass of 2.2653 g. This number
has five significant figures or digits because all five numbers are observed. However,
you will learn from experience that the final digit on the right (3) is somewhat uncertain because you may notice the balance readings can change slightly and give
masses of 2.2652, 2.2653, and 2.2654, with the mass of 2.2653 observed most of
the time. In general, in a number representing a scientific measurement, the last digit
to the right is taken to be inexact. The uncertainty in the last digit is best determined
by taking multiple measurements and calculating the standard deviation of the measurements. However, when the standard deviation is not determined experimentally,
it is common practice to assign an uncertainty of ±1 to the last significant figure.
Suppose you want to calculate the density of a piece of metal (Figure 1R.7). The
mass and dimensions were determined by standard laboratory techniques. Most of
these data have two digits to the right of the decimal, but they have different numbers of significant figures.
Data Collected
Mass of metal
13.56 g
4
Length
6.45 cm
3
Width
2.50 cm
3
Thickness
3.1 mm 5 0.31 cm
2
Figure 1R.6 Analytical
laboratory balance and
significant figures. Such
balances can determine the mass
of an object to the nearest tenth
of a milligram.
Significant Figures
The quantity 0.31 cm has two significant figures. The 3 in 0.31 is exactly known, but
the 1 is uncertain. That is, the thickness of the metal piece may have been as small
as 0.30 cm or as large as 0.32 cm. The width of the piece is reported as 2.50 cm,
where 2.5 is known with certainty, but the final 0 is uncertain. There are three significant figures in 2.50.
When you read a number in a problem or collect data in the laboratory (­ Figure 1R.8),
how do you determine which digits are significant?
First, is the number an exact number or a measured quantity? If it is an exact
number, you don’t have to worry about the number of significant figures. For example, there are exactly 100 cm in 1 m. You could add as many zeros after the decimal
place as you want, and the expression would still be true. Using this relationship in a
calculation does not affect how many significant figures you report in your answer.
If, however, the number is a measured value, you must take into account
­significant figures. In the data given in Figure 1R.7, the values 13.56 g and 6.45 cm
contain only nonzero digits, which are always significant in a reported measurement.
Thus, 13.56 g has four significant figures, and 6.45 mm has three. But how many
significant figures are in the values 0.31 cm and 2.50 cm? Are the zeroes significant?
1. Zeroes between two other significant digits are significant. For example, the zero
in 103 is significant.
2. Zeroes to the right of a nonzero number, and also to the right of a decimal place,
are significant. For example, in the number 2.50 cm, the zero is significant.
2.50 cm
13.56 g
6.45 cm
© Charles D. Winters/Cengage
Measurement
© Charles D. Winters/Cengage
Determining Significant Figures
3.1 mm
Figure 1R.7 Data used to
determine the density of a
metal.
1R.3 Mathematics of Chemistry
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43
250-mL flask
contains 250.0
± 0.1 mL when
full to the mark
50-mL buret
marked in 0.10-mL
increments
20-mL pipet
volume known
to the nearest
0.02 mL
Photos: © © Charles D. Winters/Cengage
10-mL graduated
cylinder marked in
0.1-mL increments
The 10-mL graduated cylinder is
marked in 0.1-mL increments; its
contents would normally be estimated
to 0.01 mL. However, graduated
cylinders are not precision glassware.
You can expect no more than 2
significant figures when reading a
volume with this cylinder.
A 50-mL buret is marked in
0.10-mL increments, but it may
be read with greater precision
(0.01 mL).
A volumetric flask is meant to be
filled to the mark on the neck. For a
250-mL flask, the volume is known
to the nearest 0.1 mL, so the flask
contains 250.0 ± 0.1 mL when full to
the mark (four significant figures).
A pipet is like a volumetric flask in
that it is filled to the mark on its
neck. For a 20-mL pipet the
volume is known to the nearest
0.02 mL.
Figure 1R.8 Glassware and significant figures.
Zeroes and Common Laboratory
Mistakes Students often find
the mass of a chemical on a
balance and fail to write down
zeroes that are significant. For
example, if a balance displays
a mass of 2.340 g, the final
zero is significant and must
be reported as part of the
measured value. The number
2.34 g has only three significant
figures and implies the 4 is
uncertain, when in fact the
balance reading indicated the 4
is certain.
3. Zeroes that are placeholders are not significant. There are two types of numbers
that fall under this rule.
(a) The first are decimal numbers with zeroes that occur to the left of the first
nonzero digit. For example, in 0.0013, only the 1 and the 3 are significant;
the zeroes are not. This number has two significant figures. Similarly, 0.31 cm
has only two significant figures.
(b) The second are numbers with trailing zeroes that must be there to indicate the
magnitude of the number. For example, the zeroes in the number 13,000
may or may not be significant; it depends on whether they were measured
or not. To avoid confusion with regard to such numbers, assume in this
book that trailing zeroes are significant when there is a decimal point to the
right of the last zero. Thus, you would say that 13,000 has only two significant figures but that 13,000. has five. The best way to be unambiguous when
writing numbers with trailing zeroes is to use scientific notation. For
example 1.300 3 104 indicates four significant figures, whereas 1.3 3 104
indicates two.
Using Significant Figures in Calculations
When doing calculations using measured quantities, you need to follow some basic
rules so that the results reflect the precision of all the measurements that go into the
calculations. The rules used for significant figures in this book are as follows:
Rule 1. When adding or subtracting numbers, the number of decimal places in the
answer is equal to the number of decimal places in the number with the fewest
digits after the decimal.
0.12
2 decimal places
11.9
1 decimal place
110.925
3 decimal places
12.945
answer on calculator
The sum should be reported as 12.9, a number with one decimal place, because 1.9
has only one decimal place.
44
Chapter 1R / Let’s Review: The Tools of Quantitative Chemistry
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Problem Solving Tip 1R.1 Using Your Calculator
You will be performing a number of
calculations in general chemistry,
most of them using a calculator.
Many different types of ­calculators
are available, so be sure to consult
your calculator manual for ­specific
instructions to enter scientific
­notation and to find powers and
roots of numbers. You do not want to
get questions incorrect because you
­performed the wrong operations on
your calculator!
To make sure you are using your
calculator correctly, try these sample
calculations and verify that you obtain
the correct answers:
1. (6.02 3 1023)(2.26 3 1025)/367
(Answer 5 3.71 3 1016)
2. (4.32 3 1023)3
(Answer 5 8.06 3 1028)
3. (4.32 3 1023)1/3
(Answer 5 0.163)
Rule 2. In multiplication or division, the number of significant figures in the answer
is determined by the value with the fewest significant figures.
0.01208
5 0.511864. . . (answer on calculator)
0.0236
Because 0.0236 has only three significant digits, while 0.01208 has four, the answer
should have three significant digits and be reported as 0.512, or in scientific notation as 5.12 3 1021.
Rule 3. When a number is rounded off, the last digit retained is increased by one
(rounded up) only if the following digit is 5 or greater.
Full Number
Number Rounded to Three Significant Digits
12.696
12.7
16.349
16.3
18.35
18.4
18.351
18.4
Now you can apply these rules to calculate the density of the piece of metal in
Figure 1R.7.
Length 3 width 3 thickness 5 volume
mass (g)
volume (cm3)
13.56 g
5
5 2.7 g/cm3
6.45 cm 3 2.50 cm 3 0.31 cm
Density 5
The calculated density has two significant figures because the thickness has only two
significant figures. Remember that a result calculated using only multiplication and
division is reported to the same number of significant figures as the value with the
fewest significant figures.
One last word on significant figures and calculations: When working problems,
you should do the calculation with all the digits allowed by your calculator and
round off only at the end of the calculation. Rounding off in the middle of a calculation can introduce errors. In Example problems in this book, the answer to each
intermediate step is given to the correct number of significant figures plus one extra
digit for that step, so that round-off errors do not propagate in the significant figures.
The final answers to numerical problems result from retaining several digits more
than the number required by the rules of significant figures and rounding to the
correct number of significant figures only at the end.
Who Is Right—You or the Book?
If your answer to a problem in
this book does not quite agree
with the answer in Appendix N,
check if you rounded the answer
after each step and then used
that rounded answer in the
next step. If so, performing the
calculation without rounding
between steps should resolve the
discrepancy.
1R.3 Mathematics of Chemistry
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45
E xamp le 1R.4
Using Significant Figures
Problem An example of a calculation you will do later in the book (Chapter 10) is
( 0.120 ) ( 0.08206 ) ( 273.15 1 5 )
Volume of gas (L) 5
( 230/760.0 )
Calculate the final answer to the correct number of significant figures.
What Do You Know? You know the rules for determining the number of significant figures for each number in the equation.
Strategy First decide on the number of significant figures represented by each n­ umber
and then apply Rules 123 (pages 44–45).
Solution
Number
Number of
Significant
Figures
Comments
0.120
3
The final 0 on the right is significant.
0.08206
4
Neither the 0 to the left of the decimal
place nor the first 0 to the immediate right
of the decimal is significant.
273.15 1 5 5 278
3
5 has no decimal places, so the sum
cannot either.
230/760.0 5 0.30
2
230 has two significant figures because the
trailing zero is not significant. In contrast,
there is a decimal point in 760.0, so there
are four significant digits. The quotient will
have only two significant digits.
The calculation gives 9.0506. . . L. However, an analysis of significant figures shows that one
of the pieces of information involved in multiplication and division is known to only two
significant figures. Therefore, you should report the volume of gas as 9.1 L , a number with
two significant figures.
Think about Your Answer Be especially careful when you add or subtract two
numbers because it is easy to make significant figure errors when doing so. Notice that in
the addition portion of this calculation (273.15 1 5 5 278) the sum has three significant
figures.
Check Your Understanding
What is the result of the following calculation?
x5
46
(110.7 2 64)
(0.056)(0.00216)
Chapter 1R / Let’s Review: The Tools of Quantitative Chemistry
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1R.4 Problem Solving by Dimensional Analysis
Goal for Section 1R.4
•
Solve problems using dimensional analysis.
Figure1R.7 illustrated the data that were collected to determine the density of a piece
of metal. The thickness was measured in millimeters, whereas the length and width
were measured in centimeters. To find the volume of the sample in cubic centimeters, the length, width, and thickness must all be in the same unit, centimeters, so
the thickness was converted to centimeters.
3.1 mm 3
1 cm
5 0.31 cm
10 mm
Here, the thickness in millimeters (3.1 mm) was multiplied by a conversion factor
(1 cm/10 mm) to produce the result in the desired unit (0.31 cm). Notice that units
are treated like numbers. Because the unit mm is in both the numerator and the
denominator, these units are said to cancel. This leaves the answer in centimeters,
the desired unit.
This approach to problem solving is often called dimensional analysis (or
sometimes the factor-label method). It is a general problem-solving approach that
uses the dimensions or units of each value as a guide for calculations.
A conversion factor expresses the equivalence of a measurement in two different
units (1 cm 5 10 mm; 1 g 5 1000 mg; 12 eggs 5 1 dozen; 12 inches 5 1 foot).
Because the numerator and the denominator describe the same quantity, the conversion factor is equivalent to the number 1. Therefore, multiplication by this factor does
not change the measured quantity, only its units. A conversion factor is always written
so that the original units cancel, leaving the answer expressed in the new units.
Number in original unit
Quantity to
express in
new units
new unit
= new number in new unit
original unit
Conversion factor
Quantity now
expressed in new
units
Using Conversion Factors and
Doing Calculations As you
work problems in this book
and read Example problems,
notice that proceeding from
given information to an answer
very often involves a series of
multiplications. That is, you
multiply the given data by a
conversion factor, multiply the
answer of that step by another
factor, and so on, to get the
answer.
Exam p le 1R.5
Using Conversion Factors and Dimensional
Analysis
Problem Oceanographers often express the density of sea water in units of kilograms
per cubic meter. If the density of sea water is 1.025 g/cm3 at 15 ° C, what is its density in
kilograms per cubic meter?
What Do You Know? You know the density in a unit involving mass in grams and
volume in cubic centimeters. These have to be changed to their equivalents in kilograms
and cubic meters, respectively.
Strategy To simplify this problem, break it into three steps.
Step 1. Convert the mass in grams to kilograms.
Step 2. Convert the volume in cubic centimeters to cubic meters.
Step 3. Calculate the density by dividing the mass in kilograms by the volume in cubic
meters.
1R.4 Problem Solving by Dimensional Analysis
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47
Solution
Step 1. Convert the mass in grams to a mass in kilograms.
1.025 g 3
1 kg
5 1.025 3 1023 kg
1000 g
The given information is known to four significant figures. The conversion factor is an exact
number, so using it will not affect the number of significant figures.
Step 2. Convert the volume in cm3 to m3. A conversion factor to directly change units of
cubic centimeters to cubic meters is not available in the tables in this book. You can find
one, however, by cubing (raising to the third power) the relation between the meter and
the centimeter:
3


 1m 
1 m3
1 cm3 3 
5 1 cm3 3 
5 1 3 1026 m3

6
3
 100 cm 
 1 3 10 cm 
This conversion involves only numbers that are known exactly, so you don’t need to worry
about significant figures for this step. You now know that 1 cm3 is equivalent to 1 3 1026 m3.
Step 3. Calculate the density of sea water.
Density 5
1.025 3 1023 kg
5 1025 kg/m3
1 3 1026 m3
Think about Your Answer The number of significant figures reported for the final answer is determined by the given information, 1.025 g, which has four significant figures. The final answer therefore has four significant figures. Densities in units of kg/m3 can
often be large numbers. For example, the density of platinum is 21,450 kg/m3, and dry air
has a density of 1.204 kg/m3.
Check Your Understanding
The density of gold is 19,320 kg/m3. What is this density in g/cm3?
1R.5 Graphs and Graphing
Goals for Section 1R.5
•
•
Read information from graphs.
Prepare and interpret graphs of numerical information, and, if a graph produces a
straight line, find the slope and equation of the line.
In a number of instances in this text, graphs are used when analyzing experimental
data to obtain a mathematical equation that may help predict new results. The procedure used will often result in a straight line, which has the equation
y 5 mx 1 b
In this equation, y is usually referred to as the dependent variable; its value is determined from (that is, is dependent on) the values of x, m, and b. In this equation, x is
called the independent variable, and m is the slope of the line. The parameter b is
the y-intercept—that is, the value of y when x 5 0. The following example will show
two things: (1) how to construct a graph from a set of data points and (2) how to
derive an equation for the line generated by the data.
Figure 1R.9 includes the set of data points to be graphed. First, mark off each
axis in increments of the values of x and y. Here, the x-data are within the range from
48
Chapter 1R / Let’s Review: The Tools of Quantitative Chemistry
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3
Experimental data
2.5
y-data
2
x
3.35
2.59
1.08
−1.19
y-intercept, where x = 0
and y = 1.87
y
0.0565
0.520
1.38
2.45
Figure 1R.9 Plotting data.
Using Microsoft Excel with
these data and doing a linear
regression analysis gives the
equation for the line:
y 5 20.525x 1 1.87.
1.5
1
x = 2.00, y = 0.82
0.5
Using the points marked with a square,
the slope of the line is:
Slope =
∆y
0.82 − 1.87
=
= −0.525
∆x
2.00 − 0.00
0
−2
−1
0
1
x-data
2
3
4
22 to 4, so the x-axis is marked off in increments of 1 unit. The y-data fall within
the range from 0 to 2.5, so the y-axis is marked off in increments of 0.5. Each data
point is marked as a circle on the graph.
After plotting the points on the graph (round circles), draw a straight line that
comes as close as possible to representing the trend in the data. (Do not just connect
the dots!) Because there is always some inaccuracy in experimental data, the straight
line that should be drawn is unlikely to pass exactly through every point.
To identify the specific equation corresponding to the data, the y-intercept (b) and
slope (m) for the equation y 5 mx 1 b must be determined. The y-intercept is the
point at which x 5 0 and thus is the point where the line intersects the y-axis. The
slope is determined by selecting two points on the line (marked with squares in
Figure 1R.9) and calculating the difference in values of y (∆y 5 y2 2 y1) and
­
x (∆x 5 x 2 2 x 1). The slope of the line is then the ratio of these differences, m 5 ∆y/∆x.
With the slope and intercept now known, you can write the equation for the line
y 5 20.525x 1 1.87
Determining the Slope and
y-Intercept with a Computer
Program—Least-Squares
Analysis Generally, the easiest
method of determining the slope
and y-intercept of a straight line
(and thus the line’s equation)
is to use a program such as
Microsoft Excel or Apple’s
Numbers. These programs
perform a least-squares or linear
regression analysis and give
the best straight line based on
the data. (This line is referred
to in Excel and Numbers as a
trendline.)
and you can use this equation to calculate y-values for points that are not part of
the original set of x2y data. For example, when x 5 1.50, you can calculate that
y 5 20.525(1.50) 1 1.87 5 1.08.
1R.6 Problem Solving and Chemical Arithmetic
Goals for Section 1R.6
•
•
Solve problems using a systematic approach.
Incorporate quantitative information into an algebraic expression and solve that
expression.
Many aspects of chemistry involve analyzing quantitative information, so problem
solving will be important in your success. However, as in anything you do, careful
planning is important, and students usually find it helpful to follow a definite plan
as illustrated in all of the examples in the book.
1R.6 Problem Solving and Chemical Arithmetic
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49
Step 1 State the Problem. Read it carefully—and then read it again.
Step 2 What Do You Know? Determine specifically what you are trying to calculate or
conclude and what information you are given. What key principles are involved?
What information is known or not known? What information might be present just
for context? Organize the information to see what is required and discover the relationships among the data given. Try writing the information down in table form. If
the information is numerical, be sure to include units.
Strategy Maps A number of
the Example problems in this
book are accompanied by a
Strategy Map (for instance,
Example 1R.6) that outlines a
route to a solution.
Step 3 Strategy. One of the greatest difficulties for a student in introductory chemistry is picturing what is being asked for. Try sketching a picture of the situation involved. For example, we sketched a picture of the piece of metal whose density we
wanted to calculate and put the dimensions on the drawing (Figure 1R.7).
Develop a plan. Have you done a problem of this type before? If not, perhaps
the problem is a combination of several simpler ones you have seen before. Break it
down into those simpler components. Try reasoning backward from the units of the
answer. What data do you need to find an answer in those units? Drawing a strategy
map may help you plan how to solve the problem.
Step 4 Solution. Execute the plan. Carefully write down each step of the problem, being sure to keep track of the units on each number. (Do the units cancel to give you the
answer in the desired units?) Don’t skip steps. Don’t do anything except the simplest
steps in your head. Students often say they got a problem wrong because they “made
a stupid mistake.” Your instructors—and book authors—also make them, and it is usually because they don’t take the time to write down the steps of the problem clearly.
Step 5 Think about Your Answer. Ask yourself whether the answer is reasonable and
if you obtained an answer in the correct units.
Step 6 Check Your Understanding. In this text, each Example is followed by another
problem for you to try. (The solutions to those questions are given by chapter in
Appendix N.) When doing homework Study Questions, try the Practicing Skills questions to see if you understand the basic ideas.
These steps for problem solving are ones that many students have found to be
successful, so try to conscientiously follow this scheme. But also be flexible. The “What
Do You Know?” and “Strategy” steps often blend into a single set of ideas.
Exam pl e 1 R.6
Strategy Map
Problem
How thick will an oil layer be
when a given mass covers a
given area?
Problem Solving
Problem A mineral oil has a density of 0.875 g/cm3. Suppose you spread 0.75 g of this
oil over the surface of water in a large circular dish with an inner diameter of 21.6 cm. How
thick is the oil layer? Express the thickness in centimeters.
What Do You Know? You know the mass and density of the oil and the diameter
of the surface to be covered.
Strategy It is often useful to begin solving such problems by sketching a picture of the
situation.
21.6 cm
50
Chapter 1R / Let’s Review: The Tools of Quantitative Chemistry
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This helps you recognize the relationships between the key quantities (volume, area, and
thickness) in this problem. If you know the volume and the area covered by the oil, then
you can find the thickness because
Volume of oil layer 5 (thickness of layer) 3 (area of oil layer)
So, you need two things: (1) the volume of the oil layer and (2) the area of the layer.
Step 1. Calculate the volume of the oil using the mass and density of the oil.
Data/Information
Mass and density of the oil
and diameter of the circular
surface to be covered.
Step 2. Calculate the area covered by the oil.
The area can be found because the oil forms a circle, which has an area equal to π 3 r2
(where r is the radius of the dish).
Step 3. Calculate the thickness by dividing the volume by the surface area.
Solution
Step 1
Calculate the volume of the oil.
The mass of the oil layer is known, so combining the mass of oil with its density gives the
volume of the oil used:
0.75 g 3
Step 2
1 cm3
5 0.857 cm3
0.875 g
Calculate the area covered by the oil.
The oil is spread over a circular surface, whose area is given by
Area 5 π 3 (radius)2
The diameter of the oil layer is 21.6 cm. The radius is half the diameter, 10.8 cm, so
Area of oil layer 5 (π)(10.8 cm)2 5 366.4 cm2
Step 3
Calculate the thickness of the oil layer by dividing the volume by the area.
Thickness 5
0.857 cm3
volume
5
5 0.0023 cm
area
366.4 cm2
Think about Your Answer In the volume calculation, the calculator shows
0.857143. . . . The quotient should have two significant figures because 0.75 has two significant figures, so the result of this step is reported as 0.857 cm3, containing one extra digit. In
the area calculation, the calculator shows 366.435. . . . The answer to this step should have
three significant figures because 10.8 has three; again, this value is reported to one extra
digit. When these interim results are combined to calculate the thickness, the final result can
have only two significant figures. Remember that premature rounding can lead to errors.
Check Your Understanding
A particular paint has a density of 0.914 g/cm3. You need to cover a wall that is 7.6 m long and 2.74 m high with a paint layer 0.13
mm thick. What volume of paint (in liters) is required? What is the mass (in grams) of the paint layer?
Applying Chemical Principles
1R.1 Out of Gas!
On July 23, 1983, a new Boeing 767 jet aircraft was flying at
26,000 feet from Montreal to Edmonton as Air Canada Flight
143. Warning buzzers sounded in the cockpit. One of the world’s
largest planes was now a glider—the plane had run out of fuel!
How did this happen? A simple mistake had been made in
calculating the amount of fuel required for the flight because
of a mixup of units of measurement!
Applying Chemical Principles
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51
© Wayne Glowacki/Winnipeg Free Press
The Gimli glider. After running out of fuel, Air Canada Flight 143
glided 29 minutes before landing on an abandoned airstrip at
Gimli, Manitoba, near Winnipeg.
Like all Boeing 767s, this plane had a sophisticated fuel
gauge, but it was not working properly. The plane was still allowed to fly, however, because there is an alternative method of
determining the quantity of fuel in the tanks. Mechanics can
use a stick, much like the oil dipstick in an automobile engine,
to measure the fuel level in each of the three tanks. The mechanics in Montreal read the dipsticks, which were calibrated
in centimeters, and translated those readings to a volume in
liters. According to this, the plane had a total of 7682 L of fuel.
Pilots always calculate fuel quantities in units of mass because they need to know the total mass of the plane before
take-off. Air Canada pilots had always calculated the quantity
of fuel in pounds, but the new 767’s fuel consumption was
given in kilograms. The pilots knew that 22,300 kg of fuel was
required for the trip. If 7682 L of fuel remained in the tanks,
how much had to be added? This involved using the fuel’s
density to convert 7682 L to a mass in kilograms. The mass of
fuel to be added could then be calculated, and that mass converted to a volume of fuel to be added.
The First Officer of the plane asked a mechanic for the conversion factor to do the volume-to-mass conversion, and the mechanic replied “1.77.’’ Using that number, the First Officer and
the mechanics calculated that 4917 L of fuel should be added.
But later calculations showed that this is only about one fourth
of the required amount of fuel! Why? Because no one thought
about the units of the number 1.77. They realized later that
1.77 has units of pounds per liter and not kilograms per liter.
Out of fuel, the plane could not make it to Winnipeg, so controllers directed them to the town of Gimli and to a small airport
abandoned by the Royal Canadian Air Force. After gliding for
almost 30 minutes, the plane approached the Gimli runway. The
runway, however, had been converted to a race course for cars,
and a race was underway. Furthermore, a steel barrier had been
erected across the runway. Nonetheless, the pilot managed to
touch down very near the end of the runway. The plane sped
down the concrete strip; the nose wheel collapsed; several tires
blew—and the plane skidded safely to a stop just before the
barrier. The Gimli glider had made it! And somewhere an aircraft
mechanic is paying more attention to units on numbers.
Questions
1. What is the fuel density in units of kg/L? (1 pound 5 453.6 g)
2. What mass and what volume of fuel should have been loaded?
Have you ever noticed that there are many ties in swimming
competitions? For example, in the 2016 Summer Olympics,
there was a two-way tie for the gold medal in the women’s
100-m freestyle and a three-way tie for the silver medal in the
men’s 100-m butterfly. Olympic competitions are timed to one
hundredth of a second. You might wonder why the officials
don’t simply time the events out to a thousandth of a second,
something that is technologically feasible and done in some
sports, and eliminate most of these ties. The reason relates to
the topic of how many digits in a swimming competition are
really significant.
Consider a 50-m Olympic-sized swimming pool and a 50-m
freestyle swimming contest. The current world record of 20.91
seconds for this event was set by César Cielo of Brazil in 2009.
Assuming a person is swimming at this rate, the maximum distance traveled in one thousandth of a second is 2.4 mm. The
problem arises with the necessary specifications in the dimensions of the pool. There will always be some variation in the
lengths of the different lanes due to limitations in the construction of pools. Current specifications allow a lane to be up to 3
cm longer than the stated length of 50.00 m. It would thus not
be fair to penalize a swimmer in a lane that could be 3 cm
longer for a difference in time that would amount to at most 2.4
mm, and so timing out to thousandths of a second is not done.
52
Richard Heathcote/Getty Images Sport/Getty Images
1R.2 Ties in Swimming and Significant Figures
A Tie for Gold. Simone Manuel and Penny Oleksiak tie for gold
in the 100-m freestyle event at the 2016 Summer Olympics held
in Rio de Janeiro, Brazil.
Questions
1. Confirm that a person swimming at the world record rate for
the 50-m freestyle would travel 2.4 mm in one thousandth
of a second.
2. At this world record rate, how long would it take for a
swimmer to travel 3.0 cm?
3. Consider a lane that is 3 cm longer than the stated 50.00 m.
What is the percent error in this lane’s length?
Chapter 1R / Let’s Review: The Tools of Quantitative Chemistry
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Think–Pair–Share
1. An antacid tablet is known to contain 1.000 g of calcium
carbonate. After analyzing four tablets each, three students
report the following results:
Student A: average = 0.984 g, standard deviation = 0.005 g
Student B: average = 0.975 g, standard deviation = 0.003 g
Student C: average = 0.996 g, standard deviation = 0.008 g
Which student had the most accurate results? The least
accurate results? The most precise results? The least precise
results? Explain your answers.
2. Consider the following three measurements: 150 mL,
150. mL, and 150.0 mL.
(a) How many significant figures are present in each
measurement?
(b) Given the general guideline that in many measurements,
there is an uncertainty of ±1 in the last significant digit, give
the range of values represented by each of the measurements.
3. Carry out the following analysis.
(a) Write down the measurements 4.8 m, 4.578 m, and 3.24 m.
Underline the inexact digit in each case.
(b) Add the three measurements and report the answer,
keeping all digits that would be displayed on a calculator.
Underline any digits that have some uncertainty.
(c) Based on the principle that a reported number should contain
all digits that are known exactly and one digit that is inexact,
what answer should be reported for this calculation? How does
this answer correspond to the answer required by the addition/subtraction rule for reporting significant figures requires?
4. The density of silver is 10.5 g/cm3. You have a sample of
silver that has a mass of 38.7 g.
(a) Solve the equation d = m/V for volume (V). Substitute the
given data into that equation and calculate the volume in cm3.
(b) In an alternative approach, construct a conversion factor
that could be used to convert from the mass to the volume
and carry out this calculation.
(c) Which method do you prefer and why? (Note that some
students may prefer one method and other students may
prefer the other.)
Chapter Goals Revisited
The goals for this chapter are keyed to specific Study Questions to help you
organize your review.
1R.1 Units of Measurement
•
Use the common units for measurements in chemistry and make unit
conversions (such as from liters to milliliters). 1–12, 19–22, 39–41.
1R.2 Making Measurements: Precision, Accuracy,
Experimental Error, and Standard Deviation
•
Recognize and express uncertainties in measurements. 23, 24, 50, 62, 64, 68, 73.
1R.3 Mathematics of Chemistry
•
Express and use numbers in exponential or scientific notation. 25–28.
•
Report the answer of a calculation to the correct number of significant
figures. 29, 30.
1R.4 Problem Solving by Dimensional Analysis
•
Solve problems using dimensional analysis. 13–18, 43, 44, 59.
1R.5 Graphs and Graphing
•
Read information from graphs. 32, 33.
•
Prepare and interpret graphs of numerical information, and, if a graph
produces a straight line, find the slope and equation of the line. 31, 34, 71, 72.
1R.6 Problem Solving and Chemical Arithmetic
•
Solve problems using a systematic approach. 42, 48, 51, 55–58, 60, 65, 67.
•
Incorporate quantitative information into an algebraic expression and
solve that expression. 35–38.
Chapter Goals Revisited
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53
Key Equations
Equation 1R.1 (page 33) ​Converting a temperature from ° C to K.
T (K) 5
1K
(T °C 1 273.15 °C)
1 °C
Equation 1R.2 (page 38) Error in measurement.
Error in measurement 5 experimentally determined value 2 accepted value
Equation 1R.3 (page 39) Percent error.
error in measurement
3 100%
accepted value
experimentally determined value 2 accepted value
5
accepted value
Percent error 5
3 100%
Equation 1R.4 (page 40) Standard deviation.
∑ ( xi 2 x )
N 21
2
Standard deviation 5
Study Questions
▲ denotes challenging questions. Blue-numbered questions have answers in Appendix N and fully worked
solutions in the Student Solutions Manual.
Practicing Skills
Temperature Scales
1. Many laboratories use 25 °C as a standard temperature. What is this temperature in kelvins?
2. The temperature on the surface of the Sun is
5.5 3 103 °C. What is this temperature in kelvins?
3. Make the following temperature conversions:
°CK
(a) 16
(b) 370
(c) 40
4. Make the following temperature conversions:
°CK
(a) 77
(b) 63
(c) 1450
Length, Volume, Mass, and Density
(See Example 1R.1.)
5. A marathon distance race covers a distance of 42.195
km. What is this distance in meters? In miles?
6. The average pencil, new and unused, is 19 cm long.
What is its length in millimeters? In meters?
54
7. A standard U.S. postage stamp is 2.5 cm long and
2.1 cm wide. What is the area of the stamp in
square centimeters? In square meters?
8. A particular cat food can’s lid has a diameter of
8.3 cm. What is the surface area of this lid in
square centimeters? In square meters? [Area of a
circle 5 (π)(radius)2]
9. A typical laboratory beaker has a volume of
250. mL. What is its volume in cubic centimeters?
In liters? In cubic meters? In cubic decimeters?
10. Some soft drinks are sold in bottles with a volume
of 1.5 L. What is this volume in milliliters? In cubic
centimeters? In cubic decimeters?
11. A book has a mass of 2.52 kg. What is this mass in
grams?
12. A new U.S. dime has a mass of 2.265 g. What is its
mass in kilograms? In milligrams?
13. Ethylene glycol, C2H6O2, is an ingredient of automobile antifreeze. Its density is 1.11 g/cm3 at
20 °C. If you need 500. mL of this liquid, what
mass of the compound, in grams, is required?
14. Acetone is a compound used in many nail polish
removers. Its density is 0.7845 g/cm3 at 25 °C.
What mass of acetone is present in a sample with a
volume of 100. mL?
Chapter 1R / Let’s Review: The Tools of Quantitative Chemistry
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15. A piece of silver metal has a mass of 2.365 g. If the
density of silver is 10.5 g/cm3, what is the volume
of the silver?
16. Lead has a density of 11.35 g/cm3 at 20 °C. What
volume of lead is present in a sample with a mass
of 50.0 g?
17. You can identify a metal by carefully determining its
density (d). An unknown piece of metal, with a mass
of 2.361 g, is 2.35 cm long, 1.34 cm wide, and 1.05
mm thick. Which of the following is the element?
(a) nickel, d 5 8.91 g/cm3
(b) titanium, d 5 4.50 g/cm3
(c) zinc, d 5 7.14 g/cm3
(d) tin, d 5 7.23 g/cm3
18. Which occupies a larger volume, 600 g of water
(with a density of 0.995 g/cm3) or 600 g of lead
(with a density of 11.35 g/cm3)?
Energy Units
19. You are on a diet that calls for eating no more than
1200 Cal/day. What is this energy in joules?
20. A 2-inch piece of chocolate cake with frosting
provides 1670 kJ of energy. What is this in dietary
Calories (Cal)?
21. One food product has an energy content of 170 kcal
per serving, and another has 280 kJ per serving.
Which food provides the greater energy per serving?
22. A can of soft drink (335 mL) provides 130 Calories.
A bottle of mixed berry juice (295 mL) provides
630 kJ. Which provides the greater total energy?
Which provides the greater energy per milliliter?
Accuracy, Precision, Error, and Standard Deviation
(See Examples 1R.2 and 1R.3.)
23. You and your lab partner are asked to determine the
density of an aluminum bar. The mass is known accurately (to four significant figures). You use a simple
metric ruler to measure its dimensions and obtain the
results for Method A. Your partner uses a precision
micrometer and obtains the results for Method B.
Method A ( g/cm3)
Method B (g/cm3)
2.2
2.703
2.3
2.701
2.7
2.705
2.4
2.703
The accepted density of aluminum is 2.702 g/cm3.
(a) Calculate the average density for each method.
(b) Calculate the percent error for each method’s
average value.
(c) Calculate the standard deviation for each set of
data.
(d) Which method’s average value is more precise?
Explain your answer. Which method is more
accurate? Explain your answer.
24. The accepted value of the mass of aspirin in a tablet
of aspirin is 325 mg. Trying to verify that value, you
obtain 321 mg, 326 mg, 324 mg, and 329 mg in
four separate trials. Your partner measures 327 mg,
329 mg, 328 mg, and 326 mg.
(a) Calculate the average value and percent error
for your data and your partner’s data.
(b) Which of you is more accurate? Explain your
answer.
(c) Without doing any calculations, predict whose
set of data is more precise? Explain your
answer.
(d) Calculate the standard deviation for your data
and your partner’s data. Do these values
confirm or disprove your prediction in part (c)?
Exponential Notation and Significant Figures
(See Example 1R.4.)
25. Express the following numbers in exponential or
scientific notation, and give the number of significant figures in each.
(a) 0.0830 g
(c) 0.00602 g
(b) 136 g
(d) 3000 mL
26. Express the following numbers in exponential or
scientific notation, and give the number of significant figures in each.
(a) 1356 mL
(c) 250.0 g
(b) 0.03042 L
(d) 120 g
27. Express the following numbers without using
­scientific notation (for example 1.23 3 102 5 123),
and give the number of significant figures in each.
(a) 5.43 3 102 m
(c) 6.20 3 1024 L
(b) 4.306 3 1022 L
(d) 8.42 3 103 mL
28. Express the following numbers without using scientific notation (for example, 1.23 3 102 5 123), and
give the number of significant figures in each.
(a) 3.25 3 1022 m
(c) 4.2 3 103 mL
(b) 4.02 3 1023 mL
(d) 9.305 3 104 g
29. Carry out the following operations. Provide the
answer with the correct number of significant figures.
(a) (1.52)(6.21 3 1023)
(b) (6.217 3 103)2(5.23 3 102)
(c) (6.217 3 103) ÷ (5.23 3 102)
 7.779 
(d) (0.0546)(16.0000) 
 55.85 
Study Questions
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55
30. Carry out the following operations. Provide the
answer with the correct number of significant figures.
(a) (6.25 3 102)3
(b) 2.35 3 1023
(c) (2.35 3 1023)1/3
23.56 2 2.3 
(d) (1.68) 
 1.248 3 103 


33. Use the graph below to answer the following questions.
(a) Derive the equation for the straight line,
y 5 mx 1 b.
(b) What is the value of y when x 5 6.0?
25.00
20.00
Graphing
(See Section 1R.5)
Number of Kernels
Mass (g)
5
0.836
12
2.162
35
5.801
15.00
y values
31. To determine the average mass of a popcorn kernel,
you collect the following data:
10.00
5.00
Plot the data with number of kernels on the x-axis
and mass on the y-axis. Draw the best straight
line using the points on the graph (or do a leastsquares or linear regression analysis using a computer program), and then write the equation for
the resulting straight line. What is the slope of the
line? What does the slope of the line signify about
the mass of a popcorn kernel? What is the mass of
20 popcorn kernels? How many kernels are there in
a handful of popcorn with a mass of 20.88 g?
0
6.00
y values
5.00
4.00
3.00
4.00
5.00
34. The following data were collected in an experiment
to determine how an enzyme works in a biochemical reaction.
Amount of H2O2
Reaction Speed
(amount/second)
1.96
4.75 3 1025
1.31
4.03 3 1025
0.98
3.51 3 1025
0.65
2.52 3 1025
0.33
1.44 3 1025
0.16
0.585 3 1025
Solving Equations
3.00
35. Solve the following equation for the unknown
value, C.
2.00
(0.502)(123) 5 (750.)C
1.00
0
0.10
0.20
0.30
x values
56
2.00
(a) Plot these data as 1/amount on the x-axis and
1/speed on the y-axis. Draw the best straight
line to fit these data points.
(b) Determine the equation for the data, and give
the values of the y-intercept and the slope.
(Note: In biochemistry this is known as a
­Lineweaver–Burk plot, and the y-intercept is
related to the maximum speed of the reaction.)
7.00
0
1.00
x values
32. Use the following graph to answer the following
questions:
(a) What is the value of x when y 5 4.0?
(b) What is the value of y when x 5 0.30?
(c) What are the slope and the y-intercept of the line?
(d) What is the value of y when x 5 1.0?
8.00
0
0.40
0.50
36. Solve the following equation for the unknown
value, n.
(1.0)(22.4) 5 n(0.082057)(273.15)
Chapter 1R / Let’s Review: The Tools of Quantitative Chemistry
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
37. Solve the following equation for the unknown
value, T.
(4.184)(244)(T 2 292.0)
1 (0.449)(88.5)(T 2 369.0) 5 0
38. Solve the following equation for the unknown
value, n.
1
1
2246.0 5 1312  2 2 2 
n 
2
General Questions
These questions are not designated as to type or location in
the chapter. They may combine several concepts.
43. The anesthetic procaine hydrochloride is often
used to deaden pain during dental surgery. The
compound is packaged as a 10.% solution (by
mass; d 5 1.0 g/mL) in water. If your dentist injects
0.50 mL of the solution, what mass of procaine
hydrochloride (in milligrams) is injected?
44. You need a cube of aluminum with a mass of 7.6 g.
What must be the length of the cube’s edge (in cm)?
(The density of aluminum is 2.698 g/cm3.)
45. You have a 250.0-mL graduated cylinder containing
some water. You drop three marbles with a total
mass of 95.2 g into the water. What is the average
density of a marble?
H3N
NH3
Pt
1.97Å
Cl
© Charles D. Winters/Cengage
39. Molecular distances are usually given in nanometers
(1 nm 5 1 3 1029 m) or in picometers (1 pm 5
1 3 10212 m). However, the angstrom (Å) unit is
sometimes used, where 1 Å 5 1 3 10210 m. (The
angstrom unit is not an SI unit.) If the distance
between the Pt atom and the N atom in the cancer
chemotherapy drug cisplatin is 1.97 Å, what is this
distance in nanometers? In picometers?
Cl
Cisplatin
40. The separation between carbon atoms in diamond
is 0.154 nm. What is their separation in meters? In
picometers (pm)? In Angstroms (Å)?
0.154 nm
A portion of the diamond structure
(a)
(b)
Determining density. (a) A graduated cylinder with 61 mL
of water. (b) Three marbles are added to the cylinder.
46. You have a white crystalline solid, known to be one
of the potassium compounds listed below. To determine which, you measure its density. You measure
out 18.82 g and transfer it to a graduated cylinder
containing kerosene (in which these compounds will
not dissolve). The level of liquid kerosene rises from
8.5 mL to 15.3 mL. Calculate the density of the solid,
and identify the compound from the following list.
(a) KF, d 5 2.48 g/cm3
(b) KCl, d 5 1.98 g/cm3
(c) KBr, d 5 2.75 g/cm3
(d) KI, d 5 3.13 g/cm3
47. ▲ The smallest repeating unit of a crystal of
common salt is a cube (called a unit cell) with an
edge length of 0.563 nm.
41. A red blood cell has a diameter of 7.5 μm
(micrometers). What is this dimension in
(a) meters, (b) nanometers, and (c) picometers?
42. The platinum-containing cancer drug cisplatin
(Study Question 39) contains 65.0 mass-percent
of the metal. If you have 1.53 g of the compound,
what mass of platinum (in grams) is contained in
this sample?
0.563 nm
Sodium chloride, NaCl
Study Questions
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57
(a) What is the volume of this cube in cubic nanometers? In cubic centimeters?
(b) The density of NaCl is 2.17 g/cm3. What is
the mass of this smallest repeating unit (unit
cell)?
(c) Each repeating unit is composed of four NaCl
units. What is the mass of one NaCl formula
unit?
48. Diamond has a density of 3.513 g/cm3. The mass
of diamonds is often measured in carats, where
1 carat equals 0.200 g. What is the volume (in
cubic centimeters) of a 1.50-carat diamond?
© Charles D. Winters/Cengage
49. The element gallium has a melting point of
29.8 °C. If you hold a sample of gallium in your
hand, should it melt? Explain briefly.
Gallium metal
50. ▲ The density of pure water at various temperatures is given below.
T(° C)
d (g/cm3)
4
0.99997
15
0.99913
25
0.99707
35
0.99406
Suppose your laboratory partner tells you the
density of water at 20 °C is 0.99910 g/cm3. Is this
a reasonable number? Why or why not?
51. When you heat popcorn, it pops because it loses
water explosively. Assume a kernel of corn, with a
mass of 0.125 g, has a mass of only 0.106 g after
popping.
(a) What percentage of its mass did the kernel lose
on popping?
(b) Popcorn is sold by the pound in the United
States. Using 0.125 g as the average mass of a
popcorn kernel, how many kernels are there in
a pound of popcorn? (1 lb 5 453.6 g)
58
52. ▲ The aluminum in a package containing 75 ft2 of
kitchen foil weighs approximately 12 ounces. Aluminum has a density of 2.70 g/cm3. What is the approximate thickness of the aluminum foil in millimeters?
(1 ounce 5 28.4 g; 12 in 5 1 ft; and 1 in 5 2.54 cm)
53. ▲ Fluoridation of city water supplies has been
practiced in the United States for several decades. It
is done by continuously adding sodium fluoride to
water as it comes from a reservoir. Assume you live
in a medium-sized city of 150,000 people and that
660 L (170 gal) of water is used per person per day.
What mass of sodium fluoride (in kilograms) must
be added to the water supply each year (365 days)
to have the required fluoride concentration of
1 ppm (part per million)—that is, 1 kilogram of
fluoride per 1 million kilograms of water? (Sodium
fluoride is 45.0% fluoride, and water has a density
of 1.00 g/cm3.)
54. ▲ Over two centuries ago, Benjamin Franklin
showed that 1 teaspoon of oil would cover
about 0.5 acre of still water. If you know that
1.0 3 104 m2 5 2.47 acres and that there is approximately 5 cm3 in a teaspoon, what is the thickness
of the 0.5-acre layer of oil? How might this thickness be related to the sizes of molecules?
55. ▲ Automobile batteries are filled with an aqueous
solution of sulfuric acid. What is the mass of the
acid (in grams) in 750. mL of the battery acid solution if the density of the solution is 1.285 g/cm3
and the solution is 38.08% sulfuric acid by mass?
56. ▲ Hydrochloric acid is a solution of the gas
hydrogen chloride (HCl) dissolved in water. What
mass of HCl is present in 50. mL of a concentrated
hydrochloric acid solution that is 37% HCl by mass
and has a density of 1.2 g/mL?
57. ▲ Household bleach is often a solution of the
compound sodium hypochlorite (NaClO) in water.
What volume of a bleach solution that is 5.25%
NaClO is needed to provide 8.0 g of NaClO?
Assume that the density of the bleach solution is
1.1 g/mL.
58. A 26-meter-tall statue of the Buddha in Tibet
is covered with 279 kg of gold. If the gold was
applied to a thickness of 0.0015 mm, what
surface area is covered (in square meters)?
(Gold density 5 19.3 g/cm3)
59. At 25 °C, the density of water is 0.997 g/cm3,
whereas the density of ice at 210 °C is 0.917 g/cm3.
(a) If a soft-drink can (volume 5 250. mL) is filled
completely with pure water at 25 °C and then
frozen at 210 °C, what volume does the ice
occupy?
(b) Can the ice be contained within the can?
Chapter 1R / Let’s Review: The Tools of Quantitative Chemistry
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60. Suppose your bedroom is 18 feet long and 15
feet wide, and the distance from floor to ceiling
is 8 feet 6 inches. You need to know the volume
of the room in metric units for some scientific
calculations.
(a) What is the room’s volume in cubic meters? In
liters?
(b) What is the mass of air in the room in kilograms? In pounds? (Assume the density of air
is 1.2 g/L and that the room is empty of
furniture.)
(b) The unknown is one of the seven metals listed
in the following table. Is it possible to identify
the metal based on the density you have calculated? Explain.
61. A spherical steel ball has a mass of 3.475 g and a
diameter of 9.40 mm. What is the density of the
steel? [The volume of a sphere 5 (4/3)πr3 where
r 5 radius.]
62. ▲ You are asked to identify an unknown liquid
that is one of the liquids listed below. You pipet a
3.50-mL sample into a beaker. The empty beaker
had a mass of 12.20 g, and the beaker plus the
liquid weighed 16.08 g.
Metal
Density
(g/cm3)
Metal
Density
(g/cm3)
Zinc
7.13
Nickel
8.90
Iron
7.87
Copper
8.96
Cadmium
8.65
Silver
10.50
Cobalt
8.90
64. ▲ There are five hydrocarbon compounds (compounds of C and H) that have the formula C6H14.
(Compounds that have the same formula but
differ in the way the atoms are attached are called
isomers.) All are liquids at room temperature but
have slightly different densities.
Hydrocarbon
Density (g/mL)
Substance
Density at 25 ° C (g/cm3)
Hexane
0.6600
Ethylene
glycol
1.1088 (major component of
automobile antifreeze)
2,3-Dimethylbutane
0.6616
Water
0.9971
1-Methylpentane
0.6532
Ethanol
0.7893 (alcohol in alcoholic
beverages)
2,2-Dimethylbutane
0.6485
2-Methylpentane
0.6645
Acetic acid
1.0492 (active component of vinegar)
Glycerol
1.2613 (solvent used in home care
products)
(a) Calculate the density and identify the
unknown.
(b) If you were able to measure the volume to only
two significant figures (that is, 3.5 mL, not
3.50 mL), will the results be sufficiently precise
to identify the unknown? Explain.
63. ▲ You have an irregularly shaped piece of an
unknown metal. To identify it, you determine its
density and then compare this value with known
values that you look up in the chemistry library.
The mass of the metal is 74.122 g. Because of the
irregular shape, you measure the volume by submerging the metal in water in a graduated cylinder.
When you do this, the water level in the cylinder
rises from 28.2 mL to 36.7 mL.
(a) What is the density of the metal? (Use the
correct number of significant figures in your
answer.)
(a) You have a pure sample of one of these hydrocarbons, and to identify it you decide to
measure its density. You determine that a
5.0-mL sample (measured in a graduated cylinder) has a mass of 3.2745 g (measured on an
analytical balance). Assume that the accuracy
of the values for mass and volume is plus or
minus one (61) in the last significant figure.
What is the density of the liquid?
(b) Can you identify the unknown hydrocarbon
based on your experiment?
(c) Can you eliminate any of the five possibilities
based on the data? If so, which one(s)?
(d) You need a more precise volume measurement
to solve this problem, and you redetermine the
volume to be 4.93 mL. Based on this new information, what is the unknown compound?
65. ▲ Suppose you have a cylindrical glass tube with a
thin capillary opening, and you wish to determine
the diameter of the opening. You can do this experimentally by weighing a piece of the tubing before
and after filling a portion of the capillary with
Study Questions
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59
mercury. Using the following information, calculate
the diameter of the opening.
Mass of tube before adding mercury 5 3.263 g
Mass of tube after adding mercury 5 3.416 g
Length of capillary filled with mercury 5 16.75 mm
Density of mercury 5 13.546 g/cm3
Volume of cylindrical capillary filled with mercury 5
(π)(radius)2(length)
66. Copper: Copper has a density of 8.96 g/cm3. An
ingot of copper with a mass of 57 kg (126 lb)
is drawn into wire with a diameter of 9.50 mm.
What length of wire (in meters) can be produced?
[Volume of wire 5 (π)(radius)2(length)]
67. ▲ Copper:
(a) Suppose you have a cube of copper metal that
is 0.236 cm on a side with a mass of 0.1206 g.
If you know that each copper atom
(radius 5 128 pm) has a mass of
1.055 3 10222 g, how many atoms are there in
this cube? What percentage of the volume
occupied by the cube is filled with atoms? How
much of the lattice is empty space? Why is
there empty space in the lattice?
(b) Now look at the smallest, repeating unit of the
crystal lattice of copper.
Cube of copper atoms
Smallest repeating unit
Knowing that an edge of this cube is 361.47 pm
and the density of copper is 8.960 g/cm3, calculate the number of copper atoms in this smallest, repeating unit.
In the Laboratory
69. A sample of unknown metal is placed in a graduated cylinder containing water. The mass of the
sample is 37.5 g, and the water levels before and
after adding the sample to the cylinder are as
shown in the figure. Which metal in the following
list is most likely the sample? (d is the density of
the metal.)
(a) Mg, d 5 1.74 g/cm3
(d) Al, d 5 2.70 g/cm3
(b) Fe, d 5 7.87 g/cm3
(e) Cu, d 5 8.96 g/cm3
(c) Ag, d 5 10.5 g/cm3
(f) Pb, d 5 11.3 g/cm3
25
20
20
15
15
10
10
5
5
Graduated cylinders with unknown metal (right)
70. Iron pyrite is often called fool’s gold because it
looks like gold (see page 13). Suppose you have a
solid that looks like gold, but you believe it to be
fool’s gold. The sample has a mass of 23.5 g. When
the sample is lowered into the water in a graduated
cylinder (Study Question 69), the water level rises
from 47.5 mL to 52.2 mL. Is the sample fool’s gold
(d 5 5.00 g/cm3) or pure gold (d 5 19.3 g/cm3)?
71. You can analyze for a copper compound in water
using an instrument called a spectrophotometer.
[A spectrophotometer is a scientific instrument
that measures the absorbance of light (of a given
wavelength) by the solution.] The absorbance at
a given wavelength of light (A) depends directly
on the mass of compound per liter of solution. To
calibrate the spectrophotometer, you collect the following data:
68. You set out to determine the density of lead in the
laboratory. Using a top loading balance to determine the mass and the water displacement method
(Study Question 45) to determine the volume of a
variety of pieces of lead, you calculate the following densities: 11.6 g/cm3, 11.8 g/cm3, 11.5 g/cm3,
and 12.0 g/cm3. You consult a reference book and
find that the accepted value for the density of lead
is 11.3 g/cm3. Calculate your average value, percent
error, and standard deviation of your results.
60
25
Absorbance
(A)
Mass of Copper Compound
per Liter (g/L)
0.257
1.029 3 1023
0.518
2.058 3 1023
0.771
3.087 3 1023
1.021
4.116 3 1023
Chapter 1R / Let’s Review: The Tools of Quantitative Chemistry
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Plot the absorbance (A) against the mass of copper
compound per liter (g/L), and find the slope (m)
and intercept (b) (assuming that A is y and the
mass of copper compound per liter of solution is
x in the equation for a straight line, y 5 mx 1 b).
What is the mass of copper compound in the solution in g/L and mg/mL when the absorbance is
0.635?
72. A gas chromatograph is calibrated for the analysis
of isooctane (a major gasoline component) using
the following data:
Percent Isooctane
(x-data)
Instrument Response
(y-data)
0.352
1.09
0.803
1.78
1.08
2.60
1.38
3.03
1.75
4.01
If the instrument response is 2.75, what percentage
of isooctane is present? (Data are taken from D. A.
Skoog, D. M. West, F. J. Holler, and S. R. Crouch,
Analytical Chemistry, An Introduction, 7th ed.,
Belmont, CA: Brooks/Cole, 2000.)
73. A general chemistry class carried out an experiment
to determine the percentage (by mass) of acetic
acid in vinegar. Ten students reported the following values: 5.22%, 5.28%, 5.22%, 5.30%, 5.19%,
5.23%, 5.33%, 5.26%, 5.15%, and 5.22%. Determine the average value and the standard deviation
from these data. How many of these results fell
within 6 one standard deviation of this average
value?
Study Questions
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61
2 Atoms, Molecules, and Ions
Transition Metals
Group 2B (12)
Group 2A (2)
Magnesium—Mg
Titanium—Ti
Vanadium—V
Chromium—Cr
Manganese—Mn
Iron—Fe
Cobalt—Co
Nickel—Ni
Copper—Cu
Mercury—Hg
Group 1A (1)
8A
(18)
1A
(1)
Lithium—Li
2A
(2)
1
H
2
Li Be
3A 4A 5A 6A 7A
O (17)
(13) (14) (15) (16)
He
F
Ne
B
3B
(3)
7B
(7)
8B
1B 2B
(9) (10) (11) (12)
C
N
Al Si
P
4B
(4)
5B
(5)
6B
(6)
Ca Sc Ti
V
Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr
3
Na Mg
4
K
5
Rb Sr
6
Cs Ba La Hf Ta
7
Fr Ra Ac Rf Db Sg Bh Hs Mt Ds Rg Cn Nh Fl Mc Lv
Y
(8)
S
Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te
W Re Os Ir
Zinc—Zn
Group 8A (18),
Noble Gases
Cl Ar
I
Xe
Pt Au Hg Tl Pb Bi Po At Rn
Ts Og
Neon—Ne
Potassium—K
Group 4A (14)
Group 3A (13)
Boron—B
Carbon—C
Group 5A (15)
Tin—Sn
Group 6A (16)
Group 7A (17)
Sulfur—S
Nitrogen—N2
Bromine—Br
Aluminum—Al
Silicon—Si
Lead—Pb
Selenium—Se
Phosphorus—P
© Charles D. Winters/Cengage
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
C hapt e r O ut li n e
2.1
Atomic Structure, Atomic Number, and Atomic Mass
2.2
Atomic Weight
2.3
The Periodic Table
2.4
Molecules: Formulas, Models, and Names
2.5
Ions
2.6
Ionic Compounds: Formulas, Names, and Properties
2.7
Atoms, Molecules, and the Mole
2.8
Chemical Analysis: Determining Compound Formulas
2.9
Instrumental Analysis: Determining Compound Formulas
This chapter begins an exploration of the chemistry of the elements—the building
blocks of chemistry—and the compounds they form. There are currently 118 known
elements, most of which combine to form the millions of compounds known to
chemists. You will learn about the submicroscopic particles that make up these pure
substances as well as some of the properties you can observe on the macroscopic
scale. Some important skills you will learn are how to determine the names and
formulas of many compounds and how to perform calculations that will allow you
to connect the macroscopic measurement of mass to the particulate understanding
of these substances. Throughout this chapter, you will see how the periodic table of
the elements is used to help organize much of this information and as a tool to help
you understand quantitative relationships in chemistry.
2.1 Atomic Structure, Atomic
Number, and Atomic Mass
Goals for Section 2.1
•
•
•
Describe electrons, protons, and neutrons, and the general structure of the atom.
Define the terms atomic number and mass number.
Define isotopes and give the atomic symbol for a specific isotope.
Atomic Structure
Around 1900, a series of experiments done by scientists in England, including Sir
Joseph John Thomson (1856–1940) and Ernest Rutherford (1871–1937), established a model of the atom that is still the basis of modern atomic theory.
◀ Some of the 118 known elements.
63
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Nucleus with protons (positive
electric charge) and neutrons
(no electric charge).
Atoms are made of subatomic particles: electrically positive protons, electrically negative electrons, and, in all except one type of hydrogen atom, electrically
neutral neutrons. The model places the more massive protons and neutrons in a
very small nucleus (Figure 2.1), which contains all the positive charge and almost
all the mass of an atom. Electrons, with a much smaller mass than protons or
neutrons, surround the nucleus and occupy most of the volume. In an electrically
neutral atom, the number of electrons equals the number of protons. As you will
see, the chemical properties of elements and compounds depend largely on the
electrons in their atoms.
Atomic Number
Electrons (negative electric charge).
The number of electrons and protons
is equal in an electrically neutral atom.
Figure 2.1 The structure of the
atom. This figure is not drawn
to scale. If the nucleus were
really the size depicted here,
the electron cloud would extend
over 200 m. The atom is mostly
empty space! In this illustration,
the electrons are depicted as a
cloud around the nucleus. The
most accurate model of the atom
represents electrons as waves,
not particles.
Historical Perspective on the
Development of Our Understanding
of Atomic Structure A brief
history of important experiments
and the scientists involved
in developing the modern
view of the atom is given on
pages 72–74.
Unified Atomic Mass Units Masses
of fundamental atomic particles
are sometimes expressed using
a non-SI unit called the unified
atomic mass unit (u), sometimes
referred to as the dalton (Da).
One unified atomic mass unit,
1 u, is one-twelfth the mass
of an atom of carbon with six
protons and six neutrons. Such
a carbon atom has a mass of
exactly 12 u. 1 atomic mass
unit (u) = 1.66054 × 10−24 g.
64
All atoms of a given element have the same number of protons in the nucleus.
­Hydrogen is the simplest element, with one nuclear proton. All helium atoms have
two protons, all lithium atoms have three protons, and all beryllium atoms have
four protons. The number of protons in the nucleus of an element is given by its
atomic number, which is generally indicated by the symbol Z.
The 118 known elements are listed in the periodic table inside the front cover of
this book and on the list inside the back cover. The integer number at the top of the
box for each element in the periodic table is its atomic number. A copper atom (Cu),
for example, has an atomic number of 29, so its nucleus contains 29 protons.
A ­uranium atom (U) has 92 nuclear protons and Z = 92.
Copper
29
Cu
Atomic number
Symbol
Relative Atomic Mass
As eighteenth- and nineteenth-century chemists tried to understand how the elements combine, they carried out increasingly quantitative studies aimed at learning,
for example, how much of one element combines with another. Based on this work,
they learned that substances have a constant composition, which led to the
­conclusion that chemists can define the relative masses of elements that combine to
produce a new substance. At the beginning of the nineteenth century, John Dalton
(1766–1844) suggested that the combinations of elements involve atoms, and he
proposed a relative scale of atom masses. Dalton based this scale on hydrogen having a mass of 1. Later, oxygen atoms were chosen as the standard, and oxygen was
assigned a mass of exactly 16, but there were disagreements between chemists and
physicists about the details of this definition.
In 1959–1960, the International Union of Pure and Applied Chemistry (IUPAC)
and the International Union of Pure and Applied Physics (IUPAP) agreed to a new
unified standard: a carbon atom having 6 protons and 6 neutrons in the nucleus is
assigned a relative mass of exactly 12. The masses of other atoms are determined relative to the mass for this type of carbon atom. For example, chemical experiments
and physical measurements show that the mass of an oxygen atom with 8 protons
and 8 neutrons is 1.33291 times the mass of a carbon atom with 6 protons and
6 neutrons. So, the oxygen atom has a relative mass of 1.33291 × 12 = 15.9949.
Mass Number
Protons and neutrons have relative atomic masses close to 1, while the mass of an
electron is only about 1/2000 of this value (Table 2.1). You can estimate the
­approximate mass of an atom if you know the number of neutrons and protons in
that atom. The sum of the number of protons and neutrons for an atom is called its
mass number and is given the symbol A.
Chapter 2 / Atoms, Molecules, and Ions
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Table 2.1
Properties of Subatomic Particles*
mass
Particle
Grams
Relative Atomic Mass
Charge
Symbol
Electron
9.109384 × 10
0.0005485799
1−
−01e or e−
Proton
1.672622 × 10−24
1.007276
1+
1
+
1p or p
Neutron
1.674927 × 10−24
1.008665
0
1
0n or n
−28
How Small Is an Atom? The radius
of the typical atom is between
30 and 300 pm (3 × 10−11 m
to 3 × 10−10 m). To get a
feeling for the incredible
smallness of an atom, consider
that 1 cm3 of water contains
about three times as many atoms
as the Atlantic Ocean contains
teaspoons of water.
*These values and others in the book are taken from the National Institute of Standards and Technology
website at http://physics.nist.gov/cuu/Constants/index.html
A = mass number = number of protons + number of neutrons
For example, a sodium atom, which has 11 protons and 12 neutrons in its nucleus,
has a mass number of 23 (A = 11 p + 12 n). The most common atom of uranium
has 92 protons and 146 neutrons, and a mass number of A = 238. Atoms are commonly symbolized with the following notation:
Mass number
Atomic number
A
ZX
Element symbol
The subscript Z is optional because each atomic number represents a unique
element. For example, the atoms described previously have the symbols 1213Na and
238
23
Na and 238U. In words, these are referred to as “sodium-23” and
92U, or just
“uranium-238.”
Exam p le 2.1
Atomic Composition
Problem How many protons and how many electrons are in an atom of phosphorus
with 16 neutrons? What is its mass number? What is the complete symbol for such an
atom?
What Do You Know? You know the name of the element and the number of
neutrons. The number of protons and electrons are equal in a neutral atom.
Strategy The number of protons in an atom is given by the atomic number shown on
the periodic table. The mass number is the sum of the number of protons and neutrons.
Solution A phosphorus atom has 15 protons and 15 electrons. A phosphorus atom
with 16 neutrons has a mass number of 31.
Mass number = number of protons + number of neutrons = 15 + 16 = 31
The atom’s complete symbol is 1315P.
Think about Your Answer Once you know the identity of an element, you can
determine the number of protons in the nucleus from its atomic number shown on the
periodic table. You can determine the mass number if you also know the number of neutrons in the atom.
Check Your Understanding
1.
What is the mass number of an iron atom with 30 neutrons?
2.
How many protons, neutrons, and electrons are in a 64Zn atom?
2.1 Atomic Structure, Atomic Number, and Atomic Mass
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
65
Chemistry in Your Career
Tiffany J. Carey
Tiffany J. Carey
Tiffany Carey’s first interest was biology, she reports,
“But when I learned all life as we know it would not
exist if the angle between hydrogen atoms in a
­water molecule was just 5 degrees different, I was
intrigued.” Carey went on to earn a B.S. in biology
with a chemistry minor and then an M.S. in environmental science and environmental engineering.
Today Carey is a quality assurance officer for a
major U.S. water utility that delivers 1 billion gallons per day of high-quality drinking water, where
she ensures that data generated by the labs are of
the highest integrity and accuracy. Carey enjoys
the mutual respect she shares with coworkers, saying that “we all have a part to play in the big picture of enriching the environment and protecting
public health for millions of people.”
Carey emphasizes the importance of chemistry in the environmental sector. “No one ever told
me that the deciding factor in getting a job was
going to be having enough chemistry credits, but
it is, especially for government or public utility
jobs.”
Isotopes
Solid H2O
d = 0.917 g/cm3
Solid D2O
d = 1.11 g/cm3
Figure 2.2 Ice made
from “heavy water” sinks
in “ordinary” water. Water
containing ordinary hydrogen
(11H, protium) forms a solid that
is less dense than liquid H2O, so
it floats in the liquid. D2O-ice is
denser than liquid H2O, so solid
D2O sinks in liquid H2O.
© Charles D. Winters/Cengage
Liquid H2O
d = 0.9998 g/cm3
All the atoms in a naturally occurring sample of a given element have the same mass
in only a few instances (for example, aluminum, fluorine, and phosphorus). Most
elements consist of atoms with several different mass numbers. For example, there
are two kinds of boron atoms, one with a mass of about 10 (10B) and a second with
a mass of about 11 (11B). Atoms of tin can have any of 10 different masses ranging
from 112 to 124. Atoms with the same atomic number but different mass numbers
are called isotopes. Scientists often refer to a particular isotope by giving its mass
number. For example, an atom of iron with a mass number of 58 is referred to as
iron–58 or symbolized as 58Fe.
All atoms of an element have the same number of protons. To have different
masses, isotopes must have different numbers of neutrons. The nucleus of a
10
B atom (Z = 5) contains five protons and five neutrons, whereas the nucleus of a
11
B atom contains five protons and six neutrons.
The isotopes of hydrogen are so important that they have special names and
symbols. All hydrogen atoms have one proton. When that is the only nuclear particle, the isotope is called protium, or just hydrogen. The isotope of hydrogen with
one neutron, 12H, is called deuterium, or heavy hydrogen (symbol = D). The nucleus
of radioactive hydrogen-3, 13H, or tritium (symbol = T), contains one proton and two
neutrons.
The substitution of one isotope of an element for another isotope of the same
element in a compound usually has only a small, almost negligible, effect on chemical and physical properties. An exception is the substitution of deuterium for hydrogen because the mass of deuterium is double that of hydrogen (Figure 2.2).
E xamp le 2.2
Isotope Composition
Problem Two naturally occurring isotopes of oxygen are 16O and 18O. How many
­protons, neutrons, and electrons are in the atoms of each of these isotopes?
66
Chapter 2 / Atoms, Molecules, and Ions
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
What Do You Know? You know that 16O and 18O are isotopes of oxygen. You also
know that the atomic number for each element is shown on the periodic table and that
in a neutral atom the number of protons and number of electrons are the same.
Mass number is the sum of the protons and neutrons in an atom. Isotopes have the same
number of protons but different numbers of neutrons, which results in different
­
mass numbers.
Strategy
Step 1. Determine the number of protons in the atoms from the atomic number of the
element shown on the periodic table.
Step 2. Determine the number of neutrons in each atom by subtracting the atomic
number from the mass number.
Number of neutrons = mass number (number of protons and neutrons) − atomic
number (number of protons)
Step 3. Determine the number of electrons from the number of protons and the
charge on the atom. In a neutral atom, the number of electrons equals the number of
protons.
Solution
Step 1. The atomic number of oxygen shown on the periodic table is eight, so there are
eight protons in the atoms of both 16O and 18O.
Step 2. The number of neutrons is the difference between the mass number (protons
and neutrons) and the atomic number (protons). Isotopes of the same element have
different numbers of neutrons.
For 16O: mass number − atomic number = 16 − 8 = 8 neutrons in 16O
For 18O: mass number − atomic number = 18 − 8 = 10 neutrons in 18O
Step 3. Because there are 8 protons in each isotope, 16O and 18O each contain 8
electrons.
Think about Your Answer Isotopes of the same element have the same number of protons and the same number of electrons. They differ only in the number of
neutrons.
Check Your Understanding Two naturally occurring isotopes of uranium are
U and 238U. How many protons, neutrons, and electrons are in the atoms of each of
these isotopes?
235
2.2 Atomic Weight
Goal for Section 2.2
•
Perform calculations that relate the atomic weight (atomic mass) of an element
and isotopic abundances and masses.
2.2 Atomic Weight
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
67
Determining Atomic Mass and Isotopic Abundance
The masses of isotopes and their abundances are determined experimentally by
mass spectrometry. In a mass spectrometer, the atoms or molecules of a sample are
converted into gaseous, charged species that are then separated based on their mass
and charge (Figure 2.3). Modern mass spectrometers can measure isotopic masses to
as many as nine significant figures.
Except for carbon-12, whose mass is defined to be exactly 12, isotopic masses
do not have integer values. Isotopic masses are, however, always close to the
mass numbers for the isotope. For example, the atomic mass of boron-11
(11B, 5 protons and 6 neutrons) is 11.0093, and the atomic mass of iron-58 (58Fe,
26 protons and 32 neutrons) is 57.9333.
A sample of water from a lake consists almost entirely of H2O where the
H atoms are the 1H isotope. A few molecules, however, have deuterium (2H) substituted for 1H. This is because 99.989% of all hydrogen atoms on Earth are 1H atoms.
That is, the abundance of 1H atoms is 99.989%.
Isotopic Masses and the Mass
Defect Actual masses of atoms
are always less than the sum
of the masses of the subatomic
particles composing that atom.
This is called the mass defect,
and the reason for it is discussed
in Chapter 20.
Percent abundance 5
number of atoms of a given isotope
3 100% (2.1)
total number of atoms of all isotopes of that element
The remainder of naturally occurring hydrogen is deuterium, whose abundance is
only 0.012%. Tritium, the radioactive 3H isotope, occurs naturally in only trace
amounts.
Consider the two isotopes of chlorine. The chlorine-35 isotope has an abundance of 75.76%; the abundance of chlorine-37 is 24.24%. Among 10,000 chlorine
atoms from an average natural sample, 7576 are chlorine-35 atoms and 2424 of
them are chlorine-37 atoms.
Acceleration
Deflection
Heavy ions
are deflected
too little.
e−e−e−
e−e−e−
e−e−e−
Gas
inlet
1
−
Repeller Electron
trap
plate
2
Analysis
Magnet
Electron gun
+
Detection
Accelerating
plates
3
20Ne+
Magnet
4
Light ions
are deflected
too much.
1 A sample is introduced as a
4 This chamber is in a magnetic
vapor into the ionization chamber.
field, which is perpendicular to
the direction of the beam of
2 There it is bombarded with highenergy electrons that strip
charged particles.
5
electrons from the atoms or
The magnetic field causes the
molecules of the sample.
beam to curve. The radius of
curvature depends on the mass
3 The resulting positive particles
are accelerated by a series of
and charge of the particles (as well
negatively charged accelerator
as the accelerating voltage and
plates into an analyzing chamber.
strength of the magnetic field).
To mass
analyzer
22Ne+
To vacuum pump
5
21Ne+
Detector
Here, particles of 21Ne+ are focused
on the detector, whereas beams of
ions of 20Ne+ and 22Ne+ (of lighter
or heavier mass) experience greater
and lesser curvature, respectively,
and so fail to be detected.
A mass spectrum is a plot of the
relative abundance of the charged
particles versus the ratio of
mass/charge (m/Z).
Relative Abundance
Vaporization Ionization
100
80
60
40
20
0
20
21
22
m/Z
By changing the magnetic field,
charged particles of different
masses can be focused on the
detector to generate the observed
spectrum.
Figure 2.3 Mass spectrometer. A mass spectrometer separates ions of different mass and charge in a gaseous sample of ions. The
instrument allows the researcher to determine the accurate mass of each ion.
68
Chapter 2 / Atoms, Molecules, and Ions
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Atomic Weight
Every sample of chlorine has some atoms with an atomic mass of 34.96885 and
others with an atomic mass of 36.96590. The atomic weight of the element, the
average mass of a representative sample of chlorine atoms, is somewhere between
these values. For chlorine for example, the atomic weight is 35.45. If isotope masses
and abundances are known, the atomic weight of an element can be calculated using Equation 2.2.
 % abundance isotope 1 
Atomic weight 5 
 (mass of isotope 1)

100
 % abundance isotope 2 
1
 (mass of isotope 2) 1 . . .

100
(2.2)
For chlorine with two isotopes (35Cl, abundance = 75.76%; 37Cl, abundance =
24.24%),
 75.76 
 24.24 
Atomic weight 5 
× 34.96885 1 
× 36.96590 5 35.45
 100 
 100 
Equation 2.2 gives an average mass, weighted by the abundance of each isotope for
the element. As illustrated by the data in Table 2.2, the atomic weight of an element
is often close to the mass of the most abundant isotope or isotopes.
For each stable element the atomic weight is given in the periodic table. The
atomic weight is shown for a couple of unstable (radioactive) elements, but more
often, the mass number of the most stable isotope is given in parentheses. For
­example, the isotopic masses and abundances of the radioactive element uranium
(U) are known, so an atomic weight of 238.03 is listed on the periodic table. On the
other hand, no atomic weight is shown for neptunium (Np), but the mass number
of its most ­stable isotope, 237, is given in parentheses.
Table 2.2
Isotope Abundance and Atomic Weight
Element
Symbol
Atomic
Weight
Mass
Number
Isotopic
Mass
Natural Abundance
(%)
Hydrogen
H
1.008
Boron
Neon**
Magnesium
1
1.0078
99.989
D*
2
2.0141
0.012
T†
3
3.0160
0
B
Ne
Mg
10.81
20.180
24.305
10
10.0129
19.9
11
11.0093
80.1
20
19.9924
90.48
21
20.9938
0.27
22
21.9914
9.25
24
23.9850
78.99
25
24.9858
10.00
26
25.9826
11.01
*D = deuterium; †T = tritium, radioactive; **See the mass spectrum of Ne in Figure 2.3.
2.2 Atomic Weight
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69
A Closer Look
Isotopic Abundances and Atomic Weights
Accurate atomic weights for elements are of great importance for quantitative chemistry. The international body
that establishes the values for atomic weights is the
­Commission on Isotopic Abundances and Atomic Weights
(CIAAW) of the International Union of Pure and Applied
Chemistry (IUPAC).
Modern mass spectrometers can determine isotopic
atomic masses out to nine or more decimal places. For elements with only one naturally occurring isotope, the atomic
weights are therefore known to nine or more decimal
places. For example, fluorine has only one naturally occurring isotope, 19F, and has an atomic weight of 18.998403162.
Most elements, however, have more than one isotope. Their
atomic weights depend on both the isotopic masses and the
isotopic abundances. In introductory chemistry courses, it is
often assumed that the different isotopes of an element are
uniformly distributed on the Earth. However, this is not strictly
true. Samples of an element’s atoms from different sources
may have different isotopic abundances. These differences result in the isotopic abundances being known to fewer significant figures than isotopic masses, which limits the precision of
atomic weights. For example, the current value for the atomic
weight of copper is 63.546, a value that only goes to the third
decimal place. This result follows the general rule for measured
values that the final decimal place has some uncertainty.
For some elements, the CIAAW has determined that the
variation in the abundances of the elements is so large that
the atomic weight is best represented as a range of values
(Table). In such cases, the atomic weight of the element calculated for normal materials would vary, depending upon the
source of the material, but would be somewhere within that
range.
Using ranges for atomic weights, however, can make learning
chemistry more complicated than desired. To avoid this, the periodic tables in this text show the conventional value for these elements that is often used for samples from an unspecified source.
Elements with Defined Atomic Weight Ranges
Element
Atomic Weight
Range
Conventional
Value
hydrogen
1.00784–1.00811
1.008
lithium
6.938–6.997
6.94
boron
10.806–10.821
10.81
carbon
12.0096–12.0116
12.011
nitrogen
14.00643–14.00728
14.007
oxygen
15.99903–15.99977
15.999
magnesium
24.304–24.307
24.305
silicon
28.084–28.086
28.085
sulfur
32.059–32.076
32.06
chlorine
35.446–35.457
35.45
argon
39.792–39.963
39.95
bromine
79.901–79.907
79.904
thallium
204.382–204.385
204.38
lead
206.14–207.94
207.2
E xamp le 2.3
Calculating Atomic Weight from Isotope
Abundance
Problem Bromine has two naturally occurring isotopes. One has a mass of 78.918 and
© Charles D. Winters/Cengage
an abundance of 50.7%. The other isotope has a mass of 80.916 and an abundance of
49.3%. Calculate the atomic weight of bromine.
Br2 vapor
Br2 liquid
Elemental bromine. Bromine is a
deep orange-red, volatile liquid
at room temperature. It consists
of Br2 molecules in which two
bromine atoms are chemically
bonded together. There are
two, stable, naturally occurring
isotopes of bromine atoms:
79
Br (50.7% abundance) and
81
Br (49.3% abundance).
70
What Do You Know? You know the mass and abundance of each of the two
isotopes.
Strategy The atomic weight of any element is the weighted average of the masses of
the isotopes in a representative sample. Use Equation 2.2 to calculate the atomic weight.
Solution
Atomic weight of bromine = (50.7/100)(78.918) + (49.3/100)(80.916) = 79.9
Think about Your Answer You can also estimate the atomic weight from the
data given. There are two isotopes, mass numbers of 79 and 81, in approximately equal
abundance. From this, you would expect the average mass to be about 80, midway between the two mass numbers. The calculation bears this out.
Chapter 2 / Atoms, Molecules, and Ions
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Check Your Understanding
Verify that the atomic weight of boron is 10.8, given the following information:
B mass = 10.0129; percent abundance = 19.9%
10
B mass = 11.0093; percent abundance = 80.1%
11
Calculating Isotopic Abundances
Problem Antimony, Sb, has two stable isotopes: 121Sb, atomic mass = 120.904, and
Sb, atomic mass = 122.904. What are the relative abundances of these isotopes?
123
What Do You Know? You know the masses of the two isotopes of the element and
know that their weighted average, the atomic weight, is 121.76 (see the periodic table).
Strategy To calculate the abundances recognize there are two unknown but related
quantities, the fractional abundances of 121Sb and 123Sb (where the fractional abundance of
an isotope is the percent abundance of the isotope divided by 100). These are related by
Equation 2.2 and by an equation that shows the sum of the two fractional abundances must
equal 1. These two equations can be solved for the two unknown fractional abundances.
© Charles D. Winters/Cengage
Exam p le 2.4
A sample of the metalloid
antimony. The element has two
stable isotopes, 121Sb and 123Sb.
Solution Equation 2.2 can be written for antimony as follows.
Atomic weight = 121.76 = (fractional abundance of 121Sb)(120.904) +
(fractional abundance of 123Sb)(122.904)
or
121.76 = x(120.904) + y(122.904)
where x = fractional abundance of 121Sb and y = fractional abundance of 123Sb. You know
that the sum of fractional abundances of the isotopes must equal 1 (x + y = 1).
Because y = fractional abundance of 123Sb = 1 − x, you can make a substitution for y.
121.76 = x(120.904) + (1 − x)(122.904)
Expanding this equation, you have
121.76 = 120.904x + 122.904 − 122.904x
Finally, solving for x, you find
121.76 − 122.904 = (120.904 − 122.904)x
x = 0.572
The fractional abundance of 121Sb is 0.572 and its percent abundance is 57.2%. This means
that the percent abundance of 123Sb must be 42.8%.
Think about Your Answer You might have predicted that the lighter isotope
(121Sb) must be more abundant because the atomic weight is closer to 121 than to 123.
2.2 Atomic Weight
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71
Check Your Understanding
Neon has three stable isotopes, one with a small abundance. What are the abundances of
the other two isotopes?
Ne, mass = 19.992440; percent abundance = ?
20
Ne, mass = 20.993846; percent abundance = 0.27%
21
Ne, mass = 21.991385; percent abundance = ?
22
Key Experiments
The Nature of the Atom and Its Components
The idea that atoms are the building blocks for matter was set
on the right track by the English chemist John Dalton in the
early 1800s, but little was known about atoms at that time and
for a long time after. Dalton proposed that an atom was a
“solid, massy, hard, impenetrable, moveable particle,” far
+ Slits to focus a
–
narrow beam of rays
Electrically charged
deflection plates
+
–
+
+
Negative
electrode
Positive electrodes
accelerate electrons
A beam of electrons (cathode
rays) is accelerated through two
focusing slits.
–
Fluorescent
sensitized
–
screen
Magnetic field coil
To vacuum pump
perpendicular to electric field
When passing through an
electric field the beam of
electrons is deflected.
The experiment is arranged so that
the electric field causes the beam of
electrons to be deflected in one
direction. The magnetic field deflects
the beam in the opposite direction.
Figure 1 Cathode rays: Thomson’s experiment to measure the
electron’s charge-to-mass ratio. The second half of the ­nineteenth
century saw a series of experiments involving cathode ray tubes.
First described in 1869 by William Crookes (1832–1919), a
­cathode ray tube is an evacuated container with two electrodes.
When a high voltage is applied, particles (cathode rays) flow from
the negative electrode (the cathode) to the anode. These particles
were deflected by electric and magnetic fields, and by balancing
these effects, it was possible to determine their charge-to-mass
72
from the modern description of a nuclear atom with protons,
neutrons, and electrons. Our current understanding required
ingenious experiments, carried out in the late 1800s and early
1900s. This section describes the main ideas for a few of these
experiments.
Electrically deflected
electron beam
Undeflected
electron beam
Magnetically
deflected
electron beam
By balancing the effects of the
electrical and magnetic fields, the
charge-to-mass ratio of the electron
can be determined.
­ niversity
ratio (e/m). In 1897, J. J. Thomson (1856–1940) at the U
of Cambridge in England estimated that these particles had about
three orders of magnitude less mass than a hydrogen atom. They
became known as electrons, a term already used to describe the
smallest particle of electricity. Thomson reasoned that electrons
must originate from the atoms of the cathode, and he speculated
that an atom was a uniform sphere of positively charged matter in
which negative electrons were embedded, a model that is now
known to be incorrect.
Chapter 2 / Atoms, Molecules, and Ions
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
particles
rays
Photographic film
or phosphor screen
+
Lead block
shield
particles,
attracted to
+ plate
particles
particles,
attracted to
– plate
–
Slit
Charged
plates
Radioactive
element
Emitted radiation passes
through oppositely
charged plates.
Positive particles deflect toward the – plate,
negative particles deflect toward the + plate,
and neutral rays continue undeflected.
Identification of the radiation emanating from radioactive substances soon followed. Three types of radiation were observed and
given the labels alpha, beta, and gamma. Charge-to-mass studies
revealed that alpha rays are helium nuclei (He2+) and beta rays are
electrons. Gamma rays have neither mass nor charge; they are a
highly energetic form of electromagnetic radiation.
Figure 2 Radioactivity. Evidence that atoms were made up from
smaller particles was also inferred from the discovery of radioactivity. In 1896, Henri Becquerel (1852–1908) found that uranium
emitted invisible rays that caused a covered photographic plate to
darken. Marie Curie (1867–1934) invented the term radioactivity
to describe this new phenomenon. She and her husband Pierre
Curie (1859–1906) concluded that the observed radiation is the
result of the disintegration of atoms.
Oil atomizer
Light source
to illuminate
drops for
viewing
X-ray source
A fine mist of oil drops is
introduced into one chamber.
The droplets fall one by one into
the lower chamber under the
force of gravity.
Voltage applied
to plates
Oil droplets
under observation
Positively charged plate
+
+
Telescope
–
–
Negatively charged plate
Gas molecules in the bottom
chamber are ionized (split into
electrons and a positive fragment)
by a beam of X-rays. The electrons
adhere to the oil drops, some
droplets having one electron,
some two, and so on.
Figure 3 Millikan’s experiment to determine the electron charge.
Cathode ray experiments allow measurement of the charge-to-mass
ratio of a charged particle, but not the charge or mass individually.
In 1908, the U.S. physicist Robert Millikan (1868–1953), at the
California Institute of Technology, conducted an experiment to
measure the charge on the electron. In his experiment, tiny oil
droplets were sprayed into a chamber and then subjected to X-rays,
causing them to take on a negative charge. The drops could be
These negatively charged droplets
continue to fall due to gravity.
By carefully adjusting the voltage on
the plates, the force of gravity on the
droplet is exactly counterbalanced
by the attraction of the negative
droplet to the upper, positively
charged plate.
Analysis of these forces leads to a
value for the charge on the electron.
suspended in air if the force of gravity was balanced against an
electric field, and from an analysis of these forces on the droplet
the charge could be calculated. Millikan determined that the electronic charge was 1.592 × 10−19 coulombs (C), not far from
­today’s accepted value of 1.602 × 10−19 C. Millikan correctly
assumed this was the fundamental unit of charge. Knowing this
value and the charge-to-mass ratio determined by Thomson, the
mass of an electron could be calculated.
2.2 Atomic Weight
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
73
Beam of
particles
Nucleus of
gold atoms
Atoms in Electrons occupy
gold foil space outside nucleus.
Undeflected
particles
Gold foil
Deflected
particles
particles
Some particles
are deflected
considerably.
A few particles
collide head-on
with nuclei and are
deflected back
toward the source.
Most particles
pass straight
through or are
deflected very little.
Figure 4 Rutherford’s experiment to determine the structure of
the atom. Although scientists recognized that atoms were made up
of smaller particles, it was not clear how these particles fit together.
Around 1910, Ernest Rutherford (1871–1937) established the
model accepted today. Rutherford interpreted an experiment conducted by two colleagues, Hans Geiger (1882–1945) and Ernest
Marsden (1889–1970), in which they bombarded thin gold foil
with α particles. Almost all the particles passed straight through
the gold foil as if there was nothing there. However, a few α particles were deflected sideways and some even bounced right back.
This experiment proved that an atom of gold is mostly empty space
with a tiny nucleus at its center. The electrons surround the nucleus and account for most of the volume of the atom. Rutherford
Source of narrow beam
of fast-moving particles
ZnS fluorescent
screen
calculated that the central nucleus of an atom occupied only
1/10,000th of its volume. He also estimated that a gold nucleus
had a positive charge of around 100 units and a radius of about
10−12 cm. (The values are now known to be 79+ for atomic charge
and 10−13 cm for the radius.)
The final piece of the picture of atomic structure was not established for another decade. Scientists knew that there had to be
something else in the nucleus, and it had to be a heavy particle to
account for the mass of the element. In 1932, the British physicist
James Chadwick (1891–1974) found the missing particle. These
particles, now known as neutrons, have no electric charge and a
mass of 1.675 × 10−24 g, slightly greater than the mass of a
proton.
2.3 The Periodic Table
Goals for Section 2.3
•
Know the terminology of the periodic table (periods, groups) and know how to
use the information given in the periodic table.
•
Recognize similarities and differences in properties of some of the common
elements of a group.
Features of the Periodic Table
The main organizational features of the periodic table are the following (Figure 2.4):
•
74
Elements with similar chemical and physical properties lie in vertical columns
called groups or families. The periodic table commonly used in the United
States has groups numbered 1 through 8, with each number followed by the
letter A or B. The A groups are often called the main group elements and the B
groups are the transition elements. In other parts of the world, the groups are
numbered 1–18. In this book, groups are referenced using the system commonly used in the United States, followed by the group number using the 1–18
system in parentheses. For example, the elements carbon, silicon, germanium,
tin, lead, and flerovium comprise Group 4A (14).
Chapter 2 / Atoms, Molecules, and Ions
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A
Figure 2.4 Periods and
groups in the periodic table. ​An
alternative to this labeling system
numbers the groups from 1 to
18 going from left to right. This
notation is generally used outside
the United States.
A
1 2
3 4 5 6 7 8
B
3 4 5 6 7
8
1 2
Groups or Families
1
2
3
4
5
6
7
Main Group Metals
Transition Metals
Metalloids
Nonmetals
Periods
The horizontal rows of the table are called periods, and they are numbered
beginning with 1 for the period containing only H and He. Currently, 118 elements are known, filling periods 1 through 7.
The periodic table can be divided into several regions according to the properties of the elements. In Figures 2.4 and 2.5 (and the table on the inside cover of this
book), metals are on the left and are indicated in shades of blue. Nonmetals, on the
right (with the exception of hydrogen), are indicated in orange. Metalloids, along the
metal-nonmetal boundary, appear in green. Elements gradually become less metallic from left to right across a period.
You are probably familiar with many properties of metals from your own experience. At room temperature and normal atmospheric pressure, metals are solids (except for mercury), can conduct electricity, are usually ductile (can be drawn into
wires) and malleable (can be rolled into sheets), and can form alloys (mixtures of
one or more metals with another metal). Iron (Fe) and aluminum (Al) are used in
automobile parts because of their ductility, malleability, and low cost relative to
other metals. Copper (Cu) is used in electric wiring because it conducts electricity
better than most other metals.
The nonmetals, which lie to the right of a diagonal line that stretches from B to
Te in the periodic table, have a wide variety of properties. Some are solids (carbon,
sulfur, phosphorus, and iodine). Ten elements are gases at room temperature
(­hydrogen, oxygen, nitrogen, fluorine, chlorine, helium, neon, argon, krypton, and
xenon). One nonmetal, bromine, is a liquid at room temperature. With the exception of carbon in the form of graphite, nonmetals do not conduct electricity, which
is one of the main features that distinguishes them from metals.
The elements along the diagonal line from boron (B) to tellurium (Te) have
properties that make them difficult to classify as metals or nonmetals. Chemists call
them metalloids or, sometimes, semimetals. Metalloids are elements that have
some of the physical characteristics of a metal but some of the chemical characteristics of a nonmetal. This definition reflects the ambiguity in the behavior of these
elements. Antimony (Sb), for example, conducts electricity just as well as many metals. Its chemistry, however, resembles that of phosphorus, a nonmetallic element.
We include only B, Si, Ge, As, Sb, and Te in this category, but some chemists include
one or more other elements.
Some elements can exist in several different and distinct forms called
allotropes, each of which has its own properties. For example, oxygen occurs as two
allotropes in nature: O2 (usually referred to as oxygen) and O3 (called ozone). Humans will die without a sufficient source of O2 molecules to breathe. On the other
hand, O3 is poisonous. Nonetheless, O3 plays an important role in the environment
by blocking out some of the radiation that reaches the Earth from the sun.
Placing H in the Periodic Table
Where to place H? Hydrogen is
a nonmetal, but periodic tables
often show it in Group 1A (1)
on the left side of the periodic
table even though it is not a
metal. However, in some of its
reactions, it forms a 1+ ion like
other members of Group 1A (1).
For this reason, H is often
placed in this group.
Forms of silicon
© Charles D. Winters/Cengage
•
Silicon—a metalloid. Only six
elements are generally classified
as metalloids or semimetals. This
photograph shows solid silicon
in various forms, including a
wafer that holds printed electronic
circuits.
Metalloids For more information
on which elements are
considered metalloids, see the
article “Which Elements are
Metalloids?” by René E. Vernon,
Journal of Chemical Education,
2013, 90, 1703–1707.
2.3 The Periodic Table
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75
76
Chapter 2 / Atoms, Molecules, and Ions
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H
3B
(3)
4B
(4)
20
(223)
Fr
87
132.91
Cs
55
85.468
Rb
37
39.098
21
Sc
89
(226)
Ra
(227)
Ac
137.33 138.91
88
57
La
Ba
56
Y
88.906
Sr
39
87.62
38
40.078 44.956
Ca
19
K
24.305
Mg
6B
(6)
7B
(7)
(8)
V
23
Zr
Nb
41
73
Ta
(267)
Rf
104
(268)
105
Db
178.49 180.95
72
Hf
91.224 92.906
40
47.867 50.942
22
Ti
Figure 2.5 The periodic table of the elements.
Th
Ce
25
Mn
Mo
Tc
75
Re
(98)
43
Pa
238.03
U
(237)
Np
93
92
Pm
61
(277)
108
Hs
91
231.04
Rh
77
Ir
(10)
1B
(11)
Sm
Pu
(244)
94
29
Cu
Pd
Ag
47
79
Au
Eu
63
(281)
Ds
110
Am
(243)
95
Gd
31
81
80
113
112
Tb
65
(285)
Cn
Cm
(247)
96
(247)
Bk
97
Dy
Cf
(251)
98
7
N
5A
(15)
P
15
33
As
83
82
115
Ho
67
(289)
Es
(252)
99
Er
Fm
(257)
100
9
F
7A
(17)
35
Br
35.45
17
Cl
Te
I
53
85
Tm
69
(293)
Md
(258)
101
Yb
He
Xe
Lu
71
(294)
118
Og
(222)
86
Rn
131.29
54
83.798
36
Kr
39.95
18
Ar
20.180
10
Ne
4.0026
2
8A
(18)
No
(259)
102
Lr
(262)
103
173.05 174.97
70
(294)
Ts
117
116
Lv
(210)
At
(209)
84
Po
127.60 126.90
52
78.971 79.904
34
Se
32.06
S
16
15.999 18.998
8
O
6A
(16)
167.26 168.93
68
(290)
Mc
114
Fl
208.98
Bi
207.2
Pb
118.71 121.76
Sb
51
Sn
50
72.630 74.922
32
Ge
28.085 30.974
14
Si
12.011 14.007
6
C
4A
(14)
162.50 164.93
66
(286)
Nh
200.59 204.38
Tl
Hg
112.41 114.82
In
49
48
Cd
69.723
65.38
Ga
30
Zn
26.982
2B
(12)
157.25 158.93
64
(282)
Rg
111
195.08 196.97
78
Pt
106.42 107.87
46
58.693 63.546
28
Ni
150.36 151.96
62
(276)
Mt
109
190.23 192.22
76
Os
(145)
Nd
Ru
45
101.07 102.91
44
144.24
Pr
60
(270)
107
Bh
27
Co
55.845 58.933
26
Fe
140.91
59
(269)
Sg
106
183.84 186.21
74
W
95.95
42
51.996 54.938
24
Cr
(9)
Al
13
5
B
3A
(13)
12
5B
(5)
Atomic weight
Symbol
Atomic number
10.81
8B
238.03
U
92
9.0122
22.990
Na
11
6.94
Be
NONMETALS
4
3
Li
METALLOIDS
2A
(2)
TRANSITION METALS
MAIN GROUP METALS
1A
(1)
1.008
1
Note: Atomic weights are
58
IUPAC values. For elements
Lanthanides
for which IUPAC recommends
140.12
ranges of atomic weights,
conventional values are shown.
90
Numbers in parentheses are
Actinides
mass numbers of the most
232.04
stable isotope of an element.
7
6
5
4
3
2
1
Martyn F. Chillmaid/Science Source
Group 1A (1) metals are
soft and some, like sodium
and potassium, can be cut
with a knife.
© Charles D. Winters/Cengage
Group 1A (1) metals react
vigorously with water to
give hydrogen gas and an
alkaline solution of the
metal hydroxide.
Figure 2.6
Properties of the
alkali metals.
A Brief Overview of the Periodic Table
and the Chemical Elements
Elements in the leftmost column, Group 1A (1), are known as the alkali metals
(except H). The word alkali comes from Arabic. Ancient Arabian chemists discovered
that the ashes of certain plants, which they called al-qali, gave water solutions that
felt slippery and burned the skin. Chemists now know these ashes contain compounds of Group 1A (1) elements that produce alkaline (basic) solutions.
All the alkali metals are solids at room temperature and reactive. For example,
they react with water to produce hydrogen and alkaline solutions (Figure 2.6).
­Because of their reactivity, these metals are only found in nature combined in compounds (such as NaCl), never as free elements.
Group 2A (2) is also composed entirely of metals that occur naturally only in
compounds. Except for beryllium (Be), these elements react with water to produce
alkaline solutions, and most of their oxides (such as lime, CaO) form alkaline solutions; hence, they are known as the alkaline earth metals. Magnesium (Mg) and
calcium (Ca) are the seventh and fifth most abundant elements in the Earth’s crust,
respectively (Table 2.3). Calcium, one of the important elements in teeth and bones,
occurs naturally in vast limestone deposits. Calcium carbonate (CaCO3) is the chief
constituent of limestone as well as corals, sea shells, marble, and chalk. Radium
(Ra), the heaviest alkaline earth element, is radioactive.
Table 2.3
The 10 Most Abundant Elements in
the Earth’s Crust
Rank
Element
Abundance (ppm)*
1
Oxygen
474,000
2
Silicon
277,000
3
Aluminum
82,000
4
Iron
41,000
5
Calcium
41,000
6
Sodium
23,000
7
Magnesium
23,000
8
Potassium
21,000
9
Titanium
5600
10
Hydrogen
1520
*ppm = parts per million = g per 1000 kg.
2.3 The Periodic Table
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77
A Closer Look
Mendeleev and the Periodic Table
John C. Kotz
Statue of Dmitri Mendeleev and a periodic
table. This statue and mural are at the
Institute of Metrology in St. Petersburg,
Russia.
78
column where he believed an unknown element should exist. He deduced that these
spaces would be filled by undiscovered
elements. For example, he left a space between Si (silicon) and Sn (tin) in Group 4A
(14) for an element he called eka-silicon.
Based on the progression of properties in
this group, Mendeleev was able to predict
the properties of the missing element.
With the discovery of germanium (Ge) in
1886, Mendeleev’s prediction was
confirmed.
In Mendeleev’s table the elements were
ordered by increasing atomic weight. A
glance at a modern table, however, shows
that, if listed in order of increasing atomic
weight, three pairs of elements (Ni and Co,
Ar and K, and Te and I) would be out of order. Mendeleev assumed the atomic weights
known at that time were i­naccurate—not a
bad assumption based on the analytical
methods then in use. In fact, his order is
correct and his assumption that element
“Key Experiments”) and examined the
X-rays emitted in the process. Moseley
­realized the wavelength of the X-rays emitted by a given element was related in a
precise manner to the positive charge in
the nucleus of the element and that this
provided a way to experimentally determine the atomic number of a given element. Indeed, once atomic numbers could
be determined, chemists recognized that
organizing the elements in a table by
increasing atomic number corrected the
­
inconsistencies in Mendeleev’s table. The
law of chemical periodicity now states that
the properties of the elements are periodic
functions of atomic number.
References
For more on the periodic table, see:
1. J. Emsley: Nature’s Building Blocks—An
A–Z Guide to the Elements, Second Edition,
New York, Oxford University Press, 2011.
2. E. Scerri, The Periodic Table, Second Edition,
New York, Oxford University Press, 2020.
Chapter 2 / Atoms, Molecules, and Ions
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Photos: © Charles D. Winters/Cengage
RE IH E N
Although the arrangement of elSodium
Germanium
Iodine
Copper
ements in the periodic table is
now understood on the basis of
atomic structure, the table was
originally developed from many
experimental observations of
the chemical and physical properties of elements and is the
result of the ideas of a number
TABELLE I I .
of chemists in the eighteenth
GRUPPE I. G RUPPE II. G RUPPE III. G RUPPE IV. G RUPPE V. G RUPPE VI. G RUPPE VII.
GRU PPE VI I I .
and nineteenth centuries.
—
—
—
RH 4
RH 3
RH 2
RH
—
In 1869, at the University of
R2O
RO
R2O3
R 2O 7
RO 4
RO 2
R 2O 5
RO 3
St. Petersburg in Russia, Dmitri
1
H=1
­Ivanovich Mendeleev (1834–1907)
2 Li = 7
Be = 9,4
B = 11
C = 12
N = 14
O = 16
F = 19
3
Na = 23
Mg = 24
Al = 27,3
Si = 28
P = 31
S = 32
Cl = 35,5
wrote a textbook on chemistry. As
4 K = 39
Fe = 56, Co = 59,
Ca = 40
— = 44
Ti = 48
V = 51
Cr = 52
Mn = 55
he pondered the chemical and
Ni = 59, Cu = 63.
physical properties of the elements,
5
(Cu = 63)
Zn = 65
— = 68
— = 72
As = 75
Se = 78
Br = 80
6 Rb = 85
Ru = 104, Rh = 104,
Sr = 87
?Yt = 88
Zr = 90
Nb = 94
Mo = 96
— = 100
he realized that, if the elements
Pd = 106, Ag = 108.
were arranged in order of increasing
7
(Ag = 108)
Cd = 112
In = 113
Sn = 118
Sb = 122
Te = 125
J = 127
atomic weight, elements with simi8 Cs = 133
————
Ba = 137
?Di = 138
?Ce = 140
—
—
—
lar properties appeared in a regular
9
( —)
—
—
—
—
—
—
10 —
Os = 195, Ir = 197,
—
?Er = 178
?La = 180
Ta = 182
W = 184
—
pattern. That is, he saw a ­periodicity
Pt = 198, Au = 199.
or periodic repetition of the proper11
(Au = 199)
Hg = 200
Tl = 204
Pb = 207
Bi = 208
—
—
ties of elements. Mendeleev orga12 —
————
—
—
Th = 231
—
U = 240
—
nized the known elements into a
table by lining them up in horizonThe original Mendeleev table showing the places he left for as yet-undiscovered elements.
tal rows in order of increasing
atomic weight. When he came to an ele- started a new row. As more elements were properties were a function of their atomic
ment with properties similar to one already added to the table, new rows were begun, weight was wrong.
in the row, he started a new row. For ex- and elements with similar properties (such
In 1913, H. G. J. Moseley (1887–1915),
ample, the elements Li, Be, B, C, N, O, and as Li, Na, and K) were placed in the same a young English scientist working with
F were in a row. Sodium was the next ele- vertical column.
Ernest Rutherford (1871–1937), bomment then known; because its properties
An important feature of Mendeleev’s barded many different metals with elecclosely resembled those of Li, Mendeleev table was that he left an empty space in a trons in a cathode-ray tube (see page 72,
© Charles D. Winters/Cengage
Figure 2.7 Liquid
gallium. Bromine and mercury
are the only elements that are
liquids at room temperature and
usual atmospheric pressures.
Gallium and cesium melt slightly
above room temperature.
Gallium melts (melting point =
29.8 °C) when held in the hand.
Group 3A (13) includes a metalloid—boron—and metals—aluminum, gallium
(Figure 2.7), indium, and thallium. As a metalloid, boron (B) has a different chemistry than the other elements in its group. Nonetheless, all form compounds with
analogous formulas, such as BCl3 and AlCl3, and this similarity marks them as
members of the same periodic group. Boron occurs in the mineral borax, a compound used as a cleaning agent, antiseptic, and flux for metal work.
Aluminum (Al) is the most abundant metal in the Earth’s crust at 8.2% by mass.
It is exceeded in abundance only by the nonmetal oxygen and metalloid silicon, and
is usually found in minerals and clays.
In Group 4A (14), there is a nonmetal, carbon (C), two metalloids, silicon (Si)
and germanium (Ge), and two metals, tin (Sn) and lead (Pb). Because of the change
from nonmetallic to metallic behavior, the properties of the elements of this group
have more variation than in most others. Nonetheless, there are similarities. For
example, these elements form compounds with analogous formulas such as CO2,
SiO2, GeO2, and PbO2.
Carbon has many allotropes, the best known of which are graphite and diamond. Graphite consists of flat sheets in which each carbon atom is connected to
three others (Figure 2.8a). Because the sheets of carbon atoms cling only weakly to
one another, the layers slip easily over each other. This explains why graphite is soft,
a good lubricant, and used in pencil lead. [Pencil “lead” is not the element lead (Pb)
but a composite of clay and graphite that leaves a trail of graphite on the page as
you write.]
In diamond, each carbon atom is connected to four others at the corners of a
tetrahedron, and this extends throughout the solid (Figure 2.8b). This structure
causes diamonds to be extremely hard, denser than graphite (d = 3.51 g/cm3 for diamond versus d = 2.22 g/cm3 for graphite), and chemically less reactive. Because diamonds are both hard and excellent conductors of heat, they are used on the tips of
metal- and rock-cutting tools.
In the late 1980s, another form of carbon was identified as a component of
black soot, the stuff that collects when carbon-containing materials are burned in a
deficiency of oxygen. This substance is made up of molecules with 60 carbon atoms
arranged as a spherical cage (Figure 2.8c). The surface is made up of five- and sixmember rings and resembles a hollow soccer ball. The shape also reminded its discoverers of an architectural dome conceived by the U.S. philosopher and engineer,
R. Buckminster Fuller (1895–1983). This led to the official name of the allotrope,
buckminsterfullerene, although chemists often call these molecules “buckyballs.”
Carbon’s chemistry is unique because of its ability to bond to other carbon
­atoms to form chains and rings to which atoms of other elements can be attached.
Most of the millions of chemical compounds known contain carbon.
Elements 113–118. The heaviest
elements in Groups 3A–8A
(13–18) with atomic numbers
113–118 are all laboratorycreated and have very short
existences. Their properties and
reactions are not well known so
they have been omitted from the
overview of group properties in
this chapter.
2.3 The Periodic Table
Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
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79
© Charles D. Winters/Cengage
© Charles D. Winters/Cengage
Mark A. Schneider/Science Source
(a) Graphite. Graphite consists of layers of
carbon atoms.
Each six-member ring shares an edge with
three other six-member rings and three
five-member rings.
Each C atom is connected tetrahedrally to four
other C atoms.
Each carbon atom is linked to three others to
form a sheet of six-member, hexagonal rings.
(b) Diamond. In diamond the carbon atoms
are also arranged in six-member rings, but the
rings are not planar.
Figure 2.8 The allotropes of carbon.
(c) Buckyballs. A member of the family
called buckminsterfullerenes, C60 is an
allotrope of carbon. Sixty carbon atoms are
arranged in a spherical cage that resembles a
hollow soccer ball. Chemists call this
molecule a buckyball. C60 is a black powder; it
is shown here in the tip of a pointed glass
tube.
Oxides of silicon are the basis of many minerals such as clay, quartz, and beautiful gemstones like amethyst. Tin and lead have been known for centuries because
they are easily smelted from their ores. Tin alloyed with copper makes bronze,
which was used in ancient times in utensils and weapons. Lead has been used in
water pipes and paint, even though the element is toxic to humans.
Like Group 4A (14), Group 5A (15) contains nonmetals (N and P), metalloids
(As and Sb), and a metal (Bi). In spite of these variations, they form analogous
compounds such as the oxides N2O5, P4O10, and As2O5. Nitrogen occurs naturally
in the form of the diatomic molecule N2 (Figure 2.9) and makes up about threefourths of Earth’s atmosphere. It is also found in biochemically important ­substances
such as chlorophyll, proteins, and DNA. Scientists have long studied ways to make
compounds from atmospheric nitrogen, a process called nitrogen fixation. Nature
accomplishes this easily in some prokaryotic organisms, but extreme conditions
(high temperatures, for example) must be used in the laboratory and industry for
N2 to react with other elements (such as H2 to make ammonia, NH3, which is widely
used as a fertilizer).
Phosphorus is also essential to life. It is an important constituent in bones,
teeth, and DNA. The element glows in the dark if it is in the air (owing to its reaction with O2), and its name is based on Greek words meaning “light-bearing.”
This element has several allotropes, the most important being white and red
H2
N2
O2
O3
F2
Cl 2
Br2
I2
Figure 2.9 Elements that exist as diatomic or triatomic molecules. Seven of the known
elements exist as diatomic, or two-atom, molecules. Oxygen has an additional allotrope, ozone,
with three O atoms in each molecule.
80
Chapter 2 / Atoms, Molecules, and Ions
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Iodine, I2
© Charles D. Winters/Cengage
Bromine, Br2
© Charles D. Winters/Cengage
phosphorus. White phosphorus (composed of P4 molecules) ignites spontaneously
in air, so it is normally stored under water. When it reacts with air, it forms P4O10,
which can react with water to form phosphoric acid (H3PO4), a compound used in
food products such as soft drinks. Red phosphorus is used in the striking strips on
match books. When a match is struck, potassium chlorate in the match head mixes
with some red phosphorus on the striking strip, and the friction is enough to ignite
the mixture.
In Group 6A (16) there is again a variation of properties. Oxygen, sulfur, and
­selenium are nonmetals, tellurium is a metalloid, and polonium, a radioactive
element discovered in 1898 by Marie and Pierre Curie (see “A Closer Look:
­
­Marie ­Curie”), is a metal. Nonetheless, there is a family resemblance in their chemistries. All of oxygen’s fellow group members form oxygen-containing compounds
(SO2, SeO2, and TeO2), and all form sodium-containing compounds (Na2O, Na2S,
Na2Se, and Na2Te).
Oxygen, which constitutes about 20% of Earth’s atmosphere and which combines readily with most other elements, is at the top of Group 6A (16). Most of the
energy that powers life on Earth is derived from reactions in which oxygen combines
with other substances.
Sulfur has been known in elemental form since ancient times as brimstone or
“burning stone” (Figure 2.10). Sulfur, selenium, and tellurium are often referred to
collectively as chalcogens (from the Greek word, khalkos, for copper) because most
copper ores contain these elements. Their compounds can be foul-smelling and
poisonous; nevertheless, sulfur and selenium are essential components of the human diet. By far the most important compound of sulfur is sulfuric acid (H2SO4),
which is manufactured in larger amounts than any other compound.
As in Group 5A (15), the second- and third-period elements of Group 6A (16)
have different structures. Like nitrogen, oxygen is also a diatomic molecule (see
Figure 2.9). Unlike nitrogen, however, oxygen has an allotrope, the triatomic molecule ozone, O3. Sulfur, which can be found in nature as a yellow solid, has many
allotropes, the most common of which consists of eight-member, crown-shaped
rings of sulfur atoms (see Figure 2.10).
The Group 7A (17) elements—fluorine, chlorine, bromine, iodine, and radioactive astatine—are all nonmetals and all exist as diatomic molecules. At room temperature fluorine (F2) and chlorine (Cl2) are gases. Bromine (Br2) is a liquid and
iodine (I2) is a solid, but bromine and iodine vapor are clearly visible over the liquid
or solid (Figure 2.11).
Figure 2.10 Sulfur. The most
common allotrope of sulfur consists
of S atoms arranged in eightmember, crown-shaped rings.
Special Group Names Some
groups have widely used
common names.
Group 1A (1): Alkali metals
Group 2A (2): Alkaline earth
metals
Group 7A (17): Halogens
Group 8A (18): Noble gases
Figure 2.11 Bromine
and iodine. These and other
Group 7A (17) elements are
commonly called halogens.
2.3 The Periodic Table
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81
Marie Curie, originally known as ­Maria
Skłodowska, was born in Russiancontrolled Warsaw, Poland in 1867.
The daughter of school teachers, she
hoped to attend college, but at that
time the universities in Poland (and
many other European countries) refused to admit women. The Sorbonne
in Paris, France did admit women,
but she did not have enough money to
attend. She and her sister Bronya,
however, worked out a plan to help
each other. Maria would work as a tutor
and governess to support Bronya while
Bronya studied medicine in Paris, and
then Bronya would work to support Maria.
Maria also attended what was called a flying, or floating university in Warsaw — an
illegal underground movement in which
classes were taught informally in private
homes and other locations. Of particular
importance to Maria was a secret Polish
laboratory at the Museum of Industry and
Agriculture, where she learned techniques
of chemical analysis. In 1891, Bronya fulfilled her promise, and Maria enrolled at
the Sorbonne as Marie. She lived frugally,
spending almost nothing on luxuries or
sometimes even on food. On several occasions, she fainted because she worked
constantly and had such a meager diet.
She finished first in her physics masters
program in the summer of 1893 and second in her mathematics masters the following year. She then decided to do
advanced work in physics.
In 1896, Henri Becquerel accidentally
discovered radioactivity when he placed a
uranium compound in a drawer with photographic plates that were covered with black
paper. When he developed the plates, an
image was produced even though the plates
had not been exposed to light. Becquerel
correctly surmised that the uranium was
emitting some type of radiation. (Abel
Niepce de Saint Victor carried out similar
experiments in 1857 and came to the same
conclusion, but his results were largely ignored.) Marie Curie was considering possible topics for her doctoral thesis at the
Sorbonne and had read about Becquerel’s
work. She was interested in whether other
elements would emit such radiation. Because her French physicist husband, Pierre
Curie (1859–1906), was working at the
École Supérieure de Physique et de Chimie
82
Hulton Archive/Stringer/Getty Images
A Closer Look
Marie Curie (1867–1934)
Marie and Pierre Curie in their
laboratory.
Industrielles de la Ville de Paris (ESPCI),
she was allowed to perform her research in
a laboratory there. After Pierre developed a
highly sensitive method to detect radiation
from a radioactive source, Marie tested every substance she could find for radioactivity. She determined that the degree to
which uranium ore was radioactive—a term
she invented to describe the p
­ henomenon—
depended on the percent of uranium
­present, which confirmed ­Becquerel’s hypothesis that uranium metal itself was radioactive. When she tested pitchblende, a
common ore containing uranium and other
metals, she was astonished to find that it
was even more radioactive than pure uranium. There was only one explanation:
pitchblende contained at least one element
more radioactive than uranium.
In 1898, using chemical analysis techniques, the Curies separated pitchblende
into fractions containing different materials. One fraction, which contained mainly
the non-radioactive element bismuth, was
nonetheless very radioactive; another contained mostly barium and was also very
radioactive. The Curies proposed that each
of these fractions contained a new element. They named the element in the bismuth fraction polonium after Marie’s
homeland of Poland and the element in
the barium fraction radium. Working in a
run-down shed at the ESPCI, the Curies
started with tons of pitchblende and
performed thousands of steps to successfully isolate about a tenth of a gram of
pure radium chloride in 1902.
In 1903, Marie earned her doctorate in
physics at the Sorbonne, becoming the
first women in France to earn this degree.
Also in 1903, Henri Becquerel and Marie
and Pierre Curie shared the Nobel Prize in
physics for their discovery of “spontaneous
radioactivity.” Initially, the Nobel committee nominated only Becquerel and Pierre
for this honor, but Pierre declared that he
would not accept the prize unless Marie
was also included. Marie was the first
woman to win a Nobel Prize. In 1906,
Pierre was killed when he fell beneath the
wheels of a horse-drawn vehicle, and Marie
was hired to take his place at the Sorbonne, where he had moved in 1900. She
was the first woman ever hired to teach at
that university. In 1910, working with
­André-Louis Debierne, Marie isolated pure
metallic radium. Despite her accomplishments, the French Academy of Sciences
chose to elect Édouard Branly instead of
Marie Curie to fill a vacancy in the organization in 1911. This was, in part, because
of her Polish ancestry, a false rumor that
she was Jewish, and being a woman. In
1911–the same year she was rejected by
the Academy–Marie was awarded a second
Nobel Prize, this time in chemistry for the
discovery of radium and polonium. As of
2021, only four people and two organizations have ever won multiple Nobel Prizes.
The Radium Institute in Paris, established
in 1909, opened in 1914 with ­Marie as the
Director of the Physics and Chemistry Laboratory. During World War I, Marie learned
that the entire French Army had only one
X-ray station. She proposed and helped
­establish mobile X-ray units and then taught
herself to drive and operated one of the 20
mobile units, nicknamed “petite Curies,” used
during the war. Approximately one ­million
soldiers were treated with these m
­
­ obile
units. ­Marie Curie died at the age of 66 in
1934 from aplastic pernicious a­nemia, a
condition she developed from years of
­radiation exposure. A unit of radioactivity
­(curie, Ci) and an element (curium, Cm) are
named in honor of Marie and Pierre Curie.
One of their daughters, Irène, married
­Frédéric Joliot, and they shared the 1935
Nobel Prize in chemistry for their discovery
of induced radioactivity.
Chapter 2 / Atoms, Molecules, and Ions
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The Group 7A (17) elements are among the most reactive of all elements, and
all combine violently with alkali metals to form salts such as table salt, NaCl. The
name for this group, the halogens, comes from the Greek words hals, meaning
“salt,” and genes, for “forming.”
The Group 8A (18) elements—helium, neon, argon, krypton, xenon, and radioactive radon—are all nonmetals and are the least reactive elements. All are gases,
and none is abundant on Earth or in the Earth’s atmosphere (although argon is the
third most abundant gas in dry air at 0.9%). Because of this, they were not discovered until the end of the nineteenth century (see page 118). Their general lack of
reactivity is reflected in the common name for this group, the noble gases. Just as
the Western nobility class at this time was considered to be generally aloof and hesitant to mix with commoners, these elements are very unreactive toward other
elements.
Helium, the second most abundant element in the universe after hydrogen,
was detected in the sun in 1868 by analysis of the solar spectrum but was not
found on Earth until 1895. It is now widely used, with a worldwide production of
the gas in 2019 of about 160 billion liters. The biggest single use of helium is to
cool the magnets found in magnetic resonance imaging (MRI) units in hospitals
and nuclear magnetic resonance (NMR) spectrometers in research laboratories
(Figure 2.12). These magnets must be cooled with liquid helium to 4 K because,
at this extremely low temperature, the magnets are superconductors of electricity.
They can then generate the high magnetic fields needed to produce an image of
your body. In addition, helium gas is used to fill weather balloons (and party balloons) and in the semiconductor industry. The United States supplies most of the
helium, but there are periodic shortages that seriously disrupt commerce and
research.
Stretching between Groups 2A and 3A (2 and 13) in the periodic table is a series
of elements called the transition elements. These fill the B-groups (3–12) in the
fourth through the seventh periods in the center of the periodic table. All are metals,
and 13 of them are in the top 30 elements in terms of abundance in the Earth’s crust.
Most occur naturally in combination with other elements, but a few—copper (Cu),
silver (Ag), gold (Au), and platinum (Pt)—can be found in nature as pure
elements.
Virtually all of the transition elements have commercial uses. They are used as
structural materials (iron, titanium, chromium, copper); in paints (titanium,
­chromium); in the catalytic converters in automobile exhaust systems (platinum
and rhodium); in coins (copper, nickel, zinc); and in batteries (manganese, nickel,
zinc, cadmium, mercury).
Two rows at the bottom of the table accommodate the lanthanides [the series
of elements between lanthanum (Z = 57) and hafnium (Z = 72)] and the actinides
[the series of elements between actinium (Z = 89) and rutherfordium (Z = 104)].
The lanthanides are often referred to as rare earth elements (Figure 2.13). In fact,
they are not very rare but are geologically widely dispersed. In spite of the difficulty
in mining rare earth–containing minerals, they have become very important commercially. They are used in magnets, LCD screens, smartphones, computers, hybrid
car batteries, wind turbines, and powders used to polish glass. Minerals containing
rare earth elements are mined largely in China, and there is concern that a worldwide shortage looms.
Figure 2.12 Helium, a noble
gas, and MRI units. The magnets
of MRI units need to be cooled
to 4 K with liquid helium to
generate the high magnetic field
required. This is the largest use of
this noble gas.
2.3 The Periodic Table
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83
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Bjoern Wylezich/Shutterstock.com
(a)
(b)
Figure 2.13 The rare earth elements. ​(a) A sample of the rare earth element terbium. (b) A mine for ores containing
rare earth elements. The rare earth elements are essential for many modern devices. Obtaining sufficient amounts has
environmental, economic, and geopolitical consequences.
2.4 Molecules: Formulas,
Models, and Names
Goals for Section 2.4
•
Recognize and interpret molecular formulas, condensed formulas, and structural
formulas.
•
•
•
Recognize and interpret different types of molecular models.
Remember formulas and names of common molecular compounds.
Name and write formulas for binary molecular compounds.
In Chapter 1, you learned that some elements (such as oxygen, O2) and many compounds (such as water, H2O) are composed of particles called molecules. A molecule is the smallest identifiable unit into which some pure substances like sugar and
water can be divided and still retain the composition and chemical properties of the
substance. Such substances are composed of identical molecules consisting of two
or more atoms bound firmly together. In the reaction below (and in Figure 2.14),
molecules of sulfur, S8, combine with molecules of oxygen, O2, to produce molecules of the compound sulfur dioxide, SO2.
S8(s) + 8 O2(g) → 8 SO2(g)
sulfur + oxygen → sulfur dioxide
The composition of each element and compound in the chemical change (or chemical reaction) is represented by a formula that indicates the types and numbers of
atoms present in one molecule of the substance. For example, one molecule of SO2
is composed of one S atom and two O atoms.
Formulas
There is often more than one way to write the formula of a molecular compound.
For example, the formula of ethanol (also called ethyl alcohol) can be represented
as C2H6O (Figure 2.15). This molecular formula describes the composition of ethanol molecules—two carbon atoms, six hydrogen atoms, and one oxygen atom per
84
Chapter 2 / Atoms, Molecules, and Ions
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Sulfur dioxide, SO2 (g)
Oxygen, O2 (g)
Photos: © Charles D. Winters/Cengage
Sulfur, S8 (s)
Figure 2.14 The reaction of the molecular elements sulfur and oxygen to give the molecular compound sulfur dioxide.
molecule—but it gives no structural information. Structural information—how the
atoms are connected and how the molecule fills space—is important because it
helps you to understand how a molecule can interact with other molecules.
To provide some structural information, it is useful to write a condensed
­formula, which indicates how certain atoms are grouped together. For example, the
condensed formula of ethanol, CH3CH2OH (Figure 2.15), indicates that the molecule consists of three groups: a CH3 group, a CH2 group, and an OH group. Writing
the formula as CH3CH2OH also shows that the compound is not dimethyl ether,
CH3OCH3, a compound with the same molecular formula but with a different
­structure and distinctly different properties.
That ethanol and dimethyl ether are different molecules is also clear from their
structural formulas (Figure 2.15). This type of formula gives an even higher level of
structural detail, showing how all of the atoms are attached within a molecule. The
lines between atoms represent the chemical bonds that hold atoms together in the
molecule.
Writing Formulas When writing
molecular formulas of organic
compounds (compounds with
C, H, and other elements)
the convention is to write C
first, then H, and finally other
elements in alphabetical order.
For example, acrylonitrile,
a compound used to make
consumer plastics, has the
condensed formula CH2CHCN.
Its molecular formula would be
C3H3N.
Molecular Models
The physical and chemical properties of compounds are often closely related to their
structures (which is why you will see so many molecular models in this book). For
example, two well-known features of ice are related to its underlying molecular
structure (Figure 2.16). The first is the shape of ice crystals: The sixfold symmetry of
macroscopic ice crystals also appears at the particulate level in the form of six-sided
rings of hydrogen and oxygen atoms. The second is water’s unusual property of
NAME
MOLECULAR
FORMULA
CONDENSED
FORMULA
Ethanol
C2H6O
CH3CH2OH
STRUCTURAL
FORMULA
MOLECULAR MODEL
H H
H
C
C
O
H
H H
Dimethyl
ether
C2H6O
CH3OCH3
C
H
Here the two compounds have
the same molecular formula.
Condensed or structural formulas,
or a molecular model, clearly
show these compounds are
different.
H
H
H
Figure 2.15 Four approaches
to showing molecular formulas. ​
O
C
H
H
2.4 Molecules: Formulas, Models, and Names
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85
The six-sided structure of a
snowflake is a reflection of the
underlying molecular structure
of ice.
Standard Colors for Atoms in
Molecular Models The colors
listed here are used for
molecular models in this book
and are generally used by
chemists.
carbon atoms
hydrogen atoms
oxygen atoms
nitrogen atoms
Alexey Kljatov/Shutterstock.com
Ice consists of six-sided rings
formed by water molecules, in
which each side of a ring
consists of two O atoms and
an H atom.
Figure 2.16 Ice. ​
Snowflakes reflect the
underlying structure of ice.
being less dense as a solid than as a liquid. The lower density of ice, which has enormous consequences for Earth’s climate, results from the fact that molecules of water
do not pack together tightly in ice.
Because molecules are three dimensional, it is often difficult to represent their
structures on paper. Certain conventions have been developed, however, that help
represent three-dimensional structures on two-dimensional surfaces. Perspective
drawings are often used (Figure 2.17).
Molecular models are very useful for visualizing structures. These models
show how atoms are attached to one another and show the molecule’s overall
three-dimensional structure. In ball-and-stick models, spheres of different colors
represent the atoms, and sticks represent the bonds holding them together.
­Molecules can also be represented using space-filling models, representations of
molecules in which the atoms are partial spheres that have diameters proportional
to those of the atoms and are joined directly to one another. These models are a
better representation of relative sizes of atoms and their proximity to each other.
A disadvantage of pictures of space-filling models is that atoms can often be hidden from view.
chlorine atoms
Naming Molecular Compounds
There are many molecular compounds you will encounter often. You should understand how to name many of them and, in many cases, be able to write their formulas. First consider molecules formed from combinations of two nonmetals, known
as binary compounds of nonmetals. Although there are exceptions, most binary
molecular compounds are a combination of nonmetallic elements from Groups
4A–7A (14–17) with one another or with hydrogen. The formula is generally
H
Bonds in
plane
of paper
C
H
H
H
H
John C. Kotz
Bond going
away from
H
observer
Bond coming
toward
observer
Perspective
drawing
Plastic model
Ball-and-stick
model
Space-filling model
C
H
H
The three representations
in a single drawing.
Figure 2.17 Ways of depicting a molecule, here the methane (CH4) molecule.
86
Chapter 2 / Atoms, Molecules, and Ions
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written by putting the elements in order of increasing group number. If the two elements are from the same group, then the element further down in the group is usually written first.
Hydrogen forms binary compounds with all of the nonmetals except the noble
gases. For compounds of oxygen, sulfur, and the halogens, the H atom is generally
written first in the formula and is named first. The other nonmetal is named by
­adding -ide to the stem of the name.
Compound
Name
HF
Hydrogen fluoride
HCl
Hydrogen chloride
H2S
Hydrogen sulfide
Formulas of Binary Nonmetal
Compounds Containing
Hydrogen Simple hydrocarbons
(compounds of C and H) such
as methane (CH4) and ethane
(C2H6) have formulas written
with H following C, and the
formulas of ammonia and
hydrazine have H following
N. Water and the hydrogen
halides, however, have the
H atom preceding O or the
halogen atom. Tradition is
the only explanation for such
irregularities in writing formulas.
When naming most other binary molecular compounds, the number of
­atoms of a given type in the compound is designated with a prefix.
Number of Atoms
Prefix
1
mono-
2
di-
3
tri-
4
tetra-
5
penta-
6
hexa-
7
hepta-
8
octa-
9
nona-
10
deca-
If there is only one atom of a given type and it is the second element listed in
the binary compound formula, the prefix “mono-” is used. However, “mono-” is
omitted if it is the first element in the formula.
Exam p le 2.5
Binary Molecular Compounds
Problem Provide the missing name or formula for the following binary compounds.
(a) What is the name of N2O5?
(b) What is the name of CO?
(c) What is the formula of tetraphosphorus decaoxide?
What Do You Know? You know the formulas for N2O5 and CO and the name for
tetraphosphorus decaoxide. You know that binary molecular compounds are usually
c­ omposed of two nonmetals, and you know the prefixes used in naming binary molecular
compounds.
2.4 Molecules: Formulas, Models, and Names
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87
Strategy For parts a and b, first determine whether the binary compound is likely to
be molecular by noting if both elements are nonmetals. The prefix system is used in naming most binary molecular compounds. For part c, the prefixes indicate the number of
atoms for each element present in a binary molecular compound.
Solution
(a) The two elements present in the compound are nitrogen and oxygen. These are both
nonmetals, so the compound is molecular. There are two atoms of nitrogen, so the
prefix di- is used. There are five atoms of oxygen, so the prefix penta- is used. The ending of the second element is changed to -ide. The name is dinitrogen pentaoxide.
(b) The two elements present are carbon and oxygen, which are both nonmetals, so the
compound is molecular. There is only one atom of each element present. The prefix
mono- is not used for the first element in a compound’s name, so no prefix is needed
for the carbon. The prefix mono- is used with the second element, and the ending of
the element’s name needs to be changed to -ide. Rather than say monooxide, however, the word is simplified to monoxide. The name is carbon monoxide.
(c) The prefixes indicate the number of atoms for each element present in the binary
molecular compound. Tetra- indicates that four atoms of phosphorus are present,
and deca- indicates that ten atoms of oxygen are present. The order for writing the
element symbols is the same as that in the name: the element further left on the
periodic table is written first. The formula is P4O10.
Think about Your Answer All the compounds in this example were binary molecular compounds, so the appropriate prefixes were used in the names. In the same way that
monoxide is used rather than monooxide, some people also drop a final “a” in a prefix used
before oxide. Thus, some would have given the name of N2O5 as dinitrogen pentoxide.
Check Your Understanding
(1) What are the names of (a) SO2, (b) NI3, and (c) N2O.
(2) Write the formulas for (a) dinitrogen tetraoxide, (b) sulfur hexafluoride, and (c) disulfur
decafluoride.
Hydrocarbons Compounds such
as methane, ethane, propane,
and butane belong to a class of
hydrocarbons called alkanes.
methane, CH4
ethane, C2H6
88
propane, C3H8
butane, C4H10
Finally, some binary compounds of nonmetals were discovered many years ago
and have common names.
Compound
Common Name
Compound
Common Name
CH4
Methane
N2H4
Hydrazine
C2H6
Ethane
PH3
Phosphine
C3H8
Propane
NO
Nitric oxide
C4H10
Butane
N2O
Nitrous oxide (laughing
gas)
NH3
Ammonia
H2O
Water
Chapter 2 / Atoms, Molecules, and Ions
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2.5 Ions
Goals for Section 2.5
•
Recognize that metal atoms commonly lose one or more electrons to form positive
ions, called cations, and nonmetal atoms often gain electrons to form negative
ions, called anions.
•
•
Predict the charge on monatomic cations and anions based on Group number.
Name and write the formulas of ions.
The compounds you have encountered so far in this chapter are molecular
­
compounds, that is, compounds that consist of discrete molecules at the particulate
level. Ionic compounds make up another major class of compounds, and many may
be familiar to you (Figure 2.18). Table salt, or sodium chloride (NaCl), and lime
(CaO) are just two. Ionic compounds consist of ions, that is, atoms or groups of
atoms that bear a positive or negative electric charge.
Monatomic Ions
Atoms of many elements can lose or gain electrons to form monatomic ions, ions
consisting of a single atom bearing a charge. Some commonly-encountered ions are
listed in Figure 2.19. How do you know if an atom is likely to gain or lose electrons?
It depends on whether the element is a metal or nonmetal. In reactions,
•
Metals generally lose one or more electrons.
•
Nonmetals frequently gain one or more electrons.
Monatomic Cations
If an atom loses an electron (which is transferred to an atom of another element in
the course of a reaction), the atom now has one less negative electron than it has
positive protons in the nucleus. The result is a positively charged ion called a cation
(Figure 2.20a). (The name is pronounced “cat′-i-on.”) For example, the elements of
Group 1A (1), such as lithium, lose one electron per atom, resulting in the formation of cations with a 1+ charge.
Figure 2.18 Some common ionic compounds.
Gypsum, CaSO4 ⋅ 2 H2O
Common
Name
Calcite, CaCO3
Fluorite, CaF2
Orpiment, As2S3
© Charles D. Winters/Cengage
Hematite, Fe2O3
Name
Formula
Ions
Involved
Calcite
Calcium
carbonate
CaCO3
Ca2+, CO32−
Fluorite
Calcium
fluoride
CaF2
Ca2+, F−
Gypsum
Calcium
sulfate
dihydrate
CaSO4 ∙ 2 H2O
Ca2+, SO42−
Hematite
Iron(III) oxide
Fe2O3
Fe3+, O2−
Orpiment
Arsenic(III)
sulfide
As2S3
As3+, S2−
2.5 Ions
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89
1A
(1)
Figure 2.19 Charges on some
common monatomic cations
and anions. Metals usually form
2A
(2)
H+
cations, and nonmetals usually
form anions. Ions with the same
magnitude charge (for example,
2+ and 2–) are highlighted with
the same color in this figure.
NOTE: It is important to recognize
that transition metals (and a few
main group metals) form cations of
several charges. Examples include
Cr2+ and Cr3+, Fe2+ and Fe3+,
as well as Cu+ and Cu2+. As
explained in the text, their names
must reflect this.
7A 8A
(17) (18)
1+ and 1−
3A 4A 5A 6A
−
(13) (14) (15) (16) H
2+ and 2−
3+ and 3−
4+
Li+
3B
Na+ Mg2+ (3)
K+ Ca2+
4B
(4)
5B
(5)
Ti4+
N3− O2−
(8)
8B
(9)
Cr2+ Mn2+ Fe2+
Co2+
6B
(6)
7B
(7)
Cr3+
1B 2B
3+
(10) (11) (12) Al
Fe3+ Co3+
Ni2+
Ag+ Cd2+
Cs+ Ba2+
Hg22+
Hg2+
→
e−
+
(3 protons and 3 electrons)
Writing Ion Formulas When writing
the formula of an ion, the charge
on the ion must be included.
S2− Cl−
Se2− Br−
Cu2+ Zn2+
Rb+ Sr2+
Li atom
P3−
Cu+
F−
Sn2+
Te2− I−
Pb2+ Bi3+
Li+ cation
(3 protons and 2 electrons)
Elements of Group 2A (2) lose two electrons in reactions, forming cations with a
charge of 2+,
→
Ca atom
2 e−
+
(20 protons and 20 electrons)
Ca2+ cation
(20 protons and 18 electrons)
and elements of Group 3A (13) lose three electrons, forming cations with a charge
of 3+.
→
Al atom
Group 3A (13) cations Like
aluminum, other metals in
Group 3A (13) also lose three
electrons to form cations with
a 3+ charge. But they also, on
occasion, form M+ cations with
a single positive charge.
3 e−
(13 protons and 13 electrons)
+
Al3+ cation
(13 protons and 10 electrons)
How can you predict the number of electrons gained or lost in reactions of elements in Groups 1A, 2A, and 3A (1, 2, and 13)?
•
Metals of Groups 1A, 2A, and 3A lose one or more electrons to form positive
ions having a charge equal to the group number of the metal.
(a) Formation of a lithium (Li+) cation from a neutral Li atom.
3e –
e–
2e –
3p
3n
3p
3n
Li
Li +
3p
3n
3e –
3p
3n
2e –
F
F–
9p
10n
9e –
9p
10n
10e –
Lithium ion, Li +
A lithium-6 atom is electrically neutral
because the number of positive charges
(three protons) and negative charges (three
electrons) are the same. When it loses one
electron, it has one more positive charge
than negative charge, so it has a net charge
of 1+. The resulting lithium cation is
symbolized as Li +.
Lithium, Li
(b) Formation of a fluoride (F –) anion from a neutral F atom.
9e –
9p
10n
10e –
e–
9p
10n
Fluorine, F
A fluorine-19 atom is also electrically
neutral, having nine protons and nine
electrons. A fluorine atom can acquire an
electron to produce an F− anion. This anion
has one more electron than it has protons,
so it has a net charge of 1−.
Fluoride ion, F –
Figure 2.20 Ions.
90
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•
The number of electrons remaining on the cations formed from Group 1A (1)
elements, Group 2A (2) elements, and aluminum is the same as the number of
electrons in an atom of the noble gas that precedes it in the periodic table.
Transition metals (B-group elements) also form cations, but unlike the A-group
metals, there is no easily predictable pattern of behavior. In addition, transition
metals often form several different ions. Iron, for example, may form either Fe2+ or
Fe3+ ions in its reactions. Copper may form 1+ or 2+ ions, but silver forms only a
1+ ion.
Monatomic Anions
Nonmetals can gain electrons to form negatively charged ions. If an atom gains one
or more electrons, there will now be more negatively charged electrons than protons
(Figure 2.20b). A negatively charged ion is called an anion (pronounced “an′i-on”). An oxygen atom, for example, can gain two electrons in a reaction to form
an ion with the formula O2−:
O atom
+
2 e−
→
(8 protons and 8 electrons)
O2− anion
(8 protons and 10 electrons)
A chlorine atom can add a single electron to form Cl−.
Cl atom
+
e−
→
(17 protons and 17 electrons)
Cl− anion
(17 protons and 18 electrons)
Two general observations can be made concerning the formation of anions from
nonmetals.
•
Nonmetals of Groups 5A–7A form negative ions with a charge equal to the
group number of the nonmetal minus 8.
•
The number of electrons on the anion is the same as the number of electrons in
an atom of the noble gas that follows it in the periodic table.
Notice that the nonmetal hydrogen appears at two locations in Figure 2.19. The
H atom can either lose or gain electrons, depending on the other atoms it
encounters.
Electron lost:
H (1 proton, 1 electron) → H+ (1 proton, 0 electrons) + e−
Electron gained:
H (1 proton, 1 electron) + e−
→
Monatomic anion charges Using
the 1–18 system of group
numbers, the charge on a
monatomic anion from Groups
15–17 is equal to the group
number minus 18.
H− (1 proton, 2 electrons)
Finally, the noble gases very rarely form monatomic cations and never form monatomic anions in chemical reactions.
Naming Monatomic Ions
Monatomic ions are named using the following rules:
1. For a monatomic positive ion (that is, a metal cation) from Group 1A (1),
Group 2A (2), or aluminum, the name is that of the metal plus the word
“­cation.” For example, Al31 is the aluminum cation.
2. Transition metals often form more than one type of positive ion. To specify
which ion is involved, the charge of transition metal cations is indicated by a
Roman numeral in parentheses immediately following the ion’s name. For example, Co21 is the cobalt(II) cation, and Co31 is the cobalt(III) cation.
3. A monatomic negative ion (that is, a nonmetal anion) is named by adding -­ide
to the stem of the name of the nonmetal element from which the ion is derived
(Figure 2.21). For example, the anions of the Group 7A (17) elements, the halogens, are known as fluoride, chloride, bromide, and iodide ions and as a group
are called halide ions.
Elements with Multiple Ion
Charges These occur especially
in the transition metals.
However, some main group
metals such as tin (Sn21 and
Sn41) and lead (Pb21 and Pb41)
can also have multiple ion
charges. It is our practice to
always indicate the ion charge
with a Roman numeral when
naming compounds of the
transition metal elements and
in other cases when multiple
charges are possible.
2.5 Ions
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91
1–
3–
H−
2–
hydride
ion
F−
N3− O2−
nitride
ion
oxide
ion
fluoride
ion
P3−
S2−
Cl−
phosphide sulfide
ion
ion
chloride
ion
Se2− Br−
selenide bromide
ion
ion
Te2−
I−
telluride
ion
iodide
ion
Figure 2.21 Names and
charges of some common
monatomic anions.
Polyatomic Ions
A polyatomic ion is an electrically-charged particle that consists of two or more atoms bonded together (Figure 2.22 and Table 2.4). For example, carbonate ion,
CO32−, a common polyatomic anion, consists of one C atom and three O atoms.
The ion has two units of negative charge because there are two more electrons
(a total of 32) in the ion than there are protons (a total of 30) in the nuclei of one
C atom and three O atoms.
The ammonium ion, NH4+, is a common polyatomic cation. In this case, four
H atoms surround an N atom, and the ion has a 1+ electric charge. This ion has
10 electrons, but there are 11 positively charged protons in the nuclei of the N and
H atoms (7 for N, 1 for each H).
Naming Polyatomic Ions
The polyatomic cation you will encounter most in this book and the laboratory is
the ammonium ion, NH4+. Do not confuse the ammonium cation with the ammonia molecule, NH3, which has no electric charge and one less H atom.
Polyatomic anions are common, especially those containing oxygen (called
oxoanions). The names of some of the most common oxoanions are given in T
­ able 2.4.
Although most of these names must be memorized, some guidelines can help.
1. For a series of oxoanions with two members, the oxoanion with the greater
number of oxygen atoms is given the suffix -ate, and the oxoanion with the
smaller number of oxygen atoms has the suffix -ite. For example, consider the
following pairs of ions:
NO3− is the nitrate ion, whereas NO2− is the nitrite ion.
SO42− is the sulfate ion, whereas SO32− is the sulfite ion.
Photos: © Charles D. Winters/Cengage
2. For a series of oxoanions with more than two members, the ion with the largest
number of oxygen atoms has the prefix per- and the suffix -ate. The ion with the
smallest number of oxygen atoms has the prefix hypo- and the suffix -ite. The
chlorine oxoanions are the most commonly encountered example.
Perchlorate ion
ClO3−
Chlorate ion
ClO2−
Chlorite ion
ClO−
Hypochlorite ion
CO32–
Calcite, CaCO3
Calcium carbonate
per . . . ate
increasing
oxygen content
ClO4−
. . . ate
. . . ite
hypo . . . ite
PO43–
Apatite, Ca5F(PO4)3
Calcium fluorophosphate
SO42–
Celestite, SrSO4
Strontium sulfate
Figure 2.22 Common ionic compounds containing polyatomic ions.
92
Chapter 2 / Atoms, Molecules, and Ions
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Table 2.4
Formula
Formulas and Names of Some Common Polyatomic Ions
Name
Formula
Name
Cation: Positive Ion
NH4+
Polyatomic ions To be successful
Ammonium ion
Anions: Negative Ions
Based on a Group 4A (14) element
Based on a Group 7A (17) element
CN−
Cyanide ion
ClO−
Acetate ion
−
Chlorite ion
−
Chlorate ion
−
Perchlorate ion
CH3CO2
−
C2O42−
Hypochlorite ion
ClO2
Carbonate ion
CO3
2−
HCO3
−
Hydrogen carbonate ion
(or bicarbonate ion)
ClO3
ClO4
Oxalate ion
Based on a Group 5A (15) element
Based on a transition metal
−
Nitrite ion
CrO42−
Chromate ion
−
Nitrate ion
Cr2O7
2−
Dichromate ion
−
Permanganate ion
NO2
NO3
Phosphate ion
PO4
3−
HPO4
2−
Hydrogen phosphate ion
−
Dihydrogen phosphate ion
H2PO4
in your study of chemistry,
you must know the names and
formulas (including the ion
charges) of the common ions
listed in this table.
MnO4
Based on a Group 6A (16) element
OH−
Hydroxide ion
SO32−
Sulfite ion
SO42−
Sulfate ion
HSO4−
Hydrogen sulfate ion (or bisulfate ion)
3. Oxoanions that contain hydrogen are named by adding the word “hydrogen”
before the name of the oxoanion. Two hydrogens in an anion are indicated by
“dihydrogen.” Some hydrogen-containing oxoanions also have common names.
For example, the hydrogen carbonate ion, HCO3−, is called the bicarbonate ion.
Ion
Systematic Name
HPO4
Common Name
2−
Hydrogen phosphate ion
−
Dihydrogen phosphate ion
−
Hydrogen carbonate ion
Bicarbonate ion
−
Hydrogen sulfate ion
Bisulfate ion
−
Hydrogen sulfite ion
Bisulfite ion
H2PO4
HCO3
HSO4
HSO3
2.6 Ionic Compounds: Formulas,
Names, and Properties
Goals for Section 2.6
•
•
•
Write formulas for ionic compounds by combining ions in the proper ratio to give
a zero overall charge.
Give the names and write the formulas of ionic compounds.
Understand the importance of Coulomb’s law in chemistry, which describes the
electrostatic forces of attraction and repulsion of ions.
2.6 Ionic Compounds: Formulas, Names, and Properties
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93
© Charles D. Winters/Cengage
Ionic compounds are electrically neutral combinations of cations and anions. How
can you recognize that a compound is likely to be ionic? One indicator is the presence of a metal as the first element in a compound’s formula or the presence of a
polyatomic cation such as the ammonium ion (NH4+). Thus, compounds such as
NaCl, CuCl2, AgNO3, and (NH4)2CO3 are all classified as being ionic. It is important for you to be able to name ionic compounds and to write their formulas.
Because compounds are electrically neutral, they have no net electric charge.
Thus, in an ionic compound the numbers of positive and negative ions must be
such that the positive and negative charges balance. In sodium chloride, the sodium
ion has a 1+ charge (Na+) and the chloride ion has a 1− charge (Cl−). These ions
must be present in a 1∶1 ratio, and so the formula is NaCl.
The gemstone ruby is largely the compound formed from aluminum ions
(Al3+) and oxide ions (O2−). To have a compound with the same number of positive
and negative charges, two Al3+ ions must combine with three O2− ions to give a
formula of Al2O3.
2 Al3+ ions charge = 2 × (3+) = 6+
3 O2− ions charge = 3 × (2−) = 6−
The Color of Rubies The beautiful
red color of a ruby comes from
a trace of Cr3+ ions that take the
place of a few of the Al3+ ions
in the solid.
Al2O3 overall charge = (6+) + (6−) = 0
Calcium, a Group 2A (2) metal, forms a cation having a 2+ charge. It can combine with a variety of anions to form ionic compounds such as those in the following table:
Compound
Ion Combination
CaCl2
Ca
CaCO3
Ca
Ca3(PO4)2
3 Ca2+ + 2 PO43−
2+
2+
Overall Charge on Compound
−
+ 2 Cl
(2+) + 2 × (1−) = 0
+ CO3
2−
(2+) + (2−) = 0
3 × (2+) + 2 × (3−) = 0
In writing formulas of ionic compounds, the convention is that the symbol of
the cation is given first, followed by the anion symbol. Also notice the use of
parentheses when more than one polyatomic ion of a given kind is present [as
in Ca3(PO4)2]. (None, however, are used when only one polyatomic ion is present, as in CaCO3.)
Problem Solving Tip 4.1 Formulas for Ions and Ionic Compounds
Writing formulas for ionic compounds
requires that you know the formulas
and charges of the most common
ions. The charges on monatomic ions
are often evident from the position
of the element in the periodic table,
but you simply have to remember the
formulas and charges of polyatomic
ions, especially the most common
ones such as nitrate, sulfate, carbonate, phosphate, and acetate.
94
If you cannot remember the
formula of a polyatomic ion or if you
encounter an ion you have not seen
before, you may be able to figure out
its formula. For example, suppose you
are told that the formula for sodium
formate is NaCHO2. You know that
the sodium ion is Na+, so the formate
ion must be the remaining portion of
the compound; it must have a charge
of 1− to balance the 1+ charge on
the sodium ion. Thus, the formate ion
must be CHO2−.
Finally, when writing the formulas
of ions, you must include the charge
on the ion (except in the formula of
an ionic compound). Writing Na when
you mean sodium ion is incorrect.
There is a vast difference in the properties of the element sodium (Na) and
those of its ion (Na+).
Chapter 2 / Atoms, Molecules, and Ions
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Exam p le 2.6
Ionic Compound Formulas
Problem For each of the following ionic compounds, write the symbols for the ions
present and give the relative number of each: (a) Li2CO3 and (b) Fe2(SO4)3.
What Do You Know? You know the formulas of the ionic compounds, the predicted charges on monatomic ions (see Figure 2.19), and the formulas and charges of
polyatomic ions (see Table 2.4).
Strategy Divide the formula of the compound into the cations and anions. To accomplish this you will have to recognize, and remember, the formulas of common monatomic
and polyatomic ions.
Solution
(a)
Li2C O3 is composed of two lithium ions, Li+, for each carbonate ion, CO32−. Li is a
Group 1A (1) element and always has a 1+ charge in its compounds. Because the two
1+ charges balance the negative charge of the carbonate ion, the latter must be 2−.
(b) Fe2(SO4)3 contains two iron(III) ions, Fe3+, for every three sulfate ions, SO42−. The way
to recognize this is to recall that sulfate has a 2− charge. Because three sulfate ions are
present (with a total charge of 6−), the two iron cations must have a total charge of
6+. This is possible only if each iron cation has a charge of 3+.
Think about Your Answer Remember that the formula for an ion must include
its composition and its charge. Formulas for ionic compounds are always written with the
cation first and then the anion, but ion charges are not included.
Identifying Charges on Transition
Metal Cations Because ionic
compounds are electrically
neutral, the charges on transition
metal cations can be determined
if anion charges are known.
Check Your Understanding
Give the number and identity of the constituent ions in each of the following ionic compounds: NaF, Cu(NO3)2, and NaCH3CO2.
Exam p le 2.7
Writing Ionic Compound Formulas
Problem Write formulas for ionic compounds composed of an aluminum cation and
each of the following anions: (a) fluoride ion, (b) sulfide ion, and (c) nitrate ion.
What Do You Know? You know the names of the ions involved, the predicted
charges on monatomic ions (see Figure 2.19), and the names, formulas, and charges of
polyatomic ions (see Table 2.4).
Strategy First decide on the formula of the Al cation and the formula of each anion.
Combine the Al cation with each type of anion to form electrically neutral compounds.
2.6 Ionic Compounds: Formulas, Names, and Properties
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95
Solution Aluminum is a Group 3A (13) metal and therefore has a charge of 3+.
(a) Fluorine is a Group 7A (17) element. The charge of the fluoride ion is predicted to be
1− (from 7 − 8 = 1−). Therefore, 3 F− ions are needed to combine with one Al3+. The
formula of the compound is AlF3.
(b) Sulfur is a nonmetal in Group 6A (16), so it forms a 2− anion. Thus, two Al3+ ions [total
charge is 6+ = 2 × (3+)] and three S2− ions [total charge is 6− = 3 × (2−)] are
needed. The compound has the formula Al2S3.
(c) The nitrate ion has the formula NO3− (see Table 2.4). The answer is therefore similar to
the AlF3 case, and the compound has the formula Al(NO3)3. Here parentheses are
placed around NO3 to show that three polyatomic NO3− ions are involved.
Think about Your Answer Compounds are electrically neutral; they have an
overall charge of zero. In an ionic compound, the sum of the positive charges must equal
the sum of the negative charges. The most common error students make is not knowing
the correct charge on an ion.
Check Your Understanding
(a) Write the formulas of all electrically neutral ionic compounds that can be formed by
combining the cations Na+ and Ba2+ with the anions S2− and PO43−.
(b) Iron forms ions having 2+ and 3+ charges. Write the formulas of the compounds
formed between chloride ions and these two different iron cations.
Names of Compounds Containing
Transition Metal Cations Be sure
to remember that the charge
on a transition metal cation is
indicated by a Roman numeral
and is included in the name.
Names of Ionic Compounds
The name of an ionic compound is built from the names of the positive and negative ions in the compound. The name of the positive cation is given first, followed
by the name of the negative anion.
E xamp le 2.8
Naming Ionic Compounds
Problem What is the name for each of the following compounds?
(a) CaBr2
(b) (NH4)2CO3
(c) Co2O3
What Do You Know? You know the formulas of the given compounds. You also
know the names and charges of the monatomic and polyatomic ions.
Strategy First, determine if the compound should be named as an ionic compound. If
so, the name consists of the names of the ions in the compound. If a transition metal (or
other metal with multiple possible charges) is the cation, you will need to determine its
charge and write the charge in Roman numerals after the metal’s name.
96
Chapter 2 / Atoms, Molecules, and Ions
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Solution
(a) The first element listed in the formula (calcium) is a metal, so the compound will be
named as an ionic compound. There is one calcium cation (Ca2+), and two bromide
anions (Br–) in the formula. The name of the compound is calcium bromide . Calcium
is not a transition metal or other metal that can have multiple charges, so no Roman
numerals are needed.
(b) This compound contains the ammonium (NH4+) cation, so the compound will be
named as an ionic compound. There are two ammonium cations and one carbonate
anion (CO32–) in the formula. Its name is ammonium carbonate .
(c) The first element listed in the formula (cobalt) is a transition metal. The compound
will be named as an ionic compound, and the charge of the cobalt must be determined. Keep in mind that the overall compound is electrically neutral, that is, the
sum of the ion charges must be zero. There are three oxide ions (O2–) in the formula.
This gives an overall negative charge of 3 × (2–) = 6–. The total positive charge must
be 6+. There are two cobalt ions whose charges must add up to 6+, so each must be
3+. The compound is cobalt(III) oxide .
Think about Your Answer Recall that different rules are used to name binary
molecular compounds (see Example 2.5).
Check Your Understanding
What is the name for each of the following compounds?
(a) NaHSO4
(b) Mg(OH)2
(c) TiCl2
Properties of Ionic Compounds
When a particle with a negative electric charge is brought near another particle with
a positive electric charge, there is a force of attraction between them (Figure 2.23).
In contrast, there is a repulsive force when two particles with the same charge—both
positive or both negative—are brought together. These forces are called electrostatic
forces, and the force of attraction (or repulsion) between ions is given by Coulomb’s
law (Equation 2.3):
charge on + and − ions
Force = −k
proportionality constant
charge on electron
(n+e)(n−e)
d2
(2.3)
distance between ions
where, for example, n+ is +3 for Al3+ and n− is −2 for O2−. Based on Coulomb’s
law, the force of attraction (Figure 2.23) between oppositely charged ions increases
•
as the ion charges (n+ and n−) increase. Thus, the attraction between ions having charges of 2+ and 2− is greater than that between ions having 1+ and
1− charges.
•
as the distance between the ions becomes smaller.
The Importance of Coulomb’s
Law Coulomb’s law is the
basis for understanding many
fundamental concepts of
chemistry. Among the chapters
where this is important are:
Chapter 3: dissolving
compounds in water.
Chapters 6 & 7: the interaction
of electrons and the atomic
nucleus.
Chapters 8 & 9: the interaction
of atoms to form molecules.
Chapter 11: the interactions
between molecules
(intermolecular forces).
Chapter 12: the formation of
ionic solids.
Chapter 13: the solution
process.
2.6 Ionic Compounds: Formulas, Names, and Properties
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97
+1
n+ = 1
+
Li+
d small
d
n– = –1
−
+
−1
+2
d large
−2
LiF
F−
Ions such as Li+ and F– are held together by a coulombic force
of attraction. Here a lithium ion is attracted to a fluoride ion,
and the distance between the nuclei of the two ions is d.
(a)
As ion charge increases,
force of attraction increases
As distance increases,
force of attraction decreases
(b)
Figure 2.23 Coulomb’s law and electrostatic forces.
The simplest ratio of cations to anions in an ionic compound is represented by
its formula. However, ionic compounds do not consist of simple pairs or small groups
of positive and negative ions. Instead, an ionic solid consists of millions upon millions of ions arranged in an extended three-dimensional network called a crystal
­lattice. A portion of the lattice for NaCl, illustrated in Figure 2.24, represents a
­common way of arranging ions for compounds that have a 1∶1 ratio of cations to
anions.
If ionic compounds are prepared in
water solution and then isolated as
solids, the crystals often retain molecules of water. These compounds are
called hydrated compounds. For example, crystals of the red cobalt(II)
compound in the Figure have six water
molecules per CoCl2. By convention,
the formula for this compound is written as CoCl2 ∙ 6 H2O. The dot between
CoCl2 and 6 H2O ­
indicates that
6 ­molecules of water are ­associated with
every CoCl2. The name of the compound is
cobalt(II) chloride hexahydrate.
Hydrated cobalt(II) chloride, the red
solid in the Figure, turns purple and then
deep blue as it is heated and loses water to
form anhydrous CoCl2; anhydrous means
without water. Upon on exposure to moist
air, anhydrous CoCl2 takes up water and is
converted back into the red hydrated compound. It is this property that allows crystals of the blue compound to be used as a
humidity indicator. You may have seen
them in small bags packed with a piece of
electronic equipment.
Hydrated compounds are common. The
walls of your home are probably covered
with drywall, or plaster board, which contains hydrated calcium sulfate (gypsum,
98
Photos: © Charles D. Winters/Cengage
A Closer Look
Hydrated Ionic Compounds
Cobalt(II) chloride
hexahydrate [CoCl2 • 6 H2O]
is deep red.
When it is heated,
the compound loses
some of the water of
hydration.
CaSO4 ∙ 2 H2O) and anhydrous CaSO4, sandwiched between paper. If gypsum is heated
between 120 and 180 °C, the water is
partly driven off to give CaSO4 ∙ 1/2 H2O, a
compound known as plaster of Paris. If you
have ever broken an arm or leg and had to
have a cast, the cast may have been made
of this compound. It is an effective casting
material because, when added to water, it
Heating ultimately leads to
the deep blue compounds
CoCl2 • 2 H2O and CoCl2.
Experiments show that
some also decomposes to
black CoO and HCl.
forms a thick slurry that can be poured
into a mold or spread out over a part of the
body. As it takes on more water, the material increases in volume and forms a hard,
inflexible solid. Plaster of Paris is also a
useful material for artists, because the expanding compound fills a mold completely
and makes a high-quality reproduction.
Chapter 2 / Atoms, Molecules, and Ions
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© Charles D. Winters/Cengage
Ionic compounds have characteristic properties that can be understood in terms
of the charges of the ions and their arrangement in the lattice. Because each ion is
surrounded by oppositely charged nearest neighbors, it is held tightly in its allotted
location. At room temperature, each ion can move just a bit around its average position, but considerable energy must be added before an ion can escape the attraction
of its neighboring ions. Only if enough energy is added will the lattice structure
collapse and the substance melt. Greater attractive forces mean that ever more
energy—and higher and higher temperatures—is required to cause melting. Thus,
Al2O3, composed of Al3+ and O2− ions, melts at a much higher temperature
(2072 °C) than NaCl (801 °C), composed of Na+ and Cl− ions.
Most ionic compounds are hard solids. That is, the solids are not pliable or soft.
This is again related to the lattice of ions. The nearest neighbors of a cation in a lattice are anions, and the force of attraction makes the lattice rigid. However, a blow
with a hammer can cause the lattice to break cleanly along a sharp boundary. The
hammer blow displaces layers of ions just enough to cause ions of like charge to
become nearest neighbors, and the repulsion between these like-charged ions forces
the lattice apart (Figure 2.25).
Figure 2.24 Sodium
chloride. A crystal of NaCl
2.7 Atoms, Molecules, and the Mole
consists of an extended lattice of
sodium ions and chloride ions in
a 1∶1 ratio. When melted, the
crystal lattice collapses and the
ions move freely and can conduct
an electrical current.
Goals for Section 2.7
•
•
•
Understand the mole concept and molar mass and their application.
•
Calculate the amount (5 number of moles) of a compound represented by a given
mass, and vice versa.
•
Use Avogadro’s number to calculate the number of atoms or ions in a compound.
Use the molar mass of an element and Avogadro’s number in calculations.
Calculate the molar mass of a compound from its formula and a table of atomic
weights.
© Charles D. Winters/Cengage
When two chemicals react with each other, chemists want to know how many atoms
or molecules of each are used so that formulas can be established for the reaction’s
An ionic solid is rigid owing to the
forces of attraction between oppositely
charged ions. When struck sharply,
however, the crystal can cleave cleanly.
Figure 2.25 Ionic solids.
When a crystal is struck,
layers of ions move slightly,
and ions of like charge
become nearest neighbors.
Repulsions between ions
of similar charge cause the
crystal to cleave.
2.7 Atoms, Molecules, and the Mole
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99
Amedeo Avogadro, conte di Quaregna (1776–1856), was an
Italian nobleman and a lawyer. In about 1800, he turned to
science and was the first professor of mathematical physics
in Italy.
Avogadro did not propose the notion of a fixed number of
particles in a chemical unit. Rather, the number was named
in his honor because he performed experiments in the
­nineteenth century that laid the groundwork for the concept.
Just how large is Avogadro’s number? One mole of unpopped popcorn kernels would cover the continental United
States to a depth of about 9 miles.
When the mole was first adopted as an SI base unit, it was
defined as the number of elementary entities as there are atoms
in exactly 12 g of carbon-12. Using that definition, the mass of
12 g for a mole of carbon-12 was a defined quantity (with an infinite number of significant figures). Avogadro’s number, on the
other hand, was a measured quantity that was determined experimentally and changed over time as experiments yielded more accurate and precise values.
The Mole The term mole was
introduced about 1895 by
Wilhelm Ostwald (1853–1932),
who derived the term from the
Latin word moles, meaning a
“heap” or a “pile.”
On May 20, 2019, the definition of the
mole was officially changed by the Bureau
International des Poids et Mesures that oversees the SI base units. In the new definition,
the number of particles in a mole is defined
to be exactly 6.02214076 × 1023. The value
of Avogadro’s number is now a defined quantity. The mass of carbon-12 that corresponds
to this number is now an experimentally Amedeo Avogadro
measured value with limited accuracy and
precision.
The new definition of the mole is a fundamental change and has
various implications, but most of these are beyond the level of
an introductory chemistry course. The crucial things for you to
­remember are that
1. one mole of any substance contains Avogadro’s number of
elementary entities (atoms, molecules, ...) of that substance,
and
2. one mole of any substance contains the same number of
particles as one mole of any other substance.
Edgar Fahs Smith Collection
A Closer Look
Amedeo Avogadro and His Number
products. To do this, a method of counting atoms and molecules is needed. Is there
a way to connect the macroscopic world, the world you can see with your eyes or
with ordinary scientific instruments, with the particulate world of atoms, molecules,
and ions? The solution to this problem is to define a unit of matter that contains a
known number of particles. That chemical unit is the mole.
The mole (abbreviated mol) is the SI base unit for measuring an amount of a
substance (Table 1R.1, page 32) and is defined as follows:
A mole is the amount of a substance that contains exactly 6.02214076 × 1023
elementary entities (atoms, molecules, or other particles).
1 mole = 6.02214076 × 1023 particles
This value is known as Avogadro’s number (symbolized by NA) in honor of Amedeo
Avogadro, an Italian lawyer and physicist (1776–1856) who conceived the basic
idea (but never determined the number).
The key to understanding the concept of the mole is recognizing that one mole
always contains the same number of particles, no matter what the substance. One
mole of sodium contains the same number of atoms as one mole of iron and the
same number of atoms as the number of molecules in one mole of water.
An Important Difference Between
the Terms Amount and Quantity
in Chemistry The amount of a
substance is the number of
moles of that substance. In
contrast, quantity refers, for
example, to the mass or volume
of the substance.
100
Atoms and Molar Mass
The mass in grams of one mole of any element is the molar mass of that element.
Molar mass is abbreviated with a capital italicized M and has units of grams per
mole (g/mol). An element’s molar mass is the quantity in grams numerically equal
to its atomic weight. Using copper as an example,
Molar mass of copper (Cu) =
mass of 1.000 mol of Cu atoms
= 63.55 g∙mol
= mass of 6.022 × 1023 Cu atoms
Chapter 2 / Atoms, Molecules, and Ions
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© Charles D. Winters/Cengage
Copper
63.55 g
Sulfur
32.06 g
Magnesium
24.31 g
Silicon
28.09 g
Tin
118.7 g
Figure 2.26 One mole of common elements. (Left to right) Sulfur powder, magnesium chips,
tin, and silicon. (Above) Copper beads.
Figure 2.26 shows one mole of some common elements. Although each of
these “piles of atoms” has a different volume and different mass, each contains
6.022 × 1023 atoms.
The mole concept is the cornerstone of quantitative chemistry. It is essential to
be able to convert from moles to mass and from mass to moles. Dimensional analysis, which is described in Chapter 1R, page 47, shows that this can be done in the
following way:
MASS
Moles to Mass
grams
Moles ×
= grams
1 mol
molar mass
MOLES CONVERSION
Mass to Moles
Grams ×
1 mol
= moles
grams
1/molar mass
For example, what mass, in grams, is represented by 0.35 mol of aluminum?
Using the molar mass of aluminum (27.0 g/mol), you can determine that 0.35 mol
of Al has a mass of 9.5 g.
0.35 mol Al 3
27.0 g Al
5 9.5 g Al
1 mol Al
Molar masses of the elements are usually known to four or more significant
figures. To avoid having the molar mass affect the precision of a calculation, the
convention usually followed in this book is to use a value of the molar mass with
at least one more significant figure than in any other number in the calculation. For
example, if you weigh out 16.5 g of carbon, you use 12.01 g/mol for the molar mass
of C to find the amount of carbon present.
2.7 Atoms, Molecules, and the Mole
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101
16.5 g C ×
1 mol C
12.01 g C
= 1.37 mol C
Note that four significant figures are used in the
molar mass, but there are three in the sample mass.
E xamp le 2.9
Mass, Moles, and Atoms
Problem Consider two elements in the same vertical column of the periodic table,
lead and tin.
(a) What mass of lead, in grams, is equivalent to 2.50 mol of lead (Pb)?
© Charles D. Winters/Cengage
(b) What amount of tin, in moles, is represented by 36.6 g of tin (Sn)? How many atoms
of tin are in the sample?
Lead. A 150-mL beaker
containing 2.50 mol or 518 g
of lead.
(c) What is the mass of an average atom of each of these elements?
What Do You Know? You know the amount of lead and the mass of tin. You also
know, from the periodic tables in this book, the molar masses of lead (207.2 g/mol) and tin
(118.7 g/mol). For parts (b) and (c) Avogadro’s number is needed.
Strategy
Part (a) Multiply the amount of Pb by the molar mass.
Part (b) Multiply the mass of tin by (1/molar mass). To determine the number of atoms,
multiply the amount of tin by Avogadro’s number.
© Charles D. Winters/Cengage
Part (c) Multiply the molar mass of each element by (1/Avogadro’s number).
Solution
(a) Convert the amount of lead in moles to mass in grams.
2.50 mol Pb 3
207.2 g
5 518 g Pb
1 mol Pb
(b) Convert the mass of tin to the amount in moles,
Tin. A sample of tin having a
mass of 36.6 g (or 1.86 × 1023
atoms).
36.6 g Sn 3
1 mol Sn
5 0.3083 mol Sn 5 0.308 mol Sn
118.7 g Sn
and then use Avogadro’s number to find the number of atoms in the sample.
Number of digits In the first step of
part (b), the highlighted answer
(0.308 mol Sn) is reported to
the proper number of significant
figures. The intermediate value
(0.3083 mol Sn) with one extra
digit is also shown and carried
forward to the next step so
that round-off error does not
accumulate (see page 45).
102
0.3083 mol Sn 3
6.022 3 1023 atoms Sn
5 1.86 × 1023 atoms Sn
1 mol Sn
(c) Convert the molar mass of each element to grams per atom.
1 mol Sn
118.7 g Sn
3
5 1.971 × 1022 g Sn/atoms
mol Sn
6.022 3 1023 atoms
1 mol Pb
207.2 g Pb
3
5 3.441 × 1022 g Pb/atoms
mol Pb
6.022 3 1023 atoms
Chapter 2 / Atoms, Molecules, and Ions
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Think about Your Answer Parts (a) and (b) were solved using g/mol or mol/g as
conversion factors. To be sure you have used them correctly, you should keep track of the
units of each term (page 47) and think about the numerical values of your answers. For
example, in part (b), if you had inverted the conversion factor (mol/atoms instead of
­atoms/mol), the units would not have canceled properly and you would have calculated
that there was less than one atom in 0.308 mol of Sn, an unreasonable answer. In part (c),
notice the extremely small values obtained for the average masses of the atoms. Atoms
are tiny! Finally, note that these are average masses per atom because of the presence of
more than one isotope for each element.
Check Your Understanding
What mass of gold, Au, contains 2.6 × 1024 atoms?
Molecules, Compounds, and Molar Mass
The formula of a compound tells you the type of atoms or ions in the compound
and the relative number of each. For example, one molecule of methane, CH4, is
made up of one atom of C and four atoms of H. But suppose you have Avogadro’s
number of C atoms (6.022 × 1023) combined with the proper number of H
­atoms. (For CH4 this means there are 4 × 6.022 × 1023 H atoms per mole of
C atoms.) What masses of atoms are combined, and what is the mass of this
many CH4 molecules?
C
+
4H
n
Molecular Weight and Molar
Mass The old term molecular
weight is sometimes
encountered. This is the sum
of the atomic weights of the
constituent elements.
CH4
6.022 × 1023 C atoms
4 × 6.022 × 1023 H atoms
6.022 × 1023 CH4 molecules
= 1.000 mol of C
= 4.000 mol of H atoms
= 1.000 mol of CH4 molecules
= 12.01 g of C atoms
= 4.032 g of H atoms
= 16.04 g of CH4 molecules
Students studying chemistry for the
first time are often perplexed by the
idea of the mole. But it is just a counting unit with an odd name. Pairs and
dozens are two other common counting units. For example, a pair of objects has two of the same things (two
shoes or two gloves), and there are 12
eggs or apples in a dozen. In the same
way, a mole of atoms or a mole of jelly
beans has 6.022 × 1023 objects.
The great advantage of counting
units is that if you know the number of
units you also know the number of objects.
If you know there are 3.5 dozen apples in
a box, you know there are 42 apples. And,
if you have 0.308 mol of tin (36.6 g), you
know you have 1.86 × 1023 atoms of tin.
When you carry out a chemical reaction
in the lab, you need to know how many
“chemical units” of an element are involved. Atoms obviously cannot be counted
out one by one. Instead, you can weigh a
given mass of the element, and, from the
molar mass, determine the number of
“chemical units” or moles. And, if you really wanted to know the information, you
could calculate the number of atoms
involved.
© Charles D. Winters/Cengage
A Closer Look
The Mole, a Counting Unit
Counting Units. The unit dozen, which refers
to 12 objects, is a common counting unit. Similarly, the mole is a chemical counting unit. Just as
a dozen always has 12 objects, a mole always has
6.022 × 1023 objects.
2.7 Atoms, Molecules, and the Mole
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103
Because you know the number of moles of C and H atoms in 1 mol of CH4, you
can calculate the masses of carbon and hydrogen that must be combined. It follows
that the mass of CH4 is the sum of these masses. That is, 1 mol of CH4 has a mass
equal to the mass of 1 mol of C atoms (12.01 g) plus 4 mol of H atoms (4.032 g).
Thus, the molar mass, M, of CH4 is 16.04 g/mol.
Because you know the mass of one mole of methane, 16.04 g, and that there are
6.022 × 1023 molecules present in one mole, you can also calculate the mass of an
average molecule of CH4. (This is an average mass because there are several isotopes
of carbon and hydrogen; the mass of a given molecule is determined by the isotopes
making up that molecule.)
Average molecular mass 5
C
O
O
C OH
C
H
Notice the dual meaning of the subscripts in the molecular formula of a compound. On the particulate scale, the subscripts represent the number of atoms of
each element present in one molecule of the compound. Thus, one molecule of CH4
consists of one carbon atom and four hydrogen atoms. On the macroscopic scale,
the subscripts refer to the number of moles of each element present in one mole of
the compound. One mole of CH4 consists of 1 mole of carbon atoms and 4 moles
of hydrogen atoms.
Figure 2.27 illustrates 1-mol quantities of several common compounds. To find
the molar mass of any compound, you need to add up the atomic weights for each
element in the compound, taking into account any subscripts on elements. For example, the molecular formula of aspirin is C9H8O4. In one mole of aspirin there are
9 mol of carbon atoms, 8 mol of hydrogen atoms, and 4 mol of oxygen atoms,
which add up to 180.15 g/mol of aspirin.
O
CH3
C
C
C
C
H
16.04 g
1 mol
?
5 2.664 3 1023 g/molecule
mol
6.022 × 1023 molecule
H
C
Mass of C in 1 mol C9H8O4 5 9 mol C 3
12.01 g C
5 108.09 g C
1 mol C
Mass of H in 1 mol C9H8O4 5 8 mol H 3
1.008 g H
5 8.064 g H
1 mol H
Mass of O in 1 mol C9H8O4 5 4 mol O 3
16.00 g O
5 64.00 g O
1 mol O
H
Aspirin Formula Aspirin has the
molecular formula C9H8O4 and
a molar mass of 180.15 g/mol.
Aspirin is the common name of
the compound acetylsalicylic
acid.
Total mass of 1 mol of C9H8O4 5 molar mass of C9H8O4 5 180.15 g
Aspirin, C9H8O4
180.2 g/mol
Copper(II) chloride
dihydrate, CuCl2 ∙ 2 H2O
170.5 g/mol
Iron(III) oxide, Fe2O3
159.7 g/mol
© Charles D. Winters/Cengage
H2O
18.02 g/mol
Figure 2.27 One mole of some compounds. In the second compound, CuCl2 ∙ 2 H2O, one
formula unit has one Cu2+ ion, two Cl− ions, and two water molecules. The molar mass is the
sum of the mass of 1 mol of Cu, 2 mol of Cl, and 2 mol of H2O.
104
Chapter 2 / Atoms, Molecules, and Ions
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As was the case with elements, it is important to be able to convert between
amounts (moles) and mass (grams). For example, if you take 325 mg (0.325 g) of
aspirin in one tablet, what amount of the compound have you ingested? Based on
a molar mass of 180.15 g/mol, there is 0.00180 mol of aspirin per tablet.
0.325 g aspirin 3
1 mol aspirin
5 0.001804 mol aspirin 5 0.00180 mol aspirin
180.15 g aspirin
Using the molar mass of a compound it is possible to determine the number of
molecules in any sample from the sample mass and to determine the mass of one
molecule. For example, the number of aspirin molecules in one tablet is
0.001804 mol aspirin 3
6.022 3 1023 molecules
5 1.09 3 1021 molecules
1 mol aspirin
and the average mass of one molecule is
1 mol aspirin
180.15 g aspirin
3
5 2.992 3 1022 g/molecule
1 mol aspirin
6.022 3 1023 molecules
Ex am p le 2.10
Strategy Map
Molar Mass and Moles
Problem
Calculate the molar mass of
oxalic acid. Find the amount
of oxalic acid in a given mass.
Then find the number of molecules and the number of C
atoms in the sample.
Problem What is the molar mass of oxalic acid (H2C2O4)? What amount of oxalic acid is
Data/Information
•­ Mass of sample
•­ Formula of compound
•­ Avogadro’s number
present in 16.5 g of oxalic acid? How many molecules of oxalic acid are in 16.5 g of the
acid? How many atoms of carbon are in 16.5 g of oxalic acid?
What Do You Know? You know the mass and formula of oxalic acid. The molar
mass of the compound can be calculated based on the formula.
Strategy
Step 1. Calculate the molar mass of the compound. The molar mass is the sum of the
masses of the component atoms.
Step 2. Use the molar mass to convert mass to amount (moles).
Step 3. Use Avogadro’s number to calculate the number of molecules from the amount.
Step 4. Determine the number of atoms of carbon.
Solution
Step 1
Calculate the molar mass of the compound.
The formula for oxalic acid, H2C2O4, tells you that there are 2 mol H, 2 mol C, and 4 mol O in 1
mol of oxalic acid. The molar masses for these elements are on the periodic table.
2 mol H per mol H2C2O4 3
1.008 g H
= 2.016 g H per mol H2C2O4
1 mol H
2 mol C per mol H2C2O4 3
12.01 g C
= 24.02 g C per mol H2C2O4
1 mol C
4 mol O per mol H2C2O4 3
16.00 g O
= 64.00 g O per mol H2C2O4
1 mol O
Molar mass of H2C2O4 = 90.04 g per mol H2C2O4
2.7 Atoms, Molecules, and the Mole
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105
Step 2
Use the molar mass to convert mass to amount (moles).
The molar mass (expressed here in units of 1 mol/90.04 g) is used in all mass–mole
conversions.
16.5 g H2C2O4 3
Step 3
1 mol
5 0.1833 g H2C2O4 5 0.183 mol H2C2O4
90.04 g H2C2O4
Use Avogadro’s number to calculate the number of molecules from the amount.
0.1833 mol 3
6.022 3 1023 molecules
5 1.104 3 1023 molecules
1 mol
= 1.10 × 1023 molecules
Step 4
Determine the number of carbon atoms.
From the formula, you know that there are two atoms of carbon in each molecule of oxalic acid.
1.104 3 1023 molecules 3
2 C atoms
5 2.21 × 1023 C atoms
1 molecule
Think about Your Answer The mass of oxalic acid is 16.5 g, much less than the
mass of a mole (90.04 g), so make sure your answer reflects this. The number of acid molecules should be much less than in one mole of molecules.
Check Your Understanding
If you have 454 g of citric acid (H3C6H5O7), what amount (moles) does this represent? How many molecules? How many atoms of
carbon?
Ionic compounds such as NaCl do not exist as individual molecules. Thus, for
ionic compounds chemists write the simplest formula that shows the relative number of each kind of atom in a formula unit of the compound, and the molar mass
is calculated from this formula (M for NaCl = 58.44 g/mol). To differentiate substances like NaCl that do not contain molecules, chemists sometimes refer to their
formula mass instead of their molar mass.
2.8 Chemical Analysis: Determining
Compound Formulas
Goals for Section 2.8
•
•
Express the composition of a compound in terms of percent composition.
Determine the empirical and molecular formula of a compound using percent
composition or other experimental data.
Given a sample of an unknown compound, how can its formula be determined?
The answer lies in chemical analysis, a major branch of chemistry that deals with
the determination of formulas and structures.
Percent Composition
A central principle of chemistry is that any sample of a pure compound always consists of the same elements combined in the same proportion by mass. Suppose you
have 1.0000 mol of NH3 or 17.031 g. This mass of NH3 is composed of 14.007 g of N
(1.0000 mol) and 3.024 g of H (3.000 mol). You can calculate the percent composition by mass for each element in the compound by dividing the mass of each element
in the compound by the total mass of the compound and then multiplying by 100.
106
Chapter 2 / Atoms, Molecules, and Ions
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mass of N in 1 mol NH3
3 100%
mass of 1 mol NH3
Mass percent N in NH3 5
5
14.007 g N
3 100%
17.031 g NH3
5
composition can be expressed
as a percent (mass of an
element in a 100-g sample).
82.244% of NH3 mass
is nitrogen.
5 82.244%
mass of H in 1 mol NH3
3 100%
mass of 1 mol NH3
Mass percent H in NH3 5
Molecular Composition Molecular
3.024 g H
3 100%
17.031 g NH3
5 17.76%
H
N
H
H
17.76% of NH3 mass
is hydrogen.
These values tell you that in a 100.00-g sample there are 82.24 g of N and 17.76 g
of H.
Ex am p le 2.11
Using Percent Composition
Problem
(a) What is the mass percent of each element in propane, C3H8?
(b) What mass of carbon is contained in 454 g of propane?
What Do You Know? You know the formula of propane. You will need the atomic
weights of C and H to calculate the mass percent of each element.
Strategy
(a) Mass percent of each element in propane:
Step 1. Calculate the molar mass of propane.
Step 2. The mass percent of each element is the mass of the element in one mole of the
compound divided by the molar mass of the compound and multiplied by 100.
(b) The mass of C in 454 g of C3H8 is obtained by multiplying this mass by the % C and
dividing by 100.
Solution
(a) Mass percent of each element in C3H8:
Step 1. The molar mass of C3H8 is 44.10 g/mol.
Step 2. Mass percent of C and H in C3H8:
3 mol C
1 mol C3H8
3
12.01 g C
5 36.03 g C/1 mol C3H8
1 mol C
Mass percent of C in C3H8 5
8 mol H
1 mol C3H8
3
36.03 g C
3 100% 5 81.70% C
44.10 g C3H8
1.008 g H
5 8.064 g H/1 mol C3H8
1 mol H
Mass percent of H in C3H8 5
8.064 g H
3 100% 5 18.29% H
44.10 g C3H8
2.8 Chemical Analysis: Determining Compound Formulas
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107
(b) Mass of C in 454 g of C3H8:
454 g C3H8 3
81.70 g C
5 371 g C
100.0 g C3H8
Think about Your Answer Once you know the percent C in the sample, you
could calculate the percent H from it knowing that %H = 100% − %C.
Check Your Understanding
1.
Express the composition of ammonium carbonate, (NH4)2CO3, in terms of the mass of
each element in 1.00 mol of compound and the mass percent of each element.
2.
What is the mass of carbon in 454 g of octane, C8H18?
Empirical and Molecular Formulas
from Percent Composition
Now consider the reverse of the procedure just described. That is, use relative mass
or percent composition data to find a molecular formula. Suppose you know the
identity of the elements in a sample and have determined the mass of each element in a given mass of compound by chemical analysis. You can then calculate
the relative amount (moles) of each element, which is also the relative number of
atoms of each element in the formula of the compound. For example, for a compound composed of atoms of A and B, the steps from percent composition to a
formula are as follows:
S T EP 1.
ST EP 2.
ST EP 3.
Convert
mass percent
to mass
Convert
mass to
amount
Find
mole ratio
%A
%B
gA
x mol A
gB
y mol B
ST EP 4 .
Determine the
whole-number
ratio of A to B
x mol A
y mol B
AaBb
For example, hydrazine is a compound used to remove oxygen from water in
heating and cooling systems and is a close relative of ammonia. Chemical analysis
shows that hydrazine is composed of 87.42% N and 12.58% H. What is the molecular formula for hydrazine?
Deriving a Formula Percent
composition gives the mass
of an element in 100 g of a
sample. However, in deriving a
formula, any amount of sample
is appropriate if you know the
mass of each element in that
sample mass.
108
Step 1. Convert mass percent to mass. The mass percentages of N and H in hydrazine tell you there are 87.42 g of N and 12.58 g of H in a 100.00-g sample.
Step 2. Convert the mass of each element to amount (moles). The amount of each
element in the 100.00-g sample is
87.42 g N 3
1 mol N
5 6.2412 mol N
14.007 g N
12.58 g H 3
1 mol H
5 12.480 mol H
1.008 g H
Chapter 2 / Atoms, Molecules, and Ions
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Problem Solving Tip 2.2 Finding Empirical and Molecular Formulas
significant figures can give a misleading result.
• The experimental data available
to find a formula may be in the
form of percent composition or
the masses of elements combined
in some mass of compound. No
matter what the starting point,
the first step is always to convert
masses of elements to moles.
• Determining the molecular formula
of a compound after calculating
the empirical formula requires
knowing the molar mass of the
compound.
• When finding atom ratios, always
divide the larger number of moles
by the smaller one.
• Empirical and molecular
• When both the percent composi-
f­ormulas can differ for molecular
­compounds. In contrast, there
is no molecular formula for an
ionic compound; all that can be
recorded is the empirical formula.
• Be sure to use at least three significant figures when calculating
empirical formulas. Using fewer
tion and the molar mass are known
for a compound, the alternative
method mentioned in Think about
Your Answer in Example 2.12 can
be used.
Step 3. Find the mole ratio of elements. Use the amount (moles) of each element in
the 100.00-g sample to find the amount of one element relative to the other. (To do
this it is usually best to divide the larger amount by the smaller amount.)
12.480 mol H
2.000 mol H
5
6.2412 mol N
1.000 mol N
Step 4. Determine the whole-number ratio. The mole ratio shows that there are
2 mol of H atoms for every 1 mol of N atoms in hydrazine. This is already a wholenumber ratio, so no further adjustment is needed. Thus, in one molecule, two atoms
of H occur for every atom of N; that is, the formula is
NH2
This simplest, whole-number atom ratio of atoms in a formula is called the
empirical formula.
The molecular formula, however, must convey not only this information but also
the total number of atoms in the molecule. For example, the empirical formula NH2
for hydrazine tells you the relative numbers of atoms for each element in the compound: Hydrazine has twice as many H atoms as N atoms. But the molecular formula
could be NH2, N2H4, N3H6, N4H8, or any other formula with a 1:2 ratio of N to H.
To determine the molecular formula from the empirical formula, you need to
know the molar mass of the compound. For example, experiments show that the
molar mass of hydrazine is 32.0 g/mol, twice the formula mass of NH2, which is
16.0 g/mol. Thus, the molecular formula of hydrazine is two times the empirical
formula of NH2, that is, N2H4.
Ex am p le 2.12
Calculating a Formula from Percent
Composition
Problem Many soft drinks contain sodium benzoate as a preservative. When you consume the sodium benzoate, it reacts with the amino acid glycine in your body to form
hippuric acid, which is then excreted in the urine. Hippuric acid has a molar mass of
179.17 g/mol and is 60.33% C, 5.06% H, and 7.82% N; the remainder is oxygen. What are
the empirical and molecular formulas of hippuric acid?
What Do You Know? You know the mass percent of C, H, and N. The mass percent
of oxygen is not known but is obtained by difference. You know the molar mass of
­hippuric acid but will need the atomic weights of C, H, N, and O for the calculation.
2.8 Chemical Analysis: Determining Compound Formulas
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109
Strategy Map
Strategy
Problem
Determine empirical and
molecular formulas based
on known composition and
known molar mass.
Step 1. Convert each mass percent to mass.
Data/Information
•­ Molar mass
•­ Percent composition
Step 1
Step 2. Convert the mass of each element to amount (moles).
Step 3. Find the mole ratios of the elements.
Step 4. Determine the whole-number ratios to find the empirical formula.
Step 5. Divide the known molar mass by the empirical formula mass to determine the
molecular formula.
Solution
Convert each mass percent to mass.
Assume that each mass percent is equivalent to the mass in grams in a 100-g sample. Thus,
there are 60.33 g C, 5.06 g H, and 7.82 g of N in 100 g of hippuric acid. The mass of oxygen
in the sample is
100.00 g = 60.33 g C + 5.06 g H + 7.82 g N + mass of O
Mass of O = 26.79 g O
Step 2
Convert the mass of each element to amount (moles).
Use the molar mass of each element to determine the amount (moles) of each element
present in the 100-g sample.
Step 3
60.33 g C 3
1 mol C
5 5.0229 mol C
12.011 g C
5.06 g H 3
1 mol H
5 5.020 mol H
1.008 g H
7.82 g N 3
1 mol N
5 0.5582 mol N
14.01 g N
26.79 g O 3
1 mol O
5 1.6745 mol O
15.999 g O
Find the mole ratios of the elements.
To find the mole ratios, the best approach is to base the ratios on the smallest number of
moles present—in this case, nitrogen.
mol C
5.0229 mol C
9.00 mol C
5
5
5 9 mol C/1 mol N
mol N
0.5582 mol N
1.00 mol N
mol H
5.020 mol H
8.99 mol H
5
5
5 9 mol H/1 mol N
mol N
0.5582 mol N
1.00 mol N
mol O
1.6745 mol O
3.00 mol O
5
5
5 3 mol O/1 mol N
mol N
0.5582 mol N
1.00 mol N
Step 4
Determine the whole-number ratios to find the empirical formula.
These mole ratios are already whole-number ratios, so no further adjustment is needed.
There are 9 mol of C, 9 mol of H, and 3 mol of O for each mol of N. Thus, the empirical
formula is C9H9NO3.
110
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Step 5
Divide the known molar mass by the empirical formula mass to determine the
molecular formula.
The formula mass of the empirical formula (C9H9NO3) is
(9 3 12.011 g/mol) 1 (9 3 1.008 g/mol) 1 (1 × 14.007 g/mol) 1 (3 3 15.999 g/mol)
5 179.175 g/mol
The experimentally determined molar mass of hippuric acid is 179.17 g/mol.
molar mass
179.17 g/mol
5
5 0.99997 5 1
empirical formula mass
179.175 g/mol
The ratio of the molar mass to the empirical formula mass is 1, so the molecular formula is
the same as the empirical formula. The molecular formula is C9H9NO3.
Think about Your Answer There is another approach to finding the molecular
formula if you know both the percent composition and the molar mass of a compound.
Multiply the molar mass of the compound by the percent composition of each element
divided by 100. This gives the mass of the element in 1 mol of the compound, which can
then be converted to amount (moles). For example, for carbon in hippuric acid,
179.17 g hippuric acid 3
108.09 g C 3
Hippuric Acid, C9H9NO3 This
substance, which can be isolated
as white crystals, is found in the
urine of humans and herbivorous
animals.
60.33 g C
5 108.09 g C
100 g hippuric acid
1 mol C
5 9.000 mol C
12.011 g C
Likewise, the following values are obtained for the other elements in hippuric acid: 9.07 g
of H (8.99 mol), 14.01 g N (1.000 mol), and 48.00 g of O (3.000 mol). This gives a molecular
formula of C9H9NO3. However, this approach can only be used when you know both the
percent composition and the molar mass.
Check Your Understanding
1.
What is the empirical formula of naphthalene, C10H8?
2.
The empirical formula of acetic acid is CH2O. If its molar mass is 60.05 g/mol, what is the molecular formula of acetic acid?
3.
Isoprene is a liquid compound that can be polymerized to form natural rubber. It is composed of 88.17% carbon and 11.83%
hydrogen. Its molar mass is 68.11 g/mol. What are its empirical and molecular formulas?
4.
Camphor is found in camphor wood, much prized for its distinctive odor. It is composed of 78.90% carbon and 10.59%
­hydrogen. The remainder is oxygen. What is its empirical formula?
Determining a Formula from Mass Data
The composition of a compound in terms of mass percent gives the mass of each
element in a 100.0-g sample. Sometimes, other information about the composition
of compounds is collected in the laboratory and used to determine the formulas of
compounds. Two ways of doing this are:
•
Combining known masses of elements to give a sample of the compound of
known mass. Element masses can be converted to amounts (moles), and the
ratio of amounts gives the combining ratio of atoms—that is, the empirical
formula. This approach is described in Example 2.13.
2.8 Chemical Analysis: Determining Compound Formulas
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111
•
Decomposing a known mass of an unknown compound into “pieces” of known
composition. If the masses of the “pieces” can be determined, the ratio of moles
of the “pieces” gives the formula. An example is a decomposition such as
Ni(CO)4(ℓ) → Ni(s) + 4 CO(g)
The masses of Ni and CO can be converted to moles, whose 1∶4 ratio would
reveal the formula of the compound. This approach is described in
Example 2.14.
E xamp le 2.13
Formula of a Compound from Combining Masses
Problem Oxides of virtually every element are known. Bromine, for example, forms
several oxides when treated with ozone (O3). Suppose you allow 1.250 g of bromine, Br2,
to react with ozone and obtain 1.876 g of BrxOy. What is the empirical formula of the
product?
What Do You Know? You began with a given mass of bromine and all of the
bromine became part of bromine oxide of unknown formula. You also know the mass of
the product, and because you know the mass of Br in this product, you can determine the
mass of O in the product.
Strategy
Step 1. The mass of oxygen is determined as the difference between the product mass
and the mass of bromine used.
Step 2. Calculate the amounts of Br and O from the masses of each element.
Step 3. Find the mole ratio of the elements.
Step 4. Determine the whole-number ratio to find the empirical formula.
Solution
Step 1. You already know the mass of bromine in the compound, so you can calculate the
mass of oxygen in the compound.
1.876 g product − 1.250 g Br2 = 0.626 g O
Step 2. Next, calculate the amount of each reactant. Notice that, although Br2 was the
reactant, you need to know the amount of Br atoms in the product.
1.250 g Br2 3
1 mol Br2
5 0.0078218 mol Br2
159.81 g Br2
0.0078218 mol Br2 3
0.626 g O 3
2 mol Br
5 0.015644 mol Br
1 mol Br2
1 mol O
5 0.03913 mol O
16.00 g O
Step 3. Find the ratio of moles of O to moles of Br:
Mole ratio 5
0.03913 mol O
2.50 mol O
5
0.015644 mol Br
1.00 mol Br
Step 4. T he mole ratio is 2.5 mol O/1.0 mol Br. However, atoms combine in the ratio of
small whole numbers, so you should double the mole ratio found in Step 3 to give
a whole-number ratio of 5 mol O to 2 mol Br. Thus, the product is Br2O5 (dibromine pentaoxide).
112
Chapter 2 / Atoms, Molecules, and Ions
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Think about Your Answer The whole–number ratio of 5∶2 was found by realizing that 2.5 = 2 1/2 = 5/2. The calculation gave the empirical formula for this compound. To determine whether this is also the molecular formula, the molar mass of the
compound would have to be determined.
Check Your Understanding
Gallium oxide, GaxOy, forms when gallium is combined with oxygen. Suppose you allow
1.25 g of gallium (Ga) to react with oxygen and obtain 1.68 g of GaxOy. What is the formula
of the product?
Ex am p le 2.14
Determining the Formula
of a Hydrated Compound
hydrated copper(II) sulfate, CuSO4 ∙ x H2O, that is, the
number of water molecules for each unit of CuSO4. In
the laboratory you weigh out 1.023 g of the solid. After
heating the solid thoroughly in a porcelain crucible
­(Figure), 0.654 g of nearly white, anhydrous copper(II)
sulfate, CuSO4, remains.
White CuSO4
Blue
CuSO4 ∙ x H2O
© Charles D. Winters/Cengage
Problem You want to know the value of x in blue,
1.023 g CuSO4 ∙ x H2O + heat → 0.654 g CuSO4 + ? g H2O
Strategy Map
Problem
Determine formula of
hydrated salt based on masses
of water and dehydrated salt.
What Do You Know? You know the mass of the
copper(II) sulfate sample including water (before heating) and with no water (after heating). Therefore, you
know the mass of CuSO4 and can determine the mass of water in the sample.
Strategy To find x you need to know the amount of H2O per mole CuSO4.
Data/Information
•­ Mass of sample before and
after heating to dehydrate
Step 1. Determine the mass of water released upon heating the hydrated compound.
Step 2. Calculate the amount (moles) of CuSO4 and H2O from their masses and molar
masses.
Step 3. Determine the smallest whole-number ratio (amount H2O/amount CuSO4).
Solution
Step 1
Determine the mass of water released upon heating the hydrated compound.
Mass of hydrated compound
− Mass of anhydrous compound, CuSO4
Mass of water
Step 2
1.023 g
−0.654
0.369 g
Calculate the amount (moles) of CuSO4 and H2O from their masses and
molar masses.
0.369 g H2O 3
1 mol H2O
5 0.02048 mol H2O
18.02 g H2O
0.654 g CuSO4 3
1 mol CuSO4
5 0.004098 mol CuSO4
159.6 g CuSO4
2.8 Chemical Analysis: Determining Compound Formulas
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113
Determine the mole ratio (amount H2O relative to amount CuSO4).
Step 3
0.02048 mol H2O
5.00 mol H2O
5
0.004098 mol CuSO4
1.00 mol CuSO4
The water-to-CuSO4 ratio is 5∶1, so the formula of the hydrated compound is
CuSO4 ∙ 5 H2O. Its name is copper(II) sulfate pentahydrate.
Think about Your Answer The ratio of the amount of water to the amount of
CuSO4 is a whole number. This is almost always the case with hydrated compounds.
Check Your Understanding
Hydrated nickel(II) chloride is a beautiful green, crystalline compound. When heated strongly, the compound is dehydrated. If
0.235 g of NiCl2 ∙ x H2O gives 0.128 g of NiCl2 on heating, what is the value of x?
2.9 Instrumental Analysis:
Determining Compound Formulas
Goals for Section 2.9
•
•
Determine a molecular formula from a mass spectrum.
Identify isotopes using mass spectrometry.
Determining a Formula by Mass Spectrometry
In addition to the more traditional chemical methods of determining a molecular
formula described in previous sections of this chapter, there are many instrumental
methods as well. One of them is mass spectrometry, a technique that was introduced
earlier when discussing the existence of isotopes and their relative abundance (see
Figure 2.3). If a compound can be vaporized, the vapor can be passed through an
electron beam in a mass spectrometer where high-energy electrons collide with the
gas-phase molecules. These high-energy collisions cause molecules to lose electrons
and become positive ions, which usually fragment into smaller pieces. Figure 2.28
shows the mass spectrum for the compound ethanol, CH3CH2OH. The horizontal
axis shows the mass-to-charge ratio (m/Z) of a given ion. Because almost all observed ions have a charge of Z = 1+, the value observed is the mass of the ion. As
illustrated in this figure, the cation created from ethanol itself (CH3CH2OH+,
m∙Z = 46), called the parent ion or molecular ion, fragments (losing an H atom)
to give another cation (CH3CH2O+, m∙Z = 45), which further fragments. Analysis
of the spectrum can help identify a compound and can give an accurate molar mass.
Molar Mass and Isotopes in Mass Spectrometry
Bromobenzene, C6H5Br, has a molar mass of 157.010 g∙mol. Why, then, are there
two prominent lines at mass-to-charge ratios (m∙Z) of 156 and 158 in the mass
spectrum of the compound (Figure 2.29)? The answer shows the influence of isotopes on molar mass.
Bromine has two naturally occurring isotopes, 79Br and 81Br. They are 50.7% and
49.3% abundant, respectively. Using the most abundant isotopes of C and H (12C
and 1H), the mass of the molecule having the 79Br isotope, C6H579Br, is 156. The
mass of the molecule containing the 81Br isotope, C6H581Br, is 158. The relative size
of these two peaks in the spectrum reflects the relative abundances of the two
­bromine isotopes.
114
Chapter 2 / Atoms, Molecules, and Ions
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Relative abundance of ions
100
CH2OH+
(m/Z = 31)
80
60
CH3CH2O+
(m/Z = 45)
C2H5+
(m/Z = 29)
40
CH3CH2OH+
(m/Z = 46)
CH +
3
(m/Z = 15)
20
0
One of many
fragment ion peaks
10
Parent ion peak
20
30
40
50
Mass-to-charge ratio (m/Z)
Figure 2.28 Mass spectrum of ethanol, CH3CH2OH. The peak (or line) in the spectrum at mass 46
is from an ion of ethanol that has not undergone decomposition (CH3CH2OH+). This ion is referred to
as the parent ion or molecular ion. The mass designated by the peak for the parent ion confirms the
formula of the mole­cule. Other peaks are for fragment ions. This pattern of lines can provide further,
unambiguous evidence of the formula of the compound.
The calculated molar mass of bromobenzene (157.010 g/mol) reflects the abundances of all of the isotopes. In contrast, the mass spectrum has a line for each possible combination of isotopes. This also explains why there are also small lines at
the mass-to-charge ratios of 157 and 159. They arise from various combinations of
1
H, 12C, 13C, 79Br, and 81Br atoms. In fact, careful analysis of such patterns can identify a molecule unambiguously.
100
158 = (12C)6(1H)581Br+
Relative abundance of ions
80
156 = (12C)6(1H)579Br+
60
40
20
0
0
40
80
120
160
Mass-to-charge ratio (m/Z)
Figure 2.29 Mass spectrum of bromobenzene, C6H5Br. Two parent ion peaks are present at
m/Z ratios of 156 and 158. The similar peak heights reflect the near equal abundances of the two
isotopes of bromine, 79Br and 81Br.
2.9 Instrumental Analysis: Determining Compound Formulas
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115
E xamp le 2.15
Isotopic Abundance by Mass Spectrometry
Problem The mass spectrum of phosphorus trichloride is illustrated here.
Relative abundance of ions
100
101
80
103
60
136 138
40
20
105
66
68
0
60
80
100
120
Mass-to-charge ratio (m/Z)
140
140
Phosphorus, 31P, has one stable isotope. Chlorine has two stable isotopes, 35Cl and 37Cl.
(a) What molecular species give rise to the parent ion peaks at m/Z ratios of 136, 138,
and 140?
(b) What species give rise to the peaks at m/Z ratios of 101, 103, and 105?
(c) Predict the structural formula (see page 85) of phosphorus trichloride from its mass
spectrum.
What Do You Know? You know that PCl3 molecules ionize to form positive ions.
Some of the (parent) ions fragment into smaller ions. You also know the mass numbers of
each atom. The mass spectrum shows you the mass of each ion divided by its charge
(m/Z).
Strategy Try to generate the m/Z ratios observed in the mass spectrum by combining
the mass numbers of the elements (35Cl, 37Cl, and 31P) in various ways.
Solution
(a) The parent ion peaks correspond to ions that have not fragmented. A parent ion
formed from one 31P atom and three 35Cl atoms has a m/Z ratio of 136 (if the ion charge
is 1+). One 31P atom combined with two 35Cl atoms and one 37Cl atom has a m/Z ratio
of 138. Finally, a 31P atom combined with one 35Cl and two 37Cl atoms has a m/Z ratio of
140. Thus, the molecular species are P35Cl3 (m/Z = 136), P35Cl237Cl (m/Z = 138), and
P35Cl37Cl2 (m/Z = 140).
(b) The ions with m/Z ratios of 101, 103, and 105 have the formula PCl2+. The species are
P35Cl2 (m/Z = 101), P35Cl37Cl (m/Z = 103), and P37Cl2 (m/Z = 105).
(c) The probable structure is below.
Cl
Cl
P
Cl
The mass spectrum shows fragment ions of PCl+ (m/Z = 66 and 68) and PCl2+, but no fragment ions of Cl2+ or Cl3+. The absence of Cl2+ or Cl3+ ions is evidence that the chlorine
atoms are attached to the phosphorus atom, not each other.
116
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Think about Your Answer When identifying ions in a mass spectrum, it is important to use the masses of each isotope of an element, rather than the average atomic
mass of the element.
Check Your Understanding
The mass spectrum of phosphorus tribromide is illustrated below. Bromine has two stable
isotopes, 79Br and 81Br with abundances of 50.7% and 49.3%, respectively.
Relative abundance of ions
100
191
80
60
270 272
189 193
40
20
0
180
268
200
220
240
Mass-to-charge ratio (m/Z)
260
274
280
(a) What molecular species give rise to the parent ion peaks at m/Z ratios of 270 and 272?
(b) Explain why the relative abundances of the ions at m/Z ratios of 268 and 274 are
­approximately one-third of those at 270 and 272.
Applying Chemical Principles
In 1991, a hiker in the Alps on the Austrian-Italian border
found the well-preserved remains of an approximately 46-year-old
man, now nicknamed “The Iceman,” who lived about 5300 years
ago. Studies using isotopes of oxygen, strontium, lead, and
argon, among others, have helped scientists paint a d
­ etailed
picture of the man and his life.
The abundance of the 18O isotope of oxygen in a person is
related to the latitude and altitude at which the person was
born and raised. Oxygen in biominerals such as teeth and bones
comes primarily from ingested water. The lakes and rivers on
the northern side of the Alps are known to have a lower 18O
content than those on the southern side of the mountains. The
18
O content of the teeth and bones of the Iceman was found to
be relatively high and characteristic of the watershed south of
the Alps. He had clearly been born and raised in that area.
Reuters/Alamy Stock Photo
2.1 Using Isotopes: Ötzi, the Iceman of the Alps
Ötzi the Iceman. A well-preserved mummy of a man who
lived in northern Italy about 5300 years ago.
Applying Chemical Principles
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117
The relative abundance of isotopes of many other elements
also varies slightly from place to place and in their incorporation
into different minerals. Strontium, a member of the same periodic group as calcium, is incorporated into teeth and bones. The
ratio of strontium isotopes, 87Sr/86Sr, and of lead isotopes,
206
Pb/204Pb, in the Iceman’s teeth and bones was characteristic
of soils from a narrow region of Italy south of the Alps, which
established more clearly where he was born and lived most of his
life.
The investigators also looked for food residues in the Iceman’s
intestines. Although a few grains of cereal were found, they located
tiny flakes of mica believed to have broken off stones used to grind
grain and that were therefore eaten when the man ate the grain.
They analyzed these flakes using argon isotopes, 40Ar and 39Ar, and
found their signature was like that of mica in an area south of the
Alps, thus establishing where he lived in his later years.
The many isotope studies show that the Iceman lived thousands of years ago in a small area about 10–20 kilometers
west of Merano in northern Italy.
For details of the isotope studies, see W. Müller, et al.,
­Science, 2003, 302, 862–866.
Questions
1. How many neutrons are there in atoms of 18O? In each of the
two isotopes of lead?
2. There are three stable isotopes of oxygen (16O, mass 15.9949,
99.757%, 17O, mass 16.9991, 0.038%, and 18O, 17.9992,
0.205%). Use these data to calculate the atomic weight of oxygen.
The practice in medicine for some centuries is to find compounds that are toxic to certain organisms but not so toxic that
the patient is harmed. In the early part of the twentieth century, Paul Ehrlich set out to find just such a compound that
would cure syphilis, a sexually transmitted disease that was
rampant at the time. He screened hundreds of compounds,
and found that his 606th compound was effective: an arseniccontaining drug now called salvarsan. It was used for some
years for syphilis treatment until penicillin was discovered in
the 1930s.
Salvarsan was a forerunner of the modern drug industry. Interestingly, what chemists long thought to be a single compound
was in fact discovered to be a mixture of compounds. Question
2 will lead you to the molecular formula for each of them.
Questions
1. Arsenic is found widely in the environment and is a major
problem in the ground water supply in Bangladesh. Orpiment
is one arsenic-containing mineral and enargite is another.
The latter has 19.024% As, 48.407% Cu, and 32.569% S.
What is the empirical formula of the mineral?
2. Salvarsan was long thought to be a single substance.
Recently, however, a mass spectrometry study of the compound shows it to be a mixture of two molecules with the
John C. Kotz
2.2 Arsenic, Medicine, and the Formula of Compound 606
A sample of orpiment, a common arsenic-containing mineral
(As2S3). The name of the element is thought to come from the Greek word
for this mineral, which was long favored by seventeenth century Dutch
painters as a pigment.
same empirical formula. Each has the composition 39.37%
C, 3.304% H, 8.741% O, 7.652% N, and 40.932% As. One
has a molar mass of 549 g/mol and the other has a molar
mass of 915 g/mol. What are the molecular formulas of the
compounds?
2.3 Argon—An Amazing Discovery
118
Wellcome Images CC/DIOMEDIA
Sir William Ramsay (1852–1916). Ramsay
was a Scottish chemist who discovered
several of the noble gases (for which he
received the Nobel Prize in Chemistry in
1904). Lord Rayleigh received the Nobel Prize
in Physics, also in 1904, for the discovery of
argon.
Wellcome Images CC/Diomedia
The noble gas argon was discovered by Sir William
Ramsay and John William Strutt (the third Lord
­Rayleigh) in England and reported in scientific journals
in 1895. In making this discovery, Ramsay and Lord
Rayleigh made highly accurate measurements of gas
densities. They found that gaseous nitrogen (N2)
formed by thermal decomposition of ammonia had a
density that was slightly lower than the density of the
gas that remained after O2, CO2, and H2O were removed from air. The reason for the difference is that
the sample derived from air contained a very small
amount of other gases. After removing N2 from the
sample by reacting it with red hot magnesium (to form
Chapter 2 / Atoms, Molecules, and Ions
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Mg3N2), a small quantity of gas remained that was more dense
than air. This was identified as argon.
Lord Rayleigh’s experimentally determined densities for
oxygen, nitrogen, and air are given below:
from the isotopic masses and fractional abundances.)
Assume dry air with CO2 removed is 20.96% (by volume)
oxygen, 78.11% nitrogen, and 0.930% argon. Determine the
density of argon.
3. Atmospheric argon is a mixture of three stable isotopes, 36Ar,
38
Ar, and 40Ar. Use the information in the table below to
determine the atomic mass and natural abundance of 40Ar.
Gas
Density
(g/L)
Oxygen
1.42952
Isotope
Atomic Mass
Abundance (%)
Nitrogen, derived from air
1.25718
36
Ar
35.967545
0.334
Nitrogen, derived from ammonia
1.25092
38
Ar
37.962732
0.063
Air, with water and CO2 removed
1.29327
40
Ar
?
?
Questions
1. To determine the density of atmospheric nitrogen, Lord
Rayleigh removed the oxygen, water, and carbon dioxide from
air, then filled an evacuated glass globe with the remaining
gas. He determined that a mass of 0.20389 g of nitrogen
has a density of 1.25718 g/L under standard conditions of
temperature and pressure. What is the volume of the globe
(in cm3)?
2. The density of a mixture of gases may be calculated by
summing the products of the density of each gas and the
fractional volume of space occupied by that gas. (Note the
similarity to the calculation of the molar mass of an element
4. Given that the density of argon is 1.78 g/L under standard
conditions of temperature and pressure, how many argon
atoms are present in a room with dimensions 4.0 m ×
5.0 m × 2.4 m that is filled with pure argon under these
conditions of temperature and pressure?
References
1. Lord Rayleigh and William Ramsay, Proceedings of the Royal
Society of London, 1894, 57, 265–287.
2. The Gases of the Atmosphere and Their History, 4th ed., William
Ramsay, MacMillan and Co., Limited, London, 1915.
Think–Pair–Share
1. Consider particles
compositions:
with
the
following
subatomic
Particle
Protons
Neutrons
Electrons
1
16
16
18
2
18
22
18
3
19
20
19
4
19
21
19
5
20
20
20
6
20
20
18
(a) For each particle, write the complete chemical symbol
including the element symbol, atomic number, mass number, and any nonzero charge.
(b) Which particles are neutral atoms? Which are isotopes?
Which is a cation? Which is an anion? Explain how you
arrived at these answers.
2. The figure shows a molecular model of ribose, a component
of RNA. The black spheres represent carbon atoms, the red
spheres represent oxygen atoms, and the white spheres represent hydrogen atoms.
Ribose
(a) What is the molecular formula?
(b) What is the empirical formula?
3. Write the formulas and names for the ionic compounds that
can be made by combining the cations: NH41, Mg21, and
Fe31 with the anions Cl2, SO422, and PO432.
4. Increasing amounts of carbon dioxide are implicated in
climate change.
(a) What is the formula of carbon dioxide?
(b) How many atoms of carbon and how many atoms of
­oxygen are present in one molecule of carbon dioxide?
(c) How many moles of carbon and how many moles of
­oxygen are present in one mole of carbon dioxide?
(d) Is there a relationship between the answers for parts b
and c? Explain in a few sentences why there should or
should not be a relationship.
Think–Pair–Share
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119
5. Empirical Formulas and Mole Ratios
(a) What empirical formula (XmYn) is indicated by each of the
following mole ratios of the elements X and Y?
(i) 1.50 mol Y/mol X
(ii) 1.33 mol Y/mol X
(iii) 1.67 mol Y/mol X
(iv) 2.50 mol Y/mol X
(v) 1.25 mol Y/mol X
(b) Most textbooks, including this one, instruct students to
obtain the mole ratios in an empirical formula problem by
dividing the larger number of moles by the smaller number
of moles. If you did the opposite and divided the smaller
number of moles by the larger number, could you interpret
the results to obtain the correct empirical formula? Why
or why not? Why do you think textbooks say to divide the
larger number of moles by the smaller number?
6. Two compounds have the molecular formula C2H4Cl2:
1,1-dichloroethane and 1,2-dichloroethane. These are their
structural formulas:
H H
Cl
C
C
H H
H
H
C
C
H
Cl H
Cl Cl
1,1-dichloroethane
1,2-dichloroethane
ass spectrometry was performed on samples of each comM
pound, and the following spectra were obtained.
Mass Spectrum
100
Relative Intensity
80
60
40
20
0
0
20
40
60
80
100
120
80
100
120
m/Z
Mass Spectrum
100
Relative Intensity
80
60
40
20
0
0
20
40
60
m/Z
120
Chapter 2 / Atoms, Molecules, and Ions
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(a) Both spectra have peaks at m/Z 5 98, 100, and 102. Given
that carbon is predominantly carbon-12 (99% abundance)
and that the element chlorine has two stable isotopes:
35
Cl (76% abundance) and 37Cl (24% abundance), what are
the molecular formulas for these ions? Do these ions help
identify which spectrum corresponds to which compound?
(b) Do the two spectra provide evidence that the two samples
are different compounds? Which spectrum corresponds to
which compound? How do you know?
Chapter Goals Revisited
The goals for this chapter are keyed to specific Study Questions to help you
organize your review.
2.1 Atomic Structure, Atomic Number, and Atomic Mass
•
Describe electrons, protons, and neutrons, and the general structure of
the atom. 1, 3.
•
Define the terms atomic number and mass number. 2, 5–7.
•
Define isotopes and give the atomic symbol for a specific isotope.
8–11, 111.
2.2 Atomic Weight
•
Perform calculations that relate the atomic weight (atomic mass) of an
element and isotopic abundances and masses. 17–22, 166b, 168.
2.3 The Periodic Table
•
Know the terminology of the periodic table (periods, groups) and know
how to use the information given in the periodic table. 23, 24, 27, 29, 113.
•
Recognize similarities and differences in properties of some of the
common elements of a group. 26, 30.
2.4 Molecules: Formulas, Models, and Names
•
•
Recognize and interpret molecular formulas, condensed formulas, and
structural formulas. 31, 32.
Recognize and interpret different types of molecular models. 31, 32, 85,
129, 134, 138, 139.
•
Remember formulas and names of common molecular compounds. 36.
•
Name and write formulas for binary molecular compounds. 33–36.
2.5 Ions
•
Recognize that metal atoms commonly lose one or more electrons to form
positive ions, called cations, and nonmetal atoms often gain electrons to
form negative ions, called anions. 43, 44, 126.
•
Predict the charge on monatomic cations and anions based on Group
number. 37–40.
•
Name and write the formulas of ions. 41, 42.
Chapter Goals Revisited
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121
2.6 Ionic Compounds: Formulas, Names, and Properties
•
Write formulas for ionic compounds by combining ions in the proper ratio
to give a zero overall charge. 47–54.
•
Give the names and write the formulas of ionic compounds. 55–62.
•
Understand the importance of Coulomb’s law in chemistry, which
describes the electrostatic forces of attraction and repulsion of ions.
63–64.
2.7 Atoms, Molecules, and the Mole
•
Understand the mole concept and molar mass and their application.
•
Use the molar mass of an element and Avogadro’s number in calculations.
•
Calculate the molar mass of a compound from its formula and a table of
atomic weights. 77–80.
•
Calculate the amount (= number of moles) of a compound represented by
a given mass, and vice versa. 81, 82.
•
Use Avogadro’s number to calculate the number of atoms or ions in a
compound. 83–86, 128.
65–68, 72, 73.
69–71, 115, 116.
2.8 Chemical Analysis: Determining Compound Formulas
•
Express the composition of a compound in terms of percent composition.
•
Determine the empirical and molecular formula of a compound using
percent composition or other experimental data. 95–106, 137, 143, 145.
87–89.
2.9 Instrumental Analysis: Determining Compound
Formulas
•
Determine a molecular formula from a mass spectrum. 107, 108.
•
Identify isotopes using mass spectrometry. 109, 110.
Key Equations
Equation 2.1 (page 68) Percent abundance of an isotope.
Percent abundance 5
number of atoms of a given isotope
3 100%
total number of atoms of all isotopes of that element
Equation 2.2 (page 69) Calculate the atomic weight from isotope abundances and
the atomic mass of each isotope of an element.
 % abundance isotope 1 
Atomic weight 5 
 (mass of isotope 1)

100
 % abundance isotope 2 
1
 (mass of isotope 2) 1 . . .

100
122
Chapter 2 / Atoms, Molecules, and Ions
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Equation 2.3 (page 97) Coulomb’s Law, the force of attraction between oppositely
charged ions.
charge on + and − ions
Force = −k
proportionality constant
charge on electron
(n+e)(n−e)
d2
distance between ions
Study Questions
▲ denotes challenging questions. Blue-numbered questions have answers in Appendix N and fully
worked solutions in the Student Solutions Manual.
Practicing Skills
Atoms: Their Composition and Structure
(See Section 2.1 and Example 2.1.)
1. What are the three fundamental particles from
which atoms are built? What are their electric
charges? Which of these particles constitute the
nucleus of an atom? Which is the least massive
particle of the three?
2. Define mass number. What is the difference
between the mass number and atomic mass of an
atom?
3. An atom has a very small nucleus surrounded by
an electron cloud. Figure 2.1 represents the
nucleus with a diameter of about 2 mm and
describes the electron cloud as extending over
200 m. If the diameter of an atom is 1 × 10−8 cm,
what is the approximate diameter of its nucleus?
4. A gold atom has a radius of 145 pm. If you could
string gold atoms like beads on a thread, how
many atoms would you need to have a necklace
36 cm long?
5. Give the complete symbol (AZX), including atomic
number and mass number, for each of the
following atoms: (a) potassium with 22 neutrons,
(b) vanadium with 28 neutrons, and (c) gallium
with 38 neutrons.
6. Give the complete symbol (AZX), including atomic
number and mass number, for each of the
following atoms: (a) iron with 30 neutrons,
(b) uranium with 146 neutrons, and (c) lead with
125 neutrons.
7. How many electrons, protons, and neutrons are
there in each of the following atoms?
(a) magnesium-24, 24Mg
(b) tin-119, 119Sn
(c) thorium-232, 232Th
(d) carbon-13, 13C
(e) copper-63, 63Cu
(f) bismuth-205, 205Bi
8. Atomic structure.
(a) The synthetic radioactive element technetium
is used in many medical studies. Give the
number of electrons, protons, and neutrons in
an atom of technetium-99.
(b) Radioactive americium-241 is used in household smoke detectors and in bone mineral
analysis. Give the number of electrons,
protons, and neutrons in an atom of
americium-241.
9. Cobalt has three radioactive isotopes used in
medical studies. Atoms of these isotopes have 30,
31, and 33 neutrons, respectively. Give the complete symbol for each of these isotopes.
10. Naturally occurring silver exists as two isotopes
having mass numbers 107 and 109. How many
protons, neutrons, and electrons are there in each
of these isotopes?
11. Name and describe the composition of the three
hydrogen isotopes.
12. Which of the following are isotopes of element X,
which has an atomic number of 16:
32
36
36
16
30
16 X , 16 X , 18 X , 8 X , and 14 X
Study Questions
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123
Key Experiments Developing Atomic Structure
(See pages 72–74.)
13. From cathode ray experiments, J. J. Thomson estimated that the mass of an electron was “about a
thousandth” of the mass of a proton. How accurate is that estimate? Calculate the ratio of the
mass of an electron to the mass of a proton.
14. In 1886, Eugene Goldstein observed positively
charged particles called canal rays moving in the
opposite direction to electrons in a cathode ray tube
(illustrated below). From their mass, he concluded
that these particles were formed from residual gas in
the tube. For example, if the cathode ray tube contained helium, the canal rays consisted of He+ ions.
Describe a process that could lead to these ions.
Cathode rays
Anode
+
–
–
Positive (canal) rays
Cathode with holes
(pierced disk)
+
+
–
–
+
16. Early in the 1800s, John Dalton proposed that an
atom was a “solid, massy, hard, impenetrable,
moveable particle.” Critique this description. How
does this description misrepresent atomic
structure?
Atomic Weight
(See Examples 2.3 and 2.4.)
17. Thallium has two stable isotopes, 203Tl and 205Tl.
Knowing that the atomic weight of thallium is
204.4, which isotope is the more abundant of
the two?
18. Strontium has four stable isotopes. Strontium-84
has a very low natural abundance, but 86Sr, 87Sr,
and 88Sr are all reasonably abundant. Knowing
that the atomic weight of strontium is 87.62,
which of the more abundant isotopes
predominates?
19. Verify that the atomic weight of lithium is 6.94,
given the following information:
Li, mass = 6.015123; percent abundance = 7.6%
Li, mass = 7.016003; percent abundance = 92.4%
6
+
7
Electron
Gas molecules
To vacuum pump
20. Verify that the atomic weight of magnesium is
24.31, given the following information:
Positive ion
Canal rays. In 1886, Eugene Goldstein detected a stream of
particles traveling in the direction opposite to that of the
negatively charged cathode rays (electrons). He called this
stream of positive particles “canal rays.”
15. Marie and Pierre Curie and others observed that a
radioactive substance could emit three types of
radiation: alpha (α), beta (β), and gamma (γ). If
the radiation from a radioactive source is passed
between electrically charged plates, some particles
are attracted to the positive plate, some to the
negative plate, and others feel no attraction.
Which particles are positively charged, which are
negatively charged, and which have no charge? Of
the two charged particles, which has the most
mass?
particles
rays
Photographic film
or phosphor screen
+
Lead block
shield
particles,
attracted to
+ plate
–
Slit
particles
particles,
attracted to
– plate
Charged
plates
Radioactive
element
Radioactivity. Alpha (α), beta (β), and gamma (γ) rays from a
radioactive element are separated by passing them between
electrically charged plates.
124
Mg, mass = 23.985042; percent abundance = 78.99%
Mg, mass = 24.985837; percent abundance = 10.00%
26
Mg, mass = 25.982593; percent abundance = 11.01%
24
25
21. Gallium has two naturally occurring isotopes,
69
Ga and 71Ga, with masses of 68.9256 and
70.9247, respectively. Calculate the percent abundances of these isotopes of gallium.
22. Europium has two stable isotopes, 151Eu and
153
Eu, with masses of 150.9199 and 152.9212,
respectively. Calculate the percent abundances of
these isotopes of europium.
The Periodic Table
(See Section 2.3.)
23. Titanium and thallium have symbols that are
easily confused with each other. Give the symbol,
atomic number, atomic weight, and group and
period number of each element. Are they metals,
metalloids, or nonmetals?
24. In Groups 4A–6A (14–16), there are several elements whose symbols begin with S. Name these
elements, and for each one give its symbol, atomic
number, group number, and period. Describe each
as a metal, metalloid, or nonmetal.
25. How many periods of the periodic table have
8 elements, how many have 18 elements, and how
many have 32 elements?
Chapter 2 / Atoms, Molecules, and Ions
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26. How many elements occur in the seventh period?
What is the name given to the majority of these
elements, and what well-known property characterizes them?
27. Select answers to the questions listed below from
the following list of elements whose symbols start
with the letter C: C, Ca, Cr, Co, Cd, Cl, Cs, Ce,
Cm, Cu, and Cf. (You should expect to use some
symbols more than once.)
(a) Which are nonmetals?
(b) Which are main group elements?
(c) Which are lanthanides?
(d) Which are transition elements?
(e) Which are actinides?
(f) Which are gases?
28. Give the name and chemical symbol for the
following.
(a) a nonmetal in the second period
(b) an alkali metal in the fifth period
(c) the third-period halogen
(d) an element that is a gas at 20 °C and 1 atmosphere pressure
29. Classify the following elements as metals, metalloids, or nonmetals: N, Na, Ni, Ne, and Np.
30. Here are symbols for five of the seven elements
whose names begin with the letter B: B, Ba, Bk, Bi,
and Br. Match each symbol with one of the
descriptions below.
(a) a radioactive element
(b) a liquid at room temperature
(c) a metalloid
(d) an alkaline earth element
(e) a group 5A (15) element
Molecules: Formulas, Models, and Names
(See Section 2.4 and Example 2.5)
31. A model of nitric acid is illustrated here. Write
the molecular formula for nitric acid, and draw
the structural formula. Describe the structure
of the molecule. Is it flat? That is, are all the atoms
in the plane of the paper? (Color code: nitrogen
atoms are blue; oxygen atoms are red; and hydrogen atoms are white.)
32. A model of the amino acid asparagine is illustrated here. Write the molecular formula for the
compound, and draw its structural formula.
Asparagine, an
an amino
Asparagine,
aminoacid
acid
33. Name each of the following binary molecular
compounds:
(a) NF3
(c) BI3
(b) HI
(d) PF5
34. Name each of the following binary molecular
compounds:
(a) N2O5
(c) OF2
(b) P4S3
(d) XeF4
35. Give the formula for each of the following
compounds:
(a) sulfur dichloride
(b) dinitrogen pentaoxide
(c) silicon tetrachloride
(d) diboron trioxide (commonly called boric oxide)
36. Give the formula for each of the following
compounds:
(a) bromine trifluoride
(b) xenon difluoride
(c) hydrazine
(d) diphosphorus tetrafluoride
(e) butane
Ions and Ion Charges
(See Figure 2.19 and Table 2.4.)
37. What is the charge on the common monatomic
ions of the following elements?
(a) magnesium
(c) nickel
(b) zinc
(d) gallium
38. Identify the common charges on monatomic ions
of the following elements.
(a) sodium
(c) iron
(b) aluminum
(d) copper
Nitric acid
39. What is the charge on the common monatomic
ions of the following elements?
(a) selenium
(c) oxygen
(b) fluorine
(d) nitrogen
Study Questions
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125
40. What is the charge on the common monatomic
ions of the following elements?
(a) bromine
(c) phosphorus
(b) sulfur
(d) chlorine
41. Give the symbol, including the correct charge, for
each of the following ions:
(a) barium ion
(b) titanium(IV) ion
(c) phosphate ion
(d) hydrogen carbonate ion
(e) sulfide ion
(f) perchlorate ion
(g) cobalt(II) ion
(h) sulfate ion
42. Give the symbol, including the correct charge, for
each of the following ions:
(a) permanganate ion
(b) nitrite ion
(c) dihydrogen phosphate ion
(d) ammonium ion
(e) phosphate ion
(f) sulfite ion
43. When a potassium atom becomes a monatomic
ion, how many electrons does it lose or gain?
What noble gas atom has the same number of
electrons as a potassium ion?
44. When oxygen and sulfur atoms become monatomic ions, how many electrons does each lose or
gain? Which noble gas atom has the same number
of electrons as an oxide ion? Which noble gas
atom has the same number of electrons as a
sulfide ion?
45. How many protons and electrons are present in
each of the following ions?
(a) Na1
(c) F−
21
(b) Mg
(d) O2−
46. How many protons and electrons are present in
each of the following ions?
(c) Cu21
(a) Ca21
(b) Cu1
(d) Cl−
Ionic Compounds
(See Examples 2.6–2.8)
47. What are the charges on the ions in an ionic
compound containing the elements barium
and bromine? Write the formula for the
compound.
48. What are the charges of the ions in an ionic compound containing cobalt(III) and fluoride ions?
Write the formula for the compound.
126
49. Give the name, formula, and the number of each ion
that makes up each of the following compounds:
(a) K2S
(d) (NH4)3PO4
(b) CoSO4
(e) Ca(ClO)2
(c) KMnO4
(f) NaCH3CO2
50. Give the name, formula, and the number of each
ion that makes up each of the following
compounds:
(a) Mg(CH3CO2)2
(d) Ti(SO4)2
(b) Al(OH)3
(e) KH2PO4
(c) CuCO3
(f) CaHPO4
51. Cobalt forms Co21 and Co31 ions. Write the formulas for the two cobalt oxides formed by these
transition metal ions.
52. Platinum is a transition element and forms Pt21
and Pt41 ions. Write the formulas for the compounds of each of these ions with (a) chloride
ions and (b) sulfide ions.
53. Which of the following are correct formulas for
ionic compounds? For those that are not, give the
correct formula.
(c) Ga2O3
(a) AlCl2
(b) KF2
(d) MgS
54. Which of the following are correct formulas for
ionic compounds? For those that are not, give the
correct formula.
(a) Ca2O
(c) Fe2O5
(b) SrBr2
(d) Li2O
55. Name each of the following ionic compounds:
(a) K2S
(c) (NH4)3PO4
(b) CoSO4
(d) Ca(ClO)2
56. Name each of the following ionic compounds:
(a) Ca(CH3CO2)2
(c) Al(OH)3
(b) Ni3(PO4)2
(d) KH2PO4
57. Name each of the following ionic compounds.
(a) Na2S
(c) NaHSO4
(b) Na2SO3
(d) Na2SO4
58. Name each of the following ionic compounds.
(a) Li3N
(b) LiNO2
(c) LiNO3
59. Give the formula for each of the following ionic
compounds:
(a) ammonium carbonate
(b) calcium iodide
(c) copper(II) bromide
(d) aluminum phosphate
(e) silver(I) acetate
Chapter 2 / Atoms, Molecules, and Ions
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60. Give the formula for each of the following ionic
compounds:
(a) calcium hydrogen carbonate
(b) potassium permanganate
(c) magnesium perchlorate
(d) potassium hydrogen phosphate
(e) sodium sulfite
61. Write the formulas for the four ionic compounds
that can be made by combining each of the
cations Na1 and Ba21 with the anions CO322
and I2. Name each of the compounds.
62. Write the formulas for the four ionic compounds
that can be made by combining the cations Mg21
and Fe31 with the anions PO432 and NO32. Name
each compound formed.
Coulomb’s Law
(See Equation 2.3 and Figure 2.23.)
63. Sodium ions, Na1, form ionic compounds with fluoride ions, F2, and iodide ions, I2. The radii of these
ions are as follows: Na1 5 116 pm; F2 5 119 pm;
and I2 5 206 pm. In which ionic compound, NaF
or NaI, are the forces of attraction between cation
and anion stronger? Explain your answer.
64. Consider the two ionic compounds NaCl and
CaO. In which compound are the cation–anion
attractive forces stronger? Explain your answer.
Atoms and the Mole
(See Example 2.9.)
65. Calculate the mass, in grams, of each the
following:
(a) 3.2 mol of aluminum
(b) 2.35 × 1023 mol of iron
(c) 0.12 mol of calcium
(d) 23.0 mol of neon
66. Calculate the mass, in grams, of each the following:
(a) 1.24 mol of gold
(b) 14.8 mol of He
(c) 0.43 mol of platinum
(d) 2.42 × 1024 mol of Rh
67. Calculate the amount (moles) represented by each
of the following:
(a) 87.21 g of Cu
(b) 0.024 g of sodium
(c) 2.0 mg of iridium
(d) 6.75 g of Au
68. Calculate the amount (moles) represented by each
of the following:
(a) 9.4 g of Li
(c) 0.037 g of platinum
(b) 0.942 g of tin
(d) 1.84 g of Xe
69. Calculate the average mass of an atom (in grams)
for each of the following elements.
(a) helium
(c) scandium
(b) fluorine
(d) bismuth
70. Calculate the average mass of an atom (in grams)
for each of the following elements.
(a) beryllium
(c) krypton
(b) aluminum
(d) mercury
71. You are given 1.0-g samples of He, Fe, Li, Si, and
C. Which sample contains the largest number of
atoms? Which contains the smallest?
72. You are given 0.10-g samples of K, Mo, Cr, and Al.
List the samples in order of the amount (moles),
from smallest to largest.
73. Analysis of a 10.0-g sample of apatite (a major
component of tooth enamel) showed that it
was made up of 3.99 g Ca, 1.85 g P, 4.14 g O,
and 0.020 g H. List these elements based on
relative amounts (moles), from smallest to
largest.
74. A semiconducting material is composed of 52 g
of Ga, 9.5 g of Al, and 112 g of As. Which
element has the largest number of atoms in this
material?
Molecules, Compounds, and the Mole
(See Example 2.10.)
75. The molecular formula of glucose is C6H12O6.
What amount (moles) of carbon is present in 1.0
mol of glucose? In 0.20 mol of glucose?
76. The formula of water is H2O. What amounts
(moles) of hydrogen and oxygen are present in
1.0 mol of water? In 4.0 mol of water?
77. Calculate the molar mass of each of the following
compounds:
(a) Fe2O3, iron(III) oxide
(b) BCl3, boron trichloride
(c) C6H8O6, ascorbic acid (vitamin C)
(d) Mg(NO3)2, magnesium nitrate
78. Calculate the molar mass of each of the following
compounds:
(a) CaCO3, calcium carbonate, a compound used
as an antacid
(b) Fe(C6H11O7)2, iron(II) gluconate, a dietary
supplement
(c) CH3CH2CH2CH2SH, butanethiol, has a skunklike odor
(d) C20H24N2O2, quinine, used as an antimalarial
drug
Study Questions
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127
79. Calculate the molar mass of each hydrated compound. Note that the water of hydration is
included in the molar mass. (See page 104.)
(a) Ni(NO3)2 ∙ 6 H2O
(b) CuSO4 ∙ 5 H2O
80. Calculate the molar mass of each hydrated compound. Note that the water of hydration is
included in the molar mass. (See page 104.)
(a) H2C2O4 ∙ 2 H2O
(b) MgSO4 ∙ 7 H2O, Epsom salt
84. How many ammonium ions and how many
sulfate ions are present in a 0.20 mol sample of
(NH4)2SO4? How many atoms of N, H, S and O
are contained in this sample?
85. Acetaminophen, whose structure is drawn below,
is the active ingredient in some nonprescription
pain killers. The recommended dose for an adult
is two 500‑mg caplets. How many molecules
make up one dose of this drug?
81. What mass is represented by 0.0255 mol of each
of the following compounds?
(a) C3H7OH, 2-propanol, rubbing alcohol
(b) C11H16O2, an antioxidant in foods, also known
as BHA (butylated hydroxyanisole)
(c) C9H8O4, aspirin
(d) (CH3)2CO, acetone, an important industrial
solvent
82. Assume you have 0.123 mol of each of the following compounds. What mass of each is present?
(a) C14H10O4, benzoyl peroxide, used in acne
medications
(b) Dimethylglyoxime, used in the laboratory to
test for nickel(II) ions
CH3
C
N
OH
C
N
OH
CH3
CH3 H
H
C
C
CH3
S
H
H
(d) DEET, a mosquito repellent
HC
HC
H
C
C
C
CH
O
CH2
C
N
CH2
CH3
CH3
CH3
83. Sulfur trioxide, SO3, is made industrially in enormous quantities by combining oxygen and sulfur
dioxide, SO2. What amount (moles) of SO3 is represented by 1.00 kg of sulfur trioxide? How many
molecules? How many sulfur atoms? How many
oxygen atoms?
128
86. An Alka-Seltzer tablet contains 325 mg of
aspirin (C9H8O4), 1916 mg of NaHCO3, and
1000. mg of citric acid (H3C6H5O7). (The last
two compounds react with each other to provide
the fizz, bubbles of CO2, when the tablet is put
into water.)
(a) Calculate the amount (moles) of each substance in the tablet.
(b) If you take one tablet, how many molecules of
aspirin are you consuming?
Percent Composition
(c) The compound below, responsible for the
skunky taste in beer exposed to light.
C
Acetaminophen
(See Example 2.11.)
87. Calculate the mass percent of each element in the
following compounds:
(a) PbS, lead(II) sulfide, galena
(b) C3H8, propane
(c) C10H14O, carvone, found in caraway seed oil
88. Calculate the mass percent of each element in the
following compounds:
(a) C8H10N2O2, caffeine
(b) C10H20O, menthol
(c) CoCl2 ∙ 6 H2O
89. Calculate the mass percent of copper in CuS,
copper(II) sulfide. If you wish to obtain 10.0 g of
copper metal from copper(II) sulfide, what mass
of CuS (in grams) must you use?
90. Calculate the mass percent of titanium in the
mineral ilmenite, FeTiO3. What mass of ilmenite
(in grams) is required if you wish to obtain 750 g
of titanium?
Chapter 2 / Atoms, Molecules, and Ions
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Empirical and Molecular Formulas
(See Example 2.12.)
91. Succinic acid occurs in fungi and lichens. Its
empirical formula is C2H3O2, and its molar mass
is 118.1 g/mol. What is its molecular formula?
92. An organic compound has the empirical formula
C2H4NO. If its molar mass is 116.1 g/mol, what is
the molecular formula of the compound?
93. Complete the following table:
Empirical
Formula
Molar Mass
(g/mol)
CH
26.0
(b) CHO
116.1
(a)
(c)
Molecular
Formula
C8H16
94. Complete the following table:
(a)
Empirical
Formula
Molar Mass
(g/mol)
C2H3O3
150.0
(b) C3H8
(c)
Molecular
Formula
44.1
B4H10
95. Acetylene is a colorless gas used as a fuel in
welding torches, among other things. It is 92.26%
C and 7.74% H. Its molar mass is 26.02 g/mol.
What are the empirical and molecular formulas of
acetylene?
96. A large family of boron-hydrogen compounds has
the general formula BxHy . One member of this
family contains 88.5% B; the remainder is
hydrogen. What is its empirical formula?
97. Cumene, a hydrocarbon, is a compound composed only of C and H. It is 89.94% carbon, and
its molar mass is 120.2 g/mol. What are the
empirical and molecular formulas of cumene?
98. In 2006, a Russian team discovered an interesting
molecule they called sulflower because of its
shape and because it was based on sulfur. It is
composed of 57.17% S and 42.83% C and has a
molar mass of 448.70 g/mol. Determine the
empirical and molecular formulas of sulflower.
99. Mandelic acid is an organic acid composed of
carbon (63.15%), hydrogen (5.30%), and oxygen
(31.55%). Its molar mass is 152.14 g/mol. Determine the empirical and molecular formulas of the
acid.
100. Nicotine, a poisonous compound found in
tobacco leaves, is 74.0% C, 8.65% H, and 17.35%
N. Its molar mass is 162 g/mol. What are the
empirical and molecular formulas of nicotine?
Determining Formulas from Mass Data
(See Examples 2.13 and 2.14.)
101. A compound containing xenon and fluorine
was prepared by shining sunlight on a mixture
of Xe (0.674 g) and excess F2 gas. If you isolate
0.869 g of the new compound, what is its
empirical formula?
102. Elemental sulfur (1.47 g) is combined with
fluorine, F2, to give a compound with the formula
SFx , a very stable, colorless gas. If you isolate
6.70 g of SFx , what is the value of x?
103. You weigh out 1.523 g of solid BaCl2 ∙ x H2O and
heat this compound to remove the water. The
mass of the remaining anhydrous compound is
1.298 g. What is the name and formula of the
hydrated compound?
104. Glauber’s salt, which has laxative properties, is a
hydrate of sodium sulfate. You weigh out 2.343 g
of Glauber’s salt (Na2SO4 ∙ x H2O) and heat it to
remove the water. The mass of the anhydrous
compound that remains is 1.033 g. What is the
chemical name and formula of Glauber’s salt?
105. Epsom salt is used in tanning leather and in
medicine. It is hydrated magnesium sulfate,
MgSO4 ∙ 7 H2O. The water of hydration is lost on
heating, with the number lost depending on the
temperature. Suppose you heat a 1.394-g sample
at 100 °C and obtain 0.885 g of a partially
hydrated sample, MgSO4 ∙ x H2O. What is the
value of x?
106. You combine 1.25 g of germanium, Ge, with
excess chlorine, Cl2. The mass of product, GexCly,
is 3.69 g. What is the formula of the product,
GexCly?
Mass Spectrometry
(See Section 2.9.)
107. The mass spectrum of nitrogen dioxide is illustrated in the following figure.
(a) Identify the cations present for each of the
four peaks in the mass spectrum.
(b) Does the mass spectrum provide evidence that
the two oxygen atoms are attached to a central
nitrogen atom (ONO), or that an oxygen
atom is at the center (NOO)? Explain.
Study Questions
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129
100
80
80
NO2
Relative Abundance
Relative abundance of ions
100
30
60
46
40
20
30
40
Mass-to-charge ratio (m/Z)
100
50
104
Relative abundance of ions
30
40
POF3
60
110. The highest mass peaks in the mass spectrum of
Br2 occur at m/Z 158, 160, and 162. The ratio of
intensities of these peaks is approximately 1:2:1.
Bromine has two stable isotopes, 79Br (50.7%
abundance) and 81Br (49.3% abundance).
(a) What molecular species gives rise to each of
these peaks?
(b) Explain the relative intensities of these peaks.
(Hint: Consider the probabilities of each atom
combination.)
General Questions
These questions are not designated as to type or location in
the chapter. They may combine several concepts.
58
Symbol
60
40
Ni
20
69
47 50
50
66
33
S
Number of protons
10
Number of neutrons
10
Number of electrons
in the neutral atom
0
30
50
111. Fill in the blanks in the table (one column per
element).
85
30
25
Name of element
88
70
90
Mass-to-charge ratio (m/Z)
110
109. The mass spectrum of CH3Cl is illustrated here.
You know that carbon has two stable isotopes,
12
C and 13C with relative abundances of 98.9%
and 1.1%, respectively, and chlorine has two isotopes, 35Cl and 37Cl with abundances of 75.76%
and 24.24%, respectively.
(a) What molecular species gives rise to the lines
at m/Z of 50 and 52? Why is the line at 52
about 1/3 the height of the line at 50?
(b) What species might be responsible for the line
at m/Z 5 51?
130
20
(m/Z)
108. The mass spectrum of phosphoryl chloride, POF3,
is illustrated here.
(a) Identify the cation fragment at a m/Z ratio
of 85.
(b) Identify the cation fragment at a m/Z ratio
of 69.
(c) Which two peaks in the mass spectrum
provide evidence that the oxygen atom is connected to the phosphorus atom and is not
connected to any of the three fluorine atoms?
80
40
0
10
14
0
10
60
20
16
20
CHCl3
112. Potassium has three naturally occurring isotopes
(39K, 40K, and 41K), but 40K has a very low natural
abundance. Which of the other two isotopes is
more abundant? Briefly explain your answer.
113. Crossword Puzzle: In the 2 × 2 box shown here,
each answer must be correct four ways: horizontally, vertically, diagonally, and by itself. Instead of
words, use symbols of elements. When the puzzle
is complete, the four spaces will contain the overlapping symbols of 10 elements. There is only one
correct solution.
1
2
3
4
Chapter 2 / Atoms, Molecules, and Ions
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
Horizontal
1–2: two-letter symbol for a metal used in ancient
times
3–4: two-letter symbol for a metal that burns in
air and is found in Group 5A (15)
Vertical
1–3: two-letter symbol for a metalloid
2–4: two-letter symbol for a metal used in U.S.
coins
Single squares: All one-letter symbols
1: a colorful nonmetal
2: colorless, gaseous nonmetal
3: an element that makes fireworks green
4: an element that has medicinal uses
Diagonal
1–4: two-letter symbol for an element used in
electronics
2–3: two-letter symbol for a metal used with Zr to
make wires for superconducting magnets
This puzzle first appeared in Chemical & Engineering News, p. 86, December 14, 1987 (submitted by
S. J. Cyvin) and in Chem Matters, October 1988.
114. The following chart shows a general decline in
abundance in the solar system with increasing
mass among the first 30 elements. The decline
continues beyond zinc. Notice that the scale on
the vertical axis is logarithmic, that is, it progresses
in powers of 10. The abundance of nitrogen, for
example, is 1/10,000 (1/104) of the abundance of
hydrogen. All abundances are plotted as the
number of atoms per 1012 atoms of H. (The fact
that the abundances of Li, Be, and B, as well as
those of the elements near Fe, do not follow the
general decline is a consequence of the way that
elements are synthesized in stars.)
1014
1012
Relative abundance
1010
108
106
115. Copper atoms.
(a) What is the average mass of one copper atom?
(b) Students in a college computer science class
once sued the college because they were asked
to calculate the cost of one atom and could
not do it. But you are in a chemistry course,
and you can do this. (See E. Felsenthal, Wall
Street Journal, May 9, 1995.) If the cost of
2.0-mm diameter copper wire (99.999% pure)
is currently $80.10 for 7.0 g, what is the cost
of one copper atom?
116. Which of the following is impossible?
(a) silver foil that is 1.2 × 1024 m thick
(b) a sample of potassium that contains 1.784 ×
1024 atoms
(c) a gold coin of mass 1.23 × 1023 kg
(d) 3.43 × 10227 mol of S8 molecules
117. Reviewing the periodic table.
(a) Name the element in Group 2A (2) and the
fifth period.
(b) Name the element in the fifth period and
Group 4B (4).
(c) Which element is in the second period in
Group 4A (14)?
(d) Which element is in the fourth period in
Group 5A (15)?
(e) Which halogen is in the fifth period?
(f) Which alkaline earth element is in the third
period?
(g) Which noble gas element is in the fourth
period?
(h) Name the nonmetal in Group 6A (16) and the
third period.
(i) Name a metalloid in the fourth period.
118. Identify two nonmetallic elements that have allotropes and describe the allotropes of each.
104
102
0
(a) What is the most abundant main group
metal?
(b) What is the most abundant nonmetal?
(c) What is the most abundant metalloid?
(d) Which of the transition elements is most
abundant?
(e) Which halogens are included on this plot, and
which is the most abundant?
H He Li Be B C N O F NeNaMg Al Si P S Cl Ar K Ca Sc Ti V Cr MnFe Co Ni Cu Zn
Element
The abundance of the elements in the solar system from H
to Zn
119. In each case, decide which represents more mass:
(a) 0.5 mol of Na, 0.5 mol of Si, or 0.25 mol of U
(b) 9.0 g of Na, 0.50 mol of Na, or 1.2 × 1022
atoms of Na
(c) 10 atoms of Fe or 10 atoms of K
Study Questions
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131
120. The recommended daily allowance (RDA) of iron
for women 19–50 years old is 18 mg. How many
moles is this? How many atoms?
121. Order the following samples from smallest to
largest mass:
(a) 3.79 × 1024
(d) 7.40 mol Si
atoms Fe
(e) 9.221 mol Na
(b) 19.92 mol H2
(f) 4.07 × 1024 atoms Al
(c) 8.576 mol C
(g) 9.2 mol Cl2
122. ▲ When a sample of phosphorus burns in air, the
compound P4O10 forms. One experiment showed
that 0.744 g of phosphorus formed 1.704 g of
P4O10. Use this information to determine the ratio
of the atomic weights of phosphorus and oxygen
(mass P/mass O). If the atomic weight of oxygen
is assumed to be 16.00, calculate the atomic
weight of phosphorus.
123. ▲ When a sample of carbon burns completely in
air, the compound CO2 forms. One experiment
showed that 0.876 g of carbon formed 3.210 g of
CO2. Use this information to determine the ratio
of the atomic weights of oxygen and carbon (mass
O/mass C). If the atomic weight of carbon is
assumed to be 12.01, calculate the atomic weight
of oxygen.
124. A reagent occasionally used in chemical synthesis
is sodium–potassium alloy. (Alloys are mixtures
of metals, and Na-K has the interesting property
that it is a liquid.) One formulation of the alloy
(the one that melts at the lowest temperature)
contains 68 atom percent K; that is, out of every
100 atoms, 68 are K and 32 are Na. What is the
mass percent of potassium in sodium–potassium
alloy?
125. Write formulas for all of the compounds that can
be made by combining the cations NH41 and
Ni21 with the anions CO322 and PO432.
126. How many electrons are in a strontium atom (Sr)?
Does an atom of Sr gain or lose electrons when
forming an ion? How many electrons are gained
or lost by the atom? When Sr forms an ion, the
ion has the same number of electrons as which
one of the noble gases?
127. Which of the following compounds has the
highest mass percent of chlorine?
(a) BCl3
(d) AlCl3
(b) AsCl3
(e) PCl3
(c) GaCl3
128. Which of the following samples has the largest
number of ions?
(a) 1.0 g of CaCl2
(d) 1.0 g of SrCO3
(b) 1.0 g of MgCl2
(e) 1.0 g of BaSO4
(c) 1.0 g of CaS
129. The structure of one of the bases in DNA, adenine, is
shown here. Which represents the greater mass: 40.0 g
of adenine or 3.0 × 1023 molecules of the compound?
Adenine
130. Ionic and molecular compounds of the halogens.
(a) What are the names of BaF2, SiCl4, and NiBr2?
(b) Which of the compounds in part (a) are ionic,
and which are molecular?
(c) Which has the largest mass, 0.50 mol of BaF2,
0.50 mol of SiCl4, or 1.0 mol of NiBr2?
131. A drop of water has a volume of about 0.050 mL.
How many molecules of water are in a drop of
water? (Assume water has a density of 1.00 g/cm3.)
132. Capsaicin, the spicy compound in chili peppers,
has the formula C18H27NO3.
(a) Calculate its molar mass.
(b) If you eat 55 mg of capsaicin, what amount
(moles) have you consumed?
(c) Calculate the mass percent of each element in
the compound.
(d) What mass of carbon (in milligrams) is there
in 55 mg of capsaicin?
133. Calculate the molar mass and the mass percent of
each element in the blue solid compound
Cu(NH3)4SO4 ∙ H2O. What is the mass of copper
and the mass of water in 10.5 g of the compound?
134. Write the molecular formula and calculate the molar
mass for each of the molecules shown here. Which
has the largest mass percent of carbon? Of oxygen?
(a) ethylene glycol (used in antifreeze)
H H
H
O
C
C
O H
H H
Ethylene glycol
132
Chapter 2 / Atoms, Molecules, and Ions
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
(b) dihydroxyacetone (used in artificial tanning
lotions)
H
O
H
O H
C
C
C
H
O H
H
Dihydroxyacetone
(c) ascorbic acid, commonly known as vitamin C
Ephedrine
HO
H
H
H
C
C
C
H
OH
O
C
OH
C
O
C
OH
Ascorbic acid, vitamin C
135. Malic acid, an organic acid found in apples, contains
C, H, and O in the following ratios: C1H1.50O1.25.
What is the empirical formula of malic acid?
136. Your doctor has diagnosed you as being anemic—
that is, as having too little iron in your blood. At
the drugstore, you find two iron-containing
dietary supplements: one with iron(II) sulfate,
FeSO4, and the other with iron(II) gluconate,
Fe(C6H11O7)2. If you take 100. mg of each compound, which will deliver more atoms of iron?
137. A compound composed of iron and carbon monoxide, Fex(CO)y , is 30.70% iron. What is the
empirical formula for the compound?
138. Ma huang, an extract from the ephedra species of
plants, contains ephedrine. The Chinese have used
this herb for more than 5000 years to treat
asthma. More recently, ephedrine has been used
in diet pills that can be purchased over the
counter in herbal medicine shops. However, very
serious concerns were raised regarding these pills
following reports that their use led to serious
heart problems.
(a) From the following molecular model of
ephedrine, determine the molecular formula
for ephedrine and calculate its molar mass.
(b) What is the weight percent of carbon in
ephedrine?
(c) Calculate the amount (moles) of ephedrine in
a 0.125 g sample.
(d) How many molecules of ephedrine are there
in 0.125 g? How many C atoms?
139. Saccharin, a molecular model of which is shown
below, is more than 300 times sweeter than sugar.
It was first made in 1897, when it was common
practice for chemists to record the taste of any
new substances they synthesized.
(a) Write the molecular formula for the compound, and draw its structural formula.
(S atoms are yellow.)
(b) If you ingest 125 mg of saccharin, what amount
(moles) of saccharin have you ingested?
(c) What mass of sulfur is contained in 125 mg of
saccharin?
Saccharin
140. Name each of the following compounds and indicate which ones are best described as ionic:
(a) ClF3
(f) OF2
(b) NCl3
(g) KI
(c) SrSO4
(h) Al2S3
(d) Ca(NO3)2
(i) PCl3
(e) XeF4
(j) K3PO4
141. Write the formula for each of the following compounds and indicate which ones are best
described as ionic:
(a) sodium hypochlorite
(b) boron triiodide
(c) aluminum perchlorate
(d) calcium acetate
(e) potassium permanganate
(f) ammonium sulfite
(g) potassium dihydrogen phosphate
(h) disulfur dichloride
(i) chlorine trifluoride
(j) phosphorus trifluoride
Study Questions
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133
142. Complete the table by placing symbols, formulas,
and names in the blanks.
Cation
Anion
Name
Formula
ammonium bromide
Ba2+
BaS
Cl
2
iron(II) chloride
F
PbF2
2
Al31
CO322
iron(III) oxide
143. Empirical and molecular formulas.
(a) Fluorocarbonyl hypofluorite is composed of
14.6% C, 39.0% O, and 46.3% F. The molar
mass of the compound is 82 g/mol. ­Determine
the empirical and molecular formulas of the
compound.
(b) Azulene, a beautiful blue hydrocarbon, is
93.71% C and has a molar mass of 128.16 g/
mol. What are the empirical and molecular
formulas of azulene?
144. Cacodyl, a compound containing arsenic, was
reported in 1842 by the German chemist Robert
Wilhelm Bunsen. It has an almost intolerable
garlic-like odor. Its molar mass is 210 g/mol, and
it is 22.88% C, 5.76% H, and 71.36% As. Determine its empirical and molecular formulas.
145. The action of bacteria on meat and fish produces
a compound called cadaverine. As its name and
origin imply, it stinks! (It is also present in bad
breath and adds to the odor of urine.) It is
58.77% C, 13.81% H, and 27.40% N. Its molar
mass is 102.2 g/mol. Determine the molecular
formula of cadaverine.
146. ▲ In the laboratory you combine 0.125 g of
nickel with CO and isolate 0.364 g of Ni(CO)x .
What is the value of x?
147. ▲ A compound called MMT was once used to
boost the octane rating of gasoline. What is the
empirical formula of MMT if it is 49.5% C, 3.2%
H, 22.0% O, and 25.2% Mn?
148. ▲ Elemental phosphorus is made by heating
calcium phosphate with carbon and sand in an
electric furnace. What is the mass percent of phosphorus in calcium phosphate? Use this value to
calculate the mass of calcium phosphate (in kilograms) that must be used to produce 15.0 kg of
phosphorus.
149. ▲ Chromium is obtained by heating
chromium(III) oxide with carbon. Calculate the
134
mass percent of chromium in the oxide, and then
use this value to calculate the quantity of Cr2O3
required to produce 850 kg of chromium metal.
150. ▲ Stibnite, Sb2S3, is a dark gray mineral from
which antimony metal is obtained. What is the
mass percent of antimony in the sulfide? If you
have 1.00 kg of an ore that contains 10.6%
antimony, what mass of Sb2S3 (in grams) is in
the ore?
151. ▲ Direct reaction of iodine (I2) and chlorine (Cl2)
produces an iodine chloride, Ix Cly , a bright yellow
solid. If you completely consume 0.678 g of I2 in
a reaction with excess Cl2 and produce 1.246 g of
Ix Cly , what is the empirical formula of the compound? A later experiment showed that the molar
mass of Ix Cly is 467 g/mol. What is the molecular
formula of the compound?
152. ▲ In a reaction, 2.04 g of vanadium combined
with 1.93 g of sulfur to give a pure compound.
What is the empirical formula of the product?
153. ▲ Iron pyrite, often called fool’s gold, has the
formula FeS2. If you could convert 15.8 kg of iron
pyrite to iron metal, what mass of the metal
would you obtain?
154. Which of the following statements about 57.1 g of
octane, C8H18, is (are) not true?
(a) 57.1 g is 0.500 mol of octane.
(b) The compound is 84.1% C by weight.
(c) The empirical formula of the compound is
C4H9.
(d) 57.1 g of octane contains 28.0 g of hydrogen
atoms.
155. The formula of barium molybdate is BaMoO4.
Which of the following is the formula of sodium
molybdate?
(a) Na4MoO
(d) Na2MoO4
(b) NaMoO
(e) Na4MoO4
(c) Na2MoO3
156. ▲ A metal M forms a compound with the formula
MCl4. If the compound is 74.75% chlorine, what
is the identity of M?
157. Pepto-Bismol, which can help provide relief for an
upset stomach, contains 262 mg of bismuth subsalicylate, C21H15Bi3O12, per tablet. If you take two
tablets for your stomach distress, what amount (in
moles) of this active ingredient are you taking?
What mass of Bi are you consuming in two
tablets?
158. ▲ The weight percent of oxygen in an oxide that
has the formula MO2 is 15.2%. What is the molar
mass of this compound? What element or
elements are possible for M?
Chapter 2 / Atoms, Molecules, and Ions
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159. The mass of 2.50 mol of a compound with the
formula ECl4, in which E is a nonmetallic
element, is 385 g. What is the molar mass of ECl4?
What is the identity of E?
160. ▲ The elements A and Z combine to produce two
different compounds: A2Z3 and AZ2. If 0.15 mol
of A2Z3 has a mass of 22.48 g and 0.15 mol of AZ2
has a mass of 12.44 g, what are the atomic
weights of A and Z?
161. ▲ Polystyrene can be prepared by heating
styrene with tribromobenzoyl peroxide in the
absence of air. A sample of polystyrene prepared
by this method has the empirical formula
Br3C6H3(C8H8)n, where the value of n can vary
from sample to sample. If one sample has
0.105% Br, what is the value of n?
162. A sample of hemoglobin is found to be 0.335%
iron. What is the molar mass of hemoglobin if
there are four iron atoms per molecule?
163. ▲ Consider an atom of 64Zn.
(a) Calculate the density of the nucleus in grams
per cubic centimeter, knowing that the nuclear
radius is 4.8 × 1026 nm and the mass of the
64
Zn atom is 1.06 × 10222 g. (The volume of a
sphere is [4/3]πr3.)
(b) Calculate the density of the space occupied by
the electrons in the zinc atom, given that the
atomic radius is 0.125 nm and the electron
mass is 9.11 × 10228 g.
(c) Having calculated these densities, what statement can you make about the relative densities of the parts of the atom?
164. ▲ Estimating the radius of a lead atom.
(a) You are given a cube of lead that is 1.000 cm
on each side. The density of lead is 11.35 g/cm3.
How many atoms of lead are in the sample?
(b) Atoms are spherical; therefore, the lead atoms
in this sample cannot fill all the available
space. As an approximation, assume that 60.%
of the space of the cube is filled with spherical
lead atoms. Calculate the volume of one lead
atom from this information. From the calculated volume (V) and the formula (4/3)πr3 for
the volume of a sphere, estimate the radius (r)
of a lead atom.
165. A piece of nickel foil, 0.550 mm thick and
1.25 cm square, is allowed to react with fluorine,
F2, to give nickel fluoride.
(a) How many moles of nickel foil were used?
(The density of nickel is 8.902 g/cm3.)
(b) If you isolate 1.261 g of the nickel fluoride,
what is its formula?
(c) What is its complete name?
166. ▲ Uranium is used as a fuel, primarily in the
form of uranium(IV) oxide, in nuclear power
plants. This question considers some uranium
chemistry.
(a) A small sample of uranium metal (0.169 g) is
heated to between 800 and 900 °C in air to
give 0.199 g of a dark green oxide, Ux Oy.
How many moles of uranium metal were
used? What is the empirical formula of the
oxide, Ux Oy? How many moles of Ux Oy must
have been obtained?
(b) The naturally occurring isotopes of uranium
are 234U, 235U, and 238U. Knowing that uranium’s atomic weight is 238.02 g/mol, which
isotope must be the most abundant?
(c) If the hydrated compound UO2(NO3)2 ∙
z H2O is heated gently, the water of hydration
is lost. If you have 0.865 g of the hydrated
compound and obtain 0.679 g of UO2(NO3)2
upon heating, how many waters of hydration
are in each formula unit of the original compound? (The oxide Ux Oy is obtained if the
hydrate is heated to temperatures over 800 °C
in the air.)
167. In an experiment, you need 0.125 mol of sodium
metal. Sodium can be cut easily with a knife
(Figure 2.6), so if you cut out a block of sodium,
what should the volume of the block be in cubic
centimeters? If you cut a perfect cube, what is the
length of the edge of the cube? (The density of
sodium is 0.97 g/cm3.)
168. Mass spectrometric analysis showed that there are
four isotopes of an unknown element having the
following masses and abundances:
Isotope
Mass
Number
Isotope
Mass
Abundance
(%)
1
136
135.9071
0.185
2
138
137.9060
0.251
3
140
139.9054
88.45
4
142
141.9093
11.11
Three elements in the periodic table that have
atomic weights near these values are lanthanum
(La), atomic number 57, atomic weight 138.91;
cerium (Ce), atomic number 58, atomic weight
140.12; and praseodymium (Pr), atomic number
59, atomic weight 140.91. Using the data above,
calculate the atomic weight, and identify the
element if possible.
Study Questions
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135
169. If Epsom salt, MgSO4 ∙ x H2O, is heated to
250 °C, all the water of hydration is lost. Upon
heating a 1.687-g sample of the hydrate, 0.824 g
of MgSO4 remains. How many molecules of water
occur per formula unit of MgSO4?
170. The alum used in cooking is potassium aluminum
sulfate hydrate, KAl(SO4)2 ∙ x H2O. To find the
value of x, you can heat a sample of the compound to drive off all of the water and leave only
KAl(SO4)2. Assume you heat 4.74 g of the
hydrated compound and that the sample loses
2.16 g of water. What is the value of x?
171. Tin metal (Sn) and purple iodine (I2) combine to
form orange, solid tin iodide with an unknown
formula.
Sn metal + solid I2
n solid SnxIy
Weighed quantities of Sn and I2 are combined, where
the quantity of Sn is more than is needed to react
with all of the iodine. After SnxIy has been formed, it
is isolated by filtration. The mass of excess tin is also
determined. The following data were collected:
Mass of tin (Sn) in the original mixture
1.056 g
Mass of iodine (I2) in the original mixture 1.947 g
Mass of tin (Sn) recovered after reaction
0.601 g
What is the empirical formula of the tin iodide
obtained?
172. ▲ When analyzed, an unknown compound gave
these experimental results: C, 54.0%; H, 6.00%;
and O, 40.0%. Four different students used these
values to calculate the empirical formulas shown
here. Which answer is correct? Why did some students not get the correct answer?
(a) C4H5O2
(c) C7H10O4
(b) C5H7O3
(d) C9H12O5
173. ▲ Two general chemistry students working together
in the lab weigh out 0.832 g of CaCl2 ∙ 2 H2O into
a crucible. After heating the sample for a short time
and allowing the crucible to cool, the students
determine that the sample has a mass of 0.739 g.
They then do a quick calculation. On the basis of
this calculation, what should they do next?
(a) Congratulate themselves on a job well done.
(b) Assume the bottle of CaCl2 ∙ 2 H2O was mislabeled; it actually contained something different.
(c) Heat the crucible again, and then reweigh it.
174. To find the empirical formula of tin oxide, you
first react tin metal with nitric acid in a porcelain
crucible. The metal is converted to tin nitrate, but,
upon heating the nitrate strongly, brown nitrogen
136
dioxide gas is evolved and tin oxide is formed. In
the laboratory you collect the following data:
Mass of crucible
13.457 g
Mass of crucible plus tin
14.710 g
Mass of crucible after heating 15.048 g
What is the empirical formula of tin oxide?
Summary and Conceptual Questions
The following questions may use concepts from this and the
previous chapter.
175. ▲ Identify, from the list below, the information
needed to calculate the number of atoms in
1.00 cm3 of iron. Outline the procedure used in
this calculation.
(a) the structure of solid iron
(b) the molar mass of iron
(c) Avogadro’s number
(d) the density of iron
(e) the temperature
(f) iron’s atomic number
(g) the number of iron isotopes
176. Consider the plot of relative element abundances
on page 131. Is there a relationship between abundance and atomic number? Is there any difference
between the relative abundance of an element of
even atomic number and the relative abundance
of an element of odd atomic number?
177. The photo here depicts what happens when a coil
of magnesium ribbon and a few calcium chips are
placed in water.
(a) Based on these observations, what might you
expect to see when barium, another Group 2A
(2) element, is placed in water?
(b) Give the period in which each element
(Mg, Ca, and Ba) is found. What correlation
do you think you might find between the
reactivity of these elements and their positions
in the periodic table?
© Charles D. Winters/Cengage
In the Laboratory
Magnesium (left ) and calcium (right ) in water
Chapter 2 / Atoms, Molecules, and Ions
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© Charles D. Winters/Cengage
178. A jar contains some number of jelly beans. To find
out precisely how many are in the jar, you could
dump them out and count them. How could you
estimate their number without counting each one?
(Chemists do just this kind of “bean counting”
when they work with atoms and molecules. Atoms
and molecules are too small to count one by one,
so chemists have worked out other methods to
determine the number of atoms in a sample.)
How many jelly beans are in the jar?
Study Questions
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137
3 Chemical Reactions
N
Precipitation
Adding a solution of K2CrO4 to a
solution of Pb(NO3)2 results in
formation of a yellow solid, PbCrO4.
Gaseous NH3 and HCl in open containers
diffuse in air, and when they come into
contact, a cloud of solid NH4Cl forms.
E
Acid-Base
R
A C
T
I
O
K2CrO4(aq)
E
R
A C
NH4Cl(s)
T
I
O
N
PbCrO4(s)
NH3(aq)
HCl(aq)
Pb(NO3)2(aq)
R
Redox
E
A C
T
I O
KOH(aq)
N
K(s)
Potassium reacts vigorously with water to form gaseous H2
and a solution of KOH.
© Charles D. Winters/Cengage
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C hapt e r O ut li n e
3.1
Introduction to Chemical Equations
3.2
Balancing Chemical Equations
3.3
Introduction to Chemical Equilibrium
3.4
Aqueous Solutions
3.5
Precipitation Reactions
3.6
Acids and Bases
3.7
Acid–Base Reactions
3.8
Oxidation–Reduction Reactions
3.9
Classifying Reactions in Aqueous Solution
Chemical reactions occur all around us, as well as within us. There are almost limitless numbers of ways that elements and compounds can combine to produce new
compounds. Fortunately, many chemical reactions are similar, and they can be
grouped or classified. This chapter will introduce three common types of reactions:
precipitation reactions, acid–base reactions, and oxidation–reduction reactions.
To describe chemical reactions, chemists use symbolic representations (see
Figure 1.6) that indicate both the formulas and the amount of each substance
involved in the reactions. This chapter begins with instructions for properly using
symbolic representations to describe chemical reactions.
3.1 Introduction to Chemical Equations
Goals for Section 3.1
• Understand the information conveyed by a balanced chemical equation including
the terminology used (reactants, products, stoichiometry, stoichiometric coefficients).
• Recognize that a balanced chemical equation is required by the law of conservation
of matter.
Chemistry is about atoms, molecules, and ions and the reactions they undergo. An
important way that chemists describe reactions is with balanced chemical equations.
A chemical equation shows the reacting elements and compounds and the products
of the reaction. A balanced equation has equal numbers of each type of atom on both
sides of the equation. For example, when a stream of chlorine gas, Cl2, is directed
onto solid phosphorus, P4, the mixture bursts into flame, and a chemical reaction
◀ Chemical reactions are at the heart of chemistry. Pictured here are three general
types of reactions: precipitation, acid–base, and oxidation–reduction (or redox).
139
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produces liquid phosphorus trichloride, PCl3 (Figure 3.1). The balanced chemical
equation for this reaction is
P4(s) + 6 Cl2(g)
4 PCl3(ℓ)
reactants
product
In a chemical equation, the formulas for the reactants (the substances combined in
the reaction) are written to the left of the arrow and the formulas of the products
(the substances produced) are written to the right of the arrow. The physical states
of reactants and products are usually indicated. The symbol (s) indicates a solid,
(g) a gas, and (ℓ) a liquid. A substance dissolved in water, that is, in an aqueous
­solution, is indicated by (aq).
In the eighteenth century, the French scientist Antoine Lavoisier (1743–1794)
introduced the law of conservation of matter, which states that matter can ­neither
be created nor destroyed. This means that if the total mass of reactants is 10 g,
and if the reaction completely converts reactants to products, the total mass of the
­products must be 10 g. This also means that if 1000 atoms of a particular e­ lement
are contained in the reactants, then those 1000 atoms must also appear in the
­products. Atoms, and thus mass, are conserved in chemical reactions.
When applied to the reaction of phosphorus and chlorine, the law of conservation
of matter requires that 1 molecule of phosphorus, P4 (with 4 phosphorus atoms), and
6 molecules of Cl2 (with 12 atoms of Cl) will produce 4 molecules of PCl3. Because
each PCl3 molecule contains 1 P atom and 3 Cl atoms, 4 PCl3 molecules are needed
to account for 4 P atoms and 12 Cl atoms in the product. The equation is balanced;
the same number of P and Cl atoms appear on each side of the equation.
6×2=
12 Cl atoms
4×3=
12 Cl atoms
P4(s) + 6 Cl2(g)
4 PCl3(ℓ)
4 P atoms
4 P atoms
In a chemical reaction, the relationship between the amounts of chemical reactants and products is called stoichiometry (pronounced “stoy-key-AHM-uh-tree”).
The coefficients in a balanced equation are called stoichiometric coefficients. (In
the P4 and Cl2 equation, these are 1, 6, and 4.) They can be interpreted as a number of atoms or molecules: 1 molecule of P4 and 6 molecules of Cl2 produce 4
molecules of PCl3. They can also refer to amounts of reactants and products: 1 mole
of P4 combines with 6 moles of Cl2 to produce 4 moles of PCl3.
PCl3
P4
P4(s) + 6 Cl 2(g)
4 PCl 3(ℓ)
R E AC TA N T S
PRODUCT
Photos: © Charles D. Winters/Cengage
Cl2
Figure 3.1 Reaction of solid white phosphorus with chlorine gas. The product is liquid phosphorus trichloride.
140
Chapter 3 / Chemical Reactions
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Chemistry in Your Career
Although his educational background is in environmental engineering, Nandan Prabhune uses
chemistry daily in his job as Innovation and Technology Head (for Water and Waste Solutions Division) in Pune, India. In this capacity, Prabhune
advises industries from mining to manufacturing
on how to treat, reuse, and recycle the wastewater
from their industrial processes. One of the things
he likes best about his job is witnessing how his
contributions help industries conserve water.
Prabhune recognizes two key components to his
success in generating creative solutions to wastewater
challenges: applying fundamental ­principles and
i­ntegrating knowledge from across disciplines,
including chemistry, microbiology, and physics.
Prabhune draws on his knowledge of chemistry when
determining the pH and chemical dosing required to
precipitate metal ions, fluoride, and silica from industrial wastewaters. He uses acid–base equilibria calculations to determine how to neutralize wastewaters
depending on which ions are present and how they
interact. “These calculations are used to design wastewater treatment processes [and] equipment and [to]
operate the engineered systems,” Prabhune explains.
Antoine Laurent Lavoisier (1743–1794)
On Monday, August 7, 1774, the ­English
chemist Joseph Priestley (1733–
1804) isolated oxygen. (The Swedish
chemist Carl Scheele [1742–1786]
also discovered the element, perhaps in
1773, but did not publish his results
until later.) To obtain oxygen, Priestley
heated mercury(II) oxide, HgO, causing
it to decompose to mercury and oxygen.
2 HgO(s) n 2 Hg(ℓ) + O2(g)
© Charles D. Winters/Cengage
Priestley did not immediately understand the significance of the discovery,
but he mentioned it to the French chemist Antoine Lavoisier in October 1774.
One of Lavoisier’s contributions to science
was his recognition of the importance of
exact scientific measurements and of
carefully planned experiments, and he
applied these methods to the study of oxygen. From this work, Lavoisier proposed
that oxygen was an element, that it was
The decomposition of red mercury(II)
oxide. The decomposition reaction yields
mercury metal and oxygen gas. The
mercury is seen as a film on the surface of
the test tube.
one of the constituents of the compound
water, and that burning involved a reaction with oxygen. He also mistakenly came
to believe Priestley’s gas was present in
all acids, so he named it oxygen from the
Greek words meaning “to form an acid.”
In other experiments, Lavoisier observed
that the heat produced by a guinea pig
exhaling a given amount of carbon dioxide
is similar to the quantity of heat produced
by burning carbon to give the same amount
of carbon dioxide. From these and other
experiments he concluded that, “Respiration is a combustion, slow it is true, but otherwise perfectly similar to that of charcoal.”
Although he did not understand the details
of the process, this was an important step
in the development of biochemistry.
Lavoisier was a prodigious scientist, and
the principles of naming chemical substances that he introduced are still in use
today. Furthermore, he wrote a textbook
in which he applied the principles of the
conservation of matter to chemistry, and
he used the idea to write early versions of
chemical equations.
Because Lavoisier was an aristocrat, he
came under suspicion during the Reign of
Terror of the French Revolution in 1794.
He was an investor in the Ferme Générale,
the infamous tax-collecting organization
in eighteenth-century France. Tobacco
was a monopoly product of the Ferme
Générale, and it was common to cheat the
purchaser by adding water to the tobacco,
a practice that Lavoisier opposed. Nonetheless, because of his involvement with
the Ferme, his career was cut short by the
guillotine on May 8, 1794, on the charge
of “adding water to the people’s tobacco.”
Madame Lavoisier, Marie-Anne Paulze
(1758–1836), was a crucial research
partner to Antoine. Although not quite 14
when they married, she became a key collaborator in Antoine’s work. She worked in
the laboratory, kept careful records of their
experiments, and used her artistic talents
to illustrate his texts. (She was trained by
the painter of the accompanying portrait.)
Her ability to translate chemistry texts in
English gave the team knowledge of other
scientists’ published works. She also
­ ublication of
played a major role in the p
Antoine’s revolutionary E
­ lementary Treatise
on Chemistry in 1789. Finally, in 1804,
she married another man of s­ cience,
­Benjamin Thompson (Count Rumford; see
page 257); the union was short lived.
Purchase, Mr. and Mrs. Charles Wrightsman Gift, in honor of Everett Fahy,
1977/The Metropolitan Museum of Art
A Closer Look
Nandan Prabhune
Nandan Prabhune
Lavoisier and his wife, Marie-Anne
Pierrette Paulze Lavoisier, as painted in
1788 by Jacques-Louis David. Lavoisier
was then 45, and Marie-Anne was 30.
3.1 Introduction to Chemical Equations
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141
Photos:© Charles D. Winters/Cengage
(a) Reaction of iron and oxygen to give iron(III)
oxide, Fe2O3.
(b) Reaction of sulfur (in the spoon) with oxygen
to give sulfur dioxide, SO2.
(c) Reaction of phosphorus and oxygen to give
tetraphosphorus decaoxide, P4O10.
Figure 3.2 Reactions of a metal and two nonmetals with oxygen.
3.2 Balancing Chemical Equations
Goal for Section 3.2
© Charles D. Winters/Cengage
• Balance simple chemical equations.
A balanced equation has the same number of atoms of each element on each side of
the equation. Many chemical equations can be balanced by trial and error, and this
method is often used. However, more systematic methods exist and are especially
useful if reactions are complicated.
When balancing chemical equations, there are two important things to
remember:
• Formulas for reactants and products must be correct. Once the correct formulas for
Figure 3.3 A combustion
reaction. Here, propane, C3H8,
burns to give CO2 and H2O.
These simple oxides are always
the products of the complete
combustion of a hydrocarbon
(that is, compounds containing
only hydrogen and carbon
atoms).
142
the reactants and products have been determined, the subscripts in their formulas
cannot be changed to balance an equation. Changing the subscripts changes the
identity of the substance. For example, you cannot change CO2 to CO to balance
an equation; carbon monoxide, CO, and carbon dioxide, CO2, are different
compounds.
• Chemical equations are balanced using stoichiometric coefficients. The entire
chemical formula of a substance is multiplied by the stoichiometric coefficient.
One general class of chemical reactions is the reaction of metals or nonmetals
with oxygen to give oxides of the general formula MxOy. For example, iron reacts with
­oxygen to give iron(III) oxide (Figure 3.2a).
4 Fe(s) + 3 O2(g) n 2 Fe2O3(s)
The nonmetals sulfur and oxygen react to form sulfur dioxide (Figure 3.2b),
S(s) + O2(g) n SO2(g)
and phosphorus, P4, reacts vigorously with oxygen to give tetraphosphorus decaoxide, P4O10 (Figure 3.2c).
P4(s) + 5 O2(g) n P4O10(s)
Chapter 3 / Chemical Reactions
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The equations written above are balanced. The same number of iron, sulfur, or
phosphorus atoms and oxygen atoms occurs on each side of these equations.
Every day, you encounter combustion reactions, the burning of a fuel in oxygen
accompanied by the evolution of energy as heat (Figure 3.3). The combustion of
octane, C8H18, a component of gasoline is an example.
2 C8H18(ℓ) + 25 O2(g) n 16 CO2(g) + 18 H2O(g)
In all combustion reactions, some or all of the elements in the reactants end up
as oxides, compounds containing oxygen. When the reactant is a hydrocarbon
(a ­compound that contains only C and H, such as octane), the products of complete combustion are carbon dioxide and water.
To illustrate equation balancing, consider the process used to write the ­balanced
equation for the complete combustion of propane, C3H8, a common fuel.
Step 1 Write correct formulas for the reactants and products. Here propane and oxygen are the reactants, and car-
unbalanced equation
C3H8(g) + O2(g) 88888888888888n CO2(g) + H2O(g)
bon dioxide and water are the products.
Step 2 Balance the C atoms. In combustion reactions such as
this, it is usually best to balance the C atoms first and leave the
O atoms until the end (because O atoms are often found in
more than one product). In this case, there are three C atoms
in the reactants, so three must occur in the products. Three
CO2 molecules are therefore required on the right side.
unbalanced equation
C3H8(g) + O2(g) 88888888888888n 3 CO2(g) + H2O(g)
Step 3 Balance the H atoms. A molecule of propane con-
tains eight H atoms. Each molecule of water has two H
atoms, so four molecules of water account for the required
eight H atoms on the right side.
Step 4 Balance the O atoms. Ten O atoms are on the right side
(3 × 2 = 6 in CO2 plus 4 × 1 = 4 in H2O). Five O2 molecules
are needed to supply the required ten O atoms.
unbalanced equation
C3H8(g) + O2(g) 88888888888888n 3 CO2(g) + 4 H2O(g)
balanced equation
C3H8(g) + 5 O2(g) 88888888888888n 3 CO2(g) + 4 H2O(g)
Step 5 Verify that the number of atoms of each element is
balanced. There are three C atoms, eight H atoms, and ten
O atoms on each side of the equation.
3 C atoms
3 C atoms
8 H atoms
8 H atoms
10 O atoms
10 O atoms
Exam p le 3.1
Strategy Map
Problem
Balance the equation for the
reaction of NH3 and O2.
Data/Information
The formulas of the reactants
and products are given.
Balancing an Equation for a Combustion Reaction
Problem Write the balanced equation for the combustion of ammonia gas (NH3) to
give water vapor and nitrogen monoxide gas (NO).
What Do You Know? You know the correct formulas and/or names for the reactants (NH3 and oxygen, O2) and the products (H2O and nitrogen monoxide, NO). You also
know their states.
Strategy First write the unbalanced equation. Next balance the N atoms, then the
H atoms, and finally the O atoms.
3.2 Balancing Chemical Equations
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143
Solution
Step 1
Write the equation using the correct formulas for the reactants and products.
The reactants are NH3(g) and O2(g), and the products are NO(g) and H2O(g).
unbalanced equation
NH3(g) + O2(g) 88888888888888n NO(g) + H2O(g)
Step 2
Balance the N atoms.
There is one N atom on each side of the equation. The N atoms are in balance, at least for
the moment.
unbalanced equation
NH3(g) + O2(g) 88888888888888n NO(g) + H2O(g)
Step 3
Balance the H atoms.
There are three H atoms on the left and two on the right. To have the same number on
each side (six), use two molecules of NH3 on the left and three molecules of H2O on the
right (which gives six H atoms on each side).
unbalanced equation
2 NH3(g) + O2(g) 88888888888888n NO(g) + 3 H2O(g)
Notice that after balancing the H atoms, the N atoms are no longer balanced. To bring
them into balance, use two NO molecules on the right.
unbalanced equation
2 NH3(g) + O2(g) 88888888888888n 2 NO(g) + 3 H2O(g)
Step 4
Balance the O atoms. This is best left to the final step.
After Step 3, there is an even number of O atoms (two) on the left and an odd number
(five) on the right. Because there cannot be an odd number of O atoms on the left (O
atoms are paired in O2 molecules), multiply each coefficient on the product side of the
equation by two so that an even number of oxygen atoms (ten) now occurs on the right
side. Next, multiply the coefficient of NH3 by two so that the number of nitrogen and
hydrogen atoms remain balanced with the right side:
unbalanced equation
4 NH3(g) + O2(g) 88888888888888n 4 NO(g) + 6 H2O(g)
Now the oxygen atoms can be balanced with five O2 molecules on the left side of the
equation:
balanced equation
4 NH3(g) + 5 O2(g) 88888888888888n 4 NO(g) + 6 H2O(g)
Four N atoms, 12 H atoms, and 10 O atoms are on each side of the equation.
Think about Your Answer An alternative way to write this equation is
2 NH3(g) + 5/2 O2(g) n 2 NO(g) + 3 H2O(g)
where a fractional coefficient has been used. This equation is correctly balanced and has
some uses, but chemical equations are usually written with whole-number coefficients.
Check Your Understanding
(a) Butane gas, C4H10, can burn completely in air [use O2(g) as the other reactant] to give carbon dioxide gas and water vapor.
Write a balanced equation for this combustion reaction.
(b) Write a balanced chemical equation for the complete combustion of C3H7BO3(ℓ), a gasoline additive. The products of
combustion are CO2(g), H2O(g), and B2O3(s).
144
Chapter 3 / Chemical Reactions
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3.3 Introduction to Chemical Equilibrium
Goals for Section 3.3
• Recognize that all chemical reactions are reversible and that reactions eventually
reach a dynamic equilibrium.
• Recognize the difference between reactant-favored and product-favored reactions at
Up to this point, we have treated chemical reactions as proceeding in one direction only, with reactants being converted completely to products. Nature, however,
is more complex than this. All chemical reactions are reversible, in principle, and
many reactions lead to incomplete conversion of reactants to products.
The formation of stalactites and stalagmites in a limestone cave is an example
of a system that depends on the reversibility of a chemical reaction (Figure 3.4).
Stalactites and stalagmites are made chiefly of calcium carbonate, a mineral found
in underground deposits in the form of limestone, a leftover from ancient oceans.
If water seeping through the limestone contains dissolved CO2, a reaction occurs in
which the mineral dissolves, giving an aqueous solution of Ca(HCO3)2.
CaCO3(s) + CO2(aq) + H2O(ℓ) n Ca(HCO3)2(aq)
When the mineral-laden water reaches a cave, the reverse reaction occurs; CO2 is
released into the cave and solid CaCO3 is deposited.
Ca(HCO3)2(aq) n CaCO3(s) + CO2(g) + H2O(ℓ)
As illustrated in Figure 3.5, these reactions can also be done in a laboratory.
A
Reactants:
Solutions of CaCl2
(left) and NaHCO3
(right).
© A. N. Palmer
equilibrium.
Figure 3.4 Cave chemistry. ​
Calcium carbonate stalactites
cling to the roof of a cave,
and stalagmites grow up from
the cave floor. The chemistry
producing these formations is a
good example of the reversibility
of chemical reactions.
Figure 3.5 The
reversibility of chemical
reactions. The experiments
B
here demonstrate the
reversibility of chemical
reactions. The system is
described by the following
balanced chemical equation:
FORWARD REACTION
The solutions are
mixed, forming H2O,
CO2 gas, and CaCO3
solid.
Ca2+(aq) + 2 HCO3−(aq)
uv CaCO3(s) + CO2(g)
+ H2O(ℓ)
C
The reaction can be
reversed by bubbling
CO2 gas into the
CaCO3 suspension.
Photos: © Charles D. Winters/Cengage
Solutions of CaCl2 (a source of Ca2+ ions) and NaHCO3 (a source of HCO3– ions) are mixed and produce
solid CaCO3 and CO2 gas.
REVERSE REACTION
D
The CaCO3 dissolves
when the solution
has been saturated
with CO2.
Elapsing time...
If CO2 gas is bubbled into a suspension of CaCO3, solid CaCO3 and gaseous CO2 react to produce Ca2+ and
HCO3– ions in solution.
3.3 Introduction to Chemical Equilibrium
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145
Another example of a reversible reaction is the reaction of nitrogen with hydrogen to form ammonia gas, a compound made industrially in enormous quantities
and used directly as a fertilizer and in the production of other chemicals.
N2(g) + 3 H2(g) n 2 NH3(g)
Nitrogen and hydrogen react to form ammonia, but, under the conditions of the reaction, ammonia also breaks down into nitrogen and hydrogen in the reverse reaction.
2 NH3(g) n N2(g) + 3 H2(g)
Consider what would happen if you mixed nitrogen and hydrogen in a closed container under the proper conditions for the reaction to occur. At first, N2 and H2 react to
produce some ammonia. As the ammonia is produced, however, some NH3 molecules
decompose to re-form nitrogen and hydrogen in the reverse reaction (Figure 3.6). At
the beginning of the process, the forward reaction to give NH3 predominates, but, as
the reactants are consumed, the rate (or speed) of the forward reaction progressively
slows. At the same time, the reverse reaction speeds up as the amount of ammonia
increases. Eventually, the rate of the forward reaction equals the rate of the reverse reaction. At this point, no further macroscopic change is observed; the amounts of nitrogen,
hydrogen, and ammonia stop changing (although the forward and reverse reactions
continue). The system has reached chemical equilibrium—the state at which concentrations of reactants and products do not undergo any change. The reaction vessel
contains all three substances—nitrogen, hydrogen, and ammonia. Because the forward
and reverse processes are still occurring, this state is referred to as a dynamic equilibrium. Systems in dynamic equilibrium are represented by writing a double arrow symbol (uv) connecting the reactants and products.
N2(g) + 3 H2(g) uv 2 NH3(g)
An important principle in chemistry is that chemical reactions always proceed
spontaneously toward equilibrium. A reaction will never proceed spontaneously in
a direction that takes a system further from equilibrium.
A key question is “When a reaction reaches equilibrium, will the reactants be
converted largely to products or will most of the reactants still be present?” For now it is
useful to define product-favored reactions as reactions in which reactants are completely
or largely converted to products when equilibrium is reached. The combustion reactions
covered earlier are examples of reactions that are product-favored at equilibrium. In fact,
most of the reactions you will study in the rest of this chapter are product-favored reactions at equilibrium. The equations for reactions that are very product-favored are often
written using a single arrow (n) connecting reactants and products.
N2(g) + 3H2(g)
Amounts of products
and reactants
Reaction begins with
3:1 mixture of H2 to N2.
2 NH3(g)
Equilibrium achieved
H2
Eventually, the amounts of N2, H2, and NH3
no longer change. At this point, the reaction
has reached equilibrium. Nonetheless, the
forward reaction to produce NH3 continues,
as does the reverse reaction (the
decomposition of NH3 ).
NH3
N2
As reaction proceeds H2 and
N2 produce NH3, but the NH3
also begins to decompose
back to H2 and N2.
146
Reaction proceeding toward equilibrium
Time
Figure 3.6 The reaction of N2 and H2 to produce NH3.
Chapter 3 / Chemical Reactions
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The opposite of a product-favored reaction is one that is reactant-favored at
equilibrium. Such reactions lead to the conversion of only a small amount of the
reactants to products. An example of a reactant-favored reaction is the ionization of
acetic acid in water where only a tiny fraction of the acid produces ions.
CH3CO2H(aq) + H2O(ℓ) uv CH3CO2−(aq) + H3O+(aq)
Acetic acid is an example of a large number of acids called weak acids because the
reaction with water is reactant-favored at equilibrium, and only a few percent of the
molecules react with water to form ionic products.
3.4 Aqueous Solutions
Goals for Section 3.4
• Explain the difference between electrolytes and nonelectrolytes and recognize
examples of each.
• Predict the solubility of ionic compounds in water.
Many of the reactions you will study in your chemistry course and almost all of the
reactions that occur in living things are carried out in solutions in which the r­ eacting
substances are dissolved in water. A solution is a homogeneous mixture of two or
more substances. One substance is generally considered the solvent, the medium in
which another substance—the solute—is dissolved. The remainder of this chapter
is an introduction to some of the types of reactions that occur in ­aqueous ­solutions,
solutions in which water is the solvent. First, it is important to understand some
basic concepts about the behavior of compounds dissolved in water.
Ions and Molecules in Aqueous Solutions
Dissolving an ionic solid requires separating each ion from the oppositely charged
ions that surround it in the solid state (Figure 3.7); this process is called dissociation.
Water is especially good at dissolving ionic compounds because each water molecule
(−)
(+)
A water molecule is
electrically positive on
one side (the H atoms)
and electrically negative
on the other (the O atom).
These charges enable
water to interact with
negative and positive
ions in aqueous solution.
Water molecules are attracted to both positive
cations and negative anions in aqueous solution.
+
−
Water surrounding
a cation
Water surrounding
an anion
Figure 3.7 Water as a solvent
for ionic substances. For the
sake of simplicity and clarity, the
ions in this and other figures are
shown surrounded by four or
five water molecules. However,
experiments show it is often six.
When an ionic substance
dissolves in water, each
ion is surrounded by up to
six water molecules.
2+
−
2+
© Charles D. Winters/Cengage
2+
Copper(II) chloride is added to
water. Interactions between
water and the Cu2+ and Cl–
ions allow the solid to dissolve.
2+
−
−
The ions are now sheathed in
water molecules.
3.4 Aqueous Solutions
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147
has a positively charged end and a negatively charged end. When an ionic compound
dissolves in water, each negative ion becomes surrounded by water molecules with the
positive ends of water molecules pointing toward it, and each positive ion becomes
surrounded by the negative ends of several water molecules. The forces involved are
described by Coulomb’s law (Equation 2.3, page 97).
The water-encased ions produced by dissolving an ionic compound are free to
move about in solution. Under normal conditions, the movement of ions is random, and the cations and anions from a dissolved ionic compound are dispersed
uniformly throughout the solution. However, if two electrodes (conductors of electricity such as copper wire) are placed in the solution and connected to a battery,
positive cations are drawn toward the negative electrode and negative anions are
drawn toward the positive electrode (Figure 3.8). Conduction of electricity in the
solution is a consequence of the movement of charged particles in solution.
Compounds whose aqueous solutions conduct electricity are called electrolytes.
All ionic compounds that are soluble in water are electrolytes. The extent to which a
solution conducts electricity, its conductivity, depends on the ion concentration. You
can test the conductivity of a solution by trying to pass electricity through the solution.
The greater the ion concentration, the greater the conductivity.
For example, for every mole of NaCl that dissolves, 1 mol of Na+ and 1 mol of
Cl− ions enter the solution.
NaCl(s) n Na+(aq) + Cl−(aq)
100% Dissociation n strong electrolyte
There is a significant concentration of sodium ions and chloride ions in the solution, and the solution is a good conductor of electricity. Substances whose solutions
are good electrical conductors are called strong electrolytes (Figure 3.8a). The ions
into which an ionic compound dissociates are given by the compound’s name, and
the relative amounts of these ions are given by its formula. For example, sodium
chloride yields sodium ions (Na+) and chloride ions (Cl−) in solution in a 1:1 ratio.
Nonelectrolyte
Weak Electrolyte
Bulb is lit, showing solution
conducts electricity well.
Bulb is not lit, showing
solution does not conduct.
Bulb is dimly lit, showing
solution conducts electricity poorly.
CuCl2
Ethanol
Cu2+
−
−
Cl
Acetic acid
−
© Charles D. Winters/Cengage
2+
© Charles D. Winters/Cengage
Strong Electrolyte
© Charles D. Winters/Cengage
Figure 3.8
Acetate ion
+
H+
−
+
2+
2+
2+
−
−
(a) A strong electrolyte conducts
electricity. ​CuCl2 is completely
dissociated into Cu2+ and Cl− ions.
148
−
(b) A nonelectro­lyte does not
conduct electricity because no
ions are present in solution.
(c) A weak electrolyte conducts
electricity poorly because only a
few ions are present in solution.
Chapter 3 / Chemical Reactions
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The ionic compound copper(II) chloride, CuCl2, is also a strong electrolyte. In this
case, there are two chloride ions for each copper ion in solution.
CuCl2(s) n Cu2+(aq) + 2 Cl−(aq)
Notice that the two chloride ions per formula unit are present as two separate particles
in solution. In yet another example, the ionic compound barium nitrate yields barium
ions and nitrate ions in solution. For each Ba2+ ion in solution, there are two NO3− ions.
Ba(NO3)2(s) n Ba2+(aq) + 2 NO3−(aq)
Notice that NO3−, a polyatomic ion, does not dissociate further; the ion exists as
one unit in aqueous solution.
Compounds whose aqueous solutions do not conduct electricity are called
nonelectrolytes (Figure 3.8b). The solute particles present in these aqueous solutions are molecules, not ions. For example, when the molecular compound ethanol
(C2H5OH) dissolves in water, each molecule of ethanol stays intact as a single unit.
Ions are not produced in the solution. Other examples of nonelectrolytes are sucrose
(C12H22O11) and antifreeze (ethylene glycol, HOCH2CH2OH).
Some molecular compounds (strong acids, weak acids, and weak bases) react with
water to produce ions in aqueous solutions and are thus electrolytes. One example is
gaseous hydrogen chloride, a molecular compound, which reacts with water to form
ions. The aqueous solution is referred to as hydrochloric acid.
HCl(g) + H2O(ℓ) n H3O+(aq) + Cl−(aq)
This reaction is very product-favored. Almost every molecule of HCl ionizes in
solution, so hydrochloric acid is a strong electrolyte.
Some molecular compounds are weak electrolytes (Figure 3.8c). When these compounds dissolve in water only a small fraction of the molecules ionize to form ions.
These aqueous solutions are poor conductors of electricity. Acetic acid is a weak electrolyte. In vinegar, an aqueous solution of acetic acid, only about 0.5% of the molecules of
acetic acid are ionized to form acetate (CH3CO2−) and hydronium (H3O+) ions.
CH3CO2H(aq) + H2O(ℓ) uv CH3CO2−(aq) + H3O+(aq)
Figure 3.9 summarizes whether a given type of solute will be present in aqueous solution as ions, molecules, or a combination of ions and molecules.
Solubility of Ionic Compounds in Water
Many ionic compounds are quite soluble in water. Others dissolve only to a small
extent, while many are essentially insoluble. Fortunately, it is possible to predict
Figure 3.9 Predicting the
species present in aqueous
solution. When compounds
Solute in an Aqueous Solution
Ionic
Compound
Molecular
Compound
Acids and
Weak Bases
Strong Acids
Strong Electrolyte IONS
Most Molecular
Compounds
dissolve, ions may result from
ionic or molecular compounds.
Some molecular compounds may
remain intact as molecules in
solution. (Note that hydroxidecontaining strong bases are ionic
compounds.)
Weak Acids and
Weak Bases
Weak Electrolyte
MOLECULES and IONS
Nonelectrolyte
MOLECULES
3.4 Aqueous Solutions
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149
Water-soluble compounds
Almost all salts of
Na+, K+, NH4+
Salts of
nitrate, NO3−
chlorate, ClO3−
perchlorate, ClO4−
acetate, CH3CO2−
Almost all salts of
Cl−, Br−, I−
Exceptions
(not soluble)
Halides of
Ag+, Hg22+, Pb2+
Water-insoluble compounds
Salts containing
F−
Salts of
sulfate, SO42−
Exceptions
(not soluble)
Exceptions
(not soluble)
Fluorides of
Mg2+, Ca2+, Sr2+,
Ba2+, Pb2+
Sulfates of
Ca2+, Sr2+, Ba2+,
Pb2+, Ag+
Silver compounds
Most metal hydroxides
and oxides
Exceptions (soluble)
Exceptions (soluble)
Salts of NH4+ and
the alkali metal cations,
and BaS
Alkali metal hydroxides
and Ba(OH)2 and Sr(OH)2
Hydroxides
Photos: © Charles D. Winters/Cengage
Sulfides
Most salts of
carbonate, CO32−
phosphate, PO43−
oxalate, C2O42−
chromate, CrO42−
sulfide, S2−
AgNO3
AgCl
AgOH
(NH4)2S
(a) Nitrates are generally soluble, as are
chlorides (exceptions include AgCl).
Hydroxides are generally not soluble.
CdS
Sb2S3
NaOH Ca(OH)2 Fe(OH)3 Ni(OH)2
PbS
(b) Sulfides are generally not soluble
(exceptions include salts with NH4+
and Na+).
(c) Hydroxides are generally not soluble,
except when the cation is a Group 1A (1)
metal (or Sr2+ or Ba2+).
Figure 3.10 Guidelines to predict the aqueous solubility of ionic compounds. If a compound contains one of the ions in the columns
on the left side of the chart above, it is predicted to be at least moderately soluble in water. Exceptions to the guidelines are noted.
whether many compounds are soluble. For now, solubility will be considered an
either-or question. Compounds that visibly dissolve to a certain extent are soluble,
whereas those that show no visible signs of dissolving are insoluble.
Figure 3.10 gives broad guidelines that can help you to predict whether an ionic
compound is soluble in water based on the ions that make up the compound. For
example, sodium nitrate, NaNO 3, contains an alkali metal cation, Na +, and the
nitrate anion, NO3−. The presence of either of these ions generally ensures that the
compound is soluble in water; at 20 °C, over 90 g of NaNO3 will dissolve in 100 mL
of water. In contrast, calcium hydroxide is poorly soluble in water. If a spoonful
of solid Ca(OH)2 is added to 100 mL of water, less than 1 g will dissolve at 20 °C.
Nearly all of the Ca(OH)2 remains as a solid (Figure 3.10c).
E xamp le 3.2
Solubility Guidelines
Problem Predict whether the following ionic compounds are likely to be watersoluble. For soluble compounds, list the ions present in solution.
(a) KCl
Solubility Guidelines Observations
such as those shown in
Figure 3.10 were used to create
the solubility guidelines. Note,
however, that these are general
guidelines. There are exceptions.
150
(b) MgCO3
(c) Fe(OH)3
(d) Cu(NO3)2
What Do You Know? You know the formulas of the compounds but need to be
able to identify the ions that make up each of them in order to use the solubility guidelines in Figure 3.10.
Strategy Use the solubility guidelines given in Figure 3.10. Soluble ionic compounds
will dissociate into their respective ions in solution.
Chapter 3 / Chemical Reactions
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Solution
(a) KCl is composed of K+ and Cl− ions. The presence of either of these ions means that
the compound is likely to be soluble in water. The solution contains K+ and Cl− ions
dissolved in water.
KCl(s) n K+(aq) + Cl−(aq)
(The solubility of KCl is about 35 g in 100 mL of water at 20 °C.)
(b) Magnesium carbonate is composed of Mg 2+ and CO32− ions. Salts containing the
carbonate ion usually are insoluble, unless combined with an ion like Na + or NH4+.
Therefore, MgCO3 is likely insoluble in water. (The solubility of MgCO3 is less than
0.2 g/100 mL of water.)
(c) Iron(III) hydroxide is composed of Fe 3+ and OH− ions. Hydroxides are soluble only
when OH− is combined with ions of the alkali metals, strontium, or barium; Fe3+ is a
transition metal ion, so Fe(OH)3 is insoluble.
(d) Copper(II) nitrate is composed of Cu2+ and NO3− ions. Nitrate salts are soluble, so this
compound dissolves in water, giving ions in solution as shown in the equation below.
Cu(NO3)2(s) n Cu2+(aq) + 2 NO3−(aq)
Think about Your Answer For chemists, a set of guidelines like those in
Figure 3.10 is useful. If needed, accurate solubility information is available for many compounds in chemical resource books or online databases.
Check Your Understanding
Predict whether each of the following ionic compounds is likely to be soluble in water. If it
is soluble, write the formulas of the ions present in aqueous solution.
(a) LiNO3
(b) CaCl2
(c) Cu(OH)2
(d) NaCH3CO2
3.5 Precipitation Reactions
Goals for Section 3.5
• Recognize what ions are formed when an ionic compound dissolves in water.
• Recognize exchange reactions in which there is an exchange of anions between the
cations of reactants in solution.
• Predict the products of precipitation reactions.
• Write net ionic equations for reactions in aqueous solution.
With a background on whether compounds will yield ions or molecules when dissolved in water and whether ionic compounds are soluble or insoluble in water, you
are ready to examine the types of chemical reactions that occur in aqueous solutions.
It is useful to look for patterns that can help you predict the reaction products.
Many reactions you will encounter are exchange reactions (sometimes called
double displacement, double replacement, or metathesis reactions). In these
reactions the ions of the reactants exchange partners.
A+B− + C+D−
A+D− + C+B−
Reactions in which an insoluble solid called a precipitate forms (precipitation
reactions) are exchange reactions. For example, aqueous solutions of silver
nitrate and potassium chloride react to produce solid silver chloride and aqueous
­potassium nitrate (Figure 3.11).
3.5 Precipitation Reactions
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151
Figure 3.11 ​
Precipitation of
silver chloride.
+
−
−
+
+
+
−
© Charles D. Winters/Cengage
−
+
+
+
−
−
−
+
+
+
−
(b) Initially, the Ag+ ions (silver)
and Cl− ions (yellow) are widely
separated.
(c) Ag+ and Cl− ions approach
and form ion pairs.
(d) As more and more Ag+ and
Cl− ions come together, a
precipitate of solid AgCl forms.
(a) Mixing aqueous solutions of silver nitrate
and potassium chloride produces white, insoluble
silver chloride, AgCl.
Photos: © Charles D. Winters/Cengage
AgNO3(aq) + KCl(aq) n AgCl(s) + KNO3(aq)
The dark red
precipitate PbS forms
from the reaction
of Pb(NO3)2 and
(NH4)2S.
Reactants
Products
Ag+(aq) + NO3−(aq)
Insoluble AgCl(s)
K+(aq) + Cl−(aq)
K+(aq) + NO3−(aq)
The solubility guidelines (Figure 3.10) predict that almost all metal sulfides are
insoluble in water. If a solution of a soluble metal compound comes in contact with
a source of sulfide ions, the metal sulfide precipitates.
Pb(NO3)2(aq) + (NH4)2S(aq) n PbS(s) + 2 NH4NO3(aq)
Reactants
Products
Pb2+(aq) + 2 NO3−(aq)
Insoluble PbS(s)
2 NH4+(aq) + S2−(aq)
2 NH4+(aq) + 2 NO3−(aq)
In yet another example, the solubility guidelines indicate that with the exception of the alkali metal cations (and Sr2+ and Ba2+), metal cations form insoluble
hydroxides. Thus, water-soluble iron(III) chloride and sodium hydroxide react to
give insoluble iron(III) hydroxide.
Photos: © Charles D. Winters/Cengage
FeCl3(aq) + 3 NaOH(aq) n Fe(OH)3(s) + 3 NaCl(aq)
The orange
precipitate Fe(OH)3
forms from the
reaction of FeCl3 and
NaOH.
152
Reactants
Products
Fe3+(aq) + 3 Cl−(aq)
Insoluble Fe(OH)3(s)
3 Na+(aq) + 3 OH−(aq)
3 Na+(aq) + 3 Cl−(aq)
E xamp le 3.3
Writing the Equation for a Precipitation Reaction
Problem Does a precipitate form when aqueous solutions of potassium chromate and
silver nitrate are mixed? If so, write the balanced equation.
What Do You Know? The names of the two reactants are given. Precipitation
reactions are exchange reactions in which the reactant ions switch partners and form an
insoluble solid. You will need information on molecule solubility in Figure 3.10 to determine if any of the possible exchange reaction products are insoluble in water.
Chapter 3 / Chemical Reactions
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Strategy
Step 1. Determine the formulas from the reactant names and identify the ions that
make up these compounds.
Step 3. Determine if the possible product(s) is/are insoluble using information from
Figure 3.10.
Step 4. Write and balance the equation.
Solution
Step 1. Determine the formulas of the reactants and identify the ions.
The formulas for silver nitrate and potassium chromate are AgNO3 and K2CrO4,
respectively. AgNO3 is composed of Ag+ ions and NO3− ions. K2CrO4 is composed of K+
ions and CrO42− ions.
Step 2. Write the formulas for the possible products in this reaction by exchanging
cations and anions.
Photos: © Charles D. Winters/Cengage
Step 2. Write formulas for the possible products by exchanging cations and anions.
The red precipitate
Ag2CrO4 forms
from the reaction of
AgNO3 and K2CrO4.
Ag2CrO4 can be formed by combining Ag+ ions and CrO42− ions in a 2:1 ratio. KNO3 can
be formed by combining K+ and NO3− ions in a 1:1 ratio.
Step 3. Determine whether either possible product is insoluble using information from
Figure 3.10.
Based on the solubility guidelines, silver chromate is an insoluble compound (chromates
are insoluble except for those with Group 1A (1) cations or NH4+), and potassium nitrate is
soluble in water. A precipitate of silver chromate is predicted if the reactants are mixed.
Step 4. Write and balance the equation.
2 AgNO3(aq) + K2CrO4(aq) n Ag2CrO4(s) + 2 KNO3(aq)
Think about Your Answer You can figure out that chromate ion (CrO42−) has a
charge of 2− because potassium chromate consists of two potassium ions (K+) for each
chromate ion. Likewise, the silver ion must have a 1+ charge because it was initially
paired with nitrate ion (NO3−).
Check Your Understanding
In each of the following cases, does a precipitation reaction occur when solutions of the
two water-soluble reactants are mixed? Give the formula of any precipitate that forms,
and write a balanced chemical equation for the precipitation reactions that occur.
(a) sodium carbonate and copper(II) chloride
(b) potassium carbonate and sodium nitrate
(c) nickel(II) chloride and potassium hydroxide
Net Ionic Equations
When aqueous solutions of silver nitrate and potassium chloride are mixed, insoluble silver chloride forms, leaving potassium nitrate in solution (see Figure 3.11).
The balanced chemical equation for this process is
AgNO3(aq) + KCl(aq) n AgCl(s) + KNO3(aq)
Another way to represent this reaction is by writing an equation that shows the
soluble ionic compounds present in solution as dissociated ions. An aqueous solution
of silver nitrate contains Ag+ and NO3− ions, and an aqueous solution of potassium
chloride contains K+ and Cl− ions. In the products, potassium nitrate is present in
3.5 Precipitation Reactions
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153
solution as K+ and NO3− ions. However, silver chloride is insoluble and thus is not
present in the solution as dissociated ions. It is shown in the equation as AgCl(s).
Ag+(aq) + NO3−(aq) + K+(aq) + Cl−(aq)
AgCl(s) + K+(aq) + NO3−(aq)
after reaction
before reaction
Net Ionic Equations All chemical
equations, including net ionic
equations, must be balanced.
The same number of atoms of
each kind must appear on both
the product and reactant sides.
In addition, the sum of positive
and negative charges must
be the same on both sides of
the equation.
This type of equation is called a complete ionic equation. It represents any soluble
ionic compounds (and any strong acids and strong bases, Section 3.6) as aqueous
ions.
The K+ and NO3− ions are present in solution both before and after reaction, so
they appear on both the reactant and product sides of the complete ionic equation.
Such ions are often called spectator ions because they do not participate in the net
reaction; they only “look on” from the sidelines. Little chemical information is lost
if the equation is written without them, so you can simplify the equation to
Ag+(aq) + Cl−(aq) n AgCl(s)
The balanced equation that results from leaving out spectator ions is the net ionic
equation for the reaction. The significance of net ionic equations, and the reason
that net ionic equations are commonly used, is that they focus attention on the
reaction that takes place.
Leaving out spectator ions does not mean that K+ and NO3− ions are unimportant
in the AgNO3 + KCl reaction. Indeed, Ag+ ions cannot exist alone in solution; a negative ion, in this case NO3−, must be present to balance the positive charge of Ag+. Any
anion will do, however, as long as it forms a water-soluble compound with Ag+. Thus,
you could use AgClO4 instead of AgNO3. Similarly, there must be a positive ion present to balance the negative charge of Cl−. In this case, the positive ion present is K+ in
KCl, but you could use NaCl instead of KCl. The net ionic equation would be the same.
Finally, notice that there must always be a charge balance as well as a mass
­balance in a balanced equation. In the Ag+ + Cl− net ionic equation, the cation and
anion charges on the left add together to give a net charge of zero, the same as the
zero charge on AgCl(s) on the right.
Exampl e 3 .4
Writing and Balancing Net Ionic Equations
Strategy Map
Problem
Write the balanced net ionic
equation for the reaction of
BaCl2 + Na2SO4.
Data/Information
The formulas of the reactants
are given.
Problem Write a balanced, net ionic equation for the reaction of aqueous solutions of
BaCl2 and Na2SO4.
What Do You Know? The formulas for the reactants are given. You should recognize that this is an exchange reaction, and that you will need the information on solubilities in Figure 3.10 (page 150).
Strategy
Step 1. Determine the formulas for the reaction products by exchanging cations and
anions and write a balanced equation.
Step 2. Determine the state of each reactant and product (s, ℓ, g, or aq).
Step 3. Write the complete ionic equation, which should show any soluble ionic
compounds as aqueous cations and anions.
Step 4. Write the net ionic equation by eliminating the spectator ions.
Solution
Step 1
154
Determine the products and then write the complete balanced equation.
In this exchange reaction, the Ba2+ and Na+ cations exchange anions (Cl− and SO42−) to give
BaSO4 and NaCl. Now that the reactants and products are known, you can write an equation
for the reaction. To balance the equation, you need to place a two in front of the NaCl.
Chapter 3 / Chemical Reactions
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BaCl2 + Na2SO4 n BaSO4 + 2 NaCl
Step 2
Determine the state of each reactant and product.
Decide on the solubility of each compound (see Figure 3.10). Compounds containing sodium ions
are generally water-soluble, as are those containing chloride ions (with some important exceptions).
Sulfate salts are also usually soluble, but one important exception is BaSO4. You can use this information to add the state to each reactant and product.
BaCl2(aq) + Na2SO4(aq) n BaSO4(s) + 2 NaCl(aq)
Step 3
Write the complete ionic equation.
All soluble ionic compounds dissociate to form ions in aqueous solution. Writing the soluble substances as ions in solution results in the following complete ionic equation.
Ba2+(aq) + 2 Cl−(aq) + 2 Na+(aq) + SO42−(aq) n BaSO4(s) + 2 Na+(aq) + 2 Cl−(aq)
Step 4
Write the net ionic equation.
Eliminate the spectator ions Na+ and Cl− to give the net ionic equation.
Ba2+(aq) + SO42−(aq) n BaSO4(s)
Think about Your Answer Notice that the sum of ion charges is the same on both sides of
the equation. On the left, 2+ and 2− sum to zero; on the right the charge on BaSO4 is also zero.
Check Your Understanding
In each of the following cases, aqueous solutions containing the compounds indicated are mixed. Write balanced net ionic
­equations for the reactions that occur.
(a) CaCl2 + Na3PO4
(b) iron(III) chloride and potassium hydroxide
(c) lead(II) nitrate and potassium chloride
One application for precipitation reactions is the separation of mixtures of
ionic compounds. For example, with the appropriate choice of anions, one metal
cation may be selectively precipitated from a mixture.
Exam p le 3.5
Problem Which of the following soluble ionic compounds can be used to separate a
mixture of Cu(NO3)2(aq) and Mg(NO3)2(aq) by selectively precipitating only one of the two
metal cations?
(a) KCl
(b) NaOH
(c) Na2SO4
(d) KNO3
What Do You Know? Cu(NO3)2 and Mg(NO3)2 are water soluble and provide
Cu2+(aq) and Mg2+(aq) ions, respectively. The anions to be added (Cl −, OH−, SO42−, and
NO3−) are all provided as water-soluble ionic compounds. You will need information on
solubility from Figure 3.10. You do not need to worry about the nitrate ion forming a precipitate with the added metal cations because nitrate salts are generally soluble.
Strategy Use Figure 3.10 to predict which combination of cations and anions are insoluble
© Charles D. Winters/Cengage
Separating a Mixture by Selective Precipitation
Precipitation reaction. The reaction
of barium chloride and sodium
sulfate described in Example 3.4
produces insoluble barium sulfate
and water-soluble sodium chloride.
in water. Determine which, if any, of the anions will precipitate one of the two metal cations.
Solution
(a) KCl dissolves in water to produce Cl− ions. Neither Cu2+ nor Mg2+ will precipitate with
Cl− because both CuCl2 and MgCl2 are water-soluble compounds. KCl cannot be used to
separate the cations in an aqueous solution containing both Cu2+ and Mg2+.
3.5 Precipitation Reactions
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155
(b) NaOH dissolves in water to produce OH− ions. Both Cu2+ and Mg2+ will precipitate as
insoluble hydroxides, solid Cu(OH)2 and Mg(OH)2. NaOH will not selectively precipitate one of the two metal cations.
(c)
Na2SO4 dissolves in water to produce SO42− ions. CuSO4 is soluble in water while MgSO4 is
insoluble. Na2SO4 will selectively separate Cu2+ from Mg2+ by precipitating MgSO4.
(d) KNO3 dissolves in water to produce NO3− ions. Because both Cu(NO3)2 and Mg(NO3)2
are soluble in water, KNO3 will not selectively precipitate one of the two metal cations.
Think about Your Answer This example illustrates a laboratory procedure called
qualitative chemical analysis. The technique can be used to separate mixtures containing up
to dozens of aqueous metal cations. Given a pair of metal cations to be separated, consider
what the product might be when combined with a range of anions. You can generally find
one anion that will form a precipitate with one cation but not the other, except when trying
to separate Group 1A (1) cations, which form soluble compounds with most anions.
Check Your Understanding
Is it possible to separate an aqueous mixture of Cl− and SO42− ions with the addition of
water-soluble compounds containing each of the following cations?
(a) Pb2+
(c) Na+
(b) Ca2+
(d) Cu2+
3.6 Acids and Bases
Goals for Section 3.6
• Know the names and formulas of common acids and bases and categorize them as
strong or weak.
• Define the Arrhenius and Brønsted-Lowry concepts of acids and bases.
• Recognize substances that are amphiprotic and oxides that dissolve in water to give
acidic solutions and basic solutions.
Acids and bases are two important classes of compounds. You may already be familiar with some common properties of acids. They produce bubbles of CO2 gas when
added to a metal carbonate such as CaCO3 (Figure 3.12a), and they react with many
metals to produce hydrogen gas (H2) (Figure 3.12b). Although tasting substances is
never done in a chemistry laboratory, you have probably experienced the sour taste
of acids such as acetic acid in vinegar and citric acid (commonly found in fruits and
added to candies and soft drinks).
Extract of rose petals
in alcohol and water
Add base
Photos: © Charles D. Winters/Cengage
Add acid
(a) A piece of coral (mostly CaCO3)
dissolves in acid to give CO2 gas.
(b) Zinc reacts with hydrochloric
acid to produce zinc chloride and
hydrogen gas.
(c) An extract of red rose petals turns deep red on adding acid but turns green on adding base.
Figure 3.12 Some properties of acids and bases.
156
Chapter 3 / Chemical Reactions
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Table 3.1
Oxalic acid
H2C2O4
Common Acids and Bases*
Strong Acids
(Strong Electrolytes)
Soluble Strong Bases
(Strong Electrolytes)**
HCl
Hydrochloric acid
LiOH
Lithium hydroxide
HBr
Hydrobromic acid
NaOH
Sodium hydroxide
HI
Hydroiodic acid
KOH
Potassium hydroxide
HNO3
Nitric acid
Ba(OH)2
Barium hydroxide
HClO3
Chloric acid
Sr(OH)2
Strontium hydroxide
HClO4
Perchloric acid
H2SO4
Sulfuric acid
Weak Acids
(Weak Electrolytes)
Weak Base
(Weak Electrolyte)
HF
Hydrofluoric acid
NH3
H3PO4
Phosphoric acid
H2CO3
Carbonic acid
CH3CO2H
Acetic acid
H2C2O4
Oxalic acid
H2C4H4O6
Tartaric acid
H3C6H5O7
Citric acid
HC9H7O4
Aspirin
Ammonia
Carboxyl group
Acetic acid
CH3CO2H
Weak Acids Common acids and
bases are listed in Table 3.1.
There are numerous other weak
acids and bases, and many
are natural substances. Many
naturally occurring weak acids,
such as oxalic and acetic acids,
contain CO2H or carboxyl
groups. (The H of this group is
lost as H+.)
*The electrolytic behavior refers to aqueous solutions of these acids and bases.
**Ca(OH)2 is often listed as a strong base, although it is poorly soluble.
Acids and bases have some related properties. Solutions of acids or bases, for
example, can change the colors of natural pigments (Figure 3.12c). For example,
acids change the color of litmus, a dye derived from certain lichens, from blue to
red. Adding a base reverses the effect, making the litmus blue again. Thus, acids and
bases seem to be opposites. A base can neutralize the effect of an acid, and an acid
can neutralize the effect of a base. Table 3.1 lists some common acids and bases.
Naming Common Acids
The molecular compound HCl is called hydrogen chloride when it is in the pure,
gaseous state. However, an aqueous solution of HCl is acidic, and this solution is
given the name hydrochloric acid. The same pattern, adding hydro– at the beginning
and an –ic ending, applies to other acids where the anion has an –ide ending. For
example, HF(aq) is hydrofluoric acid and H2S(aq) is hydrosulfuric acid.
Refer to Table 3.1 and notice that other common acids, such as sulfuric acid
(H2SO4) and nitric acid (HNO3), also have names ending in –ic. When anions have
names ending in –ate (such as nitrate, sulfate, chlorate, perchlorate, and acetate),
the acid associated with that anion has a name ending in –ic. Examples are nitric,
sulfuric, chloric, perchloric, and acetic acids.
You learned in Chapter 2 that there are series of anions based on chlorine, sulfur,
and nitrogen. Among them are the hypochlorite (ClO−) and chlorite (ClO2–) ions as
well as the sulfite (SO32−) and nitrite (NO2−) ions. Acids based on ions ending in –ite
have names ending in –ous. These acids are named hypochlorous, chlorous, sulfurous,
and nitrous acids. These naming conventions are illustrated in Table 3.2.
3.6 Acids and Bases
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157
Table 3.2
Names of Common Ions and Their Corresponding Acids
Common Ions
Corresponding Acids
Cl−, chloride ion
HCl, hydrochloric acid
ClO−, hypochlorite ion
HClO, hypochlorous acid
−
HClO2, chlorous acid
−
HClO3, chloric acid
−
ClO4 , perchlorate ion
HClO4, perchloric acid
S , sulfide ion
H2S, hydrosulfuric acid
SO3 , sulfite ion
H2SO3, sulfurous acid
SO42−, sulfate ion
H2SO4, sulfuric acid
NO2−, nitrite ion
HNO2, nitrous acid
NO3−, nitrate ion
HNO3, nitric acid
ClO2 , chlorite ion
ClO3 , chlorate ion
Names of Common Ions. Review
pages 92 and 93 for the system for
naming common polyatomic ions.
2−
2−
Acids and Bases: The Arrhenius Definition
Over the years, chemists have examined the properties, chemical structures, and
reactions of acids and bases and have proposed different definitions of the terms
acid and base. The two most commonly used definitions are the one proposed by
Svante Arrhenius (1859–1927) and another proposed by Johannes N. Brønsted
(1879–1947) and Thomas M. Lowry (1874–1936).
In the late 1800s, the Swedish chemist Svante Arrhenius proposed that acids
and bases dissolve in water and ultimately form ions. This theory predated any
knowledge of the composition and structure of atoms and was not well accepted
initially. With a knowledge of atomic structure, however, the presence of ions in
solution is almost taken for granted.
The Arrhenius definition for acids and bases focuses on formation of H+ and
−
OH ions in aqueous solutions.
•
An acid is a substance that, when dissolved in water, increases the concentration
of hydrogen ions, H+, in solution.
HCl(g) n H+(aq) + Cl−(aq)
•
A base is a substance that, when dissolved in water, increases the concentration
of hydroxide ions, OH−, in the solution.
NaOH(s) n Na+(aq) + OH−(aq)
•
The reaction of an acid and a base produces a salt and water. Because the characteristic properties of an acid are lost when a base is added, and vice versa,
acid–base reactions were logically described as resulting from the combination
of H+ and OH− to form water.
HCl(aq) + NaOH(aq) n NaCl(aq) + H2O(ℓ)
Arrhenius further proposed that acid strength was related to the extent to
which the acid ionized. Some acids such as hydrochloric acid (HCl) and nitric acid
(HNO3) ionize almost completely in water; they are strong electrolytes, and so are
called strong acids. Other acids such as acetic acid and hydrofluoric acid are incompletely ionized; they are weak electrolytes and are weak acids. Weak acids exist in
solution primarily as molecules, and only a small fraction of these molecules ionize
to produce H+(aq) ions along with the appropriate anion.
Water-soluble compounds that contain hydroxide ions, such as sodium hydroxide
(NaOH) and potassium hydroxide (KOH), are strong electrolytes and strong bases.
158
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Aqueous NH3 produces a very small
number of NH4+ and OH− ions per
mole of ammonia molecules
+
© Charles D. Winters/Cengage
−
OH− ions
NH3 molecules
NH4+ ions
Figure 3.13 Ammonia, a weak electrolyte. The
name on the bottle, ammonium hydroxide, is
misleading. The solution consists almost entirely of
NH3 molecules dissolved in water. It is better referred
to as aqueous ammonia.
Aqueous ammonia, NH3(aq), is a weak electrolyte. Even though OH− ions are
not part of its formula, it does produce ammonium ions and hydroxide ions from
its reaction with water and so is a base (Figure 3.13). The fact that this is a weak
electrolyte indicates that this reaction with water to form ions is reactant-favored at
equilibrium. Most of the ammonia remains in solution in molecular form.
NH3(aq) + H2O(ℓ) uv NH4+(aq) + OH−(aq)
Although the Arrhenius theory is still used to some extent and is interesting in
a historical context, modern concepts of acid–base chemistry such as the BrønstedLowry theory have gained preference among chemists.
Acids and Bases: The Brønsted-Lowry Definition
In 1923, Johannes Brønsted (1879–1947) in Copenhagen, Denmark, and
Thomas Lowry (1874–1936) in Cambridge, England, independently suggested a
new concept of acid and base behavior. They viewed acids and bases in terms of
the transfer of a proton (H+) from one species to another, and they described all
acid–base reactions in terms of equilibria. The Brønsted-Lowry theory expanded
the scope of the definition of acids and bases and helped chemists make predictions of product- or reactant-favorability based on acid and base strength.
The main concepts of the Brønsted-Lowry theory are the following:
+
•
An acid is a proton (H ) donor.
•
A base is a proton (H+) acceptor. This definition includes the OH− ion but it also
broadens the number and type of bases to include anions derived from acids as
well as neutral compounds such as water.
•
An acid–base reaction involves the transfer of a proton from an acid to a base to
form a new acid and a new base.
According to Brønsted-Lowry theory, the behavior of acids such as HCl or
CH3CO2H in water is better considered as an acid–base reaction. Both species (both
Brønsted acids) donate a proton to water (a Brønsted base) forming H3O+(aq), the
hydronium ion.
Protons The major isotope of
hydrogen has no neutrons and
consists of one proton and
one electron. To form H+, the
electron is lost, leaving only the
proton. Because of this, H+ ions
are often referred to as protons.
H3O+ versus H+ The formula for
the hydronium ion, H3O+ is
a fairly accurate description
and is often used to represent
the hydrogen ion in solution.
However, there are instances in
this textbook when, for simplicity,
the hydrogen ion is represented
as H+(aq).
Experiments show that other
forms of the ion also exist in
water, one example being
[H3O(H2O)3]+.
3.6 Acids and Bases
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159
Hydrochloric acid, HCl(aq), is a strong electrolyte that ionizes almost completely in aqueous solution. It is classified as a strong acid.
Hydrochloric acid, a strong acid, nearly 100% ionized. Equilibrium strongly f­avors
products.
+
HCl(aq)
hydrochloric acid
strong electrolyte
≈100% ionized
H2O(𝓵)
H3O+(aq)
water
hydronium ion
Cl–(aq)
+
chloride ion
In contrast, CH3CO2H, a weak electrolyte, ionizes only to a small extent. It is classified as a weak acid.
Acetic acid, a weak acid,  100% ionized. Equilibrium favors reactants.
CH3CO2H(aq)
+
acetic acid
H2O(𝓵)
H3O+(aq)
water
hydronium ion
CH3CO2−(aq)
+
acetate ion
Sulfuric acid, a diprotic acid (an acid capable of transferring two H+ ions), reacts
with water in two steps. The first step strongly favors products, whereas the second
step is reactant-favored.
H3O+(aq) +
HSO4–(aq)
sulfuric acid
≈100% ionized
hydronium ion
hydrogen
sulfate ion
HSO4–(aq) + H2O(𝓵)
H3O+(aq) +
SO42–(aq)
hydrogen sulfate ion
<100% ionized
hydronium ion
sulfate ion
Strong acid: H2SO4(aq) + H2O(𝓵)
Weak acid:
Ammonia, a weak base, reacts with water to produce OH−(aq) ions. The reaction is reactant-favored at equilibrium.
Ammonia, a weak base,  100% ionized. Equilibrium favors reactants.
+
NH3(aq)
ammonia, base
weak electrolyte
< 100% ionized
H2O(𝓵)
NH4+(aq)
water
ammonium
ion
+
OH–(aq)
hydroxide ion
According to the Brønsted-Lowry theory, anions can add a proton and are thus
classified as bases. In particular, anions of weak acids typically behave as weak bases,
and basic solutions result from dissolving a salt containing the anion of a weak acid
in water. For example, an aqueous solution of sodium acetate is basic because of the
following reaction:
160
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Acetate ion, weak base. Equilibrium favors reactants.
CH3CO2–(aq)
acetate ion, a weak base
+
H2O(𝓵)
CH3CO2H(aq)
water
acetic acid molecule
+
OH−(aq)
hydroxide ion
Some species are described as amphiprotic, that is, they can function either
as acids or as bases depending on the reaction. In the previous examples, water
functions as a base in reactions with acids (it accepts a proton) and as an acid in
its reaction with ammonia (where it donates a proton to ammonia, forming the
ammonium ion).
Exam p le 3.6
Brønsted Acids and Bases
Problem Write a balanced net ionic equation for the reaction that occurs when the
cyanide ion, CN−, accepts a proton (H+) from water to form HCN. Is CN− a Brønsted acid or
a Brønsted base?
What Do You Know? You know the formulas of the reactants (CN− and H2O) and
of one of the products, (HCN). You also know a proton transfer occurs from water to CN−.
Strategy As it is a proton transfer, you should move an H+ ion from H2O to CN− to give
the products.
Solution
H2O(ℓ) + CN−(aq) uv OH−(aq) + HCN(aq)
In this reaction, water is the Brønsted acid and the CN− ion is the Brønsted base.
Think about Your Answer The CN− ion interacts with water to produce the
OH− ion. Most anions derived from weak acids produce basic solutions in water.
Check Your Understanding
(a) Write a balanced equation for the reaction that occurs when H3PO4, phosphoric acid,
donates a proton to water to form the dihydrogen phosphate ion.
(b) Write a net ionic equation showing the dihydrogen phosphate ion acting as a B
­ rønsted
acid in a reaction with water. Write another net ionic equation showing the dihydrogen phosphate ion acting as a Brønsted base in a reaction with water. What term is
used to describe a species such as the dihydrogen phosphate ion that can act both as
an acid or as a base?
Oxides of Nonmetals and Metals
Each acid shown in Table 3.1 has one or more H atoms that ionize in water to form
H3O+ ions. However, some compounds that have no H atoms also form acidic solutions. Carbon dioxide and sulfur trioxide, oxides of nonmetals, have no H atoms,
but both react with water to produce H3O+ ions. Carbon dioxide, for example, dissolves in water to a small extent, and a few of the dissolved molecules react with
water to form the weak acid, carbonic acid. This acid then ionizes to a small extent
3.6 Acids and Bases
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161
to form the hydronium ion, H3O+, and the hydrogen carbonate (bicarbonate) ion,
HCO3−.
CO2(g)
+
H2O(𝓵)
H2CO3(aq)
H2CO3(aq)
+
H2O(ℓ)
HCO3−(aq)
+
H3O+(aq)
The HCO3− ion can also function as an acid, ionizing to produce H3O+ and the
carbonate ion, CO32−.
CO32−(aq)
H2O(ℓ)
+
H3O+(aq)
Sulfuric Acid
More sulfuric aid is produced annually in the United States, and likely
in the world, than any other chemical. Worldwide production is around
250 million metric tons. The acid
is so important to the economy of
industrialized nations that some
economists have said sulfuric acid
production is a measure of a nation’s
industrial strength.
Sulfuric acid is a colorless, s­ yrupy
liquid with a density of 1.84 g/mL and
a boiling point of 337 °C. It has several desirable properties that have led
to its widespread use: it is generally less
expensive to produce than other acids,
is a strong acid, and can be handled in
steel containers. It reacts readily with
many organic compounds to produce
useful products and reacts readily with
lime (CaO), the least expensive and most
162
+
readily available base, to give calcium sulfate, a compound used to make wall board
for the construction industry.
The first step in the industrial preparation
of sulfuric acid is the production of sulfur
dioxide from the combustion of sulfur in air,
make phosphoric acid. The remainder is
used to make pigments, explosives, pulp
and paper, detergents, and as a component in storage batteries.
S(s) + O2(g) n SO2(g)
or using the SO 2 produced in smelting
sulfur-containing copper, nickel, or other
metal ores. The SO 2 is then combined
with more oxygen, in the presence of a
catalyst (a substance that speeds up the
reaction), to give sulfur trioxide,
2 SO2(g) + O2(g) n 2 SO3(g)
which then gives sulfuric acid when dissolved in water.
SO3(g) + H2O(ℓ) n H2SO4(aq)
Currently, over two thirds of the production is used in the fertilizer industry to
Mike Hill/Stone/Getty Images
A Closer Look
HCO3−(aq)
Sulfur. Much of the sulfur used in the United
States used to be mined, but it is now
largely a by-product from natural gas and
oil-refining processes. It takes about one ton
of sulfur to make three tons of sulfuric acid.
Chapter 3 / Chemical Reactions
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These reactions are important in our environment and in the human body. Carbon
dioxide is found in small amounts in the atmosphere, so rainwater is always slightly
acidic. In the human body, carbon dioxide is dissolved in body fluids where the
HCO3− and CO32− ions play an important role in keeping the pH stable.
Oxides of nonmetals such as CO2, SO2, SO3, and NO2 that react with water to
produce an acidic solution are called acidic oxides. In contrast, oxides of metals are
called basic oxides because they produce basic solutions if they dissolve appreciably in water. Perhaps the best example of a basic oxide is calcium oxide, CaO, often
called lime, or quicklime. Almost 20 billion kg of lime are produced annually in the
United States for use in the metals and construction industries, in sewage and pollution control, in water treatment, and in agriculture. Calcium oxide reacts with water
to give calcium hydroxide, commonly called slaked lime. Although only slightly soluble in water (about 0.2 g/100 g H2O at 10 °C), Ca(OH)2 is widely used in industry
as a base because it is inexpensive.
SO2
SO3
CaO(s) + H2O(ℓ) n Ca(OH)2(s)
lime
slaked lime
NO2
Some common nonmetal oxides
that form acids in water.
3.7 Acid–Base Reactions
Goals for Section 3.7
• Identify the Brønsted-Lowry acid and base in a reaction and write equations for
Brønsted-Lowry acid–base reactions.
• Identify common acid–base reactions in which a gas is formed and write equations
for these reactions.
Brønsted-Lowry Acid–Base Reactions
The Arrhenius definition of acids focuses on the ionization of an acid in water to form
H+ ions. The Brønsted-Lowry definition shows this as a reaction of an acid with water,
where water acts as a base and accepts a proton from the acid. The result is the formation of H3O+ ions. The Brønsted-Lowry definition, however, does not restrict the
study of acids to reactions with water. An acid can transfer a proton (H+) to any base.
Acids and bases in aqueous solution usually react to produce a salt and water.
Note that these reactions are also exchange reactions, with cations and anions
changing partners. For example (Figure 3.14),
HCl(aq)
+
hydrochloric acid
NaOH(aq)
sodium hydroxide
H2O(𝓵) + NaCl(aq)
water
sodium chloride
In chemistry, the word “salt” refers to any ionic compound whose cation comes
from a base (here Na+ from NaOH) and whose anion comes from an acid (here Cl−
from HCl). The reaction of any of the acids listed in Table 3.1 with any of the listed
­hydroxide-containing bases produces a salt and water.
Hydrochloric acid and sodium hydroxide are strong electrolytes in water (see
Figure 3.14 and Table 3.1), so the complete ionic equation for the reaction of
HCl(aq) and NaOH(aq) is written as
H3O+(aq) + Cl−(aq) + Na+(aq) + OH−(aq)
from HCl(aq)
+
from NaOH(aq)
2 H2O(ℓ) + Na+(aq) + Cl−(aq)
water
from salt
−
Because Na and Cl ions appear on both sides of the equation they can be cancelled out, and the net ionic equation is just the combination of the ions H3O+ and
OH− to give water.
H3O+(aq) + OH−(aq) n 2 H2O(ℓ)
3.7 Acid–Base Reactions
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163
NaCl (salt) + H2O
+
NaOH (base)
−
+
+
Chloride ion
Cl– (aq)
+
−
H3O+(aq) + Cl−(aq)
Hydronium ion
H3O+ (aq)
−
−
−
−
−
−
+
+
−
+
+
−
Na+(aq) + OH−(aq)
+
Sodium ion
Na+(aq)
+
+
−
Hydroxide ion
OH−(aq)
+
−
Na+(aq) + Cl−(aq)
Figure 3.14 An acid–base reaction, HCl and NaOH. On mixing, the H3O+ and OH− ions combine to produce H2O, whereas the
ions Na+ and Cl− remain in solution.
This is always the net ionic equation when a strong acid reacts with a strong base.
Reactions between strong acids and strong bases are called neutralization
reactions because, upon completion of the reaction, the solution is neither
acidic nor basic if exactly the same amounts (number of moles) of the acid and
base are mixed. The other ions (the cation of the base and the anion of the acid)
remain unchanged.
If acetic acid and sodium hydroxide are mixed, the following reaction will take
place.
CH3CO2H(aq) + NaOH(aq) n NaCH3CO2(aq) + H2O(ℓ)
Because acetic acid is a weak acid (Figure 3.8c), the molecular species is the predominant form in aqueous solutions. In ionic equations, therefore, acetic acid is
shown as molecular CH3CO2H(aq). The complete ionic equation for this reaction is
CH3CO2H(aq) + Na+(aq) + OH−(aq) n Na+(aq) + CH3CO2−(aq) + H2O(ℓ)
The only spectator ions in this equation are the sodium ions, so the net ionic equation is
CH3CO2H(aq) + OH−(aq) n CH3CO2−(aq) + H2O(ℓ)
Problem Solving Tip 3.1 Writing Net Ionic Equations
Net ionic equations are commonly
written for chemical reactions in
aqueous solution because they
describe the actual chemical species
involved in a reaction. To write net
ionic equations you must know
which compounds exist as ions in
solution.
1. Strong acids, strong bases, and
soluble salts exist as ions in solution. Examples include the acids
HCl and HNO3, a base such as
NaOH, and salts such as NaCl
and CuCl2.
2. All other species should be represented by their complete formulas.
164
Weak acids such as acetic acid
(CH3CO2H) exist in aqueous
­solutions primarily as molecules.
­Insoluble salts such as CaCO3(s) or
insoluble bases such as Mg(OH)2(s)
should not be written as dissociated
ions, even though they are ionic
compounds.
bases, and soluble salts as ions.
(Consider only species labeled (aq)
in this step.)
The best way to approach writing
net ionic equations is to follow a
precise set of steps.
3. Some ions may remain unchanged
in the reaction (the ions that
appear in the equation both as
reactants and products). These
spectator ions are not part of the
chemistry that is going on, and you
can eliminate them from each side
of the equation.
1. Write a complete, balanced equation. Indicate the state of each
substance (aq, s, ℓ, g).
2. Next rewrite the whole equation,
writing all strong acids, strong
4. Net ionic equations must be balanced. The same number of atoms
appears on each side of the arrow,
and the sum of the ion charges on
the two sides must also be equal.
Chapter 3 / Chemical Reactions
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Photos: © Charles D. Winters/Cengage
HCl (acid)
Exam p le 3.7
Net Ionic Equation for an Acid–Base Reaction
What Do You Know? The reactants are NH3(aq) and HCl(aq). A proton will
transfer from the acid to the base.
Strategy Follow the general strategy for writing net ionic equations as outlined in
Problem Solving Tip 3.1.
NH4Cl(s)
© Charles D. Winters/Cengage
Problem Ammonia, NH3, is one of the most important chemicals in industrial economies. Not only is it used directly as a fertilizer but it is the raw material for the manufacture
of nitric acid, another commercially important chemical. As a base, ammonia reacts with
acids such as hydrochloric acid. Write a balanced, net ionic equation for this reaction.
NH3(aq)
Solution A proton transfers from HCl to NH3, a weak Brønsted base, to form the ammonium ion, NH4+. This positive ion must have a negative counterion from the acid, Cl −, so
the reaction product is NH4Cl, and the overall balanced equation is
NH3(aq)
ammonia
+
HCl(aq)
hydrochloric acid
n
HCl(aq)
Reaction of gaseous HCl and
NH3. Open dishes of aqueous
ammonia and hydrochloric
acid were placed side by side.
When molecules of NH3 and
HCl escape from solution to
the atmosphere and encounter
one another, a cloud of solid
ammonium chloride, NH4Cl, is
observed.
NH4Cl(aq)
ammonium chloride
Hydrochloric acid is a strong acid and produces H3O+ and Cl− ions in water. NH4Cl is quite
soluble and exists as NH4+ and Cl− ions in solution. On the other hand, ammonia is a weak
base and so is predominantly present in the solution as the molecular species, NH3. The
complete ionic equation for this reaction is
NH3(aq) + H3O+(aq) + Cl−(aq) n NH4+(aq) + Cl−(aq) + H2O(ℓ)
Eliminating the spectator ion, Cl−, gives the net ionic equation
NH3(aq) + H3O+(aq) n NH4+(aq) + H2O(ℓ)
Think about Your Answer The net ionic equation shows that the important
aspect of the reaction between the weak base ammonia and the strong acid HCl is the
transfer of an H+ ion from the acid to the NH3. Any strong acid could be used here (HBr,
HNO3, HClO4, H2SO4) and the net ionic equation would be the same. Also notice that, even
though H2O is not in the overall balanced equation, it is present in the net ionic equation.
Check Your Understanding
Gas-Forming Acid–Base Reactions
Acid–base reactions that produce a gas represent another type of exchange reaction,
and there are several commonly encountered examples in a chemical laboratory.
The odor of rotten eggs will be very noticeable when you produce hydrogen sulfide,
H2S(g), from a metal sulfide and an acid. Probably the most commonly encountered examples of gas-forming reactions, however, involve the formation of CO2(g)
when either metal carbonates or metal hydrogen carbonates are treated with acid
(Figure 3.15). Equations for several types of gas-forming reactions are given in
Table 3.3.
Although usually written as a single equation for the formation of CO2(g) in
the reaction between a metal carbonate (or hydrogen carbonate), the formation of
CO2(g) actually occurs in two distinct steps. Consider the reaction of CaCO3 and
hydrochloric acid. The first step is an exchange reaction in which hydrogen ions are
exchanged for the cation(s) in the metal carbonate.
CaCO3(s) + 2 HCl(aq) n CaCl2(aq) + H2CO3(aq)
© Charles D. Winters/Cengage
Write the balanced, overall equation and the net ionic equation for the reaction of
­magnesium hydroxide with hydrochloric acid. (Hint: Think about the solubility guidelines.)
Figure 3.15 Dissolving
limestone (calcium carbonate,
CaCO3) in vinegar. Notice the
bubbles of CO2 rising from the
surface of the limestone. This
reaction shows why vinegar
can be used as a household
cleaning agent. It can be used,
for example, to clean the calcium
carbonate deposited from hard
water.
3.7 Acid–Base Reactions
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165
Table 3.3
Gas-Forming Acid–Base Reactions
Metal carbonate or hydrogen carbonate 1 acid n metal salt 1 CO2(g) 1 H2O(ℓ)
Na2CO3(aq) + 2 HCl(aq) n 2 NaCl(aq) + CO2(g) + H2O(ℓ)
NaHCO3(aq) + HCl(aq) n NaCl(aq) + CO2(g) + H2O(ℓ)
Metal sulfide 1 acid n metal salt 1 H2S(g)
Na2S(aq) + 2 HCl(aq) n 2 NaCl(aq) + H2S(g)
Metal sulfite 1 acid n metal salt 1 SO2(g) 1 H2O(ℓ)
Na2SO3(aq) + 2 HCl(aq) n 2 NaCl(aq) + SO2(g) + H2O(ℓ)
Ammonium salt 1 strong base n metal salt 1 NH3(g) 1 H2O(ℓ)
NH4Cl(aq) + NaOH(aq) n NaCl(aq) + NH3(g) + H2O(ℓ)
The product formed in this reaction is carbonic acid, H 2CO3. This compound is
unstable, however, and decomposes to CO2 and H2O.
H2CO3(aq) n H2O(ℓ) + CO2(g)
Carbon dioxide bubbles then escape the solution because CO2 is not very soluble in
water. The overall equation is obtained by adding the two equations.
Overall reaction: CaCO3(s) + 2 HCl(aq) n CaCl2(aq) + H2O(ℓ) + CO2(g)
Calcium carbonate is a common residue from hard water in home heating systems and cooking utensils. Washing with vinegar is a good way to clean the system
or utensils because insoluble calcium carbonate is turned into water-soluble calcium acetate in the following gas-forming reaction (see Figure 3.15).
2 CH3CO2H(aq) + CaCO3(s) n Ca(CH3CO2)2(aq) + H2O(ℓ) + CO2(g)
What is the net ionic equation for this reaction? Acetic acid is a weak acid, and
­calcium carbonate is insoluble in water. Therefore, the reactants are CH 3CO2H(aq)
and CaCO3(s). On the products side, calcium acetate is water-soluble and so is present in solution as aqueous calcium and acetate ions. Water and carbon dioxide are
molecular compounds, so the net ionic equation is
2 CH3CO2H(aq) + CaCO3(s) n Ca2+(aq) + 2 CH3CO2−(aq) + H2O(ℓ) + CO2(g)
There are no spectator ions in this reaction.
Have you ever made biscuits or muffins? As you bake the dough, it rises in
the oven because a gas-forming reaction occurs between an acid and baking soda,
sodium hydrogen carbonate (bicarbonate of soda, NaHCO3). One acid used for this
purpose is tartaric acid, a weak acid found in many foods. The net ionic equation
for a typical reaction is
H2C4H4O6(aq) +
tartaric acid
HCO3−(aq)
hydrogen carbonate ion
HC4H4O6−(aq) + H2O(ℓ) + CO2(g)
hydrogen tartrate ion
Problem Solving Tip 3.2 Recognizing Gas-Forming Acid–Base Reactions
How can you recognize that a particular acid–base reaction leads to
gas formation? After you predict the
products of the exchange reaction, be
alert for certain products:
166
(a) H2CO3: This decomposes into
carbon dioxide gas and water.
(c) H2S: This is already a gaseous
product.
(b) H2SO3: This decomposes into
sulfur dioxide gas and water.
(d) If NH4+ and OH− ions are produced, they form NH3 and water.
Chapter 3 / Chemical Reactions
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Exam p le 3.8
Gas-Forming Acid–Base Reactions
Problem Write a balanced equation for the reaction that occurs when nickel(II) carbonate is treated with sulfuric acid.
What Do You Know? You know the names of the reactants and therefore their
formulas. You should recognize that the reaction of a metal carbonate with an acid is a
gas-forming reaction (CO2 is formed in these reactions).
Strategy First, write the formulas for the reactants. Next, determine the products of
the reaction and their formulas. Finally, write and balance the equation.
Solution The reactants are NiCO3 and H2SO4, and the products of the reaction are
NiSO4, CO2, and H2O. The complete, balanced equation is
NiCO3(s) + H2SO4(aq) n
NiSO4(aq) + H2O(ℓ) + CO2(g)
Think about Your Answer The products in this reaction were determined by
first exchanging cations (Ni2+ and 2 H+) and anions (CO32− and SO42−). This exchange
reaction is then followed by a second reaction in which one product, H2CO3, decomposes
to give CO2 and H2O.
Check Your Understanding
Barium carbonate, BaCO3, is used in the brick, ceramic, glass, and chemical manufacturing
industries. Write a balanced equation that shows what happens when barium carbonate is
treated with nitric acid. Give the name of each of the reaction products.
3.8 Oxidation–Reduction Reactions
Goals for Section 3.8
• Determine oxidation numbers of elements in a compound and understand that
these numbers represent the charge an atom has, or appears to have, when the
electrons of the compound are counted according to a set of guidelines.
• Recognize common oxidizing and reducing agents.
• Identify oxidation–reduction reactions (redox reactions), identify the oxidizing and
reducing agents and the substances oxidized and reduced in the reaction.
Iron ore, which is largely Fe2O3, is reduced
to metallic iron with carbon (C) or carbon
monoxide (CO) in a blast furnace. The C or
CO is oxidized to CO2.
The terms oxidation and reduction come from reactions that have been known for
­centuries. Ancient civilizations learned how to change metal oxides and sulfides
into the metal, that is, how to reduce ore to the metal. A modern example is the
reduction of iron(III) oxide with carbon monoxide to give iron metal.
Fe2O3 loses oxygen and is reduced.
2 Fe(s) + 3 CO2(g)
CO is the reducing agent. It
gains oxygen and is oxidized.
In this reaction, carbon monoxide is the agent that brings about the reduction of
iron ore to iron metal, so carbon monoxide is called the reducing agent.
Jan Halaska/Science Source
Fe2O3(s) + 3 CO(g)
3.8 Oxidation–Reduction Reactions
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167
When Fe2O3 is reduced by carbon monoxide, oxygen is removed from the iron
ore and added to the carbon monoxide. The carbon monoxide, therefore, is oxidized
by the addition of oxygen. Any process in which oxygen is added to another substance is an oxidation.
Magnesium metal and oxygen produce magnesium oxide. In this reaction,
oxygen is called the oxidizing agent because it is the substance responsible for the
oxidation of the metal.
© Charles D. Winters/Cengage
Mg combines with
oxygen and is oxidized.
Burning magnesium metal in
air produces magnesium oxide.
The magnesium is oxidized by
the oxidizing agent O2. Oxygen
is reduced by the reducing
agent Mg.
2 Mg(s) + O2(g)
2 MgO(s)
O2 is the oxidizing agent.
Oxidation–Reduction Reactions and Electron Transfer
The concept of oxidation–reduction reactions can be extended to a vast number of
other reactions that do not involve oxygen. Rather than concentrate on whether oxygen is gained or lost, focus on what happens with electrons during the course of the
reaction. All oxidation and reduction reactions can be accounted for by considering them to occur by a transfer of electrons between substances. When a substance
accepts electrons, it is said to be reduced because there is a reduction in the numerical value of the charge on an atom of the substance. In the reaction of a silver salt
with copper metal, positively charged Ag+ ions accept electrons from copper metal
and are reduced to uncharged silver atoms (Figure 3.16).
Ag+ ions accept electrons from Cu and are
reduced to Ag. Ag+ is the oxidizing agent.
Ag+(aq) + e− n Ag(s)
2 Ag+(aq) + Cu(s)
2 Ag(s) + Cu2+(aq)
Cu donates electrons to Ag+ and is oxidized to Cu2+.
Cu is the reducing agent.
Cu(s) n Cu2+(aq) + 2 e−
Photos: © Charles D. Winters/Cengage
Because copper metal supplies the electrons that cause Ag+ ions to be reduced, Cu
is the reducing agent.
Pure copper wire
Copper wire in dilute AgNO3
solution after several hours
Blue color due to
Cu2+ ions formed
in redox reaction
Silver crystals
formed after
several weeks
Figure 3.16 The oxidation of copper metal by silver ions. A clean piece of copper wire is placed in
a solution of silver nitrate, AgNO3. Over time, the copper reduces Ag+ ions, forming silver crystals, and
the copper metal is oxidized to copper ions, Cu2+. The blue color of the solution is due to the presence of
aqueous copper(II) ions formed in the reaction.
168
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When a substance loses electrons, the numerical value of the charge on an
atom of the substance increases. The substance is said to have been oxidized.
In this example, copper metal releases electrons on going to Cu2+, so the copper metal is oxidized. For this to happen, something must be available to accept
the electrons from copper. In this case, Ag+ is the electron acceptor; its charge is
reduced to zero in silver metal, so the silver metal ion is reduced. Because Ag+ is
the agent that causes Cu metal to be oxidized, Ag+ is the oxidizing agent.
Balancing Equations for Redox
Reactions The notion that a
redox reaction can be divided
into a reduction portion and
an oxidation portion will lead
to a method of balancing
more complex equations for
redox reactions described in
Chapter 19.
Exam p le 3.9
Recognizing Oxidation–Reduction Reactions
Problem Aluminum reacts with oxygen to produce aluminum oxide.
4 Al(s) + 3 O2(g) n 2 Al2O3(s)
Which species is oxidized and which is reduced? Identify the oxidizing agent and the
reducing agent.
elements have no charges. The product, aluminum oxide, is an ionic compound. The
charges on aluminum and oxide ions are always 3+ and 2−, respectively.
Strategy Determine which species gains electrons (is reduced) and which loses electrons
(is oxidized), then assign the terms oxidizing agent and reducing agent to the correct species.
Solution Each aluminum atom loses three electrons when converted from the element to a cation. Al(s) is oxidized. Each oxygen atom gains two electrons when converted
from an element to an anion. O2(g) is reduced. When oxidized, the aluminum reduces
another species. Al(s) is the reducing agent. Oxygen takes electrons from another species.
O2(g) is the oxidizing agent.
© Charles D. Winters/Cengage
What Do You Know The reactants, aluminum and oxygen, are elements, and
A redox reaction. One component of a toy sparkler is aluminum, which reacts with oxygen to
form Al2O3.
Think about Your Answer Remember that a species gains electrons when
reduced. Reducing oxygen lowers (or reduces) its charge from zero to 2−. In addition,
notice that the species that is oxidized is always the reducing agent (aluminum is oxidized
and is the reducing agent), and the reduced species is always the oxidizing agent (oxygen
is reduced and is the oxidizing agent).
Check Your Understanding
Upon heating, magnesium reacts with iodine to produce magnesium iodide.
Mg(s) + l2(s) n Mgl2(s)
Which species is oxidized and which is reduced? Identify the oxidizing agent and the
reducing agent.
The observations outlined so far lead to several important conclusions:
•
•
•
•
If one substance is oxidized, another substance in the same reaction must be
reduced. For this reason, such reactions are called oxidation–reduction reactions,
or redox reactions for short.
The reducing agent is itself oxidized, and the oxidizing agent is reduced.
Reduction involves the gain of electrons, oxidation involves the loss of electrons.
The extents of oxidation and reduction in a reaction must be the same. This
means that the number of electrons released when a substance is oxidized must
equal the number of electrons gained by the substance being reduced.
Oxidation Numbers
How can you tell an oxidation–reduction reaction when you see one? ­Oxidation–
reduction reactions are most apparent when elements react to form an ionic
3.8 Oxidation–Reduction Reactions
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169
Writing Charges and Oxidation
Numbers on Ions Convention­ally,
charges on ions are written
as (number, sign), whereas
oxidation numbers are written as
(sign, number). For example, the
oxidation number of the Cu2+
ion is +2 and its charge is 2+.
Monatomic Ions The ion charges
for most main group elements
are determined from the
element's group number.
Peroxides In hydrogen peroxide
(H2O2), each hydrogen atom
has an oxidation number of +1.
To balance this, each oxygen
must have an oxidation number
of −1. A 3% aqueous solution
of H2O2 is sometimes used as an
antiseptic.
compound, such as Mg and O2 combining to form MgO. Elemental magnesium
loses electrons to form the Mg2+ cation and O2 gains electrons to form O2− ions. For
many other oxidation–reduction reactions, however, it is difficult to decide which
species are gaining and losing electrons. The way to decide is to look for a change
in the oxidation number of an element in the course of the reaction. The oxidation
number of an atom in a molecule or ion is defined as the charge an atom has, or
appears to have, as determined by the following guidelines.
Guidelines for Assigning Oxidation Numbers
1. Each atom in a pure element has an oxidation number of zero. The oxidation
number of Cu in metallic copper is 0, and it is 0 for each atom in I2 and S8.
2. For monatomic ions, the oxidation number is equal to the charge on the ion.
Magnesium forms ions with a 2+ charge (Mg2+); the oxidation number of
magnesium in this ion is therefore +2.
3. When combined with another element, fluorine always has an oxidation
­number of 21.
4. The oxidation number of O is 22 in most compounds. Exceptions to this rule occur
(a) when oxygen is combined with fluorine (where oxygen takes on a positive
oxidation number),
(b) in compounds called peroxides (such as Na2O2) and superoxides (such as KO2)
in which oxygen has an oxidation number of −1 and −1/2, respectively.
5. Cl, Br, and I have oxidation numbers of 21 in compounds, except when combined with oxygen and fluorine. This means that Cl has an oxidation number of
−1 in NaCl (in which Na’s oxidation number is +1, as predicted by the fact that
it is an element of Group 1A). In the ion ClO−, however, the Cl atom has an oxidation number of +1 (and O has an oxidation number of −2; see Guideline 4).
6. The oxidation number of H is 11 in most compounds. The main exception to this
guideline occurs when H forms a binary compound with a metal. In such cases, the
metal forms a positive ion and H becomes a hydride ion, H−. Thus, in CaH2 the
oxidation number of Ca is +2 (equal to the group number) and that of H is −1.
7. The algebraic sum of the oxidation numbers for the atoms in a neutral compound must be zero; in a polyatomic ion, the sum must equal the ion charge. For
example, in HClO4 the H atom is assigned +1 and each O atom is assigned −2.
This means the Cl atom must be +7. In ClO4−, the sum of the oxidation states of
O (−2 × 4 = −8) and Cl (+7) is the charge on the ion, −1.
E xamp le 3.10
Determining Oxidation Numbers
Problem Determine the oxidation number of the indicated element in each of the following compounds or ions:
(a) aluminum in aluminum oxide, Al2O3
(b) phosphorus in phosphoric acid, H3PO4
(c) sulfur in the sulfate ion, SO42−
(d) each Cr atom in the dichromate ion, Cr2O72−
What Do You Know? Correct formulas for each species are given.
Strategy Follow the Guidelines for Assigning Oxidation Numbers, paying particular
attention to Guidelines 4, 6, and 7.
Solution
(a) Al2O3 is a neutral compound, so the sum of the oxidation numbers of all the elements in
Al2O3 must be zero. Assuming that oxygen has its usual oxidation number of −2, you can
solve the following algebraic equation for the oxidation number of aluminum.
170
Chapter 3 / Chemical Reactions
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Net charge on Al2O3 = sum of oxidation numbers for two Al atoms + three O atoms
0 = 2(x) + 3(−2) and so x = +3
The oxidation number of Al must be +3, in agreement with its position in Group
3A (13) of the periodic table.
(b) H3PO4 has an overall charge of 0. If each of the oxygen atoms has an oxidation number of
−2 and each of the H atoms is +1, then the oxidation number of phosphorus is +5.
Net charge on H3PO4 = sum of oxidation numbers for three H atoms
+ one P atom + four O atoms
0 = 3(+1) + (x) + 4(−2) and so x = +5
(c) The sulfate ion, SO42−, has an overall charge of 2−. Oxygen is assigned its usual
oxidation number of −2, and so sulfur in this ion has an oxidation number of +6.
Net charge on SO42− = sum of oxidation number of one S atom + four O atoms
−2 = (x) + 4(−2) and so x = +6
(d) The net charge on the Cr 2O72− ion is 2−. Oxygen is assigned its usual oxidation
number of −2.
Net charge on Cr2O72− = sum of oxidation numbers for two Cr atoms + seven O atoms
−2 = 2(x) + 7(−2) and so x = +6
The oxidation number of each chromium in this polyatomic ion is +6.
Think about Your Answer In each of these examples, the oxidation number of
Al, S, P, and Cr matched the number of the periodic group (using the A and B numbering
system) in which the element is found. This is often (but not always) the case. For example,
S, P, and Cr have a range of oxidation numbers, depending on the compound.
Check Your Understanding
Assign an oxidation number to the underlined atom in each ion or molecule.
(a) Fe2O3
(b) H2SO4
(c) CO32−
(d) NO2+
Recognizing Oxidation–Reduction Reactions
A Closer Look
It is always possible to tell whether a reaction involves oxidation and reduction by
assessing the oxidation number of each element and noting whether any of these
numbers change in the course of the reaction. In many cases, however, this will not
be necessary. For example, it will always be true that a redox reaction has occurred if
an uncombined element is converted to a compound or if a well-known oxidizing
or reducing agent is involved.
Are Oxidation Numbers Real?
Do oxidation numbers reflect the
actual electric charge on an atom in
a molecule or ion? With the exception of monatomic ions such as Cl− or
Na+, the answer is generally no.
Oxidation numbers assume that all
atoms in a molecule are positive or
negative ions, which is not true. For
example, in H2O, the H atoms are not H+
ions and the O atoms are not O2− ions. This
is not to say, however, that atoms in molecules do not bear an electric charge of any
kind. Advanced calculations indicate the
O atom in water actually has a charge of
about 0.4− (or 40% of the electron charge)
and the H atoms are each about 0.2+.
So why use oxidation numbers? Oxidation
numbers provide a convenient way of dividing
up the electrons among the atoms in a molecule or polyatomic ion. Because the distribution of electrons changes in a redox reaction,
chemists use oxidation numbers to decide
whether a redox reaction has occurred and to
distinguish the oxidizing and reducing agents.
3.8 Oxidation–Reduction Reactions
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171
NO2 gas
Table 3.4
Common Oxidizing and Reducing Agents
© Charles D. Winters/Cengage
Oxidizing
Agent
Copper metal oxidized to green Cu(NO3)2
Figure 3.17 Oxidizing and
reducing agents. The reaction
of copper with nitric acid.
Copper (a reducing agent) reacts
vigorously with concentrated nitric
acid (an oxidizing agent) to give
the brown gas NO2 and a deep
blue-green solution of copper(II)
nitrate.
Reaction Product
Reducing
Agent
Reaction Product
O2, oxygen
O2−, oxide ion or O
combined in H2O or
other molecule
H2, hydrogen
H+(aq), hydrogen ion
or H combined in H2O
or other molecule
Halogen, F2, Cl2,
Br2, or I2
Halide ion, F−, Cl−,
Br−, or I−
M, metals such as
Na, K, Fe, and Al
Mn+, metal ions such
as Na+, K+, Fe2+ or
Fe3+, and Al3+
HNO3, nitric acid
Nitrogen oxides*
such as NO and NO2
C, carbon (used
to reduce metal
oxides)
CO and CO2
Cr2O72−,
­dichromate ion
Cr3+, chromium(III)
ion (in acid solution)
MnO4−,
permanganate ion
Mn2+, manganese(II)
ion (in acid solution)
*NO is produced with dilute HNO3, whereas NO2 is a product of concentrated acid.
The halogens and oxygen (see Figures 3.1 and 3.2) are oxidizing agents, and
another common oxidizing agent is nitric acid, HNO3. In Figure 3.17 copper metal
is oxidized by the acid to give copper(II) nitrate, and the nitrate ion is reduced to
the brown gas NO2. The net ionic equation for the reaction is
Oxidation number of Cu changes from 0 to +2. Cu
is oxidized to Cu2+ and is the reducing agent.
Cu(s) + 2 NO3−(aq) + 4 H3O+(aq)
Cu2+(aq) + 2 NO2(g) + 6 H2O(𝓵)
Oxidation number of N changes from +5 in NO3− to +4 in NO2.
NO3– is reduced to NO2 and is the oxidizing agent.
Nitrogen has been reduced from +5 (in the NO3− ion) to +4 (in NO2); therefore,
the nitrate ion in acid solution is the oxidizing agent. Copper metal is the reducing
agent; each metal atom has given up two electrons to produce the Cu2+ ion.
Tables 3.4 and 3.5 may help you organize your thinking as you look for
oxidation–reduction reactions and use their terminology.
Table 3.5
172
Recognizing Oxidation–Reduction Reactions
Oxidation
Reduction
In terms of
oxidation number
Increase in oxidation
number of an atom
Decrease in oxidation number of
an atom
In terms of electrons
Loss of electrons by an atom
Gain of electrons by an atom
In terms of oxygen
Gain of one or more O atoms
Loss of one or more O atoms
Chapter 3 / Chemical Reactions
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Ex am p le 3.11
Oxidation–Reduction Reaction
5 Fe2+(aq) + MnO4−(aq) + 8 H3O+(aq) n 5 Fe3+(aq) + Mn2+(aq) + 12 H2O(ℓ)
decide which atoms are undergoing a change in oxidation number and identify the
oxidizing and reducing agents.
What Do You Know? The equation given here is balanced (but, for practice, you
KMnO4(aq)
oxidizing
agent
© Charles D. Winters/Cengage
Problem For the reaction of the iron(II) ion with permanganate ion in aqueous acid,
might verify this). The reactants and products indicate that this is an oxidation–reduction
reaction. Permanganate ion, MnO4−, is a common oxidizing agent (see Table 3.4), and iron
changes from Fe2+ to Fe3+.
Fe2+(aq)
reducing
agent
Oxidizing and reducing agents.
The reaction of iron(II) ion and
permanganate ion. The reaction
of purple permanganate ion
(MnO4−) with the iron(II) ion
(Fe2+) in acidified aqueous
solution gives the nearly colorless
manganese(II) ion (Mn2+) and the
iron(III) ion (Fe3+).
Strategy Determine the oxidation number of the atoms in each molecule or ion in the
equation and identify which atoms change oxidation number.
Solution The oxidation numbers for all of the atoms involved in this reaction were
determined and are shown below the balanced equation that follows. The Mn oxidation
number in MnO4− is +7, and it decreases to +2 in the product, the Mn2+ ion. Thus, the
MnO4− ion has been reduced and is the oxidizing agent (see Table 3.4).
5 Fe2+(aq) + MnO4−(aq) + 8 H3O+(aq) n 5 Fe3+(aq) + Mn2+(aq) + 12 H2O(ℓ)
+2
+7, −2
+1, −2
+3
+2
+1, −2
The oxidation number of iron has increased from +2 to +3, so each Fe2+ ion has lost one electron upon being oxidized to Fe3+ (see Table 3.5). This means the Fe2+ ion is the reducing agent.
Think about Your Answer Once a species has been established as having been
reduced (or oxidized), you know another species has undergone the opposite process.
Check Your Understanding
The following reaction is used in a device for testing the breath of a person for the presence of ethanol. Identify the oxidizing and reducing agents, the substance oxidized, and
the substance reduced.
3 CH3CH2OH(aq) + 2 Cr2O72−(aq) + 16 H3O+(aq) n 3 CH3CO2H(aq) + 4 Cr3+(aq) + 27 H2O(ℓ)
dichromate ion;
orange-red
acetic acid
chromium(III)
ion; green
3.9 Classifying Reactions in
Aqueous Solution
Goals for Section 3.9
• Identify reactions in aqueous solution as being either precipitation, acid–base
(including gas-forming acid–base), or oxidation–reduction reactions.
• Predict products for precipitation and acid–base (including gas-forming acid–base)
reactions and write balanced chemical equations and net ionic equations for these
reactions.
One goal of this chapter has been to explore common types of reactions that
occur in aqueous solution. This helps you decide, for example, that a gas-forming
acid–base reaction occurs when an Alka-Seltzer tablet (containing citric acid and
NaHCO3) is dropped into water (Figure 3.18).
© Charles D. Winters/Cengage
ethanol
Figure 3.18 A gas-forming
acid–base reaction. An AlkaSeltzer tablet contains an acid
(citric acid) and sodium hydrogen
carbonate (NaHCO3), the
reactants in a gas-forming acid–
base reaction.
3.9 Classifying Reactions in Aqueous Solution
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173
H3C6H5O7(aq)
citric acid
+
HCO3−(aq)
hydrogen carbonate ion
H2C6H5O7−(aq) + H2O(ℓ) + CO2(g)
dihydrogen citrate ion
Of the three types of reactions in aqueous solution studied in this chapter, two
of the three (precipitation and acid–base) fall into the category of exchange reactions.
Precipitation Reactions: Ions combine in solution to form an insoluble reaction
product.
Overall Equation
Pb(NO3)2(aq) + 2 KI(aq) n PbI2(s) + 2 KNO3(aq)
Net Ionic Equation
Pb2+(aq) + 2 I−(aq) n PbI2(s)
Acid–Base Reactions: Water is a product of many acid–base reactions, and the
cation of the base and the anion of the acid form a salt.
Overall Equation for the Reaction of a Strong Acid and a Strong Base
HNO3(aq) + KOH(aq) n HOH(ℓ) + KNO3(aq)
Net Ionic Equation for the Reaction of a Strong Acid and a Strong Base
H3O+(aq) + OH−(aq) n 2 H2O(ℓ)
Overall Equation for the Reaction of a Weak Acid and a Strong Base
Alternative Organizations of Reaction Types
Chemistry is about the transformation
of one or more substances into other
substances. This is done by chemical
reactions, and thousands upon thousands of reactions have been carried
out by chemists. Although students
beginning their study of chemistry
can be bewildered by the apparent
infinite variety of these reactions,
there are some common reaction types.
We have classified them as o­ xidation–
reduction reactions and exchange reactions. The latter include precipitation and
acid–base reactions. Classifying reactions
is useful because it helps to see their
common features and to predict what
might happen if you see a new set of
reactants.
There are other ways of classifying reactions. Here are four that are commonly
used.
Synthesis describes the preparation of
a compound from elements or other compounds. You have already seen synthesis
reactions such as the preparation of ammonium chloride, which is widely used in
174
fertilizers, in medicines, in consumer products such as shampoo, and in explosives.
The synthesis of ammonium chloride can
be carried out using an acid–base reaction.
NH3(aq) + HCl(aq) n NH4Cl(aq)
Decomposition describes a reaction in
which a compound is broken apart into
smaller constituents. One such reaction is
the decomposition of hydrogen peroxide to
water and oxygen, an oxidation–­reduction
reaction seen in the figure.
2 H2O2(aq) n 2 H2O(ℓ) + O2(g)
Double displacement describes a reaction in which cations and anions in two
compounds exchange places. At least
one of the products will be a precipitate
or a molecular compound. For example,
the reaction of NaF(aq) with a strong
acid, such as HCl(aq), produces the weak
acid HF.
NaF(aq) + HCl(aq) n HF(aq) + NaCl(aq)
You might recognize this type of reaction
as an exchange reaction.
© Charles D. Winters/Cengage
A Closer Look
CH3CO2H(aq) + NaOH(aq) n NaCH3CO2(aq) + HOH(ℓ)
The decomposition of hydrogen peroxide,
H2O2. This can also be classified as an
oxidation–reduction reaction.
Single displacement describes a reaction in which one element is substituted
for another element in a compound. An
example is the oxidation–reduction reaction of CuCl2(aq) with metallic zinc.
CuCl2(aq) + Zn(s) n Cu(s) + ZnCl2(aq)
Chapter 3 / Chemical Reactions
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Net Ionic Equation for the Reaction of a Weak Acid and a Strong Base
CH3CO2H(aq) + OH−(aq) n CH3CO2−(aq) + H2O(ℓ)
Gas-Forming Acid–Base Reactions: The most common examples involve metal
­carbonates and acids but others exist. The reaction of a metal carbonate and acid
produces carbonic acid, H2CO 3, which decomposes to H2O and CO 2. Carbon
­dioxide is the gas in the bubbles you see during these reactions.
Overall Equation: CuCO3(s) + 2 HNO3(aq) n Cu(NO3)2(aq) + CO2(g) + H2O(ℓ)
Net Ionic Equation: CuCO3(s) + 2 H3O+(aq) n Cu2+(aq) + CO2(g) + 3 H2O(ℓ)
Oxidation–Reduction Reactions: These reactions are not ion-exchange reactions.
Rather, electrons are transferred from one material to another.
Overall Equation: Cu(s) + 2 AgNO3(aq) n Cu(NO3)2(aq) + 2 Ag(s)
Net Ionic Equation: Cu(s) + 2 Ag+(aq) n Cu2+(aq) + 2 Ag(s)
These reaction types are most often unique, but keep in mind that a reaction
can fall into more than one category. For example, barium hydroxide reacts readily
with sulfuric acid to give barium sulfate and water, a reaction that is both a precipitation and an acid–base reaction.
Ba(OH)2(aq) + H2SO4(aq) n BaSO4(s) + 2 H2O(ℓ)
Ex am p le 3.12
Types of Reactions
Problem Complete and balance each of the following equations for these exchange reactions and classify each as a precipitation, acid–base, or gas-forming acid–base reaction.
(a) Na2S(aq) + Cu(NO3)2(aq) n
(b) Na2SO3(aq) + HCl(aq) n
(c) HClO4(aq) + NaOH(aq) n
What Do You Know? You know the formulas and states for the reactants and that
these are all exchange reactions.
Strategy
Step 1. Recognize that these are exchange reactions. The products of each reaction can
be found by exchanging cations and anions between the two reactants.
Step 2. Determine if each product is solid, liquid, gas, or dissolved in water.
Step 3. Determine the reaction type. Look specifically for common acids and bases, for
a product that is insoluble in water, and for ions that react with acid or base to give a gas
(CO32−, S2−, SO32−, NH4+).
Step 4. Write the balanced chemical equation.
Solution
(a)
Step 1. Determine the products of the exchange reaction.
Na2S dissociates in water into Na+ and S2− ions; Cu(NO3)2 dissociates to Cu2+ and NO3−
ions. The products of the exchange reaction are NaNO3 and CuS.
3.9 Classifying Reactions in Aqueous Solution
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175
Step 2. Determine if each product is solid, liquid, gas, or dissolved in water.
NaNO3 is water-soluble. CuS is insoluble in water.
Step 3. Determine the reaction type.
The exchange reaction produces solid CuS. It is a precipitation reaction.
Step 4. Write the balanced chemical equation.
Na2S(aq) + Cu(NO3)2(aq) n 2 NaNO3(aq) + CuS(s)
(b)
Step 1. Determine the products of the exchange reaction.
Na2SO3 dissociates in water into Na+ and SO32− ions, and HCl ionizes to H+ (or H3O+)
and Cl− ions. The products of the exchange reaction are NaCl and H2SO3. The H2SO3 will
decompose into SO2 gas and H2O (see Table 3.3).
Step 2. Determine if each product is solid, liquid, gas, or dissolved in water.
NaCl is water-soluble; SO2 is a gas, and H2O is a liquid.
Step 3. Determine the reaction type.
The reactants are a strong acid (HCl) and an ionic compound containing a basic anion
(SO32−). The decomposition of H2SO3 into SO2 gas and H2O alerts you to the fact that this is
a gas-forming acid–base reaction.
Step 4. Write the balanced chemical equation.
Na2SO3(aq) + 2 HCl(aq) n 2 NaCl(aq) + SO2(g) + H2O(ℓ)
(c)
Step 1. Determine the products of the exchange reaction.
HClO4 is a strong acid that ionizes in water into H+ (or H3O+) and ClO4− ions; NaOH
dissociates into Na+ and OH− ions. The products of the exchange reaction are NaClO4 and
H2O.
Step 2. Determine if each product is solid, liquid, gas, or dissolved in water.
NaClO4 is water soluble. H2O is a liquid.
Step 3. Determine the reaction type.
HCl is a strong acid, and NaOH is a strong base. The reaction produces a water-soluble
ionic compound (NaClO4) and water. This is an acid–base reaction.
Step 4. Write the balanced chemical equation.
HClO4(aq) + NaOH(aq) n NaClO4(aq) + H2O(ℓ)
Think about Your Answer As practice, try writing the net ionic equations for
each of the preceding reactions. The answers are:
(a) S2−(aq) + Cu2+(aq) n CuS(s)
(b) SO32−(aq) + 2 H3O+(aq) n 3 H2O(ℓ) + SO2(g)
(c) H3O+(aq) + OH−(aq) n 2 H2O(ℓ)
Check Your Understanding
Classify each of the following reactions as a precipitation, acid–base, gas-forming acid–
base, or oxidation–reduction reaction. Predict the products of the reaction, and then balance the completed equation. Write the net ionic equation for each.
(a) CuCO3(s) + H2SO4(aq) n
(b) Ga(s) + O2(g) n
(c) Ba(OH)2(s) + HNO3(aq) n
(d) CuCl2(aq) + (NH4)2S(aq) n
176
Chapter 3 / Chemical Reactions
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Applying Chemical Principles
In 1987, the Nobel Prize in Physics was awarded to Georg
Bednorz and Karl Müller (IBM Labs, Zurich, Switzerland)
for their pioneering work in superconductivity, including
their discovery of a new class of superconductors based on a
compound containing lanthanum, barium, copper, and oxygen. The superconductor was identified as La 2−xBa xCuO 4,
where the value of x varies from 0.10 to 0.20. In the same
year, researchers at the University of Alabama at Huntsville
synthesized YBa2Cu3O7−x (or YBCO), where x varies from 0
to 0.50. YBCO was the first material discovered to superconduct at temperatures above the boiling point of liquid
nitrogen (77 K). Further research determined that the critical temperature for superconductivity of YBCO varies with
changes to the ratios of its components. The highest superconducting temperature, 95 K, is found for YBa2Cu3O6.93.
Superconductors are important because these materials have no resistance to the flow of electric current. Once
a current (that is, a flow of electrons) is induced in a superconductor, it will continue indefinitely with no energy loss. A
potential application for superconductors is electricity storage,
an increasingly important need in society. Currently, electricity
produced at power plants must be used as it is produced. If
unused electricity could be fed into a superconducting storage
ring, the current could be stored indefinitely. Unfortunately,
no superconductor has been discovered that can carry large
currents at temperatures greater than 77 K.
Compound formulas containing subscripts that are not
whole numbers are common for a variety of compounds, including high-temperature superconductors. Answer the following
questions concerning a couple of these superconductors.
Questions
1. Use the following mass percentages to determine the
value of x in a sample of La 2−xBa xCuO 4: %La = 63.43,
%Ba = 5.085, %Cu = 15.69, and %O = 15.80.
2. What is the percent by mass of each element in YBCO when
x = 0.07?
3. Assuming the charges on the yttrium and barium ions are
3+ and 2+, respectively, what charges are present on the
copper ions in YBa 2Cu3O7? (Note: Although most copper
Phil Degginger/Science Source
3.1 Superconductors
Superconductivity. When a superconducting material is cooled
to a low temperature, say in liquid nitrogen (boiling point 77 K),
and is placed in a magnetic field, the field does not penetrate
the super­conductor but rather is expelled from the superconductor. The effect is seen here where a magnet floats above cooled
super­conductors.
compounds are based on Cu2+, charges of 1+ and 3+ are
also possible. Assume at least one copper ion in this substance is Cu2+.)
4. The reaction of Y2O3, BaCO3, and CuO produces YBa2Cu3O7−x
with CO2(g) as the only by-product. Write a balanced chemical
equation of this reaction and determine the value of x.
5. The percentage of oxygen in YBCO is adjusted by heating it
in the presence of elemental oxygen. What mass of oxygen
is required to convert 1.00 g YBa2Cu3O6.50 to YBa2Cu3O6.93?
References
1. M. K. Wu, et al., Phys. Rev. Lett., 1987, 58, 908–910.
2. J. W. Cochrane and G. J. Russell, Supercond. Sci. Technol.,
1998, 11, 1105–1111.
3.2 Sequestering Carbon Dioxide
Scientific evidence strongly indicates that the rising concentration of CO2 in the atmosphere contributes to warming of our
planet. These concerns have led to a variety of studies that
attempt to limit the CO2 entering the atmosphere. The best
known of these involves pumping CO2 into deep wells underground. However, there are concerns about how well CO2 can
be stored this way; in particular, there is concern that this gas
will escape containment and leak to the surface.
In 2016, the results of a study conducted in Iceland to
address this problem of leakage from stored CO 2 were published. Here CO2 was dissolved in water. Then a chemical to
serve as an indicator was added, and the solution was pumped
2000 meters underground into the basalt rock strata that
underlies most of Iceland. Basalt is an igneous aluminosilicate rock, widely distributed on Earth. It consists of a matrix
of aluminum and silicon oxides in which metal ions including
Applying Chemical Principles
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177
Questions
1. Write a balanced net ionic equation for the reaction of Ca2+
ion with H2CO3.
2. One of the indicators used was CO2 labeled with the radioactive carbon isotope carbon-14. The researchers detected
Arka38/Shutterstock.com
Ca2+ are dispersed. It is somewhat porous so that water under
high pressure can be forced into the rock.
Surrounding the CO2 injection site are eight monitoring
holes 150–1300 meters deep. The injected water-CO2-­indicator
mixture slowly diffuses through the basalt and after about 60
days reaches the monitoring holes. Researchers have followed
changes in dissolved carbon (CO2) and the acidity in the mixture over time, observing an initial significant increase in CO2
and acidity, both of which quickly diminished as the flow continued. Tests of the material from the monitoring hole showed
that most of the CO2 was not being released to the atmosphere
but instead was being converted to calcite, CaCO3, within the
rock formation. The hydrogen ions, presumably, remained in
the basalt lattice.
The significance of this is that the CO2 now is bound up in
a stable solid material and no longer free to escape confinement. Will this idea catch on and be used on a large scale?
This has yet to be determined, but a new plant using similar technology was opened in Iceland in 2021 that should be
able to remove 4000 metric tons of carbon dioxide from the
atmosphere per year.
Basalt, an igneous rock found in volcanic regions. A project in
Iceland has found that the mineral is effective in sequestering CO2.
that H 2CO 3 was moving through the rock matrix by measuring the radioactivity of the water at the detection well.
Give the number of protons, electrons, and neutrons in a
carbon-14 atom.
References
1. www.smithsonianmag.com/smart-news/worlds-largest-carbon
-capture-plant-opens-iceland-180978620/
2. www.washingtonpost.com/climate-solutions/2021/09/08/co2
-capture-plan-iceland-climeworks/
In 1977, scientists were exploring the junction of two of the
tectonic plates that form the floor of the Pacific Ocean. There
they found thermal springs gushing a hot, black soup of
minerals. Seawater seeps into cracks in the ocean floor and,
as it sinks deeper into Earth’s crust, the water is superheated
to between 300 °C and 400 °C by the hot magma just below
Earth’s crust. This superhot water dissolves minerals in the crust
and is pushed back to the surface. When this hot water, now
laden with dissolved metal cations and rich in anions such as sulfide and sulfate, gushes through the surface, it cools, and metal
sulfates—such as calcium sulfate—and metal sulfides—such as
those of copper, manganese, iron, zinc, and nickel—precipitate.
Many metal sulfides are black, and the plume of material coming from the sea bottom looks like black smoke; thus, the vents
have been called black smokers. The solid sulfides and other
minerals settle around the edges of the vent on the sea floor and
eventually form a chimney of precipitated minerals. You can see
the same deposits of metal sulfides around steam vents from
volcanoes on Earth’s surface and in the water flowing away from
a volcano or steam vent.
Scientists were amazed to discover that the deep sea
vents were surrounded by peculiar animals living in the hot,
­sulfide-rich environment. Because black smokers are under
hundreds of feet of water and sunlight does not penetrate to
these depths, the animals have developed a way to live without
the energy from sunlight. In a terrestrial environment, plants
178
Ralph White/Corbis Documentary
/Getty Images
3.3 Black Smokers and Volcanoes
Metal sulfides from a black smoker. A black smoker
photographed deep in the Pacific Ocean along the East Pacific
Rise. The smoke is a cloud of insoluble metal sulfides formed
when the molten material is forced from the Earth’s interior.
use the energy of the sun to synthesize organic molecules by
the process of photosynthesis. In the lightless ecosystem deep
in the ocean, energy is derived from the oxidation of sulfides.
With this source of energy, microbes are able to make the
organic molecules that are the basis of life.
Questions
1. When the superheated water that gushes from vents in the
sea floor cools, compounds such as CaSO4, MnS, FeS, and
NiS precipitate from solution. What are the formulas and
names for the ions making up these compounds?
2. The oxidation of sulfide ion to sulfate ion by oxygen can be
carried out in the lab. What are the oxidation numbers of
sulfur in these two ions?
Chapter 3 / Chemical Reactions
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Think–Pair–Share
1. Answer the following questions about each of the following
water-soluble compounds: acetone (C3H6O), ammonia, barium bromide, barium hydroxide, nitric acid, and nitrous acid.
(a) Is the compound a strong electrolyte, weak electrolyte, or
nonelectrolyte?
(b) What type of species will the compound produce when
dissolved in water: ions, molecules, or predominantly molecules with some ions?
(c) Imagine that 1 mol of the compound appears as a reactant
in a balanced equation for a reaction in aqueous solution.
How should it be represented in the complete ionic equation for the reaction? [For example, the compound sodium
chloride (NaCl) would be represented as Na+(aq) + Cl−(aq).]
2. The ionization of hydroiodic acid in water can be represented
by the following chemical equation:
HI(aq)  H2O(ℓ) n H3O(aq)  I(aq)
(a) Based on this equation, is HI an Arrhenius acid or base? Explain.
(b) Is HI acting as a Brønsted-Lowry acid or base in this reaction? Explain.
(c) Is H2O acting as a Brønsted-Lowry acid or base in this reaction? Explain.
3. Propose a precipitation reaction to form lead(II) bromide.
(a) Write the reaction’s overall chemical equation.
(b) Write the reaction’s complete ionic equation.
(c) Write the reaction’s net ionic equation.
(d) Compare your answers to parts a–c with those of some
classmates. Do they have the same reaction and equations
as you do? How are they similar or different?
4. There are often multiple ways to prepare a particular
compound.
(a) Develop a laboratory procedure to obtain solid copper(II)
chloride beginning with one of the following reactants:
(i) copper(II) sulfide, (ii) copper(II) sulfate, or (iii) copper(II)
hydroxide. Be specific in your description of the procedure
and write the overall balanced equation for the chemical
reaction you propose.
(b) Compare your proposed experimental procedure and
chemical reaction with that of a student who chose a
­different reactant.
(i) Check each other’s work and make any modifications
needed.
(ii) Discuss how your procedures and reactions are similar
and different.
5. Zinc metal reacts with hydrochloric acid in an oxidation–
reduction reaction to form aqueous zinc chloride and hydrogen gas.
(a) Write a balanced chemical equation for this reaction.
(b) Which reactant is:
(i) oxidized?
(ii) reduced?
(iii) the oxidizing agent?
(iv) the reducing agent?
(c) Explain why this reaction is not an acid–base reaction or a
gas-forming acid–base reaction.
Chapter Goals Revisited
The goals for this chapter are keyed to specific Study Questions to help you organize
your review.
3.1 Introduction to Chemical Equations
•
Understand the information conveyed by a balanced chemical equation
including the terminology used (reactants, products, stoichiometry,
stoichiometric coefficients). 1, 2.
•
Recognize that a balanced chemical equation is required by the law of
conservation of matter. 3–6.
3.2 Balancing Chemical Equations
•
Balance simple chemical equations. 7–12, 77, 78.
3.3 Introduction to Chemical Equilibrium
•
Recognize that all chemical reactions are reversible and that reactions
eventually reach a dynamic equilibrium. 13, 14, 98.
•
Recognize the difference between reactant-favored and product-favored
reactions at equilibrium. 15, 16.
Chapter Goals Revisited
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
179
3.4 Aqueous Solutions
•
•
Explain the difference between electrolytes and nonelectrolytes and
recognize examples of each. 17, 18, 94, 95.
Predict the solubility of ionic compounds in water (Figure 3.10). 19, 20, 23,
24, 79, 80.
3.5 Precipitation Reactions
•
Recognize what ions are formed when an ionic compound dissolves in
water. 21, 22.
•
Recognize exchange reactions in which there is an exchange of anions
between the cations of reactants in solution. 27, 28, 67, 68.
•
Predict the products of precipitation reactions. 27, 28, 92.
•
Write net ionic equations for reactions in aqueous solution. 53–56.
3.6 Acids and Bases
•
Know the names and formulas of common acids and bases and categorize
them as strong or weak. 29, 30.
•
Define the Arrhenius and Brønsted-Lowry concepts of acids and bases. 45–48.
•
Recognize substances that are amphiprotic and oxides that dissolve in
water to give acidic solutions and basic solutions. 37, 39, 43.
3.7 Acid–Base Reactions
•
Identify the Brønsted-Lowry acid and base in a reaction and write
equations for Brønsted-Lowry acid–base reactions. 41, 42.
•
Identify common acid–base reactions in which a gas is formed and write
equations for these reactions (Table 3.3). 49–52.
3.8 Oxidation–Reduction Reactions
• Determine oxidation numbers of elements in a compound and understand that
these numbers represent the charge an atom has, or appears to have, when the
electrons of the compound are counted according to a set of guidelines. 61, 62.
• Recognize common oxidizing and reducing agents. 65, 66.
• Identify oxidation–reduction reactions (redox reactions), identify the
oxidizing and reducing agents and the substances oxidized and reduced in
the reaction (Tables 3.4 and 3.5). 63–66, 85, 101.
3.9 Classifying Reactions in Aqueous Solution
• Identify reactions in aqueous solution as being either precipitation, acid–base
(including gas-forming acid–base), or oxidation–reduction reactions. 67–76.
Reaction Type
Key Characteristic
Precipitation
Formation of an insoluble compound
Acid–base
Transfer of H+ (often producing a salt and water)
Gas-forming
acid–base
Evolution of a water-insoluble gas such as CO2
Oxidation–reduction
Transfer of electrons (with changes in oxidation numbers)
• Predict products for precipitation and acid–base (including gas-forming
180
acid–base) reactions and write balanced chemical equations and net ionic
equations for these reactions. 69–72.
Chapter 3 / Chemical Reactions
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Study Questions
▲ denotes challenging questions. Blue-numbered questions have answers in Appendix N and fully
worked ­solutions in the Student Solutions Manual.
Practicing Skills
Introduction to Chemical Equations
(See Section 3.1.)
© Charles D. Winters/Cengage
1. The equation for the oxidation of phosphorus in air
is P4(s) + 5 O2(g) n P4O10(s). Identify the reactants
and products and the stoichiometric coefficients. To
what do the designations s and g refer?
2. Write an equation from the following description:
reactants are gaseous NH3 and O2, products
are gaseous NO2 and liquid H2O, and the
stoichiometric coefficients are 4, 7, 4, and 6,
respectively.
3. The equation for the reaction of phosphorus and
chlorine is P4(s) + 6 Cl2(g) n 4 PCl3(ℓ). If you use
8000 molecules of P4 in this reaction how many
molecules of Cl2 are required to consume the P4
completely?
4. The equation for the reaction of aluminum and
bromine is 2 Al(s) + 3 Br2(ℓ) n Al2Br6(s). If you
use 6.0 × 1023 molecules of Br2 in a reaction how
many atoms of Al will be consumed?
5. Oxidation of 1.00 g of carbon monoxide, CO,
produces 1.57 g of carbon dioxide, CO2. How
many grams of oxygen were required in this
reaction?
6. A 0.20 mol sample of magnesium burns in air
to form 0.20 mol of solid MgO. What amount
(moles) of oxygen (O2) is required for a complete
reaction?
Balancing Equations
(See Example 3.1.)
7. Write balanced chemical equations for the following reactions.
(a) The reaction of aluminum and iron(III) oxide
to form iron and aluminum oxide (known as
the thermite reaction).
(b) The reaction of carbon and water at high temperature to form a mixture of gaseous CO and
H2 (known as water gas and once used as a
fuel).
(c) The reaction of liquid silicon tetrachloride and
magnesium forming silicon and magnesium
chloride. This is one step in the preparation of
ultrapure silicon used in the semiconductor
industry.
Thermite reaction
8. Write balanced chemical equations for the
following reactions:
(a) production of ammonia, NH3(g), by combining N2(g) and H2(g)
(b) production of methanol, CH3OH(ℓ) by combining H2(g) and CO(g)
(c) production of sulfuric acid by combining
sulfur, oxygen, and water
9. Balance the following equations:
(a) Cr(s) + O2(g) n Cr2O3(s)
(b) Cu2S(s) + O2(g) n Cu(s) + SO2(g)
(c) C6H5CH3(ℓ)+ O2(g) n H2O(ℓ) + CO2(g)
10. Balance the following equations:
(a) Cr(s) + Cl2(g) n CrCl3(s)
(b) SiO2(s) + C(s) n Si(s) + CO(g)
(c) Fe(s) + H2O(g) n Fe3O4(s) + H2(g)
11. Balance the following equations, and name each
reactant and product:
(a) Fe2O3(s) + Mg(s) n MgO(s) + Fe(s)
(b) AlCl3(s) + NaOH(aq) n
Al(OH)3(s) + NaCl(aq)
(c) Ba(NO3)2(aq) + H2SO4(aq) n
BaSO4(s) + HNO3(aq)
(d) NiCO3(s) + HNO3(aq) n
Ni(NO3)2(aq) + CO2(g) + H2O(ℓ)
12. Balance the following equations, and name each
reactant and product:
(a) SF4(g) + H2O(ℓ) n SO2(g) + HF(aq)
(b) NH3(aq) + O2(aq) n NO(g) + H2O(ℓ)
(c) BF3(g) + H2O(ℓ) n HF(aq) + H3BO3(aq)
Study Questions
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
181
Chemical Equilibrium
(See Section 3.3.)
13. Identify each of the following statements as either
true or false. Explain why the false statements are
incorrect.
(a) At equilibrium the rates of the forward and
reverse reactions are equal.
(b) When a reaction reaches equilibrium the
forward and reverse reactions cease to occur.
(c) Chemical reactions always proceed toward
equilibrium.
14. Identify each of the following statements as either
true or false. Explain why the false statements are
incorrect.
(a) All chemical reactions are product-favored at
equilibrium.
(b) There is no observable change in a chemical
system at equilibrium.
(c) An equilibrium involving a weak acid in water
is product-favored.
15. Equal amounts of two weak acids—CH3CO2H
(acetic acid) and HCO2H (formic acid)—are placed
in aqueous solution. When equilibrium has been
achieved, the HCO2H solution has a slightly greater
electrical conductivity than the CH3CO2H solution.
Which reaction is more product-favored at equilibrium? Explain your answer.
CH3CO2H(aq) + H2O(ℓ) uv H3O+(aq) + CH3CO2−(aq)
HCO2H(aq) + H2O(ℓ) uv H3O
(aq) + HCO2−(aq)
+
16. Two aqueous solutions were prepared, one
containing 0.10 mol of boric acid (H3BO3) in
200 mL and the second containing 0.10 mol
phosphoric acid (H3PO4) in 200 mL. Both were
weak conductors of electricity, but the H3PO4
solution was a noticeably stronger conductor. Write
chemical equations to describe the equilibrium in
each solution, and explain the observed difference
in conductivity.
Ions and Molecules in Aqueous Solution
(See Section 3.4 and Example 3.2.)
17. What is an electrolyte? How can you differentiate
experimentally between a weak electrolyte
and a strong electrolyte? Give an example
of each.
18. Name and give the formulas of two acids that are
strong electrolytes and one acid that is a weak
electrolyte. Name and give formulas of two bases
that are strong electrolytes and one base that is a
weak electrolyte.
182
19. Which compound or compounds in each of the following groups is (are) soluble in water?
(a) CuO, CuCl2, FeCO3
(b) AgI, Ag3PO4, AgNO3
(c) K2CO3, KI, KMnO4
20. Which compound or compounds in each of the
following groups is (are) soluble in water?
(a) BaSO4, Ba(NO3)2, BaCO3
(b) Na2SO4, NaClO4, NaCH3CO2
(c) AgBr, KBr, Al2Br6
21. The following compounds are water-soluble. What
ions are produced by each compound in aqueous
solution?
(a) KOH
(c) LiNO3
(b) K2SO4
(d) (NH4)2SO4
22. The following compounds are water-soluble. What
ions are produced by each compound in aqueous
solution?
(a) KI
(c) K2HPO4
(d) NaCN
(b) Mg(CH3CO2)2
23. Decide whether each of the following is watersoluble. If soluble, tell what ions are produced
when the compound dissolves in water.
(a) Na2CO3
(c) NiS
(b) CuSO4
(d) BaBr2
24. Decide whether each of the following is
water-soluble. If soluble, tell what ions are
produced when the compound dissolves in water.
(a) NiCl2
(c) Pb(NO3)2
(b) Cr(NO3)3
(d) BaSO4
Precipitation Reactions and Net Ionic Equations
(See Section 3.5 and Examples 3.3 and 3.4.)
25. Balance the equation for the following precipitation reaction and then write the net ionic equation.
Indicate the state of each product (s, ℓ, aq, or g).
ZnBr2(aq) + NaOH(aq) n Zn(OH)2 + NaBr
26. Balance the equation for the following precipitation
reaction, and then write the net ionic equation.
Indicate the state of each product (s, ℓ, aq, or g).
K2CO3(aq) + Fe(ClO4)2(aq) n FeCO3 + KClO4
27. Predict the products of each precipitation reaction.
Balance the equation, and then write the net ionic
equation.
(a) NiCl2(aq) + (NH4)2S(aq) n
(b) Mn(NO3)2(aq) + Na3PO4(aq) n
Chapter 3 / Chemical Reactions
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
28. Predict the products of each precipitation reaction.
Balance the equation, and then write the net ionic
equation.
(a) Pb(NO3)2(aq) + KBr(aq) n
(b) Ca(NO3)2(aq) + KF(aq) n
(c) Ca(NO3)2(aq) + Na2C2O4(aq) n
Reactions of Acids and Bases with Water
(See Section 3.6 and Example 3.6.)
29. Name the following acids and specify whether each
is a strong or weak acid.
(a) HNO3
(c) HClO4
(b) HNO2
(d) HClO2
30. Name the following acids and specify whether each
is a strong or weak acid.
(a) H2SO3
(c) HClO3
(b) H2SO4
(d) HClO
31. Write a balanced equation for the ionization of
nitric acid in water.
32. Write a balanced equation for the ionization of
perchloric acid in water.
33. Oxalic acid, H2C2O4, which is found in certain
plants, can provide two hydronium ions in water.
Write balanced equations (like those for sulfuric
acid on page 160) to show how oxalic acid can
supply one and then a second H3O+ ion.
34. Phosphoric acid can supply one, two, or three H3O+
ions in aqueous solution. Write balanced equations
(like those for sulfuric acid on page 160) to show
this successive loss of hydrogen ions.
35. Write a balanced equation for the reaction of the
weak base, methylamine, CH3NH2, with water.
36. Formate ion, HCO2−, acts as a weak base in water.
Write a balanced chemical equation for the reaction
of formate ion with water.
37. Write a balanced equation for the reaction of the
basic oxide, magnesium oxide, with water.
42. Write an equation that describes the equilibrium
that exists when the weak acid benzoic acid
(C6H5CO2H) dissolves in water. Identify each of
the four species in solution as either Brønsted acids
or Brønsted bases. Does the equilibrium favor the
products or the reactants? (In acting as an acid, the
OCO2H group supplies H+ to form H3O+.)
43. Write two chemical equations, one that shows H2O
reacting (with HBr) as a Brønsted base and a second
that shows H2O reacting (with NH3) as a Brønsted acid.
44. Write two chemical equations, one in which H2PO4−
is a Brønsted acid (in reaction with the carbonate ion,
CO32−), and a second in which HPO42− is a Brønsted
base (in reaction with acetic acid, CH3CO2H).
Reactions of Acids and Bases
(See Section 3.7 and Examples 3.7 and 3.8)
45. Complete and balance the equations for the following
acid–base reactions. Name the reactants and products.
(a) CH3CO2H(aq) + Mg(OH)2(s) n
(b) HClO4(aq) + NH3(aq) n
46. Complete and balance the equations for the following
acid–base reactions. Name the reactants and products.
(a) H3PO4(aq) + KOH(aq) n
(b) H2C2O4(aq) + Ca(OH)2(s) n
(H2C2O4 is oxalic acid, an acid capable of
donating two H+ ions. See Study Question 33.)
47. Write a balanced equation for the reaction of
­barium hydroxide with nitric acid.
48. Write a balanced equation for the reaction of
aluminum hydroxide with sulfuric acid.
49. Siderite is a mineral consisting largely of iron(II)
carbonate. Write an overall, balanced equation for
its reaction with nitric acid, and name the products.
50. The mineral rhodochrosite is manganese(II)
carbonate. Write an overall, balanced equation for
the reaction of the mineral with hydrochloric acid,
and name the products.
38. Write a balanced equation for the reaction of the
basic oxide, lithium oxide (Li2O), with water.
© Charles D. Winters/Cengage
39. Write a balanced equation for the reaction of sulfur
trioxide gas with water.
40. Write a balanced chemical equation for the reaction
of tetraphosphorus decaoxide (P4O10) with water to
form phosphoric acid.
41. Write an equation that describes the equilibrium
that exists when nitric acid dissolves in water.
­Identify each of the four species in solution as
either Brønsted acids or Brønsted bases. Does the
equilibrium favor the products or the reactants?
Rhodochrosite, a mineral consisting largely of MnCO3
51. Write an overall, balanced equation for the reaction
of (NH4)2S with HBr, and name the reactants and
products.
Study Questions
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183
52. Write an overall, balanced equation for the reaction
of Na2SO3 with CH3CO2H, and name the reactants
and products.
Writing Net Ionic Equations
(See Examples 3.4 and 3.7.)
53. Balance each of the following equations, and then
write the net ionic equation.
(a) (NH4)2CO3(aq) + Cu(NO3)2(aq) n
CuCO3(s) + NH4NO3(aq)
(b) Pb(OH)2(s) + HCl(aq) n PbCl2(s) + H2O(ℓ)
(c) BaCO3(s) + HCl(aq) n
BaCl2(aq) + H2O(ℓ) + CO2(g)
(d) CH3CO2H(aq) + Ni(OH)2(s) n
Ni(CH3CO2)2(aq) + H2O(ℓ)
54. Balance each of the following equations, and then
write the net ionic equation:
(a) Zn(s) + HCl(aq) n H2(g) + ZnCl2(aq)
(b) Mg(OH)2(s) + HCl(aq) n
MgCl2(aq) + H2O(ℓ)
(c) HNO3(aq) + CaCO3(s) n
Ca(NO3)2(aq) + H2O(ℓ) + CO2(g)
(d) (NH4)2S(aq) + FeCl2(aq) n
NH4Cl(aq) + FeS(s)
55. Balance the following equations, and then write the
net ionic equation. Show states for all reactants and
products (s, ℓ, g, aq).
(a) the reaction of aqueous solutions of silver
nitrate and potassium iodide to give silver
iodide and potassium nitrate
(b) the reaction of aqueous solutions of barium
hydroxide and nitric acid to give barium nitrate
and water
(c) the reaction of aqueous solutions of sodium
phosphate and nickel(II) nitrate to give
nickel(II) phosphate and sodium nitrate
56. Balance each of the following equations, and then
write the net ionic equation. Show states for all
reactants and products (s, ℓ, g, aq).
(a) the reaction of aqueous solutions of sodium
hydroxide and iron(II) chloride to give iron(II)
hydroxide and sodium chloride
(b) the reaction of aqueous solutions of barium
chloride with sodium carbonate to give barium
carbonate and sodium chloride
(c) the reaction of aqueous solutions of ammonia
with phosphoric acid
184
57. Write balanced net ionic equations for the
­following reactions:
(a) the reaction of nitrous acid (a weak acid) and
sodium hydroxide in aqueous solution
(b) the reaction of calcium hydroxide and hydrochloric acid
58. Write balanced net ionic equations for the
following reactions:
(a) the reaction of aqueous solutions of silver
nitrate and sodium iodide
(b) the reaction of aqueous solutions of barium
chloride and potassium carbonate
59. Write a net ionic equation for the gas-forming acid–base
reaction of aqueous solutions of (NH4)2S and HCl.
60. Write a net ionic equation for the gas-forming acid–base
reaction of aqueous solutions of NH4NO3 and KOH.
Oxidation Numbers
(See Section 3.8 and Example 3.10.)
61. Determine the oxidation number of each element
in the following ions or compounds.
(d) CaH2
(a) BrO3−
(b) C2O42−
(e) H4SiO4
−
(c) F
(f) HSO4−
62. Determine the oxidation number of each element
in the following ions or compounds.
(a) PF6−
(d) N2O5
−
(b) H2AsO4
(e) POCl3
2+
(c) UO2
(f) XeO42−
Oxidation–Reduction Reactions
(See Section 3.8 and Example 3.11.)
63. Which two of the following reactions are
oxidation–reduction reactions? Explain your answer
in each case. Classify the remaining reaction.
(a) Zn(s) + 2 NO3−(aq) + 4 H3O+(aq) n
Zn2+(aq) + 2 NO2(g) + 6 H2O(ℓ)
(b) Zn(OH)2(s) + H2SO4(aq) n
ZnSO4(aq) + 2 H2O(ℓ)
(c) Ca(s) + 2 H2O(ℓ) n Ca(OH)2(s) + H2(g)
64. Which two of the following reactions are oxidation–
reduction reactions? Explain your answer briefly.
Classify the remaining reaction.
(a) CdCl2(aq) + Na2S(aq) n
CdS(s) + 2 NaCl(aq)
(b) 2 Ca(s) + O2(g) n 2 CaO(s)
(c) 4 Fe(OH)2(s) + 2 H2O(ℓ) + O2(g) n
4 Fe(OH)3(s)
Chapter 3 / Chemical Reactions
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65. In the following reactions, decide which reactant is
oxidized and which is reduced. Designate the oxidizing agent and the reducing agent.
(a) C2H4(g) + 3 O2(g) n 2 CO2(g) + 2 H2O(ℓ)
(b) Si(s) + 2 Cl2(g) n SiCl4(ℓ)
66. In the following reactions, decide which reactant
is oxidized and which is reduced. Designate the
oxidizing agent and the reducing agent.
(a) Cr2O72− (aq) + 3 Sn2+(aq) + 14 H3O+(aq) n
2 Cr3+(aq) + 3 Sn4+(aq) + 21 H2O(ℓ)
(b) FeS(s) + 3 NO3−(aq) + 4 H3O+(aq) n
3 NO(g) + SO42−(aq) + Fe3+(aq) + 6 H2O(ℓ)
acid–base, or oxidation–reduction reactions. Show
states for reactants and products (s, ℓ, g, aq).
(a) CuCl2 + H2S n CuS + HCl
(b) H3PO4 + KOH n H2O + K3PO4
(c) Ca + HBr n H2 + CaBr2
(d) MgCl2 + NaOH n Mg(OH)2 + NaCl
(See Section 3.9 and Example 3.12.)
72. ▲ Complete and balance the equations below, and
classify them as precipitation, acid–base, gas-forming
acid–base, or oxidation–reduction reactions. Show
states for reactants and products (s, ℓ, g, aq).
(a) NiCO3 + H2SO4 n
(b) Co(OH)2 + HBr n
(c) AgCH3CO2 + NaCl n
(d) NiO + CO n
67. Balance the following equations, and then classify
each as a precipitation, acid–base, or gas-forming
acid–base reaction.
(a) Ba(OH)2(aq) + HCl(aq) n
BaCl2(aq) + H2O(ℓ)
(b) HNO3(aq) + CoCO3(s) n
Co(NO3)2(aq) + H2O(ℓ) + CO2(g)
(c) Na3PO4(aq) + Cu(NO3)2(aq) n
Cu3(PO4)2(s) + NaNO3(aq)
73. The products formed in several reactions are given
below. Identify the reactants (labeled x and y) and
write the complete balanced equation for each
reaction.
(a) x + y n H2O(ℓ) + CaBr2(aq)
(b) x + y n Mg(NO3)2(aq) + CO2(g) + H2O(ℓ)
(c) x + y n BaSO4(s) + NaCl(aq)
(d) x + y n NH4+(aq) + OH−(aq)
68. Balance the following equations, and then classify
each as a precipitation, acid–base, or gas-forming
acid–base reaction.
(a) K2CO3(aq) + Cu(NO3)2(aq) n
CuCO3(s) + KNO3(aq)
(b) Pb(NO3)2(aq) + HCl(aq) n
PbCl2(s) + HNO3(aq)
(c) MgCO3(s) + HCl(aq) n
MgCl2(aq) + H2O(ℓ) + CO2(g)
74. The products formed in several reactions are given
below. Identify the reactants (labeled x and y) and
write the complete balanced equation for each
reaction.
(a) x + y n (NH4)2SO4(aq)
(b) x + y n CaCl2(aq) + CO2(g) + H2O(ℓ)
(c) x + y n Ba(NO3)2(aq) + AgCl(s)
(d) x + y n H3O+(aq) + ClO4−(aq)
Types of Reactions in Aqueous Solution
69. Classify each of the following reactions as a precipitation, acid–base, or gas-forming acid–base r­ eaction.
Show states for the products (s, ℓ, g, aq), and then
balance the completed equation. Write the net ionic
equation.
(a) MnCl2(aq) + Na2S(aq) n MnS + NaCl
(b) K2CO3(aq) + ZnCl2(aq) n ZnCO3 + KCl
70. Classify each of the following reactions as a
precipitation, acid–base, or gas-forming acid–base
reaction. Show states for the products (s, ℓ, g, aq),
and then balance the completed equation. Write the
net ionic equation.
(a) Fe(OH)3(s) + HNO3(aq) n Fe(NO3)3 + H2O
(b) FeCO3(s) + HNO3(aq) n
Fe(NO3)2 + CO2 + H2O
71. Balance each of the following equations, and classify them as precipitation, acid–base, gas-forming
75. The products formed in several reactions are given
below. Identify the reactants (labeled x and y)
and write a complete balanced equation for each
reaction. Then write a net ionic equation for each
reaction.
(a) x + y n H2O(ℓ) + NaNO3(aq)
(b) x + y n CaCO3(s) + NaCl(aq)
(c) x + y n Sr(NO3)2(aq) + CO2(g) + H2O(ℓ)
(d) x + y n Zn3(PO4)2(s) + NaCl(aq)
76. The products formed in several reactions are given
below. Identify the reactants (labeled x and y)
and write a complete balanced equation for each
reaction. Then write a net ionic equation for each
reaction.
(a) x + y n H2O(ℓ) + NaF(aq)
(b) x + y n SrSO4(s) + KNO3(aq)
(c) x + y n KCl(aq) + H2S(g)
(d) x + y n Fe(OH)2(s) + NaI(aq)
Study Questions
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185
General Questions
These questions are not designated as to type or location in
the chapter. They may combine concepts.
77. Balance the following equations:
(a) for the synthesis of urea, a common fertilizer
CO2(g) + NH3(g) n NH2CONH2(s) + H2O(ℓ)
(b) for the reactions used to make uranium(VI)
fluoride for the enrichment of natural
uranium
UO2(s) + HF(aq) n UF4(s) + H2O(ℓ)
UF4(s) + F2(g) n UF6(s)
(c) for the reaction to make titanium(IV) chloride,
which is then converted to titanium metal
TiO2(s) + Cl2(g) + C(s) n TiCl4(ℓ) + CO(g)
TiCl4(ℓ) + Mg(s) n Ti(s) + MgCl2(s)
78. Balance the following equations:
(a) for the reaction to produce “superphosphate”
fertilizer
Ca3(PO4)2(s) + H2SO4(aq) n
Ca(H2PO4)2(aq) + CaSO4(s)
(b) for the reaction to produce diborane, B2H6
NaBH4(s) + H2SO4(aq) n
B2H6(g) + H2(g) + Na2SO4(aq)
(c) for the reaction to produce tungsten metal
from tungsten(VI) oxide
WO3(s) + H2(g) n W(s) + H2O(ℓ)
(d) for the decomposition of ammonium
dichromate
(NH4)2Cr2O7(s) n N2(g) + H2O(ℓ) + Cr2O3(s)
79. Give a formula for each of the following
compounds:
(a) a soluble compound containing the bromide
ion
(b) an insoluble hydroxide
(c) an insoluble carbonate
(d) a soluble nitrate-containing compound
(e) a weak Brønsted acid
80. Give a formula for each of the following
compounds:
(a) a soluble compound containing the acetate ion
(b) an insoluble sulfide
(c) a soluble hydroxide
(d) an insoluble chloride
(e) a strong Brønsted base
81. Indicate which of the following copper(II) salts
are soluble in water and which are insoluble:
Cu(NO3)2, CuCO3, Cu3(PO4)2, CuCl2.
186
82. Name two anions that combine with Al3+ ion to
produce water-soluble compounds.
83. Write the net ionic equation and identify the spectator ion or ions in the reaction of nitric acid and
magnesium hydroxide. What type of reaction is this?
2 H3O+(aq) + 2 NO3−(aq) + Mg(OH)2(s) n
4 H2O(ℓ) + Mg2+(aq) + 2 NO3−(aq)
84. Identify and name the water-insoluble product in
each reaction and write the net ionic equation:
(a) CuCl2(aq) + H2S(aq) n CuS + 2 HCl
(b) CaCl2(aq) + K2CO3(aq) n 2 KCl + CaCO3
(c) AgNO3(aq) + NaI(aq) n AgI + NaNO3
85. Bromine is obtained from sea water by the following redox reaction:
Cl2(g) + 2 NaBr(aq) n 2 NaCl(aq) + Br2(ℓ)
(a) What has been oxidized? What has been
reduced?
(b) Identify the oxidizing and reducing agents.
86. Identify each of the following substances as a likely
oxidizing or reducing agent: HNO3, Na, Cl2, O2,
KMnO4.
87. The mineral dolomite contains magnesium carbonate. This reacts with hydrochloric acid.
MgCO3(s) + 2 HCl(aq) n
CO2(g) + MgCl2(aq) + H2O(ℓ)
(a) Write the net ionic equation for this reaction
and identify the spectator ions.
(b) What type of reaction is this?
88. Aqueous solutions of ammonium sulfide, (NH4)2S,
and Hg(NO3)2 react to produce HgS and NH4NO3.
(a) Write the overall, balanced equation for the
reaction. Indicate the state (s, aq) for each
compound.
(b) Name each compound.
(c) What type of reaction is this?
89. Identify the primary species (atoms, molecules, or ions)
present in an aqueous solution of each of the following
compounds. Decide which compounds are Brønsted
acids or bases and whether they are strong or weak.
(c) NaOH
(a) NH3
(b) CH3CO2H
(d) HBr
90. (a)Name and give formulas for two water-soluble
compounds containing the Cu2+ ion. Name
two water-insoluble compounds containing the
Cu2+ ion.
(b) Name and give formulas for two water-soluble
compounds containing the Ba2+ ion. Name
two water-insoluble compounds containing the
Ba2+ ion.
Chapter 3 / Chemical Reactions
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91. Balance equations for these reactions that occur in
aqueous solution, and then classify each as a precipitation, acid–base, or gas-forming acid–base reaction.
Show states for the reactants and products (s, ℓ, g, aq),
give their names, and write the net ionic equation.
(a) K2CO3 + HClO4 n KClO4 + CO2 + H2O
(b) FeCl2 + (NH4)2S n FeS + NH4Cl
(c) Fe(NO3)2 + Na2CO3(aq) n FeCO3 + NaNO3
(d) NaOH + FeCl3 n NaCl + Fe(OH)3
92. For each reaction, write an overall, balanced
equation and the net ionic equation.
(a) the reaction of aqueous lead(II) nitrate and
aqueous potassium hydroxide
(b) the reaction of aqueous copper(II) nitrate and
aqueous sodium carbonate
93. You are given mixtures containing the following compounds. Which compound in each pair would remain
largely undissolved if the mixture is stirred with water?
(a) NaOH and Ca(OH)2
(c) AgI and KI
(d) NH4Cl and PbCl2
(b) MgCl2 and MgF2
94. Identify, from each list below, the compound or
compounds that will dissolve in water to give a
solution that strongly conducts electricity.
(a) CuCO3, Cu(OH)2, CuCl2, CuO
(b) HCl, HClO, HNO3, H2SO4
(a) What is the formula of the sulfur oxide produced by the combustion of sulfur?
(b) Write a balanced chemical equation for the
combustion reaction.
(c) Write a balanced chemical equation for the
reaction of the oxide with water that produces
sulfurous acid.
100. White phosphorus, P4, spontaneously combusts,
producing a solid compound with the empirical
formula P2O5. A separate experiment determines
the molar mass of the compound as 284 g/mol. It
reacts with water, producing phosphoric acid.
(a) What is the molecular formula of the oxide of
phosphorus produced by the combustion of
phosphorus?
(b) Write a balanced chemical equation for the
combustion reaction.
(c) Write a balanced chemical equation for the
reaction of the oxide with water that produces
phosphoric acid.
In the Laboratory
101. The following reaction can be used to prepare
iodine in the laboratory.
2 NaI(s) + 2 H2SO4(aq) + MnO2(s) n
Na2SO4(aq) + MnSO4(aq) + I2(g) + 2 H2O(ℓ)
96. Write net ionic equations for the following reactions:
(a) The reaction of acetic acid, a weak acid, and
Sr(OH)2(aq).
(b) The reaction of zinc and hydrochloric acid to
form zinc(II) chloride and hydrogen gas.
97. Gas evolution was observed when a solution of Na2S
was treated with acid. The gas was bubbled into a
solution containing Pb(NO3)2, and a black precipitate
formed. Write net ionic equations for the two reactions.
98. Heating HI(g) at 425 °C causes some of this
compound to decompose, forming H2(g) and I2(g).
Eventually, the amounts of the three species do not
change further; the system has reached equilibrium.
(At this point, approximately 22% of the HI has
decomposed.) Describe what is happening in this
system at the molecular level.
99. A sample of sulfur is combusted, producing an
unknown gas, SxOy. The gas reacts with water, producing sulfurous acid.
Photos: © Charles D. Winters/Cengage
95. Identify, from each list below, the compound or compounds that will dissolve in water to give a solution
that is only a very weak conductor of electricity.
(a) NH3, NaOH, Ba(OH)2, CuSO4
(b) CH3CO2H, Na3PO4, HF, HNO3
Preparation of iodine. A mixture of NaI and MnO2 was placed
in a flask (left). On adding concentrated H2SO4 (right), I2 was
evolved as a brown vapor.
(a) Determine the oxidation number of each atom
in the equation.
(b) What is the oxidizing agent, and what has
been oxidized? What is the reducing agent,
and what has been reduced?
(c) Is the reaction product-favored or
reactant-favored?
(d) Name the reactants and products.
Study Questions
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187
vessel if a reducing sugar is present. Using glucose,
C6H12O6, to illustrate this test, the oxidation–
reduction reaction occurring is
102. ▲ Some eating utensils contain silver. Over time
they may oxidize or tarnish. This is due to sulfurcontaining compounds in the air or food, which
react with the silver and form a black coating of
Ag2S. To remove the tarnish, you can warm the tarnished object with some aluminum foil in water
with a small amount of baking soda. Silver sulfide
reacts with aluminum to produce silver as well as
aluminum oxide and hydrogen sulfide.
C6H12O6 (aq) + 2 Ag+(aq) + 2 OH−(aq) n
C6H12O7(aq) + 2 Ag(s) + H2O(ℓ)
What has been oxidized, and what has been
reduced? What is the oxidizing agent, and what is
the reducing agent?
3 Ag2S(s) + 2 Al(s) + 3 H2O(ℓ) n
6 Ag(s) + Al2O3(s) + 3 H2S(aq)
Photos: © Charles D. Winters/Cengage
Hydrogen sulfide is foul smelling, but it is removed
by reaction with the baking soda.
NaHCO3(aq) + H2S(aq) n
NaHS(aq) + H2O(ℓ) + CO2(g)
Classify the two reactions, and identify any acids,
bases, oxidizing agents, or reducing agents.
(a)
(b)
Photos: © Charles D. Winters/Cengage
Tollen’s test. The reaction of silver ions with a sugar such as
glucose produces metallic silver. (a) The set-up for the reaction.
(b) The silvered test tube.
(a)
(b)
Removing silver tarnish. (a) A badly tarnished piece of silver is
placed in a dish with aluminum foil and aqueous sodium hydrogen carbonate. (b) The portion of the silver in contact with the
solution is now free of tarnish.
103. ▲ Suppose you wish to prepare a sample of magnesium chloride. One way to do this is to use an
acid–base reaction, the reaction of magnesium
hydroxide with hydrochloric acid.
Mg(OH)2(s) + 2 HCl(aq) n MgCl2(aq) + 2 H2O(ℓ)
When the reaction is complete, evaporating the
water will give solid magnesium chloride. Suggest
another way to prepare MgCl2.
104. ▲ Suggest a laboratory method for preparing
barium phosphate.
105. The Tollen’s test for the presence of reducing sugars
(say, in a urine sample) involves treating the sample
with silver ions in aqueous ammonia. The result is
the formation of a silver mirror within the reaction
188
Summary and Conceptual Questions
The following questions may use concepts from this and previous chapters.
106. There are many ionic compounds that dissolve in
water to a very small extent. One example is lead(II)
chloride. When it dissolves, an equilibrium is established between the solid salt and its component
ions. Suppose you stir some solid PbCl2 into water.
Explain how you would prove that the compound
dissolves but only to a small extent? Is the dissolving
process product-favored or reactant-​favored?
PbCl2(s) uv Pb2+(aq) + 2 Cl−(aq)
107. ▲ Most naturally occurring acids are weak acids.
Lactic acid is one example.
CH3CH(OH)CO2H(s) + H2O(ℓ) uv
H3O+(aq) + CH3CH(OH)CO2−(aq)
If you place some lactic acid in water, it will ionize
to a small extent, and an equilibrium will be established. Suggest some experiments to prove that this
is a weak acid and that the establishment of equilibrium is a reversible process.
H
H
OH O
C
C
C
OH
H H
Lactic acid
Chapter 3 / Chemical Reactions
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
114. You have a bottle of solid barium hydroxide and
some dilute sulfuric acid. You place some of the
barium hydroxide in water and slowly add sulfuric
acid to the mixture. While adding the sulfuric acid,
you measure the conductivity of the mixture.
(a) Write the complete, balanced equation for the
reaction occurring when barium hydroxide
and sulfuric acid are mixed.
(b) Write the net ionic equation for the barium
hydroxide and sulfuric acid reaction.
(c) Which diagram represents the change in conductivity as the acid is added to the aqueous
barium hydroxide? Explain briefly.
Derive the empirical formula for the red solid
based on the following composition: Ni, 20.32%;
C, 33.26%; H, 4.88%; O, 22.15%; and N, 19.39%.
(a)
112. The lanthanide elements react with oxygen to give,
generally, compounds of the type Ln2O3 (where
Ln stands for a lanthanide element). However,
there are interesting exceptions, such as a common
oxide of terbium, TbxOy. Given that the compound
is 73.95% Tb, what is its formula? What is the
oxidation number of terbium in this compound?
Write a balanced equation for the reaction of terbium and oxygen to give this oxide.
Electrical conductivity
© Charles D. Winters/Cengage
111. A common method for analyzing for the nickel
content of a sample is to use a precipitation reaction. Adding the organic compound
­dimethylglyoxime to a solution containing Ni2+
ions precipitates a red solid.
Finally, the presence of arsenic is confirmed by
adding AgNO3 to the solution of H3AsO4 to precipitate a reddish brown solid AgxAsOy. The composition of this solid is As, 16.20% and Ag, 69.96%.
(a) What are the oxidation numbers of As, S, and
N in the reaction of As2S3 with nitric acid?
(b) What is the formula of the reddish brown
solid AgxAsOy?
(c)
Volume of added
sulfuric acid
Electrical conductivity
110. ▲ Describe how to prepare zinc chloride by (a)
an acid–base reaction, (b) a gas-forming acid–base
reaction, and (c) an oxidation–reduction reaction.
The available starting materials are ZnCO3, HCl,
Cl2, HNO3, Zn(OH)2, NaCl, Zn(NO3)2, and
Zn. Write complete, balanced equations for the
reactions chosen.
3 As2S3(s) + 10 HNO3(aq) + 4 H2O(ℓ) n
6 H3AsO4(aq) + 10 NO(g) + 9 S(s)
Volume of added
sulfuric acid
(b)
Volume of added
sulfuric acid
Electrical conductivity
109. ▲ Describe how to prepare BaSO4, barium sulfate,
by (a) a precipitation reaction and (b) a gasforming acid–base reaction. The available starting
materials are BaCl2, BaCO3, Ba(OH)2, H2SO4, and
Na2SO4. Write complete, balanced equations for
the reactions chosen. (See page 155 for an illustration of the preparation of the compound.)
113. The presence of arsenic in a sample that may
also contain another Group 5A (15) element,
antimony, can be confirmed by first precipitating
the As3+ and Sb3+ ions as yellow solid As2S3 and
orange solid Sb2S3. If aqueous HCl is then added,
only Sb2S3 dissolves, leaving behind solid As2S3.
The As2S3 can then be dissolved using aqueous
HNO3.
Electrical conductivity
108. ▲ You want to prepare barium chloride, BaCl2,
using an exchange reaction of some type. To do
so, you have the following reagents from which
to select the reactants: BaSO4, BaBr2, BaCO3,
Ba(OH)2, HCl, HgSO4, AgNO3, and HNO3. Write a
complete, balanced equation for the reaction chosen. (Note: There are several possibilities.)
Volume of added
sulfuric acid
(d)
Study Questions
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189
4
Stoichiometry: Quantitative Information
about Chemical Reactions
4 PCl3(ℓ)
© Charles D. Winters/Cengage
P4(s) + 6 Cl2(g)
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C hapt e r O ut li n e
4.1
Mass Relationships in Chemical Reactions: Stoichiometry
4.2
Reactions in Which One Reactant Is Present in Limited Supply
4.3
Percent Yield
4.4
Chemical Equations and Chemical Analysis
4.5
Measuring Concentrations of Compounds in Solution
4.6
pH, a Concentration Scale for Acids and Bases
4.7
Stoichiometry of Reactions in Aqueous Solution—Fundamentals
4.8
Stoichiometry of Reactions in Aqueous Solution—Titrations
4.9
Spectrophotometry
In the last chapter, you learned how to write balanced chemical equations to
­describe chemical reactions. This chapter shows how these equations can be used to
predict how much of a reactant is required to react with a given quantity of another
reactant, or how much of a product can be produced. These calculations form the
basis of reaction stoichiometry, which deals with the amounts of reactants and
products in a chemical reaction. Chemists use these types of calculations to help
them design experiments and evaluate their results.
4.1 Mass Relationships in Chemical
Reactions: Stoichiometry
Goals for Section 4.1
•
Understand the principle of conservation of matter, which forms the basis of
chemical stoichiometry.
•
Calculate the mass of one reactant or product in a reaction knowing the balanced
equation and the mass of another reactant or product in that reaction.
•
Use amounts tables to organize chemical information.
The reaction of elemental phosphorus and chlorine produces the compound PCl3
(chapter opening photograph), and the following balanced equation shows the
quantitative relationship between reactants and products in this reaction.
◀ Reaction of solid white phosphorus with chlorine gas to produce phosphorus
trichloride. The balanced chemical equation for this reaction is
P4(s) + 6 Cl2(g) n 4 PCl3(ℓ)
This chemical equation allows chemists to predict the mass of chlorine required to react with
a given mass of phosphorus and the mass of phosphorus trichloride produced.
191
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P4(s) + 6 Cl2(g) n 4 PCl3(ℓ)
1 mol
124 g
6 mol
425 g
4 mol
549 g
At the molecular level this balanced equation tells you that one molecule of phosphorus reacts with six molecules of chlorine to produce four molecules of phosphorus
trichloride. At the macroscopic level where reactions are observed, the coefficients refer to the number of moles of each reactant and product. For example, the equation
tells you that
1 mol (124 g) of solid phosphorus (P4) + 6 mol (425 g) chlorine gas (Cl2) n
4 mol (549 g) of liquid phosphorus trichloride (PCl3)
Now, suppose only 1.45 g of P4 is used in the reaction. (a) What mass of Cl2 gas
is required and (b) what mass of PCl3 could be produced? This is an example of a
situation common in chemistry.
Part (a): Calculate the mass of Cl2 required by 1.45 g of P4
Step 1. Write the balanced equation (using correct
f­ ormulas for reactants and products). This is always
the first step when dealing with chemical reactions.
Step 2. Calculate amount (moles) from mass (grams).
Recall that chemical equations reflect the relative
amounts of reactants and products, not their masses.
Therefore, calculate the amount (moles) of P4 available.
P4(s) + 6 Cl2(g) n 4 PCl3(ℓ)
1.45 g P4 
1 mol P4
5 0.01170 mol P4
123.9 g P4
h
1
molar mass of P4
Step 3a. Use a stoichiometric factor. Use the balanced
equation to relate the amount of P4 available to the
amount of Cl2 required to completely consume the
P4. This relationship is a stoichiometric factor, a mole
ratio based on the stoichiometric coefficients in the
balanced equation. Here the balanced equation specifies that 6 mol of Cl2 is required for each mole of P4,
so the stoichiometric factor is (6 mol Cl2/1 mol P4).
Step 4a. Calculate mass from amount. Convert
the amount (moles) of Cl 2 calculated in Step 3 to
the mass of Cl2 required.
0.01170 mol P4 
6 mol Cl 2 required
5 0.07022 mol Cl 2 required
1 mol P4 available
h
stoichiometric factor from balanced equation
0.07022 mol Cl 2 
70.90 g Cl 2
5 4.98 g Cl 2
1 mol Cl 2
h
molar mass of Cl2
Part (b): Calculate mass of PCl3 produced from 1.45 g of P4 and 4.98 g of Cl2
From part (a), you know that 1.45 g of P4 and 4.98 g of Cl2 are the correct quantities needed for
complete reaction. Because mass is always conserved, the answer can be obtained by adding the
masses of P4 and Cl2 used (1.45 g P4 + 4.98 g Cl2 = 6.43 g PCl3 produced). Alternatively, Steps 3
and 4 can be repeated, but with the appropriate stoichiometric factor and molar mass.
192
Chapter 4 / Stoichiometry: Quantitative Information about Chemical Reactions
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Step 3b. Use a stoichiometric factor. Convert the
4 mol PCl 3 produced
amount of available P4 to the amount of PCl3 pro- 0.01170 mol P4  1 mol P4 available 5 0.04681 mol PCl 3 produced
duced. Here the balanced equation specifies that 4
h
mol of PCl3 is produced for each mole of P4 used, so
the stoichiometric factor is (4 mol PCl3/1 mol P4).
stoichiometric factor from balanced equation
Step 4b. Calculate the mass of product from its
amount. Convert the amount of PCl3 produced to its
0.04681 mol PCl 3 
mass in grams.
137.3 g PCl 3
5 6.43 g PCl 3
1 mol PCl 3
The total mass of reactants consumed (1.45 g of P4 and 4.98 g of Cl2) is equal to the
total mass of products formed (6.43 g of PCl3); mass is always conserved in chemical reactions.
The mole relationships of reactants and products for a reaction can be summarized in an amounts table.
+
Equation
P4(s)
6 Cl2(g)
Initial amount (mol)
0.01170
0.07022
0
Change in amount upon reaction (mol)
−0.01170
−0.07022
+0.04681
Amount after complete reaction (mol)
0
0
0.04681
n
4 PCl3(ℓ)
Amounts Tables The mole (and
mass) relationships of reactants
and products in a reaction can
be summarized in an amounts
table.
The balanced chemical equation is written across the top of the amounts table. The
next three lines contain the following information:
•
Initial amount (moles) of each reactant and product present.
•
Change in amount that occurs during the reaction.
•
Final amount of each reactant and product present after the reaction.
The completed amounts table indicates that the reactants P4 and Cl2 were initially
present in the correct stoichiometric ratio but no PCl3 was present.
Problem Solving Tip 4.1 Stoichiometry Calculations
You are asked to determine what
mass of product can be formed from
a given mass of reactant. It is not
possible to calculate the mass of
product in a single step. Instead, you
must follow a route such as that illustrated in the strategy map here for
the reaction of reactant A to give the
product B.
1. Write the balanced equation for
the chemical reaction.
xA n yB
2. Convert the mass (g) of reactant A
to the amount (moles) of A using
the molar mass of A.
grams reactant A
×
grams product B
direct calculation
not possible
1 mol A
gA
×
moles reactant A
gB
mol B
moles product B
×
y mol product B
x mol reactant A
× stoichiometric factor
3. Next, use the stoichiometric factor
to find the amount (moles) of B.
4. Finally, obtain the mass (g) of B by
multiplying the amount of B by its
molar mass.
When solving a stoichiometry
problem, remember that you will
always use a stoichiometric factor at
some point.
4.1 Mass Relationships in Chemical Reactions: Stoichiometry
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193
Chemistry in Your Career
Dr. Jessica N. Isaac
Dr. Jessica N. Isaac
Jessica N. Isaac is a Clinical Assistant Professor at
Binghamton University’s School of Pharmacy and
Pharmaceutical Sciences, where her work is “a combination of administration, teaching, ­research, clinical pharmacy, and mentorship.” As an instructor,
one of Dr. Isaac’s goals is to help “student pharmacists develop the skills necessary to safely and
­accurately prepare medications.”
“The study of chemistry is integral to the study
of pharmacy,” explains Dr. Isaac, including “stoichiometry, dimensional analysis, acid–base properties, and physiochemical properties.” The study
of pharmacy has three core components: looking
at the chemical interactions between medications
and the chemicals in the human body; creating
and refining molecules for drug development; and
combining, mixing, or altering chemicals to create
a medication tailored to the needs of a patient.
Dr. Isaac describes how her identity as a an
African American/Jamaican American woman has
shaped her both personally and professionally. “Being
a Black woman and a first-generation student in
pharmacy school presented many obstacles; however,
these experiences also helped me to develop certain
strengths. . .including social intelligence, perseverance/
resilience, creativity, gratitude, and critical thinking.”
Exam pl e 4 .1
Mass Relations in Chemical Reactions
Strategy Map
Problem
Calculate mass of O2 required
for combustion of 25.0 g of
glucose.
Data/Information
Formulas for reactants and
products and the mass of one
reactant (glucose)
Problem Glucose, C6H12O6, reacts with oxygen to give CO2 and H2O. What mass of
oxygen (in grams) is required to completely react with 25.0 g of glucose? What masses
of ­carbon dioxide and water (in grams) are formed?
What Do You Know? You are given the mass of one of the reactants (glucose)
and are asked to determine the masses of the other substances in the reaction. You know
formulas for the reactants and products and need to calculate their molar masses.
Strategy Write the balanced chemical equation for this reaction. Then, follow the
scheme outlined in Problem Solving Tip 4.1 and in the Strategy Map for this example.
Solution
Step 1
Write the balanced equation.
C6H12O6(s) + 6 O2(g) n 6 CO2(g) + 6 H2O(ℓ)
Step 2
Convert the given mass (grams) to amount (moles).
Use the molar mass of glucose to convert its mass (25.0 g) to the amount of glucose.
25.0 g glucose 
Step 3
1 mol glucose
5 0.1387 mol glucose
180.2 g glucose
Use the stoichiometric factor to convert the amount (moles) of glucose to the
amount (moles) of O2.
Use the coefficients of the balanced equation to obtain the stoichiometric factor of 6 mol
O2 per 1 mol glucose, and then convert the amount of glucose to the amount of O2.
0.1387 mol glucose 
Step 4
6 mol O2
5 0.8324 mol O2
1 mol glucose
Convert the amount (moles) of the requested substance to its mass (grams).
Use the molar mass of O2 to convert from the amount of O2 to the mass of O2.
0.8324 mol O2 
194
32.00 g O2
5 26.6 g O2
1 mol O2
Chapter 4 / Stoichiometry: Quantitative Information about Chemical Reactions
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Repeat Steps 3 and 4 to find the mass of CO2 produced in the combustion. First, relate
the amount (moles) of glucose available to the amount of CO 2 produced using a
stoichiometric factor. Then, convert the amount of CO2 to its mass in grams.
0.1387 mol glucose 
6 mol CO2
44.01 g CO2

5 36.6 g CO2
1 mol glucose
1 mol CO2
Now, how can you find the mass of H2O produced? You could repeat Steps 3 and 4. However, the total mass of reactants
25.0 g C6H12O6 + 26.6 g O2 = 51.6 g reactants
must equal the total mass of products. The mass of water that can be produced is therefore
Total mass of products = 51.6 g = 36.6 g CO2 produced + ? g H2O
Mass of H2O produced = 15.0 g
Think about Your Answer The results of this calculation can be summarized in
an amounts table.
Equation
C6H12O6(s) +
Initial amount (mol)
0.1387
6(0.1387)
= 0.8324
0
0
Change in amount
upon reaction (mol)
−0.1387
−0.8324
+0.8324
+0.8324
0
0.8324
0.8324
Amount after complete   0
reaction (mol)
6 O2(g)
Significant Figures As discussed
in Chapter 1R, we show one
more than the number of
required significant figures in
each step until the final step
when the answer is rounded
to the correct number of
significant figures.
n 6 CO2(g) + 6 H2O(ℓ)
When you know the mass of all but one of the chemicals in a reaction, you can find the
unknown mass using the law of mass conservation (the total mass of reactants must equal
the total mass of products; page 140).
Check Your Understanding
What mass of oxygen, O2, is required to completely combust 454 g of propane, C3H8? What masses of CO2 and H2O are produced?
4.2 Reactions in Which One Reactant
Is Present in Limited Supply
•
Determine which reactant is in limited supply in a reaction involving several
reactants.
•
Determine the yield of a product based on the limiting reactant.
Balanced chemical equations give the ideal stoichiometric relationship between reactants and products. However, reactions are often conducted using more of one
reactant than is called for by an exact stoichiometric ratio. This is usually done to
make sure that one of the reactants is consumed completely, even though some of
another reactant remains unused.
Suppose you burn a toy sparkler, a wire coated with a mixture of aluminum or iron
powder and potassium chlorate (Figure 4.1). The aluminum or iron burns, consuming
oxygen from the air or from the potassium salt and producing a metal oxide.
4 Al(s) + 3 O2(g) n 2 Al2O3(s)
© Charles D. Winters/Cengage
Goals for Section 4.2
Figure 4.1 Burning aluminum
and iron powder. A toy sparkler
contains a metal powder such as
Al or Fe and other chemicals such
as KClO3. When ignited, the metal
burns with a brilliant white light.
4.2 Reactions in Which One Reactant Is Present in Limited Supply
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195
The sparkler burns until the metal powder is consumed completely. What about the
oxygen? Four moles of aluminum require three moles of oxygen, but there is much,
much more O2 available in the air than is needed to consume the metal in a sparkler.
How much metal oxide is produced? That depends on the quantity of metal powder
in the sparkler, not on the quantity of O2 in the atmosphere. The metal powder in
this example is called the limiting reactant because its amount determines, or limits,
the amount of product formed. The O2 in this example is called the excess reactant
because it is present in a greater amount than required by stoichiometry.
Consider now another example, but this time from a particulate view of matter.
The balanced equation for the reaction of oxygen and carbon monoxide to give
carbon dioxide is
2 CO(g) + O2(g) n 2 CO2(g)
Suppose you have a mixture of four CO molecules and three O2 molecules. The four
CO molecules require only two O2 molecules (and produce four CO2 molecules).
This means that one O2 molecule remains after reaction is complete.
Products: 4 CO2 and 1 O2
Reactants: 4 CO and 3 O2
+
+
Because more O2 molecules are available than are required, the number of CO2
molecules produced is determined by the number of CO molecules available.
­Carbon monoxide, CO, is therefore the limiting reactant in this case.
A Stoichiometry Calculation with a Limiting Reactant
The first step in the manufacture of nitric acid is the oxidation of ammonia to NO
over a platinum-wire gauze (Figure 4.2).
4 NH3(g) + 5 O2(g) n 4 NO(g) + 6 H2O(ℓ)
Suppose equal masses of NH3 and O2 are mixed (750. g of each). Are these reactants
mixed in the correct stoichiometric ratio or is one of them in short supply? That is,
Burning ammonia
on the surface of a
platinum wire
produces so much
energy that the wire
glows bright red.
NH3(aq)
© Johnson-Matthey
NH3(g)
© Charles D. Winters/Cengage
Platinum wire mesh used
in the industrial oxidation
of ammonia.
Figure 4.2 Oxidation of ammonia. Billions of kilograms of HNO3 are made annually starting with the
oxidation of ammonia over a wire gauze containing platinum.
196
Chapter 4 / Stoichiometry: Quantitative Information about Chemical Reactions
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will one of them limit the quantity of NO that can be produced? How much NO
can be formed if the reaction using this reactant mixture goes to completion? And
how much of the excess reactant is left over when the maximum amount of NO has
been formed?
Step 1. Calculate the amount (moles) of each
r­ eactant.
750. g NH3 
750. g O2 
Step 2. Calculate the mass of product. ­Calculate
the expected mass of product, NO, based on the
amount of each reactant, NH3 and O2.
1 mol NH3
5 44.04 mol NH3 available
17.03 g NH3
1 mol O2
5 23.44 mol O2 available
32.00 g O2
44.04 mol NH3 
23.44 mol O2 
4 mol NO
30.01 g NO

5 1320. g NO
4 mol NH3
1 mol NO
4 mol NO
30.01 g NO

5 563 g NO
5 mol O2
1 mol NO
Here O2 is the limiting reactant because the amount of O2
available limits the amount of product (NO) formed to 563 g.
Step 3. Determine the limiting reactant and the
maximum mass of product that can be obtained.
Step 4. Calculate the mass of excess reactant.
4 mol NH3
17.03 g NH3

5 mol O2
1 mol NH3
Ammonia is the excess reactant because more
than enough NH 3 is available to react with
23.4 mol of O 2. To calculate the mass of NH 3
remaining after all the O2 has been used, you first
need to know the mass of NH3 required to consume all the limiting reactant, O2.
23.44 mol O2 available 
This required mass of NH3 is subtracted from the
mass of NH3 present to obtain the mass of excess
reactant left over at the end of the r­eaction. Because 431 g of NH3 is left over, this means that
319 g of the initial 750. g of NH3 has been
consumed.
Excess NH3 5 750. g NH3 present 2 319.3 g NH3 required
5 431 g NH3 left
5 319.3 g NH3 required
The information from the preceding steps can be organized in an amounts table.
Equation
4 NH3(g)
+
5 O2(g)
Initial amount (mol)
44.04
23.44
Change in amount
upon reaction (mol)
−(4/5)(23.44)
= −18.75
−23.44
Amount after complete
reaction (mol)
44.04 − 18.75
= 25.29
0
n
4 NO(g)
+
0
6 H2O(g)
0
+(4/5)(23.44)
= +18.75
+(6/5)(23.44)
= +28.13
18.75
28.13
All of the limiting reactant, O2, is consumed. Of the original 44.04 mol of NH3,
18.75 mol is consumed and 25.29 mol remains. The balanced equation indicates
that the amount of NO produced is equal to the amount of NH3 consumed, so
18.75 mol of NO is produced from 18.75 mol of NH3. In addition, 28.13 mol of
H2O is produced.
4.2 Reactions in Which One Reactant Is Present in Limited Supply
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197
Exam pl e 4 .2
A Reaction with a Limiting Reactant
Strategy Map
Problem
Calculate mass of product
from a reaction.
Problem Methanol, CH3OH, which can be used as a fuel in racing cars and in fuel cells,
can be made by the reaction of carbon monoxide and hydrogen.
CO(g) + 2 H2(g) n CH3OH(ℓ)
methanol
Suppose 356 g of CO and 65.0 g of H2 are mixed and allowed to react.
(a) What mass of methanol can be produced?
(b) What mass of the excess reactant remains after the limiting reactant has been consumed?
Data/Information
• Masses of the reactants
• Balanced equation
What Do You Know? Any problem in which masses of two or more reactants are
given is likely a limiting reactant problem. Here the balanced equation for the reaction
is also given, and you know the masses of CO and H2 available. You will need the molar
masses of the two reactants and the product to solve the problem.
Strategy See the Strategy Map for this example. After calculating the amount of each
reactant, ­calculate the mass of product expected based on the amount of each reactant
(Problem S­ olving Tip 4.1). From that, decide which reactant is limiting. Knowing that, you
now know the maximum possible mass of product. The mass of excess reactant is the
­difference between its starting mass and what was required by the limiting reactant.
Solution
(a) What is the maximum mass of product expected?
Step 1
Calculate the amount (moles) of each reactant.
Begin by calculating the amount of each reactant using its corresponding mass and
molar mass.
Amount of CO 5 356 g CO 
1 mol CO
5 12.71 mol CO
28.01 g CO
Amount of H2 5 65.0 g H2 
Step 2
Step 3
1 mol H2
5 32.24 mol H2
2.016 g H2
Calculate the mass of product that could be produced based on the amount of
each reactant.
For each reactant, use a stoichiometric factor to determine the amount of product that
could be produced, and then convert it to mass using the molar mass of the product.
12.71 mol CO 
1 mol CH3OH formed 32.04 g CH3OH

5 407 g CH3OH
1 mol CO available
1 mol CH3OH
32.24 mol H2 
1 mol CH3OH formed 32.04 g CH3OH

5 517 g CH3OH
2 mol H2 available
1 mol CH3OH
Decide which reactant is limiting. The limiting reactant is the reactant that
produces the smaller mass of product.
The amount of CO available produces less product than the amount of H2 available.
Carbon monoxide, CO, is the limiting reactant, so the maximum mass of CH3OH that can
be produced is 407 g .
(b) What mass of H2 remains when all the CO has been converted to product?
First, you must find the mass of H2 required to react with all the limiting reagent, CO.
12.71 mol CO 
198
2 mol H2
2.016 g H2

5 51.25 g H2 required
1 mol CO
1 mol H2
Chapter 4 / Stoichiometry: Quantitative Information about Chemical Reactions
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Auto-Data.Net/Shutterstock.com
You began with 65.0 g of H2, but only 51.25 g is required by the limiting reactant; thus, the
excess mass is
65.0 g H2 present − 51.25 g H2 required = 13.8 g H2 left
Think about Your Answer The amounts table for this reaction is as follows.
Equation
Initial amount (mol)
CO(g)
12.71
Change in amount upon
reaction (mol)
−12.71
Amount after complete
reaction (mol)
0
+
2 H2(g)
32.24
−2(12.71)
6.82
n
CH3OH(ℓ)
0
Methanol fuel cell car. Methanol
is widely used, and one use is as
a fuel. It can be burned directly
or used as the fuel in a battery for
an electric car.
+12.71
12.71
The mass of product formed plus the mass of H2 remaining after reaction (407.2 g CH3OH
produced + 13.8 g H2 remaining = 421 g) is equal to the mass of reactants present before
reaction (356 g CO + 65.0 g H2 = 421 g).
Check Your Understanding
The thermite reaction produces iron metal and aluminum oxide from a mixture of powdered aluminum metal and iron(III) oxide.
Fe2O3(s) + 2 Al(s) n 2 Fe(ℓ) + Al2O3(s)
Phil Degginger/Alamy Stock Photo
A mixture of 50.0 g each of Fe2O3 and Al is used. Which is the limiting reactant? What mass of iron metal can be produced?
Thermite reaction. When ignited, iron(III) oxide is
reduced by aluminum to produce iron and aluminum
oxide. The reaction generates an enormous amount of
energy, sufficient to produce iron in the molten state.
4.2 Reactions in Which One Reactant Is Present in Limited Supply
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
199
Problem Solving Tip 4.2 Moles of Reaction and Limiting Reactants
There is another method for solving
stoichiometry problems that applies
especially well to limiting reactant
problems. This involves the useful
concept of moles of reaction.
One mole of reaction occurs when
the reaction has taken place according
to the number of moles given by the
coefficients in the balanced equation.
For example, for the reaction of CO
and O2,
2 CO(g) + O2(g) n 2 CO2(g)
1 mole of reaction occurs when 2 mol
of CO and 1 mol of O2 produce 2 mol
of CO2. The unit moles of reaction
(mol-rxn) counts how many times the
chemical reaction itself occurs.
Now, suppose 9.5 g of CO and
excess O2 are combined. What
amount of CO2 (moles) can be
produced?
1 mol CO
1 mol-rxn
9.5 g CO 

28.0 g CO
2 mol CO
5 0.170 mol-rxn
0.170 mol-rxn 
2 mol CO2
1 mol-rxn
5 0.34 mol CO2
All reactants and products involved
in a chemical reaction undergo the
same number of moles of reaction
because the reaction can only occur a
certain number of times before one or
more of the reactants are consumed
and the reaction reaches completion.
If one of the reactants is in limited
supply, the actual number of times
a reaction can be carried out—the
number of moles of reaction—will be
determined by the limiting reactant.
Using an approach similar to Example
4.2, you first calculate the amount
of each reactant initially present and
then calculate the moles of reaction
that could occur with each amount of
reactant. [This is equivalent to dividing
the amount (moles) of each reactant
by its stoichiometric coefficient.] The
reactant producing the smaller number
of moles of reaction is the limiting
reactant. Once the limiting reactant is
known, you proceed as before.
As an example, consider again the
NH3/O2 reaction on page 196:
44.04 mol NH3 
1 mol-rxn
4 mol NH3
5 11.01 mol-rxn
Based on the amount of O2 available,
4.688 mol of reaction could occur.
23.44 mol O2 
1 mol-rxn
5 mol O2
5 4.688 mol-rxn
Fewer moles of reaction can occur
with the amount of O2 available, so O2
is the limiting reactant.
2. Calculate the change in amount and
the amount upon completion of the
reaction for each reactant and
product.
4 NH3(g) + 5 O2(g) n
4 NO(g) + 6 H2O(ℓ)
1. Calculate the moles of reaction
predicted for each reactant and
decide on the limiting reactant.
In the case of the NH3/O2 reaction,
1 mole of reaction uses 4 mol of
NH3 and 5 mol of O2 and produces 4
mol of NO and 6 mol of H2O. In the
example on page 197, we started with
44.04 mol of NH3, so 11.01 mol of
reaction could result.
The number of moles of reaction
predicted by the limiting reactant corresponds to the number of moles of
reaction that can actually occur. Each
reactant and product will undergo
this number of moles of reaction,
4.688 mol-rxn in this case. To calculate
the change in amount for a given reactant or product, multiply this number of
moles of reaction by the stoichiometric
coefficient of the reactant or product.
To illustrate, for NH3 this corresponds
to the following calculation:
 4 mole NH3 
4.688 mol-rxn  
 1 mol-rxn 
5 18.75 mol NH3
The amount of each reactant and product
after reaction is calculated as usual.
Amounts Table
+
5 O2(g)
4 NO(g)
+
Equation
4 NH3(g)
Initial amount (mol)
44.04
23.44
0
0
Moles of reaction based on limiting
reactant (mol)
4.688
4.688
4.688
4.688
Change in amount (mol)*
−4.688(4)
= −18.75
−4.688(5)
= −23.44
+4.688(4)
= +18.75
+4.688(6)
= +28.13
Amount after complete reaction (mol)
25.29
18.75
28.13
n
0
6 H2O(g)
*Moles of reaction are multiplied by the stoichiometric coefficient for each reactant and product.
4.3 Percent Yield
Goal for Section 4.3
•
200
Explain the differences among actual yield, theoretical yield, and percent yield, and
calculate percent yield for a reaction.
Chapter 4 / Stoichiometry: Quantitative Information about Chemical Reactions
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12 kernels popped
4 unpopped
75% yield
Photos: © Charles D. Winters/Cengage
16 kernels
Figure 4.3 Percent yield. Although not a chemical reaction, popping corn is a good analogy
to the difference between a theoretical yield and an actual yield. Here, you began with 16
popcorn kernels. If all of them popped, you would have obtained 16 pieces of popped corn (the
theoretical yield). In this case, only 12 of them popped (the actual yield). The percent yield from
the “reaction” was (12/16) × 100%, or 75%.
The maximum mass of product that can be obtained from a chemical reaction is the
theoretical yield. The theoretical yield is obtained from a stoichiometry calculation.
The actual yield of the product—the mass of material that is actually obtained in the
laboratory or a chemical plant—is almost always less than the theoretical yield.
­Product loss almost always occurs when you are trying to isolate and purify the
­compound. In addition, some reactions do not go completely to products, and other
reactions are sometimes complicated by giving more than one set of products.
To provide information to other chemists who might want to carry out a reaction, it is customary to report a percent yield (Figure 4.3). Percent yield, which
specifies how much of the theoretical yield was obtained, is defined as
Percent yield 5
actual yield
 100%
theoretical yield
(4.1)
Exam p le 4.3
Calculating Percent Yield
Problem ​Aspirin can be made in the laboratory by the following reaction:
C4H6O3(ℓ) + C7H6O3(s) n CH3CO2H(ℓ) + C9H8O4(s)
acetic
salicylic acid acetic acid aspirin
anhydride
C9H8O4(s)
aspirin
What is the percent yield of the reaction if 6.26 g of aspirin is obtained from 14.4 g of
salicylic acid (C7H6O3) and an excess of acetic anhydride?
What Do You Know? You know the initial mass of salicylic acid (14.4 g) and that it
is the limiting reactant. You also know the actual yield of the reaction (6.26 g).
Strategy ​
Step 1. The theoretical yield is the quantity of product (aspirin) that could be obtained
from the complete conversion of the limiting reactant (salicylic acid). Determine the
theoretical yield of aspirin by using a stoichiometry calculation to find the mass of aspirin
that could be obtained from the given mass of the salicylic acid.
Step 2. Calculate the percent yield of aspirin by dividing the actual yield by the
­theoretical yield and multiplying by 100.
Solution ​
Step 1. Calculate the theoretical yield from the given mass of salicylic acid:
14.4 g C7H6O3 
1 mol C7H6O3
138.1 g C7H6O3

1 mol aspirin
1 mol C7H6O3

180.2 g aspirin
5 18.79 g aspirin
1 mol aspirin
4.3 Percent Yield
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201
Step 2. Calculate the percent yield using Equation 4.1.
actual yield
 100%
theoretical yield
6.26 g aspirin obtained
5
 100%
18.79 g aspirin expected
Percent yield 5
5 33.3% yield
Think about Your Answer It is not unusual for reactions to not go to completion, produce unwanted byproducts, and/or lose products during purification steps like
filtration. For these reasons, percent yields are often less than 100%.
Check Your Understanding
A 1.50 g piece of copper wire is immersed in a solution containing an excess of silver
nitrate, and the following reaction occurs:
Cu(s) + 2 AgNO3(aq) n 2 Ag(s) + Cu(NO3)2(aq)
The resulting silver metal has a mass of 4.78 g. What is the percent yield for this reaction?
4.4 Chemical Equations and
Chemical Analysis
Goals for Section 4.4
•
•
Use stoichiometry principles to analyze a mixture of compounds.
Find the empirical formula of an unknown compound using chemical stoichiometry.
Quantitative Analysis of a Mixture
Quantitative chemical analysis involves determining the amount or concentration
of a substance in a sample. For example, the quantity of a drug of interest in a pill
of the drug is crucial information and can be determined using methods of quantitative chemical analysis. In such analyses, the principles of chemical stoichiometry
allow chemists to conduct chemical reactions and relate the quantities of products
obtained to the quantity of the material of interest. You will learn about many experimental techniques that can be used to carry out quantitative analyses such as
gravimetric analysis, combustion analysis, and titration.
Quantitative chemical analysis generally depends on one of the following basic ideas:
•
A substance (A), present in an unknown amount, can be allowed to react with
a known amount of another substance (B). If the stoichiometric ratio for their
reaction is known (A/B), the unknown amount (of A) can be determined.
•
A material of unknown composition can be converted to one or more substances of
known composition. Those substances can be identified, their amounts determined,
and these amounts related to the amount of the original, unknown substance.
An example of the first type of analysis is determining the amount of acetic acid
in vinegar. (Acetic acid is the ingredient that makes vinegar acidic.) The acid reacts
readily and completely with sodium hydroxide.
CH3CO2H(aq) + NaOH(aq) n NaCH3CO2(aq) + H2O(ℓ)
acetic acid
If the exact amount of sodium hydroxide used in the reaction can be measured, the
amount of acetic acid present can be calculated. This type of analysis is discussed in
Section 4.8.
202
Chapter 4 / Stoichiometry: Quantitative Information about Chemical Reactions
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An example of the second type of analysis is gravimetric analysis, in which the mass
of a precipitate is used to determine the composition of an unknown sample. This technique is used in the following example to analyze a sample of the mineral thenardite.
Exam p le 4.4
Analysis of a Mineral
Problem Sodium sulfate, Na2SO4, occurs naturally as the mineral thenardite. To ana© Charles D. Winters/Cengage
lyze an impure mineral sample for the quantity of Na2SO4, the sample is crushed, then
dissolved in water to form a solution of Na2SO4. Next, the aqueous solution is treated with
aqueous barium chloride, BaCl2, to give solid BaSO4 (Figure 4.4).
Na2SO4(aq) + BaCl2(aq) n BaSO4(s) + 2 NaCl(aq)
Suppose a 0.498-g sample containing thenardite produces 0.541 g of solid BaSO4. What is
the mass percent of Na2SO4 in the sample?
What Do You Know? You know the mass of the impure thenardite (Na2SO4) sample
and the mass of BaSO4 produced in the reaction. You also know the balanced equation for the
reaction leading to the formation of BaSO4. You will need molar masses of Na2SO4 and BaSO4.
Strategy First calculate the amount of BaSO4 from its mass. Because 1 mol of Na2SO4
was present in the sample for each mole of BaSO 4 isolated, you therefore know the
amount of Na2SO4 and can then calculate its mass and mass percent in the sample.
Thenardite. The mineral thenardite
is sodium sulfate, Na2SO4. It is
named after the French chemist
Louis Thenard (1777–1857).
Sodium sulfate is used in making
detergents, glass, and paper.
Solution The molar mass of BaSO4 is 233.4 g/mol. The amount of this solid is
0.541 g BaSO4 
1 mol BaSO4
5 2.318  1023 mol BaSO4
233.4 g BaSO4
Because 1 mol of BaSO4 is produced from 1 mol of Na2SO4, the amount of Na2SO4 in the
sample must also have been 2.318 × 10−3 mol.
2.318  1023 mol BaSO4 
1 mol Na2SO4
5 2.318  1023 mol Na2SO4
1 mol BaSO4
With the amount of Na2SO4 known, the mass of Na2SO4 can be calculated.
2.318  1023 mol Na2SO4 
142.0 g Na2SO4
5 0.3291 g Na2SO4
1 mol Na2SO4
Finally, the mass percent of Na2SO4 in the 0.498-g sample is
Mass percent Na2SO4 5
0.3291 g Na2SO4
 100% 5 66.1% Na2SO4
0.498 g sample
Think about Your Answer For an analytical procedure to be used, the reactants
must be completely converted to product and the product being measured must be isolated without losses in handling. Very careful experimental techniques are needed.
Check Your Understanding
One method for determining the purity of a sample of titanium(IV) oxide, TiO2, an important industrial chemical, is to react the sample with bromine trifluoride.
3 TiO2(s) + 4 BrF3(ℓ) n 3 TiF4(s) + 2 Br2(ℓ) + 3 O2(g)
This reaction occurs completely and quantitatively. That is, all the oxygen in TiO2 is evolved
as O2. Suppose 2.367 g of a TiO2-containing sample evolves 0.143 g of O2. What is the mass
percent of TiO2 in the sample?
4.4 Chemical Equations and Chemical Analysis
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203
1 A sample containing an
2 When BaCl2 is added to the
unknown amount of the sulfate ion
(here Na2SO4) can be analyzed by
adding barium chloride.
solution containing the sulfate ion,
insoluble BaSO4 is precipitated.
Sufficient Ba2+ ions are added to
ensure complete precipitation.
3 Solid BaSO4 is collected in a
4 After drying the BaSO4 in the
(a)
(b)
(c)
(d)
filter paper, the paper and solid are
weighed and the mass of BaSO4
determined.
© Charles D. Winters/Cengage
weighed filter paper. For an
accurate analysis all of the solid
must be carefully collected.
Na2SO4(aq),
clear solution
BaSO4,
white solid
BaCl2(aq),
clear solution
NaCl(aq),
clear solution
BaSO4, white solid
NaCl(aq),
clear solution caught in filter
Mass of dry BaSO4 determined
Figure 4.4 The procedure for analyzing a solution for the sulfate ion content by precipitation. This procedure is an
example of a gravimetric analysis.
Determining the Formula of a Compound
by Combustion
The empirical formula of a compound can be determined if the percent composition of the compound is known (see Section 2.8). But where do the percent composition data come from? One chemical method that works well for compounds that
burn in oxygen is combustion analysis. In this technique, the compound is completely burned in O2, and each element in the compound combines with oxygen to
produce the appropriate oxide. Based on the masses and identities of the product
oxides, the percent composition or the empirical formula of the original compound
can be determined.
Take methane, CH4, as an example. A balanced equation for its combustion
shows that every carbon atom in the original compound appears as CO2 and every
hydrogen atom appears in the form of water. In other words, for every mole of CO2
observed, there must have been one mole of carbon in the unknown compound.
Similarly, for every mole of H2O observed from combustion, there must have been
two moles of H atoms in the unknown compound.
Problem Solving Tip 4.3 General Approach to Finding an Empirical Formula by
Chemical Analysis
1. The unknown but pure compound
is converted by a chemical reaction into known products.
2. The reaction products are isolated,
and the amount (moles) of each
product is determined.
204
3. The amount (moles) of each
product is related to the amount
(moles) of each element in the
original compound.
4. The empirical formula is determined from the relative amounts
(moles) of elements in the original
compound.
Chapter 4 / Stoichiometry: Quantitative Information about Chemical Reactions
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Furnace
O2
CxHy
Sample containing
hydrogen and carbon
CH4(g)
+
H2O absorber
CO2 absorber
H2O
CO2
H2O is absorbed by
magnesium perchlorate,
CO2 passes through
CO2 is absorbed by finely
divided NaOH supported
on an inert solid support
CO2(g)
2 O2(g)
+
+
2 H2O(ℓ)
Figure 4.5 Combustion
analysis of a hydrocarbon. If a
compound containing C and H is
burned in oxygen, CO2 and H2O
are formed. These products pass
into tubes containing materials that
absorb them, called absorbents.
The difference between the mass
of each absorbent before and
after combustion gives the masses
of CO2 and H2O produced. Only
a few milligrams of a combustible
compound are needed for
analysis.
+
In the combustion experiment, the product gases carbon dioxide and water are
separated (as illustrated in Figure 4.5) and their masses determined. From these
masses it is possible to calculate the amounts (moles) of C and H in CO2 and H2O,
respectively, and then determine the ratio of the amounts of C and H in a sample of
the original compound. This ratio gives the empirical formula. If the molar mass is
known from a separate experiment, the molecular formula can also be determined.
Exam p le 4.5
Using Combustion Analysis to Determine
the Empirical Formula of a Hydrocarbon
Problem When 1.125 g of a liquid hydrocarbon, CxHy, is burned (Figure 4.5), 3.447 g
of CO2 and 1.647 g of H2O are produced. In a separate experiment, the molar mass of the
compound is found to be 86.2 g/mol. Determine the empirical and molecular formulas for
the unknown hydrocarbon, CxHy.
What Do You Know? You know the mass of the hydrocarbon, the fact that it contains only C and H, and the molar mass of this compound. You are also given the masses of
H2O and CO2 formed when the hydrocarbon is burned.
Strategy The strategy to solve this problem is outlined in the diagram below.
burn
in O2
×
1 mol H2O
18.02 g
×
g CO2
2 mol H
1 mol H2O
×
1 mol C
1 mol CO2
mol H2O
g H2O
CxHy
×
1 mol CO2
44.01 g
mol H
mol C
empirical
formula
mol CO2
The steps in this strategy are as follows:
Step 1. Calculate the amounts of CO2 and H2O from the given masses.
Step 2. Calculate the amounts of C and H from the amounts of CO2 and H2O,
respectively.
Step 3. Determine the lowest whole-number ratio of amounts of C and H. This gives the
subscripts for C and H in the empirical formula.
Step 4. Compare the experimental molar mass of the hydrocarbon with that of the
empirical formula to determine the molecular formula.
4.4 Chemical Equations and Chemical Analysis
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205
Solution
Step 1. The amounts of CO2 and H2O isolated from the combustion are
3.447 g CO2 
1 mol CO2
5 0.078325 mol CO2
44.009 g CO2
1.647 g H2O 
1 mol H2O
5 0.091424 mol H2O
18.015 g H2O
Step 2. For every mole of CO2 isolated, 1 mol of C must have been present in the
unknown compound.
0.078325 mol CO2 
1 mol C in unknown
5 0.078325 mol C
1 mol CO2
For every mole of H2O isolated, 2 mol of H must have been present in the unknown.
0.091424 mol H2O 
2 mol H in unknown
5 0.18285 mol H
1 mol H2O
Step 3. The original 1.125-g sample of compound therefore contained 0.078325 mol of
C and 0.18285 mol of H. To determine the empirical formula of the unknown,
find the ratio of moles of H to moles of C (Section 2.8).
2.3345 mol H
0.18285 mol H
5
0.078325 mol C
1.0000 mol C
The empirical formula gives the simplest whole-number ratio. The translation of this ratio
(2.335/1) to a whole-number ratio can usually be done by trial and error. Multiplying the
numerator and denominator by 3 gives 7/3. So, we know the ratio is 7 mol H to 3 mol C,
which means the empirical formula of the hydrocarbon is C3H7.
Step 4. Comparing the experimental molar mass with the molar mass calculated for the
empirical formula,
86.2 g/mol
2
Experimental molar mass
5
5
Molar mass of C3H7
43.1 g/mol
1
the molecular formula is determined to be twice the empirical formula, that is, (C3H7)2 or
C6H14.
Think about Your Answer As noted in Problem Solving Tip 2.2 (page 109), for problems of this type be sure to use data with enough significant figures to give accurate atom ratios.
Check Your Understanding
A 0.523-g sample of the unknown compound CxHy is burned in air to give 1.612 g of CO2
and 0.7425 g of H2O. A separate experiment gives a molar mass for CxHy of 114 g/mol.
Determine the empirical and molecular formulas for the hydrocarbon.
You can also determine the empirical formula of an oxygenated hydrocarbon,
such as ethanol (C2H6O), by combustion analysis. You first find the amount of carbon and hydrogen from the masses of collected carbon dioxide and water. Then, by
calculating the mass of carbon and hydrogen from the amounts of those elements
and subtracting these masses from the mass of the combusted compound, the mass
of oxygen, and ultimately the amount of oxygen, may be determined (Example 4.6).
206
Chapter 4 / Stoichiometry: Quantitative Information about Chemical Reactions
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Exam p le 4.6
Using Combustion Analysis to Determine the Empirical Formula of a Compound Containing C, H, and O
Problem You have isolated an acid from clover leaves and know it contains only the
elements C, H, and O. Burning 0.513 g of the acid in oxygen produces 0.501 g of CO2 and
0.103 g of H2O. What is the empirical formula of the acid, CxHyOz?
What Do You Know? You know the mass of the compound and that it contains
only C, H, and O. You also know the masses of H2O and CO2 formed when the compound
is burned.
Strategy To determine the empirical formula, you need to determine the amounts
(moles) of C, H, and O in the unknown compound. Follow the steps outlined below.
Step 1. Determine the amounts of C and H following the procedure in Example 4.5.
Step 2. Determine the masses of C and H from the amounts of C and H.
Step 3. The mass of O is the mass of the sample minus the masses of C and H.
Step 4. From the mass of O determine the amount of O.
Step 5. Finally, determine the smallest whole-number ratio between the amounts
of the three elements. This determines the subscripts for the elements in the
empirical formula.
Solution
Step 1. Determine the amounts of C and H in the sample.
0.501 g CO2 
1 mol CO2
1 mol C

5 0.01138 mol C
44.01 g CO2
1 mol CO2
0.103 g H2O 
1 mol H2O
2 mol H

5 0.01143 mol H
18.02 g H2O
1 mol H2O
Step 2. From these amounts, determine the mass of C and the mass of H in the sample.
12.01 g C
5 0.1367 g C
1 mol C
1.008 g H
0.01143 mol H 
5 0.01152 g H
1 mol H
0.01138 mol C 
Step 3. Using the mass of the original sample and the masses of C and H in the sample, now
determine the mass of O in the sample.
Mass of sample = 0.513 g = 0.1367 g C + 0.01152 g H + x g O
Mass of O = 0.3648 g O
Step 4. Determine the amount of O corresponding to its mass.
0.3648 g O 
1 mol O
5 0.02280 mol O
16.00 g O
Step 5. To find the mole ratios of elements, divide the amount of each element by the
smallest amount present. Because both C and H are present in nearly the same
amount in the sample, you know their ratio is 1 C∶1 H. What about O?
2 mol O
0.02280 mol O
5
0.01143 mol H
1 mol H
The mole ratios show that, for every C atom in the molecule, there is one H atom and two
O atoms. The empirical formula of the acid is therefore CHO2.
Think about Your Answer If the molar mass of the unknown compound were
known, you could derive the molecular formula of the compound.
4.4 Chemical Equations and Chemical Analysis
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207
Check Your Understanding
A Closer Look
A 0.1342-g sample of a compound composed of C, H, and O was burned in oxygen and
0.240 g of CO2 and 0.0982 g of H2O was isolated. What is the empirical formula of the compound? If the experimentally determined molar mass is 74.1 g/mol, what is the molecular
formula of the compound?
Nuclear Magnetic Resonance (NMR) Spectroscopy
More nuclei are in the lower energy state
than the higher one. When the nuclei are
exposed to electromagnetic radiation that
corresponds to the difference in energy between these two states, more in the lower
energy state change to the higher energy
state and are said to be in resonance with
the electromagnetic radiation.
The exact energy of electromagnetic radiation required to achieve this change ­depends
on the applied magnetic field strength, the
type of nucleus involved, and the electronic
structure that surrounds the nuclei. Nuclei
with different electronic environments in a
molecule will have different responses. The
responses are then measured and converted
into an NMR spectrum that displays the
responses as peaks on a graph. For example, the 1H NMR spectrum for CH3CH2Br
(Figure B) shows two regions of signals: a
triplet (a three-line group) of peaks centered at 1.68 ppm and a quartet (a f­ our-line
group) of peaks centered at 3.42 ppm. The
two sets of peaks correspond to hydrogen
atoms in two distinct types of electronic
environments in this compound.
The area under a set of peaks is proportional to the number of hydrogen atoms
involved in that type of signal. In the spectrum for CH3CH2Br, the ratio of the areas
for these two regions is 3 (for the triplet):2
(for the quartet). This indicates that the
CH3 group is responsible for the peaks
centered at 1.68 ppm. But why is a triplet
present in the spectrum rather than a single peak? It turns out that a triplet results
when two hydrogen atoms are bonded to
an adjacent atom, in this case the CH2
group adjacent to the CH3 group. Furthermore, a quartet results from the presence
of three hydrogen atoms on an adjacent
atom. Thus, the spectrum provides very
good evidence that the structure of this
compound contains a CH3 group followed
by a CH2 group. Combining this information with the 13C NMR and the mass spectrum (page 114) for the compound provides
conclusive evidence that the molecular formula for the compound is C2H5Br and its
molecular structure is CH3CH2Br.
Intensity
Determining compound formulas by
combustion analysis is a classical
technique that used to be performed
on almost every new carbon-containing
compound synthesized. Today, techniques using nuclear magnetic resonance (NMR) spectroscopy (­Figure A)
are much more common. In addition
to providing information about the
identities and relative numbers of atoms
involved in a compound, NMR also provides information about how these ­atoms
are connected in the molecules.
The nuclei of some isotopes, most notably 1H and 13C, behave as if they are spinning and so act like tiny magnets. In the
absence of an external magnetic field, the
nuclei are randomly oriented. In the presence of a strong magnetic field, however,
the nuclei align with the applied magnetic
field in one of either two orientations: 1) a
lower energy state in which their own magnetic fields are in the same direction as
the applied field or 2) a higher energy
state in which their magnetic fields are in
the opposite direction as the applied field.
10
9
8
7
6
5
4
3
2
1
0 ppm
© Charles D. Winters/Cengage
Chemical shift (δ)
Figure A An NMR spectrometer.
208
Figure B The 1H NMR spectrum and molecular model of
CH3CH2Br. The spectrum shows two sets of peaks (one centered at
1.68 ppm and one centered at 3.42 ppm) that correspond to two different
electronic environments for 1H atoms in this compound. The small peak
at 0 ppm is from another compound that was added to the sample as a
reference standard. The ppm (or parts per million) unit on the x-axis is a
measure of the (­extremely small) changes to the external magnetic field
experienced by nuclei in differing electronic environments within the molecule.
Chapter 4 / Stoichiometry: Quantitative Information about Chemical Reactions
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4.5 Measuring Concentrations
of Compounds in Solution
Goals for Section 4.5
•
Calculate the concentration of a solute in a solution in units of moles per liter
(molarity) and use solution concentrations in calculations.
•
Describe how to prepare a solution of a given concentration from the solute and
solvent or by dilution of a more concentrated solution.
Solution Concentration: Molarity
Most chemical studies require quantitative measurements, including experiments involving solutions. When doing these experiments, you will continue to use balanced
equations and moles, but measure volumes of solutions rather than masses of solids,
liquids, or gases.
Molarity, c, is the amount of solute per liter of solution.
Molarity of x (c x ) 5
amount of solute x (mol)
volume of solution (L)
(4.2)
For example, if 58.4 g (1.00 mol) of NaCl is dissolved in enough water to give a total
solution volume of 1.00 L, the molarity, c, is 1.00 mol/L. This is often abbreviated as
1.00 M, where capital M is the symbol for “moles per liter.” Another common notation is to place the formula of the compound in square brackets (for example,
[NaCl]); this notation indicates that the concentration of the solute in moles per liter
of solution is being specified.
cNaCl = [NaCl] = 1.00 mol/L = 1.00 M
© Charles D. Winters/Cengage
It is important to notice that molarity refers to the amount of solute per ­liter of
solution and not per liter of solvent. If one liter of water is added to one mole of a
solid compound, the final volume will not be exactly one liter, and the final concentration will not be exactly one mol/L (Figure 4.6). When making solutions of a given
For this photo, we measured out exactly
1.00 L of water, which was slowly added to
the volumetric flask containing 25.0 g of
CuSO4 · 5 H2O. When enough water was
added so that the solution volume was
exactly 1.00 L, approximately 8 mL was left
over from the original 1.00 L of water.
Figure 4.6 Volume of solution
versus volume of solvent. This
figure emphasizes that molar
concentrations are defined
as moles of solute per liter of
solution and not per liter of water
or other solvent.
1.00 L of 0.100 M
CuS04
To make a 0.100 M solution of CuSO4, 25.0 g
(0.100 mol) of CuSO4 · 5 H2O (the blue
crystalline solid) was placed in a 1.00-L
volumetric flask.
4.5 Measuring Concentrations of Compounds in Solution
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209
molarity, it is always the case that you should dissolve the solute in a volume of
solvent smaller than the desired volume of solution, then add solvent until the final
solution volume is reached.
Potassium permanganate, KMnO4, which was used as a germicide for treating
burns, is a shiny, purple-black solid that dissolves readily in water to give a deep
purple solution. Suppose you dissolve 0.435 g of KMnO4 in enough water to give
250. mL of solution (Figure 4.7). What is the concentration of KMnO4? First, convert the mass of KMnO4 to an amount.
0.435 g KMnO4 
Concentration of KMnO4 5 cKMnO4 5 [KMnO4 ] 5
2+
2+
2+
Next, combine the volume of solution—which must be in liters—with the amount
of KMnO4 to give the concentration. Because 250. mL is equivalent to 0.250 L,
2+
−
1 mol KMnO4
5 0.002753 mol KMnO4
158.0 g KMnO4
−
0.002753 mol KMnO4
5 0.0110 M
0.250 L
The KMnO4 concentration is 0.0110 mol/L, or 0.0110 M. This is useful information, but it is often useful to know the concentration of each type of ion in a solution. Like all soluble ionic compounds, KMnO4 dissociates into its ions, K+ and
MnO4−, when dissolved in water.
−
KMnO4(aq)
K+(aq) + MnO4−(aq)
© Charles D. Winters/Cengage
100% dissociation
One mole of KMnO4 provides 1 mol of K+ ions and 1 mol of MnO4− ions. Accordingly, 0.0110 M KMnO4 gives a concentration of K+ in the solution of 0.0110 M;
similarly, the concentration of MnO4− is also 0.0110 M.
Finally, consider an aqueous solution of CuCl2.
CuCl2(aq)
Ion concentrations for a soluble
ionic compound. Here, 1 mol
of CuCl2 dissociates to 1 mol of
Cu2+ ions and 2 mol of Cl− ions.
Therefore, the Cl− concentration
is twice the concentration
calculated for CuCl2.
Cu2+(aq) + 2 Cl−(aq)
100% dissociation
If 0.10 mol of CuCl2 is dissolved in enough water to make 1.0 L of solution, the
concentration of the copper(II) ion, [Cu2+], is 0.10 M. However, the concentration
of chloride ions, [Cl−], is 0.20 M because the compound dissociates in water to
provide 2 mol of Cl− ions for each mole of CuCl2.
Distilled water
A mark on the neck
of a volumetric
flask indicates a
volume of exactly
250. mL at 20 °C.
250-mL
volumetric flask
0.435 g KMn04
The KMn04 is first dissolved in a small
amount of water.
© Charles D. Winters/Cengage
250. mL
mark
Distilled water is added to fill the flask
with solution just to the mark on the flask.
Figure 4.7 Making a solution. A 0.0110 M solution of KMnO4 is made by adding enough water to 0.435 g of
KMnO4 to make 0.250 L of solution.
210
Chapter 4 / Stoichiometry: Quantitative Information about Chemical Reactions
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Exam p le 4.7
Concentration
Problem If 25.3 g of sodium carbonate, Na2CO3, is dissolved in enough water to make
250. mL of solution, what is the concentration of Na2CO3? What are the concentrations of
the Na+ and CO32− ions?
What Do You Know? You know the mass of the solute, Na2CO3, and the volume
of the solution. You will need the molar mass of Na2CO3 to calculate the amount of this
compound.
Strategy The concentration (moles/L) of Na2CO3 is the amount of Na2CO3 (moles)
divided by the volume (in liters). To determine the concentrations of the ions, recognize that one mole of this ionic compound contains two moles of Na+ and one mole of
CO32− ions.
Na2CO3(s) n 2 Na+(aq) + CO32−(aq)
Solution First find the amount of Na2CO3.
25.3 g Na2CO3 
1 mol Na2CO3
5 0.2387 mol Na2CO3
106.0 g Na2CO3
and then the concentration of Na2CO3 using Equation 4.2,
Concentration of Na2CO3 5
0.2387 mol Na2CO3
5 0.9548 M 5 0.955 mol/L
0.250 L
The ion concentrations follow from knowing that each mole of Na2CO3 produces 2 mol of
Na+ ions and 1 mol of CO32− ions.
[Na+] = 2 × 0.9548 M Na+(aq) = 1.91 M
[CO32−] = 0.955 M
Think about Your Answer While we refer to this solution as 0.955 M Na2CO3,
there are no actual particles of Na2CO3 present. This soluble ionic compound is present in
solution as dissociated sodium and carbonate ions.
Check Your Understanding
Sodium bicarbonate, NaHCO3, is used in baking powder formulations and in the manufacture of plastics and ceramics, among other things. If 26.3 g of the compound is dissolved
in enough water to make 200. mL of solution, what is the concentration of NaHCO3? What
are the concentrations of the ions in solution?
Equation 4.2 states that the molarity of a solution is equal to the amount of the
solute in moles divided by the volume of the solution in liters. If two of the three
quantities involved in this equation are known, then the third can be determined.
You have already seen that dividing the amount of solute in moles by the volume
of solution in liters yields the molarity. If this equation is solved for the amount of
solute in moles, it becomes
 mol 
Amount of solute x (mol) 5 c x 
 volume of solution (L)
 L 
(4.3)
You will find this equation very useful in solving stoichiometry problems that involve solutions.
4.5 Measuring Concentrations of Compounds in Solution
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211
E xamp le 4.8
Relating Amount and Molarity
Problem You transfer 50.0 mL of a 0.250 M solution of sodium chloride into a graduated cylinder. What amount (mol) of sodium chloride is in the cylinder?
What Do You Know? You know the volume of the solution in mL, the conversion
factor to convert from mL to L, and the molarity of the solution.
Strategy
Step 1. Convert the volume in mL to L using the relationship that there are 1000 mL in 1 L.
Step 2. Multiply the volume in L by the molarity to obtain the amount of solute in moles.
Solution
Step 1. The volume must first be converted to L because molarity is the amount of
­solute per L of solution.
50.0 mL 
1L
5 0.0500 L
1000 mL
Step 2. Using Equation 4.3, multiply the molarity of the solution by the volume in liters
to obtain the amount (mol) of solute delivered.
amount of NaCl 5 cNaCl  V 5
0.250 mol NaCl
 0.0500 L 5 0.0125 mol NaCl
L
Think about Your Answer One liter of a 0.250 M NaCl solution contains
0.250 mol of NaCl. In this problem, only 50.0 mL is measured out, so the amount of solute
is correspondingly less.
Check Your Understanding
What amount (mol) of glucose (C6H12O6) is present in 10.0 mL of a 0.0132 M glucose solution?
Preparing Solutions of Known Concentration
Chemists often need to prepare a given volume of solution of known concentration.
There are two commonly used methods to do this.
Combining a Weighed Solute with the Solvent
Volumetric Flask A volumetric
flask is a special flask with
a line marked on its neck
(see page 44 and Figures
4.6–4.8). If the flask is filled
with a solution to this line (at a
given temperature), it contains
precisely the volume of solution
specified.
212
Suppose you need to prepare 2.00 L of a 1.50 M solution of aqueous Na2CO3. You
have some solid Na2CO3, distilled water, and a 2.00-L volumetric flask. To make the
solution, you must weigh the required quantity of Na2CO3 as accurately as possible
considering the number of significant figures desired for the concentration, carefully
place all the solid in the volumetric flask, and then add some water to dissolve the
solid. After the solid has dissolved completely, more water is added to bring the
solution volume to 2.00 L. After thorough mixing, the solution then has the desired
concentration and volume.
But what mass of Na2CO3 is required to make 2.00 L of 1.50 M Na2CO3? First,
calculate the amount of Na2CO3 required,
Chapter 4 / Stoichiometry: Quantitative Information about Chemical Reactions
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2.00 L 
1.50 mol Na2CO3
5 3.000 mol Na2CO3 required
1.00 L solution
and then the mass in grams.
3.000 mol Na2CO3 
106.0 g Na2CO3
5 318 g Na2CO3
1 mol Na2CO3
Thus, to prepare the desired solution, you should dissolve 318 g of Na2CO3 in
enough water to make 2.00 L of solution.
Diluting a More Concentrated Solution
Another method for making a solution of a given concentration is to begin with a
concentrated solution of known concentration and add more solvent (usually water) until the desired, lower concentration is reached. Many of the solutions prepared for your laboratory course are probably made by this dilution method. It is
more efficient to store a small volume of a concentrated solution and then, when
needed, add water to make a much larger volume of a dilute solution.
Suppose you need 500. mL of aqueous 0.00100 M potassium dichromate,
K2Cr2O7, for use in chemical analysis. You have some 0.100 M K2Cr2O7 solution
available. To make the required 0.00100 M solution, place a measured volume of the
more concentrated K2Cr2O7 solution in a flask and then add water until the K2Cr2O7
is contained in the appropriate, larger volume of water (Figure 4.8).
What volume of a 0.100 M K2Cr2O7 solution must be diluted to make 500. mL
of 0.00100 M solution? The amount of solute in the dilute solution can be calculated from its volume and concentration.
 0.00100 mol 
Amount of K 2Cr2O7 in dilute solution 5 cK2Cr2O7  VK2Cr2O7 5 
  (0.500 L )

L
5 0.0005000 mol K 2Cr2O7
500-mL
volumetric
flask
© Charles D. Winters/Cengage
5.00-mL
pipet
0.100 M K2Cr2O7
Use a 5.00-mL pipet to withdraw
5.00 mL of 0.100 M
K2Cr2O7 solution.
Add the 5.00-mL sample of 0.100 M
K2Cr2O7 solution to a 500-mL
volumetric flask.
Fill the flask to the mark with
distilled water to give 0.00100 M
K2Cr2O7 solution.
Figure 4.8 Making a solution by dilution. Here, 5.00 mL of a 0.100 M K2Cr2O7 solution is diluted to 500. mL. This means the
solution is diluted by a factor of 100, from 0.100 M to 0.00100 M.
4.5 Measuring Concentrations of Compounds in Solution
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213
The more concentrated solution containing this amount of K2Cr2O7 is placed in a
500.-mL flask and then diluted to the final volume. The volume of 0.100 M K2Cr2O7
required is 5.00 mL.
0.0005000 mol K 2Cr2O7 
1.00 L
5 0.00500 L or 5.00 mL
0.100 mol K 2Cr2O7
Thus, to prepare 500. mL of 0.0010 M K2Cr2O7, place 5.00 mL of 0.100 M K2Cr2O7 in
a 500.-mL flask and add water until a volume of 500. mL is reached.
E xamp le 4.9
Preparing a Solution by Dilution
Problem What is the concentration of iron(III) ions in a solution prepared by diluting
1.00 mL of a 0.236 M solution of iron(III) nitrate, Fe(NO3)3, to a volume of 100.0 mL?
What Do You Know? You know the initial concentration and volume of the ironcontaining solution and the final volume required after dilution.
Strategy First calculate the amount of iron(III) ions in the 1.00-mL sample (amount =
concentration × volume). The concentration of the iron(III) ions in the final, dilute solution
is equal to this amount of iron(III) ions divided by the new volume.
Solution The amount of iron(III) ion in the 1.00 mL sample is
Amount of Fe31 5 cFe31  VFe31 5
0.236 mol Fe31
 1.00  1023 L 5 2.360  1024 mol Fe31
L
This amount of iron(III) ion is contained in the new volume of 100.0 mL, so the final concentration of the diluted solution is
cFe31 5 [Fe31] 5
2.360  1024 mol Fe31
5 2.36 × 10−3 M
0.1000 L
Think about Your Answer The sample has been diluted 100-fold, so the final
concentration should be 1/100th of the initial value. See also Problem Solving Tip 4.4,
which gives another method for this calculation.
Check Your Understanding
An experiment calls for you to use 250. mL of 1.00 M NaOH, but you are given a large
bottle of 2.00 M NaOH. Describe how to make the desired volume of 1.00 M NaOH.
Problem Solving Tip 4.4 Preparing a Solution by Dilution
There is a straightforward method for
determining how to dilute a solution.
The central idea is that the amount
of solute in the final, dilute solution
must equal the amount of solute
taken from the more concentrated
solution. If c is the concentration
(molarity) and V is the volume (where
the subscripts d and c identify the
dilute and concentrated solutions,
respectively), then the amount of
214
solute in either solution (in the case
of the K2Cr2O7 example in the text)
can be calculated as follows:
(a) Amount of K2Cr2O7 in the final
dilute solution is
cd × Vd = 0.000500 mol
(b) Amount of K2Cr2O7 taken from the
more concentrated solution is
cc × Vc = 0.000500 mol
Both the concentrated and dilute
solutions contain the same amount of
solute. Therefore, you can use the following equation:
Amount in concentrated solution = Amount in dilute solution
cc  Vc 5 cd  Vd
One parameter can be calculated
if the other three are known.
Chapter 4 / Stoichiometry: Quantitative Information about Chemical Reactions
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A Closer Look
Serial Dilutions
0.01000 L × 0.550 mol/L = 5.500 × 10−3 mol NaCl
Suppose you have 100.0 mL of a sea­water
sample that has an NaCl concentration of
0.550 mol/L. You transfer 10.00 mL of that
sample to a 100.0-mL volumetric flask and
fill to the mark with distilled water. After
thoroughly mixing the diluted sample, you
then transfer 5.00 mL of that sample to another 100.0-mL flask and fill to the mark
with distilled water. What is the NaCl concentration in the final 100.0-mL sample?
The original solution contains 0.550 mol/L
of NaCl. If you remove 10.00 mL, you have
removed
You often will find that a solution you
initially made in a laboratory is too concentrated for the analytical technique
you want to use. Suppose you want to
analyze a seawater sample for its chloride ion content. To obtain a solution
with a chloride concentration of the
proper magnitude for analysis by the
Mohr method (“Applying Chemical
­Principles 4.3: How Much Salt Is There
in Seawater?,” page 233), for example,
you might want to dilute the sample, not once
but several times.
and the concentration in 100.0 mL of the
diluted solution is
cNaCl = 5.500 × 10−3 mol/0.1000 L = 5.500 × 10−2 M
or 1/10 of the concentration of the original
solution.
Now, take 5.00 mL of the diluted solution and dilute that once again to
100.0 mL. The final concentration is
0.00500 L × 5.500 × 10−2 mol/L = 2.750 × 10−4 mol NaCl
1
Transfer 5.00 mL
Transfer 10.0 mL
2
NaCl concentration
0.550 mol/L
100 mL
Original Solution
100.0-mL seawater sample
cNaCl = 2.750 × 10−4 mol/0.1000 L
= 2.75 × 10−3 M
3
4
Fill to mark with
distilled water
100 mL
1/10 original
concentration
10.0-mL sample
diluted to 100.0 mL
This is 1/200 of the concentration of the
original solution.
A fair question at this point is why not just
take 1 mL of the original solution and dilute
to 200 mL? The answer is that there is less
error introduced using larger pipets such as
5.00- or 10.00-mL pipets rather than a
1.00-mL pipet. There is also a limitation in
available glassware. A 200.00-mL volumetric flask is unlikely to be available.
Fill to mark with
distilled water
100 mL
1/200 original
concentration
5.00-mL sample
diluted to 100.0 mL
4.6 pH, a Concentration Scale
for Acids and Bases
Goals for Section 4.6
•
•
Understand the pH scale.
Calculate the pH of a solution from the concentration of hydronium ions in the
solution. Calculate the hydronium ion concentration in a solution from its pH.
The concentration of hydronium ions in different aqueous solutions can differ by
many powers of 10. For example, a sample of vinegar, which contains the weak acid
acetic acid, has a hydronium ion concentration of 1.6 × 10−3 M, and pure rainwater
has [H3O+] = 2.5 × 10−6 M. These small values can be expressed using scientific notation, but a more convenient way to express such numbers is the logarithmic pH scale.
The pH of a solution is the negative of the base-10 logarithm of the hydronium
ion concentration.
pH 5 2log[H3O1]
(4.4)
Taking vinegar, pure water, blood, and ammonia as examples,
= −log (1.6 × 10−3 M) = −(−2.80) = 2.80
acidic
pH of pure water (at 25 °C) = −log (1.0 × 10−7 M) = −(−7.00) = 7.00
neutral
pH of vinegar
pH of blood
= −log (4.0 × 10
M) = −(−7.40) = 7.40
basic
pH of household ammonia
= −log (4.3 × 10−12 M) = −(−11.37) = 11.37
basic
−8
Logarithms Numbers less than
1 have negative logs. Defining
pH as −log[H3O+] produces
a positive number if the H3O+
concentration is less than
1 molar. See Appendix A
for a discussion of logs.
4.6 pH, a Concentration Scale for Acids and Bases
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215
Vinegar
Soda
Orange
pH = 2.8 pH = 2.9 pH = 3.8
Ammonia
pH = 11.4
7
Oven cleaner
pH = 11.7
14
© Charles D. Winters/Cengage
0
Blood
pH = 7.4
Figure 4.9 pH values of some common substances. Here, the bar is colored red at 0, and gradually changes color to blue at 14.
These are the colors of litmus paper, commonly used in the laboratory to decide whether a solution is acidic (litmus is red) or basic
(litmus is blue).
Logs and Your Calculator All
scientific calculators have a
key marked “log.” To find an
antilog, use the key marked
“10x” or the inverse log. In
determining [H3O+] from a pH,
when you enter the value of
x for 10x, make sure it has a
negative sign.
These pH values are more convenient numbers to use than the numbers in scientific notation. You can see that something you recognize as acidic has a relatively
low pH, whereas ammonia, a common base, has a very low hydronium ion concentration and a high pH. For aqueous solutions at 25 °C, acids have pH values less
than 7, bases have values greater than 7, and a pH of 7 represents a neutral solution
(Figure 4.9). Blood is close to neutral; its pH is slightly greater than 7.
To find the hydronium ion concentration of a solution you calculate the antilog,
or inverse log, of the pH. That is,
[H3O1] 5 102pH
(4.5)
For example, the pH of a diet soda is 2.92, and the hydronium ion concentration of
the solution is
[H3O+] = 10−2.92 = 1.2 × 10−3 M
The approximate pH of a solution can also be determined using a variety of
dyes. Litmus paper, for example, contains a dye extracted from a type of lichen, but
many other dyes are also available (Figure 4.10a). A more accurate measurement of
pH uses a pH meter (such as the one in Figure 4.10b). Here, a pH electrode is immersed in the solution to be tested, and the pH is read from the instrument.
Photos: © Charles D. Winters/Cengage
Figure 4.10 Determining pH.
(a) Some household products. Each
solution contains a few drops of an acid–
base indicator. A color of yellow (left flask)
or red (middle flask) indicates a pH less
than 7. A green (right flask) to purple color
indicates a pH greater than 7.
216
(b) The pH of a soda is measured with a pH
meter. Soft drinks are often quite acidic, owing
to dissolved CO2 and other ingredients.
Chapter 4 / Stoichiometry: Quantitative Information about Chemical Reactions
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Ex am p le 4.10
pH of Solutions
Weak and Strong Acids and
Hydronium Ion Concentration
Problem ​
(a) Lemon juice has [H3O+] = 0.0032 M. What is its pH?
(b) Seawater has a pH of 8.07. What is the hydronium ion concentration of this solution?
(c) A solution of nitric acid has a concentration of 0.0056 mol/L. What is the pH of this
solution?
What Do You Know? ​In part (a) you are given a concentration and asked to calculate the pH, whereas the opposite is true in (b). For part (c), however, you first must
recognize that HNO3 is a monoprotic acid with only one ionizable hydrogen atom per
molecule and is also a strong acid so it is approximately 100% ionized in water.
Because a weak acid (such as
acetic acid) does not ionize
completely in water, the
hydronium ion concentration
in an aqueous solution of a
weak monoprotic acid is less
than the concentration of the
acid. In contrast, the hydronium
ion concentration in strong
monoprotic acid solutions is the
same as the acid concentration.
Strategy ​Use Equation 4.4 to calculate pH from the H3O + concentration and
­Equation 4.5 to find [H3O+] from the pH.
Solution ​
(a) Lemon juice: When the hydronium ion concentration is known, the pH is found using
Equation 4.4.
pH = −log [H3O+] = −log (3.2 × 10−3) = −(−2.49) = 2.49
(b) Seawater: Here pH = 8.07. Therefore,
[H3O+] = 10−pH = 10−8.07 = 8.5 × 10−9 M
(c) Nitric acid: Nitric acid, a monoprotic strong acid (Table 3.1, page 157), is assumed to be
completely ionized in aqueous solution. Because the concentration of HNO3 is 0.0056
mol/L, the ion concentrations are also 0.0056 mol/L.
[H3O+] = [NO3−] = 0.0056 M
pH = −log [H3O+] = −log (0.0056 M) = 2.25
Think about Your Answer ​A comment on logarithms and significant figures
(Appendix A) is useful. The number to the left of the decimal point in a logarithm is called
the characteristic, and the number to the right is the mantissa. The mantissa has as many
significant figures as the number whose log was found. For example, the logarithm of
3.2 × 10−3 (two significant figures) is –2.49. (The significant figures are the two numbers
to the right of the decimal point.)
Check Your Understanding
(a) What is the pH of a solution of HCl in which [HCl] = 2.6 × 10−2 M?
(b) What is the hydronium ion concentration in orange juice with a pH of 3.80?
4.7 Stoichiometry of Reactions
in Aqueous Solution—Fundamentals
Goal for Section 4.7
• Use stoichiometry principles for reactions occurring in solution.
In the laboratory, chemists carry out most reactions in solutions, usually in water.
Even in nature, water is a common solvent. For example, the reactions in our bodies
occur in an aqueous environment.
4.7 Stoichiometry of Reactions in Aqueous Solution—Fundamentals
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217
One example of a reaction occurring with at least one reactant in solution is a
gas-forming exchange reaction involving a metal carbonate and an aqueous acid
(Figure 4.11)
CaCO3(s)
+ 2 HCl(aq) n CaCl2(aq) + H2O(ℓ) +
© Charles D. Winters/Cengage
metal carbonate +
n
salt
+
water
CO2(g)
+ carbon dioxide
What mass of CaCO3 is required to react completely with 25 mL of 0.750 M HCl?
This problem differs from the previous stoichiometry problems in that the quantity of one reactant (HCl) is given as the volume of a solution with a known concentration instead of as a mass in grams. Because the balanced equation is written
in terms of amounts (moles), the first step is to determine the amount of HCl
present so the amount of HCl available can be related to the amount of CaCO3
required.
Figure 4.11 A commercial
remedy for excess stomach
acid. The tablet contains calcium
carbonate, which reacts with
hydrochloric acid, the acid
present in the digestive system.
The most visible product is
CO2 gas, but CaCl2(aq) is also
produced.
acid
Amount of HCl 5 cHCl  VHCl 5
0.750 mol HCl
 0.025 L 5 0.0188 mol HCl
1 L
This is then related to the amount of CaCO3 required using the stoichiometric factor
from the balanced equation.
0.0188 mol HCl 
1 mol CaCO3
5 0.00938 mol CaCO3
2 mol HCl
Finally, the amount of CaCO3 is converted to a mass in grams using the molar mass
of CaCO3 as the conversion factor.
0.00938 mol CaCO3 
100. g CaCO3
5 0.94 g CaCO3
1 mol CaCO3
If you follow the general scheme outlined in Problem-Solving Tip 4.5 and pay
attention to the units on the numbers, you can successfully carry out any kind of
stoichiometry calculation that involves concentrations.
Problem Solving Tip 4.5 Stoichiometry Calculations Involving Solutions
In Problem Solving Tip 4.1, you
learned about a general approach
to stoichiometry problems. The
scheme can be modified for a reaction involving solutions such as
x A(aq) + y B(aq) n products.
grams reactant A
×
1 mol A
gA
grams reactant B
direct calculation
not possible
moles reactant A
× c molarity A
cA =
mol A
L soln.
gB
1 mol B
moles reactant B
×
mol reactant B
mol reactant A
stoichiometric factor
Volume of soln. A
218
×
×
1
cmolarity B
L soln.
1
=
cB
mol B
Volume of soln. B
Chapter 4 / Stoichiometry: Quantitative Information about Chemical Reactions
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Ex am p le 4.11
Stoichiometry of a Reaction in Solution
Strategy Map
Problem
Calculate volume of HCl
solution required to consume
given mass of a reactant (Zn)
Problem ​Metallic zinc reacts with aqueous HCl (Figure 3.12b, page 156).
Zn(s) + 2 HCl(aq) n ZnCl2(aq) + H2(g)
What volume of 2.50 M HCl, in milliliters, is required to completely consume 11.8 g of Zn?
What Do You Know? You have the balanced equation for the reaction of Zn and
HCl(aq) and know the mass of zinc and the concentration of HCl(aq).
Strategy ​
Data/Information
• Mass of Zn
• Concentration of HCl
• Balanced equation
Step 1. Calculate the amount of zinc.
Step 2. Use a stoichiometric factor (= 2 mol HCl/1 mol Zn) to relate the amount (moles)
of Zn available to the amount (moles) of HCl required.
Step 3. Calculate the volume of HCl solution from the amount of HCl and its concentration.
Solution ​
Step 1
Calculate the amount (moles) of Zn.
Begin by calculating the amount of Zn from the mass and molar mass of Zn.
11.8 g Zn 
Step 2
1 mol Zn
5 0.1805 mol Zn
65.38 g Zn
Use a stoichiometric factor to relate the amount of Zn to the amount of HCl
required.
The balanced chemical equation tells you that 2 moles of HCl are required per mole of Zn.
0.1805 mol Zn 
Step 3
2 mol HCl
5 0.3610 mol HCl
1 mol Zn
Calculate the volume of HCl solution required.
Use the concentration (mol/L) of HCl to convert the amount (moles) of HCl to the volume
(L) of HCl required.
0.3610 mol HCl 
1.00 L solution
5 0.144 L of 2.50 M HCl
2.50 mol HCl
The answer is requested in units of milliliters, so the calculated volume is converted to
milliliters. That is, 144 mL of 2.50 M HCl is required to convert 11.8 g of Zn completely to
products.
Think about Your Answer You began with 0.180 mol of zinc. Because the concentration of the HCl solution is 2.50 mol/L, it makes sense that significantly less than 1 L
of HCl solution is needed. Notice also that this is a redox reaction in which zinc is oxidized
(oxidation number changes from 0 to +2) and hydrogen, in HCl(aq), is reduced (its oxidation number changes from +1 to 0).
Check Your Understanding
​If you combine 75.0 mL of 0.350 M HCl and an excess of Na2CO3, what mass of CO2, in grams, is produced?
Na2CO3(s) + 2 HCl(aq) n 2 NaCl(aq) + H2O(ℓ) + CO2(g)
4.7 Stoichiometry of Reactions in Aqueous Solution—Fundamentals
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219
4.8 Stoichiometry of Reactions in
Aqueous Solution—Titrations
Goal for Section 4.8
•
Explain how a titration is conducted, explain the procedure for standardization of a
solution, and calculate concentrations or amounts of reactants from titration data.
Titration: A Method of Chemical Analysis
Suppose you are asked to determine the concentration of hydrochloric acid, HCl, in
a solution. Because the compound is an acid, it reacts with a base such as sodium
hydroxide.
HCl(aq) + NaOH(aq) n NaCl(aq) + H2O(ℓ)
You can use this reaction to determine the concentration of hydrochloric acid present
in a given volume of hydrochloric acid solution if the following conditions are met:
•
You can determine when the amount of sodium hydroxide added is exactly the
amount required to react with all the hydrochloric acid present in solution.
•
You know the concentration of the sodium hydroxide solution and volume
added at exactly the point of complete reaction.
These conditions are fulfilled in a titration, a procedure illustrated in ­Figure 4.12.
The solution containing hydrochloric acid is placed in a flask along with an
acid–base indicator, a dye that changes color when the pH of the reaction solution
reaches a certain value. An aqueous sodium hydroxide solution of accurately
known concentration is placed in a buret. The sodium hydroxide in the buret is
added slowly to the acid solution in the flask. As long as some acid is present in
Buret containing
a base of known
concentration
Photos: © Charles D. Winters/Cengage
Base added
from buret
Flask containing aqueous solution of sample
being analyzed and
an indicator
1 A buret, a volumetric measuring device
calibrated in divisions of 0.1 mL (and
consequently read to the nearest 0.01
mL), is filled with an aqueous solution
of a base of known concentration.
2 Base is added slowly from the buret
3 When the amount of NaOH added
to the solution containing the acid
being analyzed and an indicator.
from the buret equals the amount of
H3O+ supplied by the acid (the
equivalence point), the dye (the
indicator) changes color. (The indicator
used here is phenolphthalein.)
Figure 4.12 Titration of an acid in aqueous solution with a base.
220
Chapter 4 / Stoichiometry: Quantitative Information about Chemical Reactions
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solution, all the base supplied from the buret is consumed, the solution remains
acidic, and the indicator color is unchanged. At some point—the equivalence
point—the amount of OH− supplied by the base exactly equals the amount of
H3O+ supplied by the acid. In the vicinity of the equivalence point, the pH rises
rapidly with each additional drop of NaOH. The titration is finished when one
­final drop of NaOH raises the pH sufficiently to change the color of the acid–base
indicator. If the indicator is chosen properly, its color will change at, or very near,
the equivalence point (see Figure 4.12). Example 4.12 shows how to use the equivalence point to determine the concentration of HCl in the original solution.
Ex am p le 4.12
Acid–Base Titration
Problem ​A 15.00-mL sample of an aqueous hydrochloric acid solution with an
unknown concentration is transferred to a flask, and an acid–base indicator is added. After
adding 29.27 mL of 0.09977 M NaOH, the acid–base indicator changes color, indicating
that the equivalence point has been reached. What is the concentration of hydrochloric
acid in the original solution?
What Do You Know? ​You know the volume of the solution of hydrochloric acid
used and the volume and concentration of NaOH used in the titration.
Strategy ​
Step 1. Write a balanced chemical equation for the acid–base reaction.
Step 2. Calculate the amount of NaOH used in the titration from its volume and
concentration.
Step 3. Use the stoichiometric factor defined by the balanced equation to determine
the amount of HCl.
Step 4. Determine the molarity of the original solution using the amount of HCl and the
initial volume used.
Solution
Step 1. ​The balanced equation for the reaction of HCl and NaOH is
HCl(aq) + NaOH(aq) n NaCl(aq) + H2O(ℓ)
Step 2. The amount of NaOH used in the titration is given by
Amount of NaOH 5 cNaOH  VNaOH
5


0.09977 mol NaOH
1L
  29.27 mL 
5 0.0029203 mol NaOH
L
1000 mL 

Step 3. The balanced equation for the reaction shows that 1 mol of hydrochloric acid
reacts for every 1 mol of sodium hydroxide.
0.0029203 mol NaOH 
1 mol HCl
5 0.0029203 mol HCl
1 mol NaOH
Step 4. The concentration of HCl in the initial solution is determined by dividing the
amount of HCl by the volume (in L) of the HCl used for the titration.
15.00 mL 
1L
5 0.01500 L
1000 mL
0.0029203 mol HCl
cHCl 5
= 0.1947 M HCl
0.01500 L
4.8 Stoichiometry of Reactions in Aqueous Solution—Titrations
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221
Think about Your Answer ​HCl and NaOH react in a 1:1 ratio. The volume of
NaOH solution required is close to twice the volume of HCl solution used. It makes sense
that the concentration of HCl is about twice that of the NaOH solution.
Check Your Understanding
What is the concentration of sulfuric acid (H2SO4) in a solution if a 10.00-mL sample of this
solution requires 47.32 mL of 0.1122 M NaOH to convert the H2SO4 to SO42–? The balanced
equation for the reaction is
H2SO4(aq) + 2 NaOH(aq) n Na2SO4(aq) + 2 H2O(ℓ)
Exampl e 4 .1 3
Determining Purity by Acid–Base Titration
Strategy Map
Problem
Calculate the mass percent
of acid in an impure sample.
Determine the acid content
using an acid–base titration.
Problem ​A 1.034-g sample of impure oxalic acid, H2C2O4 (a diprotic acid), is dissolved
in water and an acid–base indicator added. The sample requires 34.47 mL of 0.485 M
NaOH to titrate the acid. One mole of acid requires two moles of NaOH for a complete
reaction. What is the mass of oxalic acid in the sample, and what is its mass percent?
What Do You Know? ​You know the mass of the impure oxalic acid sample and
the volume and concentration of NaOH solution used in the titration.
Strategy
Data/Information
• Mass of impure sample
containing acid.
• Volume and concentration
of base used in titration.
Step 1. Write a balanced chemical equation for this acid–base reaction.
Step 2. Calculate the amount of NaOH used in the titration from its volume and
concentration.
Step 3. Use the stoichiometric factor defined by the balanced equation to determine
the amount of H2C2O4.
Step 4. Calculate the mass of H2C2O4 from the amount and its molar mass.
Step 5. Determine the percent by mass of H2C2O4 in the sample.
Solution ​
Step 1
Write the balanced equation.
The balanced equation for the reaction of NaOH and H2C2O4 is
H2C2O4(aq) + 2 NaOH(aq) n Na2C2O4(aq) + 2 H2O(ℓ)
Step 2
Calculate the amount (moles) of NaOH used in the reaction.
Multiply the concentration (mol/L) of the NaOH solution by the volume (L) delivered
(Equation 4.3) to determine the amount of NaOH used in the reaction.
Amount of NaOH 5 cNaOH  VNaOH 5
Step 3
0.485 mol NaOH
 0.03447 L 5 0.01672 mol NaOH
L
Use a stoichiometric factor to relate the amount of NaOH to the amount of
oxalic acid.
The balanced equation for the reaction shows that 1 mol of oxalic acid requires 2 mol of
sodium hydroxide.
1 mol H2C2O4
0.01672 mol NaOH 
5 0.008359 mol H2C2O4
2 mol NaOH
222
Chapter 4 / Stoichiometry: Quantitative Information about Chemical Reactions
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Step 4
Calculate the mass of oxalic acid in the sample.
The mass of oxalic acid is found using the amount of the acid and its molar mass.
0.008359 mol H2C2O4 
Step 5
90.03 g H2C2O4
5 0.7526 g H2C2O4 5 0.753 g H2C2O4
1 mol H2C2O4
Calculate the mass percent of oxalic acid in the original sample.
Calculate the mass percent by dividing the mass of H2C2O4 by the original sample mass,
then multiplying by 100.
0.7526 g H2C2O4
 100% 5 72.8% H2C2O4
1.034 g sample
Think about Your Answer ​Problem Solving Tip 4.5 outlines the procedure used
to solve this problem.
Check Your Understanding
A 25.0-mL sample of vinegar (which contains the weak acid acetic acid, CH3CO2H) requires 28.33 mL of a 0.953 M solution of
NaOH for titration to the equivalence point. What is the mass of acetic acid (molar mass 5 60.05 g/mol), in grams, in the vinegar
sample, and what is the concentration of acetic acid in the vinegar?
CH3CO2H(aq) + NaOH(aq) n NaCH3CO2(aq) + H2O(ℓ)
Standardizing an Acid or Base
In Examples 4.12 and 4.13, the concentration of the base used in the titration was
given. In practice, this is usually found in a prior experiment. The procedure by
which the concentration of an analytical reagent is determined accurately is called
standardization, and there are two general approaches.
One approach is to weigh accurately a sample of a pure, solid acid or base
(known as a primary standard) and then titrate this sample with a solution of the
base or acid to be standardized (Example 4.14). An alternative approach to
standardizing a solution is to titrate it with another solution that is already
­
­standardized (see “Check Your Understanding” in Example 4.14). This is often done
using standard solutions purchased from chemical supply companies.
Exam p le 4.14
Standardizing an Acid by Titration
Problem ​Sodium carbonate, Na2CO3, is a base, and an accurately weighed sample can
be used to standardize an acid. A sample of sodium carbonate (0.263 g) requires 28.35 mL
of aqueous HCl for titration to the equivalence point, where the carbonate ion is converted
to carbon dioxide. What is the concentration of the HCl?
What Do You Know? ​The concentration of the HCl(aq) solution is the unknown
in this problem. You know the mass of Na2CO3 and the volume of HCl(aq) solution needed
to react completely with the Na2CO3. You need the molar mass of Na2CO3 and a balanced
equation for the reaction.
4.8 Stoichiometry of Reactions in Aqueous Solution—Titrations
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223
Strategy ​
Step 1. Write a balanced equation for this gas-forming acid–base reaction.
Step 2. Calculate the amount of Na2CO3 from its mass and molar mass.
Step 3. Use the stoichiometric factor (from the balanced equation) to find the amount
of HCl(aq).
Step 4. Calculate the concentration of HCl by dividing the amount of HCl by the ­volume
of the solution (in liters).
Solution ​
Step 1. The balanced equation for the reaction is written first.
Na2CO3(aq) + 2 HCl(aq) n 2 NaCl(aq) + H2O(ℓ) + CO2(g)
Step 2. Calculate the amount of the base, Na2CO3, from its mass and molar mass.
0.263 g Na2CO3 
1 mol Na2CO3
5 0.002481 mol Na2CO3
106.0 g Na2CO3
Step 3. Use the stoichiometric factor to calculate the amount of HCl in 28.35 mL.
0.002481 mol Na2CO3 
2 mol HCl required
5 0.004962 mol HCl
1 mol Na2CO3 available
Step 4. Calculate the concentration of the HCl solution by dividing the amount of HCl
by the volume of HCl used in the titration.
[HCl] 5
0.004962 mol HCl
5 0.175 M HCl
0.02835 L
Think about Your Answer ​Sodium carbonate is commonly used as a primary
standard. It can be obtained in pure form, can be weighed accurately, and it reacts completely with strong acids.
Check Your Understanding
​Hydrochloric acid, HCl, with a concentration of 0.100 M can be purchased from chemical
supply houses, and this solution can be used to standardize the solution of a base. If titrating 25.00 mL of a sodium hydroxide solution to the equivalence point requires 29.67 mL
of 0.100 M HCl, what is the concentration of the base?
Determining Molar Mass by Titration
In Chapter 2 and in this chapter we used analytical data to determine the empirical
formula of a compound. The molecular formula could then be derived if the molar
mass were known. If the unknown substance is an acid or a base, it is possible to
determine the molar mass by titration.
Exam pl e 4 .1 5
Determining the Molar Mass of an Acid by Titration
Problem ​A 1.056-g sample of an unknown organic acid, HA, is titrated with standardized NaOH (that is, with a NaOH solution whose concentration is accurately known).
­Calculate the molar mass of HA knowing the acid sample reacts with 33.78 mL of 0.256 M
NaOH according to the equation
HA(aq) + NaOH(aq) n NaA(aq) + H2O(ℓ)
224
Chapter 4 / Stoichiometry: Quantitative Information about Chemical Reactions
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Strategy Map
Problem
Calculate the molar mass of
an acid, HA, using an acid–
base titration.
Data/Information
• Mass of acid sample
• Volume and concentration
of base used in titration.
• Balanced equation
What Do You Know? ​You know the mass of the sample of unknown acid, and the
volume and concentration of NaOH(aq). From the balanced chemical equation given, you
know that the acid and base react in a 1∶1 ratio.
Strategy ​The key to this problem is to recognize that the molar mass of a substance
is the ratio of the mass of the sample (g) to its amount (mol). You know the mass, but you
need to determine the amount equivalent to that mass. The balanced chemical equation
informs you that 1 mol of HA reacts with 1 mol of NaOH, so the amount of HA equals the
amount of NaOH used in the titration. The latter can be calculated from its concentration
and volume.
Solution ​
Step 1
Calculate the amount (moles) of NaOH used in the titration.
Multiply the concentration (mol/L) by the volume (L) of NaOH (Equation 4.3) to calculate
the amount (moles) of NaOH used in the titration.
Amount of NaOH 5 cNaOH  VNaOH 5
Step 2
0.256 mol
 0.03378 L 5 8.648  1023 mol NaOH
L
Use a stoichiometric factor to relate the amount of base to the amount of acid.
The balanced chemical equation tells you that the amount of NaOH used in the titration is
the same as the amount of acid titrated.
8.648  1023 mol NaOH 
Step 3
1 mol HA
5 8.648  1023 mol HA
1 mol NaOH
Calculate the molar mass of the acid.
The ratio of the mass of HA to its amount is the molar mass.
Molar mass of acid 5
1.056 g HA
5 122 g/mol
8.648  1023 mol HA
Think about Your Answer ​The molar masses of common water-soluble acids
range from 20 g/mol (for HF) to a few hundred grams per mole.
Check Your Understanding
An unknown monoprotic acid reacts with NaOH according to the net ionic equation
HA(aq) + OH−(aq) n A−(aq) + H2O(ℓ)
Calculate the molar mass of HA if 0.856 g of the acid requires 30.08 mL of 0.323 M NaOH.
Titrations Using Oxidation–Reduction Reactions
Many oxidation–reduction reactions can be used in chemical analysis because the
reactions go rapidly to completion in aqueous solution, and methods exist to determine their equivalence points.
4.8 Stoichiometry of Reactions in Aqueous Solution—Titrations
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225
E xamp le 4.16
KMnO4 in
buret
Using an Oxidation–Reduction Reaction
in a Titration
Problem ​The iron in a sample of an iron ore can be converted quantitatively to the
iron(II) ion, Fe2+, in aqueous solution, and this solution can then be titrated with aqueous
potassium permanganate, KMnO4. The balanced, net ionic equation for the reaction occurring in this titration is
MnO4−(aq) + 5 Fe2+(aq) + 8 H3O+(aq) n Mn2+(aq) + 5 Fe3+(aq) + 12 H2O(ℓ)
Using an oxidation–reduction
reaction for analysis by titration.
Purple, aqueous KMnO4 is added
to a solution containing Fe2+. As
KMnO4 drops into the solution,
colorless Mn2+ and pale yellow
Fe3+ form.
purple
© Charles D. Winters/Cengage
Fe2+(aq)
in flask
colorless
colorless
pale yellow
A 1.026-g sample of iron-containing ore requires 24.35 mL of 0.0195 M KMnO4 to reach the
equivalence point. What is the mass percent of iron in the ore?
What Do You Know? ​You know the concentration and volume of the KMnO4 solution used to titrate Fe2+(aq) to the equivalence point. The stoichiometric factor relating
amounts of KMnO4 and Fe2+(aq) is derived from the balanced equation.
Strategy
Step 1. Use the volume and concentration of the KMnO4 solution to calculate the
amount of KMnO4 used in the titration.
Step 2. Use the stoichiometric factor to determine the amount of Fe2+ from the amount
of KMnO4.
Step 3. Convert the amount of Fe2+ to mass of iron using the molar mass of iron.
Step 4. Calculate the mass percent of iron in the sample.
Solution ​
Step 1. Calculate the amount of KMnO4.
Amount of KMnO4 5 cKMnO4  VKMnO4 5
0.0195 mol KMnO4
 0.02435 L 5 0.0004748 mol
L
Step 2. Use the stoichiometric factor to calculate the amount of iron(II) ion.
0.0004748 mol KMnO4 
5 mol Fe21
5 0.002374 mol Fe21
1 mol KMnO4
Step 3. Next, calculate the mass of iron.
0.002374 mol Fe21 ×
55.85 g Fe21
5 0.1326 g Fe21
1 mol Fe21
Step 4. Finally, determine the mass percent.
0.1326 g Fe21
 100% 5 12.9% iron
1.026 g sample
Think about Your Answer The reaction of iron(II) ions with KMnO4 is well-suited
for use in a titration because it is easy to detect when all the iron(II) ion has reacted. The
MnO4− ion is deep purple, but the reaction product, Mn2+, is colorless. Therefore, KMnO4
solution is added from a buret until the initially colorless, Fe2+-containing solution just
turns a faint purple color (due to a trace of unreacted KMnO4), the signal that the equivalence point has been reached. Because of this color change at the equivalence point, there
is no need to use a separate indicator for this titration.
226
Chapter 4 / Stoichiometry: Quantitative Information about Chemical Reactions
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Check Your Understanding
​Vitamin C, ascorbic acid (C6H8O6) (molar mass = 176.1 g/mol), is a reducing agent. One
way to determine the ascorbic acid content of a sample is to mix the acid with an excess
of iodine,
C6H8O6(aq) + I2(aq) + 2 H2O(ℓ) n C6H6O6(aq) + 2 H3O+(aq) + 2 I−(aq)
and then titrate the iodine that did not react with the ascorbic acid with sodium thiosulfate. The balanced, net ionic equation for the reaction occurring in this titration is
I2(aq) + 2 S2O32−(aq) n 2 I−(aq) + S4O62−(aq)
Suppose 50.00 mL of 0.0520 M I 2 was added to the sample containing ascorbic acid.
After the ascorbic acid/I2 reaction was complete, the I2 not used in this reaction required
20.30 mL of 0.196 M Na2S2O3 for titration to the equivalence point. Calculate the mass of
ascorbic acid in the unknown sample.
4.9 Spectrophotometry
Goal for Section 4.9
•
Understand and use the principles of spectrophotometry to determine the
concentration of a colored compound or ion in solution.
Solutions of many compounds are colored, a consequence of the absorption of light
(Figure 4.13). It is possible to measure, quantitatively, the extent of light absorption
and to relate this to the concentration of the dissolved solute. This is an example of
the use of spectrophotometry, an important analytical method and one you may
use in your laboratory course. Spectrophotometry is one of the most frequently used
methods of quantitative analysis. It is applicable to many industrial, clinical, and
forensic problems involving the quantitative determination of compounds that are
colored or that react to form a colored product.
Figure 4.13 Light absorption
and color.
The light that emerges is green.
A beam of white light
shines on a solution of
nickel(II) ions in water.
© Charles D. Winters/Cengage
The color of a solution is due to the color
of the light not absorbed by the solution.
Here, red and blue/violet light are
absorbed, and green light is transmitted.
4.9 Spectrophotometry
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227
Glowing
filament
A beam of white light
passes through a prism,
which splits the light into
its component wavelengths.
Prism or
diffraction
grating
Transmitted
light
The intensity of the transmitted
light versus wavelength is recorded
by the detector. These intensities
are then converted to absorbances.
Absorbance
Selected
wavelength
A solution
of a colored
compound
700
600
500
400
Wavelength of incident light (nm)
Figure 4.14 An absorption spectrophotometer. A spectrophotometer scans all wavelengths
of light and determines the absorbance at each wavelength. The output is an absorption
spectrum, a plot of absorbance as a function of the wavelength or frequency of the incoming or
incident light. Here, the sample absorbs light in the green-blue part of the spectrum and transmits
light in the remaining wavelengths. The sample would appear red to orange to your eye.
Every substance absorbs or transmits certain wavelengths of radiant energy but
not others (Figures 4.13 and 4.14). For example, nickel(II) ions (and chlorophyll)
absorb red and blue/violet light while they transmit green light. Your eyes see the
transmitted or reflected light as the color green. Furthermore, the specific wavelengths of light absorbed and transmitted are characteristic for a substance and can
help with the identification of an unknown.
Now look at two solutions containing copper(II) ions in test tubes of equal
diameter (Figure 4.15a). One is more intensely colored than the other. The intensity
of the color in containers of equal diameter is a measure of the concentration of the
color-producing material in the solution.
Transmittance, Absorbance, and the Beer–Lambert Law
To understand the exact relationship of light absorption and solution concentration, several terms must be defined. Transmittance (T) is the ratio of the amount of
light transmitted by or passing through the sample (P) relative to the amount of
light that initially fell on the sample (the incident light, P0).
Po
Incident light
Transmittance (T ) 5
P
Sample
Transmitted light
P
intensity of transmitted light
5
Po
intensity of incident light
The absorbance of a sample is defined as the negative logarithm of its transmittance. That is, absorbance and transmittance have an inverse relationship. As the
transmittance of a solution increases, the absorbance decreases
Absorbance = −log T = −log P/Po
The solutions in Figure 4.15 illustrate transmittance and absorbance. Figure
4.15a shows solutions with different concentrations of copper(II) sulfate in test
tubes of the same diameter. Here you may deduce that the bluer solution appears
more blue because this solution has a greater concentration of copper(II) sulfate.
That is, the absorbance, A, of a sample increases as the concentration increases.
Next, suppose that there are two test tubes of different diameter, both containing the same solution at the same concentration (Figure 4.15b). Light of the same
intensity (P0) is shined on both test tubes. In the narrower-diameter tube, the light
has to travel only a short distance through the sample before its strikes your eyes,
whereas in the other tube it has to pass through more of the sample. In the widerdiameter tube more of the light will be absorbed because the path length is longer.
In other words, absorbance increases as path length increases.
228
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0.05 M
CuSO4
1.0 M
CuSO4
1.0 M
CuSO4
(a) Test tubes of the same diameter
contain copper(II) sulfate solutions of
different concentrations. More light is
absorbed by the more concentrated
solution, and it appears darker blue.
1.0 M
CuSO4
Photos: © Charles D. Winters/Cengage
Figure 4.15 Light absorption,
concentration, and path length.
(b) Here the test tubes have copper(II)
sulfate solutions of the same concentration.
However, the distance the light travels is
longer in one than in the other.
The two observations described above constitute the Beer–Lambert law:
Absorbance (A) ∝ path length (ℓ) × concentration (c)
(4.6)
A=ε×ℓ×c
where
•
A, the absorbance of the sample, is a dimensionless number.
•
ε, a proportionality constant, is called the molar absorptivity (L/mol ⋅ cm). For a given
substance the molar absorptivity varies with wavelength and temperature, so when
doing spectrophotometric experiments these parameters must be kept constant.
•
ℓ and c have the units of length (cm) and concentration (mol/L), respectively.
Beer–Lambert Law The Beer–
Lambert law applies strictly to
relatively dilute solutions. At
higher solute concentrations, the
dependence of absorbance on
concentration may not be linear.
The Beer–Lambert law shows there is a linear relationship between a sample’s
­absorbance and its concentration for a given path length.
Spectrophotometric Analysis
There are usually four steps in carrying out a spectrophotometric analysis.
1.
Record the absorption spectrum of the substance to be analyzed. In introductory
chemistry laboratories, this is often done using instruments such as the one shown
in Figure 4.16. The result is a spectrum such as that for aqueous permanganate
Figure 4.16 ​
Spectrophotometer. ​Such
Martyn F. Chillmaid/Science Source
instruments are often found
in introductory chemistry
laboratories.
4.9 Spectrophotometry
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229
0.8
Curve 1
Absorbance
0.6
0.4
Curve 2
0.2
0
400
450
500
550
600
650
700
λ,nm
Figure 4.17 The absorption
spectrum of solutions of potassium
permanganate (KMnO4) at different concentrations. The solution for
curve 1 has a higher concentration
than that for curve 2.
ions (MnO4−) in Figure 4.17. The spectrum is a plot of the absorbance of the
sample as a function of the wavelength of the incident light. Here, the maximum
absorbance is at about 525 nm.
2. Choose the wavelength for the measurement. The absorbance at each wavelength is
proportional to concentration. Therefore, in theory you could choose any wavelength where light is absorbed for quantitative estimations of concentration.
­However, the magnitude of the absorbance is important, especially when you are
trying to detect very small amounts of material. In the spectra of permanganate ions
in Figure 4.17, note that the difference in absorbance between curves 1 and 2 is
largest at about 525 nm, and at this wavelength the change in absorbance is greatest
for a given change in concentration. That is, the measurement of absorbance as a
function of concentration is most sensitive at this wavelength. For this reason, the
wavelength of maximum absorbance is usually selected for the measurements.
3. Prepare a calibration plot. Once you have chosen the wavelength, the next step
is to construct a calibration curve or calibration plot at this wavelength. This
consists of a plot of absorbance as a function of concentration for a series of
standard solutions whose concentrations are accurately known. Because of the
linear relation between concentration and absorbance (at a given wavelength
and path length), this plot is a straight line with a positive slope. (You will prepare a calibration plot in Example 4.17.)
4. Determine the concentration of the species of interest in other solutions. Once the
calibration plot is made, and the equation for the line determined, you can find
the concentration of an unknown sample from its absorbance.
E xamp le 4.17
Using Spectrophotometry in Chemical Analysis
Problem A solution of KMnO4 has an absorbance of 0.539 when measured at 540 nm
in a 1.0-cm cell. What is the concentration of the KMnO4? Prior to determining the absorbance for the unknown solution, the following calibration data were collected.
Concentration of KMnO4 (M)
Absorbance
0.0300
0.162
0.0600
0.330
0.0900
0.499
0.120
0.670
0.150
0.840
What Do You Know? The table relates concentration and absorbance (at 540 nm
in a 1.0-cm cell) for aqueous solutions of KMnO4. The absorbance of the unknown sample
under the same conditions is given.
Strategy Prepare a calibration plot from the data given above, and then use this plot
to estimate the concentration of the unknown from its absorbance. A more accurate value
of the concentration can be obtained if you find the equation for the straight line in the
calibration plot and calculate the unknown concentration using this equation.
230
Chapter 4 / Stoichiometry: Quantitative Information about Chemical Reactions
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Solution Using Microsoft Excel (or equivalent software) or a calculator, prepare a calibration plot from the experimental data.
0.900
0.800
0.700
Absorbance
0.600
0.500
0.400
0.300
0.200
0.100
0.000
0.0000
0.0500
0.1000
0.1500
0.2000
Concentration (M)
The equation for the straight line (as determined using Excel) is
y = 5.653x − 0.009
Absorbance = 5.653 c − 0.009
If you put in the absorbance for the unknown solution,
0.539 = 5.653 c − 0.009
Unknown concentration (c) = 0.0969 M
Think about Your Answer The absorbance of the unknown was 0.539. L­ ooking
back at the calibration data, notice that this absorbance falls between the data points
for absorbances of 0.499 and 0.670. The answer determined for the concentration of
KMnO 4 in the unknown (0.0969 M) falls between the concentrations for these two
data points of 0.0900 M and 0.120 M, as it should. (See pages 48–49 for information on
graphing.)
Check Your Understanding
A solution of copper(II) ions has an absorbance of 0.418 when measured at 645 nm in a
1.0-cm cell. Using the following data, calculate the concentration of copper(II) ions in the
unknown solution.
Calibration Data
2+
Concentration of Cu
(M)
Absorbance
0.0562
0.720
0.0337
0.434
0.0281
0.332
0.0169
0.219
4.9 Spectrophotometry
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231
Applying Chemical Principles
4.1 Atom Economy
Chemists and chemical industries are increasingly following
the principles of green chemistry. One of these principles is to
convert the atoms of the reactants into the product as efficiently as possible, and one way to evaluate the efficiency of a
reaction is to calculate its atom economy.
% atom economy 5
molar mass of atoms utilized
 100%
molar mass of reactants
Gado Images/Gado/Alamy Stock Photo
An example of this concept is the reaction of methanol and
carbon monoxide to produce acetic acid. The atom economy is
100% because all of the atoms of the reactants appear in the
product.
CH3OH + CO n CH3CO2H
Atom economy is often used to develop synthetic methods for
producing commonly used chemical compounds. One such
compound is methyl methacrylate.
A protective barrier, probably made of
PMMA, to separate people from each
other during the COVID-19 pandemic.
CH2 O
H3C
C
C
O
CH3
Methyl methacrylate
Methyl methacrylate is the building block for a type of plastic
called polymethyl methacrylate (PMMA), which was first
introduced in 1933. PPMA is more commonly known as
­
acrylic, Lucite™, Plexiglas™, or another trademarked name.
Transparent sheets of PMMA can have optical clarity similar to
glass with greater impact resistance and roughly half the
weight. These properties make it a good choice for shatterproof
windows, automobile taillights, and the lenses used in cameras and stoplights. Because PMMA is biocompatible (it is not
toxic and is not rejected by the body), it is also used as a bone
cement in joint replacement surgery, in dentures, and in hard
contact lenses. During the COVID-19 pandemic, PMMA sheets
found a new use as protective barriers between people in close
proximity, like in businesses and schools.
Because of its use in PMMA and other materials, there is a
large worldwide demand for methyl methacrylate, and chemists have long tried to find better and less costly production
methods. In 2021, the companies Röhm and OQ Chemicals
announced plans to build a methyl methacrylate plant in Texas
using a method called LiMA (Leading in Methacrylates) that
was developed when Röhm was part of the company Evonik.
This plant should eventually be able to produce 250 million
kilograms of methyl methacrylate per year.
Question
1. In the LiMA process, ethylene (C2H4), methanol (CH3OH),
CO, formaldehyde (CH2O), H2, and O2 combine in three steps
to yield methyl methacrylate (and water). What is the atom
economy of this new process?
H2C
CH2 CH3OH CO
CH2O H2 1⁄2 O2
CH2 O
H3C
C
C
O
CH3 (+ 2 H2O)
Methyl methacrylate
4.2 Bleach
A bleach is a substance that removes color from something.
Household bleach is often an aqueous solution of sodium hypochlorite (NaClO), an oxidizing agent, with a concentration of
5.25 to 8.25%. These solutions are used in the home to
whiten clothing, remove stains, and disinfect surfaces. ­Sodium
hypochlorite is sometimes used to chlorinate swimming pools;
232
the desired concentration of what is called free chlorine in a
pool is around 3 mg/L. It is also used to disinfect some drinking water supplies, but at a much lower concentration. Municipal water supplies have free chlorine concentrations of 0.1 to
0.5 mg/L. The key to these uses is the very low concentrations
needed to kill microorganisms in water. The ingestion of
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household bleach often leads to a sore throat, nausea, vomiting, and difficulty swallowing. Though rare for the concentrations used in household bleach, it can be fatal.
Bleach was involved in several cases of food tampering,
including its addition to soup, infant formula, and soft drinks.
In 2017, a man working at a convenience store in Illinois was
arrested for adding bleach to a bottle of barbecue sauce used
to make sandwiches at the store. Fortunately, another worker
discovered the issue before any of the tainted food was sold.
In addition to these cases of malicious food tampering, a
group of scientists at the U.S. Centers for Disease Control
and ­Prevention (CDC) reported that an internet survey of 520
people conducted in May 2020, shortly after the beginning
of the COVID-19 pandemic, showed that 19% of the
­people surveyed had applied bleach to food and that 4%
­reported either drinking or gargling diluted bleach solutions,
which the scientists classified as “nonrecommended highrisk practices.”
One method of detecting bleach uses starch-iodide paper.
The bleach oxidizes the iodide ion to iodine in an acid solution,
© Charles D. Winters/Cengage
Starch-iodine. A distinctive
blue color is generated
when iodine reacts with
water-soluble starch.
Question
1. To determine the mass percent of sodium hypochlorite in a
sample of household bleach, you dilute a 10.0-mL sample of
the bleach solution to a volume of 100.0 mL in a volumetric
flask. You then transfer 25.0 mL of this dilute solution into
an Erlenmeyer flask and add an excess of KI and acid. The
iodine (I2) generated by the subsequent reaction is then
titrated with 0.151 M Na2S2O3; 29.34 mL of the Na2S2O3
solution is required to reach the equivalence point.
(a) Calculate the amount (mol) of NaClO present in the
25.0-mL sample of dilute bleach solution that was titrated.
(b) What is the molarity of NaClO in the dilute solution?
(c) What is the molarity of NaClO in the original bleach
solution?
(d) What mass of NaClO is present in the original 10.0 mL of
bleach solution used?
(e) What is the mass of the original 10.0 mL of bleach solution?
Assume the density of this bleach solution is 1.1 g/mL.
(f ) What is the mass percent of NaClO in the original bleach
solution?
2 I−(aq) + HClO(aq) + H3O+(aq) n I2(aq) + 2 H2O(ℓ) + Cl−(aq)
and the presence of I2 is detected by a deep blue color in the
presence of starch.
This reaction is also used in the quantitative analysis of
solutions containing bleach. Excess iodide ion (in the form of
KI) is added to the sample. The bleach in the sample (which
forms HClO in acid solution) oxidizes iodide ions to iodine, I2.
The iodine formed in the reaction is then titrated with sodium
thiosulfate, Na2S2O3 in another oxidation–reduction reaction
(as in “Check Your Understanding” in Example 4.16).
I2(aq) + 2 S2O32−(aq) n 2 I−(aq) + S4O62−(aq)
The amount of Na2S2O3 used in the titration can then be used
to determine the amount of NaClO in the original sample.
4.3 How Much Salt Is There in Seawater?
Saltiness is one of the basic taste sensations, and a taste of
seawater quickly reveals it is salty. How did the oceans become
salty?
Dissolved CO2 reacts with water to produce H2CO3, a weak
acid that partially ionizes to form hydronium and bicarbonate
ions.
CO2(g) + H2O(ℓ) n H2CO3(aq)
Indeed, this is why rain is normally acidic, and this slightly
acidic rainwater can cause substances such as limestone or
corals to dissolve, producing calcium ions and more bicarbonate ions.
CaCO3(s) + H3O+(aq) n Ca2+(aq) + HCO3−(aq) + H2O(ℓ)
John C. Kotz
H2CO3(aq) + H2O(ℓ) uv H3O+(aq) + HCO3−(aq)
Salt in seawater. Every kilogram of seawater contains about 35 g
of dissolved salts, predominantly NaCl.
Applying Chemical Principles
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233
Sodium ions arrive in the oceans by a similar reaction with
sodium-bearing minerals such as albite, NaAlSi3O6. Acidic
rain falling on the land extracts sodium ions that are then carried by rivers to the ocean.
The average chloride content of rocks in Earth’s crust is only
0.01%, so only a minute proportion of the chloride ion in the
oceans can come from the weathering of rocks and minerals.
What then is the origin of the chloride ions in seawater? The
answer is volcanoes: Hydrogen chloride gas, HCl, is a constituent of volcanic gases. Early in Earth’s history, the planet was
much hotter, and volcanoes were much more widespread. The
HCl gas emitted from these volcanoes is very soluble in water
and quickly dissolves to give a dilute solution of hydrochloric
acid. The chloride ions from dissolved HCl gas and sodium
ions from weathered rocks are the source of the salt in the sea.
Suppose you are an oceanographer, and you want to determine the concentration of chloride ions in a sample of seawater. How can you do this? And what results might you find?
There are several ways to analyze a solution for its chloride
ion content; among them is the classic Mohr method in which
a solution containing chloride ions is titrated with standardized
silver nitrate. The following reaction will occur:
Ag+(aq) + Cl−(aq) n AgCl(s)
The reaction will continue until the chloride ions are precipitated completely. To detect the equivalence point of the titration of Cl− with Ag+, the Mohr method involves the addition of
a few drops of a solution of potassium chromate. This indicator
works because silver chromate is slightly more soluble than
AgCl, so the red Ag2CrO4 precipitates only after all of the AgCl
is precipitated.
2 Ag+(aq) + CrO42−(aq) n Ag2CrO4(s)
The appearance of the red color of Ag2CrO4 (page 153) signals
the equivalence point.
Question
1. Calculate the chloride ion concentration in a sample of seawater given the following experimental information: The volume
of original seawater sample was 100.0 mL. A 10.00-mL
sample of the seawater was diluted to 100.0 mL with
distilled water, and 10.00 mL of the diluted sample was
­
again diluted to 100.0 mL. A Mohr titration was performed
on 50.00 mL from the second dilution. and this sample
required 26.25 mL of 0.100 M AgNO3 to reach the equivalence point.
4.4 The Martian
234
The Martian. The hero
of this popular book and
movie used chemistry to
survive.
Photos 12/Alamy Stock Photo
In the book and movie, The Martian (Andy Weir, Crown Publishers, 2011), astronaut Mark Watney is faced with the problem of surviving on the bleak surface of Mars. One of his first
and greatest needs is water, and he proposes to synthesize it
by reacting hydrogen and oxygen. He has a suitable oxygen
source: deriving it from atmospheric CO2. Decomposition of
hydrazine, N2H4, leftover rocket fuel from the Mars Descent
Vehicle (MDV), would produce N2 and H2.
In developing plans, he guesses that from 50 liters of liquid
oxygen he could make 100 liters of water (“50 liters of O2
make 100 liters of molecules that have one oxygen each.”). He
uses the same thinking to estimate that one liter of hydrazine,
N2H4, should yield two liters of water (“Each molecule of hydrazine has 4 hydrogen atoms in it. So each liter of hydrazine
has enough hydrogen for 2 liters of water.”).
On Feb. 18, 2021 (ten years after The Martian was published), the Mars Perseverance rover landed on Mars. Included
among its many instruments is the Mars Oxygen In-situ Resource Utilization Experiment, better known as MOXIE. The
instrument is designed to test the feasibility of converting carbon dioxide to oxygen on Mars. In its first experiment, MOXIE
created 5.4 grams of O2 by simultaneously heating and running an electrical current through CO2. The process creates
carbon monoxide, CO, and O atoms. The O atoms are unstable
and quickly bond with each other to form O2 molecules.
Questions
1. Predict whether the assumptions about liquid volumes based
upon the number of oxygen atoms in liquid O2 and hydrogen
atoms in liquid N2H4 are excellent, good, fair, or poor? If you
think the logic is flawed, explain why.
2. To calculate the volume of water produced from a known
volume of liquid oxygen requires the following conversions:
vol O2 n mass O2 n mol O2 n mol H2O n mass H2O n vol H2O
Densities: O2(ℓ) = 1.14 g/mL, H2O(ℓ) = 1.00 g/mL
Calculate the volume of water produced from 50. L of liquid
oxygen.
3. Perform a similar calculation to determine the volume of
water that can be produced from 1.0 L of liquid hydrazine
(density of hydrazine = 1.02 g/mL).
4. Are Watney’s predictions in The Martian correct?
Chapter 4 / Stoichiometry: Quantitative Information about Chemical Reactions
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re
5. Hydrazine is a liquid fuel that explosively decomposes into
hydrogen and nitrogen gases upon heating. The rapid expansion of these gases propels a space craft.
(a) Write the balanced chemical equation for the decomposition.
(b) If the decomposition of 125 g N2H4 produces 12.7 g H2, what
is the percent yield for this reaction.
6. MOXIE uses the decomposition of carbon dioxide gas into
carbon monoxide and oxygen gases.
(a) Write an overall balanced chemical equation for this chemical
reaction.
(b) Assign oxidation numbers to each atom in the balanced
equation. Carbon dioxide is simultaneously oxidized and
reduced. Which element in CO2 is oxidized? Which is
reduced?
(c) If MOXIE can produce 5.4 grams of O2 in 32 minutes, how
long will it take MOXIE to produce the roughly 750 grams of
oxygen per day necessary to sustain an astronaut?
(d) Calculate the theoretical yield of oxygen if 1200 g CO2 is
decomposed by MOXIE.
Think–Pair–Share
1. An aqueous solution of silver nitrate reacts with copper metal
to produce silver metal and an aqueous solution of copper(II)
nitrate as shown in the following figure.
You add 50.0 mL of 0.100 M silver nitrate to a 1.00-g piece of
copper metal and allow the reaction to occur. You eventually
isolate 0.964 g of silver. Without performing the calculations,
construct a detailed strategy map of the steps you would follow to determine the percent yield of silver in this reaction.
2. What volume of an aqueous 6.0 M NaOH solution should
be diluted to obtain 10. L of 0.10 M NaOH? What amount
of NaOH (in moles) is in that volume of 6.0 M NaOH? What
amount of NaOH (in moles) of NaOH is in 10. L of 0.10 M
NaOH? Explain how and why these two amounts are
related.
3. You have a solution that is approximately 0.10 M NaOH. To
determine a more precise concentration, you perform a standardization titration against the standard reagent, potassium
hydrogen phthalate (KHC8H4O4, molar mass 5 204.22 g/mol).
The net ionic equation for the reaction is
HC8H4O42(aq) 1 OH2(aq) n C8H4O422(aq) 1 H2O(ℓ)
Pure copper wire
You would like your titration to use about 25 mL of the s­ odium
hydroxide solution. About what mass of the standard reagent
should you use in the titration?
4. You conduct a combustion analysis of a known mass of an
unknown compound with the formula CxHyOz, which yields a
known mass of CO2 and a known mass of H2O. What are the
steps to obtain the number of moles of C in the original
sample? The number of moles of H? The number of moles
Copper wire in dilute AgNO3 Blue color due toof O?
Silver crystals
formed after
several weeks
© Charles D. Winters/Cengage
solution after several hours Cu2+ ions formed
in redox reaction
Copper wire in dilute AgNO3 Blue color due to
solution after several hours Cu2+ ions formed
in redox reaction
Silver crystals
formed after
several weeks
Think–Pair–Share
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235
Chapter Goals Revisited
The Goals for this chapter are keyed to specific Study Questions to help you
organize your review.
4.1 Mass Relationships in Chemical Reactions:
Stoichiometry
•
Understand the principle of conservation of matter, which forms the basis
of chemical stoichiometry. 7, 153.
•
Calculate the mass of one reactant or product in a reaction knowing the
balanced equation and the mass of another reactant or product in that
reaction. 4–8, 93, 95, 111, 113.
• Use amounts tables to organize chemical information. 9–12.
4.2 Reactions in Which One Reactant Is Present
in Limited Supply
• Determine which reactant is in limited supply in a reaction involving
several reactants. 13–16, 112.
• Determine the yield of a product based on the limiting reactant. 13–20, 112.
4.3 Percent Yield
• Explain the differences among actual yield, theoretical yield, and percent
yield, and calculate percent yield for a reaction. 23–26.
4.4 Chemical Equations and Chemical Analysis
• Use stoichiometry principles to analyze a mixture of compounds. 29–32,
145, 146.
• Find the empirical formula of an unknown compound using chemical
stoichiometry. 35–42, 104, 105.
4.5 Measuring Concentrations of Compounds
in Solution
• Calculate the concentration of a solute in a solution in units of moles per
liter (molarity) and use solution concentrations in calculations. 45–52.
• Describe how to prepare a solution of a given concentration from the solute
and solvent or by dilution of a more concentrated solution. 55–64, 137.
4.6 pH, A Concentration Scale for Acids and Bases
• Understand the pH scale. 65, 66.
• Calculate the pH of a solution from the concentration of hydronium ions
in the solution. Calculate the hydronium ion concentration in a solution
from its pH. 67–70.
4.7 Stoichiometry of Reactions in Aqueous
Solution—Fundamentals
• Use stoichiometry principles for reactions occurring in solution. 71–82,
122, 123.
236
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4.8 Stoichiometry of Reactions in Aqueous
Solution—Titrations
•
Explain how a titration is conducted, explain the procedure for
standardization of a solution, and calculate concentrations or amounts of
reactants from titration data. 83–88, 141.
4.9 Spectrophotometry
•
Understand and use the principles of spectrophotometry to determine the
concentration of a colored compound or ion in solution. 91, 92, 149.
Key Equations
Equation 4.1 (page 201) Percent yield.
Percent yield 5
actual yield
 100%
theoretical yield
Equation 4.2 (page 209) Definition of molarity, a measure of the concentration of
a solute in a solution.
Molarity of x (c x ) 5
amount of solute x (mol)
volume of solution (L)
Equation 4.3 (page 211) The amount of solute in a given volume of a solution with
a known molarity.
Amount of solute x (mol) = cx (mol/L) × volume of solution (L)
Dilution Equation (page 214) This is a shortcut to find, for example, the concentration of a solution (cd) after diluting some volume (Vc) of a more concentrated solution (cc) to a new volume (Vd).
cc × Vc = cd × Vd
Equation 4.4 (page 215) pH. The pH of a solution is the negative logarithm of the
hydronium ion concentration.
pH = −log[H3O+]
Equation 4.5 (page 216) Calculating [H3O+] from pH. The equation for calculating
the hydronium ion concentration of a solution from the pH of the solution.
[H3O+] = 10−pH
Equation 4.6 (page 229) Beer–Lambert law. The absorbance of light (A) by a substance in solution is equal to the product of the molar absorptivity of the substance
(ε), the path length of the cell (ℓ), and the concentration of the solute (c).
Absorbance (A) ∝ path length (ℓ) × concentration (c)
A=ε×ℓ×c
Key Equations
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237
Study Questions
▲ denotes challenging questions. Blue-numbered questions have answers in Appendix N and fully
worked solutions in the Student Solutions Manual.
Practicing Skills
2 Al(s) + 3 Br2(ℓ) n Al2Br6(s)
Mass Relationships in Chemical Reactions:
Basic Stoichiometry
What mass of Br2, in grams, is required for complete reaction with 2.56 g of Al? What mass of
white, solid Al2Br6 is expected?
(See Example 4.1.)
6. The balanced equation for the reduction of iron
ore to the metal using CO is
1. The reaction of iron(III) oxide with aluminum to
give molten iron is known as the thermite reaction (page 199).
Fe2O3(s) + 3 CO(g) n 2 Fe(s) + 3 CO2(g)
(a) What is the maximum mass of iron, in grams,
that can be obtained from 454 g (1.00 pound)
of iron(III) oxide?
(b) What mass of CO is required to react with
454 g of Fe2O3?
Fe2O3(s) + 2 Al(s) n 2 Fe(ℓ) + Al2O3(s)
What amount of Al, in moles, is needed for complete reaction with 3.0 mol of Fe2O3? What mass
of Fe, in grams, can be produced?
2. Potassium chlorate decomposes upon heating into
potassium chloride and oxygen gas.
7. Methane, CH4, burns in oxygen.
(a) What are the products of the reaction?
(b) Write the balanced equation for the reaction.
(c) What mass of O2, in grams, is required for
complete combustion of 25.5 g of methane?
(d) What is the total mass of products expected
from the combustion of 25.5 g of methane?
2 KClO3(s) n 2 KCl(s) + 3 O2(g)
What amount of O2, in moles, is produced from
the complete reaction of 1.0 mol of KClO3? What
mass (in grams) of O2 is produced?
3. One of the components of gasoline is the compound octane, C8H18. The chemical equation for
its complete combustion in air is
8. The formation of water-insoluble silver chloride is
useful in the analysis of chloride-containing substances. Consider the following unbalanced
equation:
2 C8H18(ℓ) + 25 O2(g) n 16 CO2(g) + 18 H2O(g)
What mass of O2 is consumed in the complete
combustion of 5.00 g of C8H18? What mass of
CO2 is produced? What mass of H2O is produced?
BaCl2(aq) + AgNO3(aq) n AgCl(s) + Ba(NO3)2(aq)
(a) Write the balanced equation.
(b) What mass of AgNO3, in grams, is required for
complete reaction with 0.156 g of BaCl2?
What mass of AgCl is produced?
4. What mass of HCl, in grams, is required to react
with 0.750 g of Al(OH)3? What mass of water, in
grams, is produced?
Amounts Tables and Chemical Stoichiometry
Al(OH)3(s) + 3 HCl(aq) n AlCl3(aq) + 3 H2O(ℓ)
For each question below, make an amounts table that
lists the initial amount or amounts of reactants, the
changes in amounts of reactants and products, and the
amounts of reactants and products after reaction (see
page 193 and Example 4.1).
Photos: © Charles D. Winters/Cengage
5. Like many metals, aluminum reacts with a
halogen (here the orange-brown liquid Br2) to
give a metal halide, aluminum bromide. (The
white solid on the lip of the beaker at the end of
the reaction is Al2Br6.)
Before reaction
After reaction
9. The metals industry was a major source of air pollution years ago. One common process involved
roasting metal sulfides in the air:
2 PbS(s) + 3 O2(g) n 2 PbO(s) + 2 SO2(g)
If 2.50 mol of PbS is heated in air, what amount
of O2 is required for complete reaction? What
amounts of PbO and SO2 are expected?
10. Iron ore is converted to iron metal in a reaction
with carbon.
2 Fe2O3(s) + 3 C(s) n 4 Fe(s) + 3 CO2(g)
238
Chapter 4 / Stoichiometry: Quantitative Information about Chemical Reactions
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If 6.2 mol of Fe2O3(s) is used, what amount of
C(s) is needed, and what amounts of Fe and CO2
are produced?
11. Chromium metal reacts with oxygen to give
chromium(III) oxide, Cr2O3.
(a) Write a balanced equation for the reaction.
(b) What mass (in grams) of Cr2O3 is produced if
0.175 g of chromium metal is converted completely to the oxide?
(c) What mass of O2 (in grams) is required for
the reaction?
12. Ethane, C2H6, burns in oxygen.
(a) What are the products of the reaction?
(b) Write the balanced equation for the reaction.
(c) What mass of O2, in grams, is required for
complete combustion of 13.6 of ethane?
(d) What is the total mass of products expected
from the combustion of 13.6 g of ethane?
Limiting Reactants
(See Example 4.2.)
13. Sodium sulfide, Na2S, is used in the leather industry to remove hair from animal hides. The Na2S is
made by the reaction
Na2SO4(s) + 4 C(s) n Na2S(s) + 4 CO(g)
Suppose you mix 25.0 g of Na2SO4 and 12.5 g of
C. Which is the limiting reactant? What mass of
Na2S is produced?
14. Ammonia gas can be prepared by the reaction
of a metal oxide such as calcium oxide with
­ammonium chloride.
CaO(s) + 2 NH4Cl(s) n
2 NH3(g) + H2O(g) + CaCl2(s)
If 112 g of CaO and 224 g of NH4Cl are mixed,
what is the limiting reactant, and what mass of
NH3 can be produced?
15. The compound SF6 is made by burning sulfur in
an atmosphere of fluorine. The balanced equation
is
S8(s) + 24 F2(g) n 8 SF6(g)
Starting with a mixture of 1.6 mol of sulfur, S8,
and 35 mol of F2,
(a) which is the limiting reagent?
(b) what amount of SF6 is produced?
16. Disulfur dichloride, S2Cl2, is used to vulcanize
rubber. It can be made by treating molten sulfur
with gaseous chlorine:
S8(ℓ) + 4 Cl2(g) n 4 S2Cl2(ℓ)
Starting with a mixture of 32.0 g of sulfur and
71.0 g of Cl2,
(a) which is the limiting reactant?
(b) what is the theoretical yield of S2Cl2?
(c) what mass of the excess reactant remains
when the reaction is completed?
17. The reaction of methane and water is one way to
prepare hydrogen for use as a fuel:
CH4(g) + H2O(g) n CO(g) + 3 H2(g)
If you begin with 995 g of CH4 and 2510 g of
water,
(a) which reactant is the limiting reactant?
(b) what is the maximum mass of H2 that can be
prepared?
(c) what mass of the excess reactant remains
when the reaction is completed?
18. Aluminum chloride, AlCl3, is made by treating
scrap aluminum with chlorine.
2 Al(s) + 3 Cl2(g) n 2 AlCl3(s)
If you begin with 2.70 g of Al and 4.05 g of Cl2,
(a) which reactant is limiting?
(b) what mass of AlCl3 can be produced?
(c) what mass of the excess reactant remains
when the reaction is completed?
(d) make an amounts table for this problem.
19. In the thermite reaction, iron(III) oxide is reduced
by aluminum to give molten iron.
Fe2O3(s) + 2 Al(s) n 2 Fe(ℓ) + Al2O3(s)
If you begin with 15.0 g of Fe2O3 and 30.0 g of Al,
(a) which reactant is limiting?
(b) what mass of Fe can be produced?
(c) what mass of the excess reactant remains after
the limiting reactant is consumed?
(d) make an amounts table for this problem.
20. Aspirin, C6H4(OCOCH3)CO2H, is produced by
the reaction of salicylic acid, C6H4(OH)CO2H,
and acetic anhydride, (CH3CO)2O (page 201).
C6H4(OH)CO2H(s) + (CH3CO)2O(ℓ) n
C6H4(OCOCH3)CO2H(s) + CH3CO2H(ℓ)
If you mix 100. g of each of the reactants, what is
the maximum mass of aspirin that can be
obtained?
Percent Yield
(See Example 4.3.)
21. In Example 4.2, you found that a particular
mixture of CO and H2 could produce 407 g
CH3OH.
CO(g) + 2 H2(g) n CH3OH(ℓ)
If only 332 g of CH3OH is actually produced,
what is the percent yield of the compound?
Study Questions
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239
22. Ammonia gas can be prepared by the following
reaction:
recovered is 1.79 g. What is the percent yield
for this reaction?
CaO(s) + 2 NH4Cl(s) n
2 NH3(g) + H2O(g) + CaCl2(s)
If 112 g of CaO and 224 g of NH4Cl are mixed,
the theoretical yield of NH3 is 68.0 g (Study
­Question 14). If only 16.3 g of NH3 is actually
obtained, what is its percent yield?
26. The compound lead(II) chloride forms when
aqueous solutions of lead(II) nitrate and sodium
chloride are mixed.
Pb(NO3)2(aq) + 2 NaCl(aq) n
PbCl2(s) + 2 NaNO3(aq)
23. The deep blue compound Cu(NH3)4SO4 is made
by the reaction of copper(II) sulfate and
ammonia.
CuSO4(aq) + 4 NH3(aq) n Cu(NH3)4SO4(aq)
(a) If you use 10.0 g of CuSO4 and excess NH3,
what is the theoretical yield of Cu(NH3)4SO4?
(b) If you isolate 12.6 g of Cu(NH3)4SO4, what is
the percent yield of Cu(NH3)4SO4?
24. Black smokers are found in the depths of the
oceans. Thinking that the conditions in these
smokers might be conducive to the formation of
organic compounds, two chemists in Germany
found the following reaction could occur in
similar conditions.
2 CH3SH + CO n CH3COSCH3 + H2S
If you begin with 10.0 g of CH3SH and excess CO,
(a) what is the theoretical yield of CH3COSCH3?
(b) if 8.65 g of CH3COSCH3 is isolated, what is its
percent yield?
(a) If 1.50 g of lead(II) nitrate and 1.00 g of
sodium chloride are each dissolved in water
and then mixed, what mass of lead(II) chloride could be obtained?
(b) The lead(II) chloride formed is obtained by
filtration and dried. The mass of lead(II) chloride recovered is 1.05 g. What is the percent
yield for this reaction?
27. The reaction of methane and water is one way to
prepare hydrogen for use as a fuel:
CH4(g) + H2O(g) n CO(g) + 3 H2(g)
If this reaction has a 37% yield under certain conditions, what mass of CH4 is required to produce
15 g of H2?
28. Methanol, CH3OH, can be prepared from carbon
monoxide and hydrogen.
CO(g) + 2 H2(g) n CH3OH(ℓ)
What mass of hydrogen is required to produce
1.0 L of CH3OH (d = 0.791 g/mL) if this reaction
has a 74% yield under certain conditions?
Analysis of Mixtures
Ralph White/Corbis Documentary/Getty Images
(See Example 4.4.)
A black smoker, deep in the Pacific Ocean.
25. The deep-red compound silver chromate forms
when aqueous solutions of silver nitrate and
potassium chromate are mixed.
2 AgNO3(aq) + K2CrO4(aq) n
Ag2CrO4(s) + 2 KNO3(aq)
(a) If 2.00 g of silver nitrate and 2.00 g of potassium chromate are each dissolved in water
and then mixed, what mass of silver chromate
could be obtained?
(b) The silver chromate formed is obtained by filtration and dried. The mass of silver chromate
240
29. A mixture of CuSO4 and CuSO4 ⋅ 5 H2O has a mass
of 1.245 g. After heating to drive off all the water,
the mass is only 0.832 g. What is the mass percent
of CuSO4 ⋅ 5 H2O in the mixture? (See page 113.)
30. A 2.634-g sample containing impure CuCl2 ⋅ 2
H2O was heated. The sample mass after heating to
drive off the water was 2.125 g. What was the mass
percent of CuCl2 ⋅ 2 H2O in the original sample?
31. A sample of limestone and other soil materials
was heated, and the limestone decomposed to
give calcium oxide and carbon dioxide.
CaCO3(s) n CaO(s) + CO2(g)
A 1.624-g sample of limestone-containing material gave 0.638 g of CO2, in addition to CaO, after
being heated at a high temperature. What was the
mass percent of CaCO3 in the original sample?
32. At higher temperatures, NaHCO3 is converted
quantitatively to Na2CO3.
2 NaHCO3(s) n Na2CO3(s) + CO2(g) + H2O(g)
Chapter 4 / Stoichiometry: Quantitative Information about Chemical Reactions
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Heating a 1.7184-g sample of impure NaHCO3
gives 0.196 g of CO2. What was the mass percent
of NaHCO3 in the original 1.7184-g sample?
33. Nickel(II) sulfide, NiS, occurs naturally as the relatively rare mineral millerite. One of its occurrences
is in meteorites. To analyze a mineral sample for
the quantity of NiS, the sample is dissolved in
nitric acid to form a solution of Ni(NO3)2.
NiS(s) + 4 HNO3(aq) n
Ni(NO3)2(aq) + 2 NO2(g) + 2 H2O(ℓ) + S(s)
The aqueous solution of Ni(NO3)2 is then reacted
with the organic compound dimethylglyoxime
(C4H8N2O2) to give the red solid Ni(C4H7N2O2)2.
Ni(NO3)2(aq) + 2 C4H8N2O2(aq) n
Ni(C4H7N2O2)2(s) + 2 HNO3(aq)
© Charles D. Winters/Cengage
Suppose a 0.468-g sample containing millerite
produces 0.206 g of red, solid Ni(C4H7N2O2)2.
What is the mass percent of NiS in the sample?
A precipitate of nickel with dimethylglyoxime, Ni(C4H7N2O2)2
34. ▲ The aluminum in a 0.764-g sample of an
unknown material was precipitated as aluminum
hydroxide, Al(OH)3, which was then converted to
Al2O3 by heating strongly. If 0.127 g of Al2O3 is
obtained from the 0.764-g sample, what is the
mass percent of aluminum in the sample?
Using Stoichiometry to Determine Empirical
and Molecular Formulas
(See Examples 4.5 and 4.6.)
35. Styrene, the building block of polystyrene, consists
of only C and H. If 0.438 g of styrene is burned in
oxygen and produces 1.481 g of CO2 and 0.303 g
of H2O, what is the empirical formula of styrene?
36. Mesitylene is a liquid hydrocarbon. Burning
0.115 g of the compound in oxygen gives 0.379 g
of CO2 and 0.1035 g of H2O. What is the empirical formula of mesitylene?
37. Naphthalene is a hydrocarbon that once was used
in mothballs. If 0.3093 g of the compound is
burned in oxygen, 1.0620 g of CO2 and 0.1739 g
of H2O are isolated.
(a) What is the empirical formula of naphthalene?
(b) If a separate experiment gave 128.2 g/mol as
the molar mass of the compound, what is its
molecular formula?
38. Azulene is a beautiful blue hydrocarbon. If
0.106 g of the compound is burned in oxygen,
0.364 g of CO2 and 0.0596 g of H2O are isolated.
(a) What is the empirical formula of azulene?
(b) If a separate experiment gave 128.2 g/mol as
the molar mass of the compound, what is its
molecular formula?
39. An unknown compound has the formula CxHyOz.
You burn 0.0956 g of the compound and isolate
0.1356 g of CO2 and 0.0833 g of H2O. What is
the empirical formula of the compound? If the
molar mass is 62.1 g/mol, what is the molecular
formula?
40. An unknown compound has the formula CxHyOz.
You burn 0.1523 g of the compound and isolate
0.3718 g of CO2 and 0.1522 g of H2O. What is the
empirical formula of the compound? If the molar
mass is 72.1 g/mol, what is the molecular
formula?
41. An unknown compound has the formula CxHyOz.
You burn 0.2427 g of the compound and isolate
0.5517 g of CO2 and 0.2258 g H2O. What is the
empirical formula of the compound? If the molar
mass is 58.1 g/mol, what is the molecular
formula?
42. An unknown compound has the formula CxHyOz.
You burn 0.1425 g of the compound and isolate
0.2089 g of CO2 and 0.0855 g H2O. What is the
empirical formula of the compound? If the molar
mass is 150.1 g/mol, what is the molecular
formula?
43. Nickel forms a compound with carbon monoxide,
Nix(CO)y. To determine its formula, you heat a
sample under conditions where the metal is converted to the oxide, NiO, and the CO is lost as
CO2. What is the empirical formula of this compound, Nix(CO)y, if a 0.0973-g sample produces
0.0426 g of NiO and 0.100 g of CO2?
44. To find the formula of a compound composed of
iron and carbon monoxide, Fex(CO)y, the compound is burned in pure oxygen to give Fe2O3 and
CO2. If you burn 1.959 g of Fex(CO)y and obtain
0.799 g of Fe2O3 and 2.200 g of CO2, what is the
empirical formula of Fex(CO)y?
Solution Concentration
(See Examples 4.7 and 4.8.)
45. If 6.73 g of Na2CO3 is dissolved in enough water
to make 250. mL of solution, what is the molar
Study Questions
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Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
241
concentration of the sodium carbonate? What are
the molar concentrations of the Na+ and
CO32− ions?
59. What volume of a 6.0 M NaOH solution should
be diluted with water to prepare a 500. mL solution with a concentration of 0.10 M NaOH?
46. Some potassium sulfate (K2SO4), 2.335 g, is dissolved in enough water to make exactly 500. mL
of solution. What is the molar concentration of
the potassium sulfate? What are the molar concentrations of the K+ and SO42− ions?
60. A fresh bottle of concentrated hydrochloric acid
has a concentration of 12 M HCl. What volume of
this solution should be diluted with water to
prepare a 250. mL solution with a concentration
of 3.0 M HCl?
47. What amount (mol) of hydrochloric acid is delivered if 500. mL of 0.200 M HCl is measured out?
What mass of HCl is delivered?
61. Which of the following methods would you use to
prepare 1.00 L of 0.125 M H2SO4?
(a) Dilute 20.8 mL of 6.00 M H2SO4 to a volume
of 1.00 L.
(b) Add 950. mL of water to 50.0 mL of 3.00 M
H2SO4.
48. What amount (mol) of acetone (CH3COCH3) is
delivered if 250. mL of 4.00 M acetone is measured out? What mass of acetone is delivered?
49. What is the mass of solute, in grams, in 250. mL
of a 0.0125 M solution of KMnO4?
50. What is the mass of solute, in grams, in 125 mL of a
1.023 × 10−3 M solution of Na3PO4? What are the
molar concentrations of the Na+ and PO43− ions?
51. What volume of 0.123 M NaOH, in milliliters,
contains 25.0 g of NaOH?
52. What volume of 2.06 M KMnO4, in liters, contains
322 g of solute?
53. Identify the ions that exist in each aqueous solution, and specify the concentration of each ion.
(a) 0.25 M (NH4)2SO4
(b) 0.123 M Na2CO3
(c) 0.056 M HNO3
54. Identify the ions that exist in each aqueous solution, and specify the concentration of each ion.
(a) 0.12 M BaCl2
(b) 0.0125 M CuSO4
(c) 0.500 M K2Cr2O7
Preparing Solutions
(See Examples 4.7 and 4.9.)
55. An experiment in your laboratory requires
500. mL of a 0.0200 M solution of Na2CO3. You
are given solid Na2CO3, distilled water, and a
500.-mL volumetric flask. Describe how to prepare
the required solution.
56. What mass of oxalic acid, H2C2O4, is required to
prepare 250. mL of a solution that has a concentration of 0.15 M H2C2O4?
57. If you dilute 25.0 mL of 1.50 M hydrochloric acid
to 500. mL, what is the molar concentration of
the dilute acid?
58. If 4.00 mL of 0.0250 M CuSO4 is diluted to
10.0 mL with pure water, what is the molar concentration of copper(II) sulfate in the diluted solution?
242
62. Which of the following methods would you use to
prepare 300. mL of 0.500 M K2Cr2O7?
(a) Add 30.0 mL of 1.50 M K2Cr2O7 to 270. mL of
water.
(b) Dilute 250. mL of 0.600 M K2Cr2O7 to a
volume of 300. mL.
Serial Dilutions
(See A Closer Look: Serial Dilutions, page 215.)
63. You have 250. mL of 0.136 M HCl. Using a volumetric pipet, you take 25.00 mL of that solution
and dilute it to 100.00 mL in a volumetric flask.
Now you take 10.00 mL of that solution, using a
volumetric pipet, and dilute it to 100.00 mL in a
volumetric flask. What is the concentration of
hydrochloric acid in the final solution?
64. ▲ Suppose you have 100.00 mL of a solution of
a dye and transfer 2.00 mL of the solution to a
100.00-mL volumetric flask. After adding water to
the 100.00 mL mark, you take 5.00 mL of that
solution and again dilute to 100.00 mL. If you
find the dye concentration in the final diluted
sample is 0.000158 M, what was the dye concentration in the original solution?
Calculating and Using pH
(See Example 4.10.)
65. A table wine has a pH of 3.40. What is the hydronium ion concentration of the wine? Is it acidic or
basic?
66. A saturated solution of milk of magnesia,
Mg(OH)2, has a pH of 10.5. What is the hydronium ion concentration of the solution? Is the
solution acidic or basic?
67. What is the hydronium ion concentration of a
0.0013 M solution of HNO3? What is its pH?
68. What is the hydronium ion concentration of a
1.2 × 10−4 M solution of HClO4? What is its pH?
Chapter 4 / Stoichiometry: Quantitative Information about Chemical Reactions
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69. Make the following conversions. In each case, tell
whether the solution is acidic or basic.
(a)
1.00
(b)
10.50
(c)
1.3 × 10−5 M
(d)
2.3 × 10−8 M
Photos: © Charles D. Winters/Cengage
[H3O+]
pH
70. Make the following conversions. In each case, tell
whether the solution is acidic or basic.
(a)
[H3O+]
pH
(a)
6.7 × 10−10 M
(b)
2.2 × 10−6 M
(c)
(d)
5.25
2.5 × 10−2 M
Stoichiometry of Reactions in Solution
(See Example 4.11.)
71. What volume of 0.109 M HNO3, in milliliters, is
required to react completely with 2.50 g of Ba(OH)2?
2 HNO3(aq) + Ba(OH)2(s) n
2 H2O(ℓ) + Ba(NO3)2(aq)
72. What mass of Na2CO3, in grams, is required for
complete reaction with 50.0 mL of 0.125 M HNO3?
Na2CO3(aq) + 2 HNO3(aq) n
2 NaNO3(aq) + CO2(g) + H2O(ℓ)
73. When an electric current is passed through an
aqueous solution of NaCl, the valuable industrial
chemicals H2(g), Cl2(g), and NaOH are produced.
2 NaCl(aq) + 2 H2O(ℓ) n
H2(g) + Cl2(g) + 2 NaOH(aq)
What mass of NaOH can be formed from 15.0 L of
0.35 M NaCl? What mass of chlorine is obtained?
74. Hydrazine, N2H4, a base like ammonia, can react
with sulfuric acid.
2 N2H4(aq) + H2SO4(aq) n 2 N2H5+(aq) + SO42−(aq)
What mass of hydrazine reacts with 250. mL of
0.146 M H2SO4?
75. In the photographic developing process, silver
bromide is dissolved by adding sodium thiosulfate.
AgBr(s) + 2 Na2S2O3(aq) n
Na3Ag(S2O3)2(aq) + NaBr(aq)
If you want to dissolve 0.225 g of AgBr (as seen in
the following photo), what volume of 0.0138 M
Na2S2O3, in milliliters, should be used?
(b)
Silver chemistry. (a) A precipitate of AgBr formed by adding
AgNO3(aq) to KBr(aq). (b) Upon adding Na2S2O3(aq), sodium
thiosulfate, the solid AgBr dissolves.
76. You can dissolve an aluminum soft drink can in
an aqueous base such as potassium hydroxide.
2 Al(s) + 2 KOH(aq) + 6 H2O(ℓ) n
2 KAl(OH)4(aq) + 3 H2(g)
If you place 2.05 g of aluminum in a beaker with
185 mL of 1.35 M KOH, will any aluminum
remain? What mass of KAl(OH)4 is produced?
77. What volume of 0.750 M Pb(NO3)2, in milliliters, is required to react completely with 1.00 L
of 2.25 M NaCl solution? The balanced equation is
Pb(NO3)2(aq) + 2 NaCl(aq) n PbCl2(s) + 2 NaNO3(aq)
78. What volume of 0.125 M oxalic acid, H2C2O4,
is required to react with 35.2 mL of 0.546 M
NaOH?
H2C2O4(aq) + 2 NaOH(aq) n
Na2C2O4(aq) + 2 H2O(ℓ)
79. A precipitate of lead(II) iodide results from the
reaction of aqueous lead(II) nitrate and potassium
iodide solutions.
(a) Write the balanced chemical equation for this
reaction.
(b) What mass of lead(II) iodide can be obtained
from the reaction of 50.0 mL of 0.0500 M
lead(II) nitrate with 50.0 mL of 0.150 M
potassium iodide?
80. A precipitate of silver chloride results from the
reaction of aqueous silver nitrate and copper(II)
chloride solutions.
(a) Write the balanced chemical equation for this
reaction.
(b) What mass of silver chloride can be obtained
from the reaction of 100. mL of 0.100 M silver
nitrate with 100. mL of 0.100 M copper(II)
chloride?
Study Questions
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243
81. What mass of carbon dioxide gas can be obtained
from the reaction of 1.00 g of sodium carbonate
with 150. mL of 0.750 M hydrochloric acid?
Tartaric acid:
82. What mass of carbon dioxide can be obtained
from the reaction of 1.50 g of calcium carbonate
with 100. mL of 0.500 M nitric acid?
A 0.956-g sample requires 29.1 mL of 0.513 M
NaOH to consume the acid completely. What is
the unknown acid?
Titrations
(See Examples 4.12–4.16.)
83. What volume of 0.812 M HCl, in milliliters, is
required to titrate 1.45 g of NaOH to the equivalence point?
NaOH(aq) + HCl(aq) n H2O(ℓ) + NaCl(aq)
84. What volume of 0.955 M HCl, in milliliters, is
required to titrate 2.152 g of Na2CO3 to the equivalence point?
Na2CO3(aq) + 2 HCl(aq) n
H2O(ℓ) + CO2(g) + 2 NaCl(aq)
85. If 38.55 mL of HCl is required to titrate 2.150 g of
Na2CO3 according to the following equation,
what is the concentration (mol/L) of the HCl
solution?
Na2CO3(aq) + 2 HCl(aq) n
2 NaCl(aq) + CO2(g) + H2O(ℓ)
86. Potassium hydrogen phthalate, KHC8H4O4, is
used to standardize solutions of bases. The acidic
anion reacts with strong bases according to the
following net ionic equation:
HC8H4O4 (aq) + OH (aq) n
C8H4O42−(aq) + H2O(ℓ)
−
−
If a 0.902-g sample of potassium hydrogen
phthalate is dissolved in water and titrated to the
equivalence point with 26.45 mL of NaOH(aq),
what is the molar concentration of the NaOH?
87. You have 0.954 g of an unknown acid, H2A, which
reacts with NaOH according to the balanced
equation
H2A(aq) + 2 NaOH(aq) n Na2A(aq) + 2 H2O(ℓ)
If 36.04 mL of 0.509 M NaOH is required to
titrate the acid to the second equivalence point,
what is the molar mass of the acid?
88. An unknown solid acid is either citric acid or
­tartaric acid. To determine which acid you have,
you dissolve a sample of the solid in water, titrate
the solution with aqueous NaOH, and from this
determine the molar mass of the unknown acid.
The appropriate equations are as follows:
Citric acid:
H3C6H5O7(aq) + 3 NaOH(aq) n
3 H2O(ℓ) + Na3C6H5O7(aq)
244
H2C4H4O6(aq) + 2 NaOH(aq) n
2 H2O(ℓ) + Na2C4H4O6(aq)
89. To analyze an iron-containing compound, you
convert all the iron to Fe2+ in aqueous solution
and then titrate the solution with standardized
KMnO4. The balanced, net ionic equation is
MnO4−(aq) + 5 Fe2+(aq) + 8 H3O+(aq) n
Mn2+(aq) + 5 Fe3+(aq) + 12 H2O(ℓ)
A 0.598-g sample of the iron-containing compound requires 22.25 mL of 0.0123 M KMnO4 for
titration to the equivalence point. What is the
mass percent of iron in the sample?
90. Vitamin C has the formula C6H8O6. Besides being
an acid, it is a reducing agent. One method for
determining the amount of vitamin C in a sample
is to titrate it with a solution of bromine, Br2, an
oxidizing agent.
C6H8O6(aq) + Br2(aq) n 2 HBr(aq) + C6H6O6(aq)
A 1.00-g chewable vitamin C tablet requires 27.85
mL of 0.102 M Br2 for titration to the equivalence
point. What is the mass of vitamin C in the tablet?
Spectrophotometry
(See Section 4.9 and Example 4.17. The problems below
are adapted from Fundamentals of Analytical Chemistry, 8th ed., by D. A. Skoog, D. M. West, F. J. Holler,
and S. R. Crouch, Thomson/Brooks-Cole, Belmont, CA
2004.)
91. A solution of a dye was analyzed by spectrophotometry, and the following calibration data were
collected.
Dye Concentration
Absorbance (A)
at 475 nm
0.50 × 10−6 M
0.24
1.5 × 10−6 M
0.36
2.5 × 10−6 M
0.44
3.5 × 10−6 M
0.59
4.5 × 10−6 M
0.70
(a) Construct a calibration plot, and determine
the slope and intercept.
(b) What is the dye concentration in a solution
with A = 0.52?
92. The nitrite ion is involved in the biochemical
nitrogen cycle. You can determine the nitrite ion
Chapter 4 / Stoichiometry: Quantitative Information about Chemical Reactions
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NO2− Ion
Concentration
Absorbance of
Solution at 550 nm
2.00 × 10−6 M
0.065
6.00 × 10
−6
M
0.205
10.00 × 10
−6
M
0.338
14.00 × 10−6 M
0.474
18.00 × 10−6 M
0.598
Unknown solution
0.402
(a) Construct a calibration plot, and determine
the slope and intercept.
(b) What is the nitrite ion concentration in the
unknown solution?
General Questions
These questions are not designated as to type or location in
the chapter. They may combine several concepts.
93. Suppose 16.04 g of benzene, C6H6, is burned in
oxygen.
(a) What are the products of the reaction?
(b) Write a balanced equation for the reaction.
(c) What mass of O2, in grams, is required for
complete combustion of benzene?
(d) What is the total mass of products expected
from the combustion of 16.04 g of benzene?
94. The metabolic disorder diabetes causes a buildup of
acetone, CH3COCH3, in the blood. Acetone, a volatile compound, is exhaled, giving the breath of
untreated diabetics a distinctive odor. The acetone is
produced by a breakdown of fats in a series of reactions. The equation for the last step, the breakdown
of acetoacetic acid to give acetone and CO2, is
producing it is the combination of arginine
(C6H14N4O2) with water to give urea and ornithine (C5H12N2O2).
C6H14N4O2 + H2O n NH2CONH2 + C5H12N2O2
arginine
urea
ornithine
If you excrete 95 mg of urea, what mass of arginine was used? What mass of ornithine was
produced?
96. The reaction of iron metal and chlorine gas to give
iron(III) chloride is illustrated below.
© Charles D. Winters/Cengage
content of a sample using spectrophotometry by
first using several organic compounds to form a
colored compound from the ion. The following
data were collected.
The reaction of iron and chlorine gas
(a) Write the balanced chemical equation for the
reaction.
(b) Beginning with 10.0 g of iron, what mass of
Cl2, in grams, is required for complete reaction? What mass of FeCl3 can be produced?
(c) If only 18.5 g of FeCl3 is obtained from 10.0 g
of iron and excess Cl2, what is the percent yield?
(d) If 10.0 g each of iron and chlorine are combined, what is the theoretical yield of iron(III)
chloride?
97. Some metal halides react with water to produce
the metal oxide and the appropriate hydrogen
halide (see photo). For example,
TiCl4(ℓ) + 2 H2O(ℓ) n TiO2(s) + 4 HCl(g)
Acetone, CH3COCH3
What mass of acetone can be produced from
125 mg of acetoacetic acid?
95. Your body deals with excess nitrogen by excreting
it in the form of urea, NH2CONH2. The reaction
© Charles D. Winters/Cengage
CH3COCH2CO2H n CH3COCH3 + CO2
The reaction of TiCl4 with the water in moist air
Study Questions
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245
(a) Name the four compounds involved in this
reaction.
(b) If you begin with 14.0 mL of TiCl4 (d =
1.73 g/mL), what mass of water, in grams,
is required for complete reaction?
(c) What mass of each product is expected?
98. The reaction of 750. g each of NH3 and O2 was
found to produce 562 g of NO (see pages 196–197).
4 NH3(g) + 5 O2(g) n 4 NO(g) + 6 H2O(ℓ)
(a) What mass of water is produced by this
reaction?
(b) What mass of O2 is required to consume
750. g of NH3?
99. Sodium azide, an explosive chemical once used in
automobile airbags, is made by the reaction:
NaNO3 + 3 NaNH2 n NaN3 + 3 NaOH + NH3
If you combine 15.0 g of NaNO3 with 15.0 g of
NaNH2, what mass of NaN3 is produced?
100. Iodine is made by the following reaction:
2 NaIO3(aq) + 5 NaHSO3(aq) n
3 NaHSO4(aq)+ 2 Na2SO4(aq) + H2O(ℓ)+ I2(aq)
(a) Name the two reactants.
(b) If you wish to prepare 1.00 kg of I2, what
masses of NaIO3 and NaHSO3 are required?
(c) What is the theoretical yield of I2 if you mixed
15.0 g of NaIO3 with 125 mL of 0.853 M
NaHSO3?
101. Saccharin, an artificial sweetener, has the formula
C7H5NO3S. Suppose you have a sample of a
saccharin-containing sweetener with a mass of
0.2140 g. After decomposition to free the sulfur
and convert it to the SO42− ion, the sulfate ion is
trapped as water-insoluble BaSO4 (Figure 4.4).
The quantity of BaSO4 obtained is 0.2070 g. What
is the mass percent of saccharin in the sample of
sweetener?
102. ▲ Boron forms a series of compounds with
hydrogen, all with the general formula BxHy.
Bx H y (s) 1 excess O2(g) →
x
y
B2O3(s) 1 H2O(g)
2
2
If 0.148 g of one of these compounds gives
0.422 g of B2O3 when burned in excess O2, what
is its empirical formula?
103. ▲ Silicon and hydrogen form a series of compounds with the general formula SixHy. To find
the formula of one of them, a 6.22-g sample of
the compound is burned in oxygen. All of the Si is
converted to 11.64 g of SiO2, and all of the H is
converted to 6.980 g of H2O. What is the
­empirical formula of the silicon compound?
246
104. ▲ Menthol, from oil of mint, has a characteristic
odor. The compound contains only C, H, and O.
If 95.6 mg of menthol burns completely in O2,
and gives 269 mg of CO2 and 111 mg of H2O,
what is the empirical formula of menthol?
105. ▲ Benzoquinone, a chemical used in the dye
industry and in photography, is an organic compound containing only C, H, and O. What is the
empirical formula of the compound if 0.105 g of
the compound gives 0.257 g of CO2 and 0.0350 g
of H2O when burned completely in oxygen?
106. ▲ Aqueous solutions of iron(II) chloride and
sodium sulfide react to form iron(II)sulfide and
sodium chloride. You combine 40. g of sodium
sulfide and 40. g of iron(II) chloride in an
aqueous solution.
(a) Write the balanced equation for the reaction.
(b) Which is the limiting reactant?
(c) What mass of FeS is produced?
(d) What mass of the excess reactant does not react?
(e) What mass of FeCl2 is required to react completely with 40. g of Na2S?
107. Sulfuric acid can be prepared starting with the
sulfide ore, cuprite (Cu2S). If each S atom in Cu2S
leads to one molecule of H2SO4, what is the theoretical yield of H2SO4 from 3.00 kg of Cu2S?
108. ▲ In an experiment, 1.056 g of a metal ­carbonate,
containing an unknown metal M, is heated to give
the metal oxide and 0.376 g CO2.
MCO3(s) + heat n MO(s) + CO2(g)
What is the identity of the metal M?
(a) M = Ni
(b) M = Cu
(c) M = Zn
(d) M = Ba
109. ▲ An unknown metal reacts with oxygen to give
the metal oxide, MO2. Identify the metal if a
0.356-g sample of the metal produces 0.452 g of
the metal oxide.
110. ▲ Titanium(IV) oxide, TiO2, is heated in hydrogen gas to give water and a new titanium oxide,
TixOy. If 1.598 g of TiO2 produces 1.438 g of TixOy,
what is the empirical formula of the new oxide?
111. ▲ Potassium perchlorate is prepared by the following sequence of reactions:
Cl2(g) + 2 KOH(aq) n KCl(aq) + KClO(aq) + H2O(ℓ)
3 KClO(aq) n 2 KCl(aq) + KClO3(aq)
4 KClO3(aq) n 3 KClO4(aq) + KCl(aq)
What mass of Cl2(g) is required to produce 234 kg
of KClO4?
Chapter 4 / Stoichiometry: Quantitative Information about Chemical Reactions
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112. ▲ Commercial sodium hydrosulfite is 90.1%
Na2S2O4. The sequence of reactions used to
prepare the compound is
Zn(s) + 2 SO2(g) n ZnS2O4(s)
ZnS2O4(s) + Na2CO3(aq) n ZnCO3(s) + Na2S2O4(aq)
(a) What mass of pure Na2S2O4 can be prepared
from 125 kg of Zn, 500. g of SO2, and an
excess of Na2CO3?
(b) What mass of the commercial product would
contain the Na2S2O4 produced using the
amounts of reactants in part (a)?
113. What mass of lime, CaO, can be obtained by
heating 125 kg of limestone that is 95.0% by mass
CaCO3?
CaCO3(s) n CaO(s) + CO2(g)
114. ▲ The elements silver, molybdenum, and sulfur
combine to form Ag2MoS4. What is the maximum
mass of Ag2MoS4 that can be obtained if 8.63 g of
silver, 3.36 g of molybdenum, and 4.81 g of sulfur
are combined? (Hint: What is the limiting
reactant?)
115. ▲ What mass of ammonium nitrate can be produced in the following reaction if 6.00 g of each
of the reactants is allowed to react?
2 N2(g) + 4 H2O(g) + O2(g) n 2 NH4NO3(s)
116. ▲ Titanium(IV) chloride can be produced from
the reaction of ilmenite ore (FeTiO3) with chlorine gas and carbon:
2 FeTiO3(s) + 7 Cl2(g) + 6 C(s) n
2 TiCl4(ℓ) + 2 FeCl3(s) + 6 CO(g)
What mass of titanium(IV) chloride can be
obtained from 100. g of FeTiO3, 175 g of Cl2, and
30.0 g of C?
117. ▲ A mixture of butene, C4H8, and butane, C4H10,
is burned in air to give CO2 and water. Suppose you
burn 2.860 g of the mixture and obtain 8.800 g of
CO2 and 4.095 g of H2O. What are the mass percentages of butene and butane in the mixture?
118. ▲ Cloth can be waterproofed by coating it with a
silicone layer. This is done by exposing the cloth
to (CH3)2SiCl2 vapor. The silicon compound
reacts with OH groups on the cloth to form a
waterproofing film (density = 1.0 g/cm3) of
[(CH3)2SiO]n, where n is a large integer number.
n (CH3)2SiCl2 + 2n OH− n
2n Cl− + n H2O + [(CH3)2SiO]n
The coating is added layer by layer, with each layer
of [(CH3)2SiO]n being 0.60 nm thick. Suppose
you want to waterproof a piece of cloth that is
3.00 square meters, and you want 250 layers of
waterproofing compound on the cloth. What
mass of (CH3)2SiCl2 do you need?
119. ▲ Copper metal can be prepared by roasting
copper ore, which can contain cuprite (Cu2S) and
copper(II) sulfide.
Cu2S(s) + O2(g) n 2 Cu(s) + SO2(g)
CuS(s) + O2(g) n Cu(s) + SO2(g)
Suppose an ore sample contains 11.00% impurity
in addition to a mixture of CuS and Cu2S. Heating
100.00 g of the mixture produces 75.40 g of
copper metal with a purity of 89.50%. What is the
weight percent of CuS in the ore? The weight
percent of Cu2S?
120. An Alka-Seltzer tablet contains exactly 100. mg of
citric acid, H3C6H5O7, plus some sodium bicarbonate. What mass of sodium bicarbonate is
required to consume 100. mg of citric acid by the
following reaction?
H3C6H5O7(aq) + 3 NaHCO3(aq) n
3 H2O(ℓ) + 3 CO2(g) + Na3C6H5O7(aq)
121. ▲ Sodium bicarbonate and acetic acid react
according to the equation
NaHCO3(aq) + CH3CO2H(aq) n
NaCH3CO2(aq) + CO2(g) + H2O(ℓ)
What mass of sodium acetate can be obtained
from mixing 15.0 g of NaHCO3 with 125 mL of
0.15 M acetic acid?
122. A noncarbonated soft drink contains an unknown
amount of citric acid, H3C6H5O7. If 100. mL of the
soft drink requires 33.51 mL of 0.0102 M NaOH
to neutralize the citric acid completely, what mass
of citric acid does the soft drink contain per
100. mL? The reaction of citric acid and NaOH is
H3C6H5O7(aq) + 3 NaOH(aq) n
Na3C6H5O7(aq) + 3 H2O(ℓ)
123. Sodium thiosulfate, Na2S2O3, is used as a fixer in
black-and-white photography. Suppose you have a
bottle of sodium thiosulfate and want to determine its purity. The thiosulfate ion can be oxidized with I2 according to the balanced, net ionic
equation
I2(aq) + 2 S2O32−(aq) n 2 I−(aq) + S4O62−(aq)
If you use 40.21 mL of 0.246 M I2 in a titration,
what is the weight percent of Na2S2O3 in a 3.232-g
sample of impure material?
124. You have a mixture of oxalic acid, H2C2O4, and
another solid that does not react with sodium
hydroxide. If 29.58 mL of 0.550 M NaOH is
required to titrate the oxalic acid in the 4.554-g
sample to the second equivalence point, what is
the mass percent of oxalic acid in the mixture?
Study Questions
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247
Oxalic acid and NaOH react according to the
equation
H2C2O4(aq) + 2 NaOH(aq) n
Na2C2O4(aq) + 2 H2O(ℓ)
125. (a) What is the pH of a 0.105 M HCl solution?
(b) What is the hydronium ion concentration in a
solution with a pH of 2.56? Is the solution
acidic or basic?
(c) A solution has a pH of 9.67. What is the
hydronium ion concentration in the solution?
Is the solution acidic or basic?
(d) A 10.0-mL sample of 2.56 M HCl is diluted
with water to 250. mL. What is the pH of the
dilute solution?
126. A solution of hydrochloric acid has a volume of
125 mL and a pH of 2.56. What mass of NaHCO3
must be added to completely consume the HCl?
127. ▲ One half liter (500. mL) of 2.50 M HCl is
mixed with 250. mL of 3.75 M HCl. Assuming the
total solution volume after mixing is 750. mL,
what is the concentration of hydrochloric acid in
the resulting solution? What is its pH?
128. A solution of hydrochloric acid has a volume of
250. mL and a pH of 1.92. Exactly 250. mL of
0.0105 M NaOH is added. What is the pH of the
resulting solution?
129. ▲ You place 2.56 g of CaCO3 in a beaker containing 250. mL of 0.125 M HCl. When the reaction has ceased, does any calcium carbonate
remain? What mass of CaCl2 can be produced?
CaCO3(s) + 2 HCl(aq) n
CaCl2(aq) + CO2(g) + H2O(ℓ)
130. The cancer drug cisplatin, Pt(NH3)2Cl2, can be
made by reacting (NH4)2PtCl4 with ammonia in
aqueous solution. Besides cisplatin, the other
product is NH4Cl.
(a) Write a balanced equation for this reaction.
(b) To obtain 12.50 g of cisplatin, what mass of
(NH4)2PtCl4 is required? What volume of
0.125 M NH3 is required?
(c) ▲ Cisplatin can react with the organic compound pyridine, C5H5N, to form a new
compound.
Pt(NH3)2Cl2(aq) + x C5H5N(aq) n
Pt(NH3)2Cl2(C5H5N)x(s)
Suppose you treat 0.150 g of cisplatin with what
you believe is an excess of liquid pyridine
(1.50 mL; d = 0.979 g/mL). When the reaction
is complete, you can find out how much pyridine
was not used by titrating the solution with standardized HCl. If 37.0 mL of 0.475 M HCl is
required to titrate the excess pyridine,
248
C5H5N(aq) + HCl(aq) n C5H5NH+(aq) + Cl−(aq)
what is the formula of the unknown compound
Pt(NH3)2Cl2(C5H5N)x?
131. ▲ You need to know the volume of water in a
small irregularly shaped swimming pool. To solve
the problem, you stir in a solution of a dye (1.0 g
of methylene blue, C16H18ClN3S, in 50.0 mL of
water). After the dye has mixed with the water in
the pool, you take a sample of the water. Using a
spectrophotometer, you determine that the concentration of the dye in the pool is 4.1 × 10−8 M.
What is the volume of water in the pool?
132. ▲ Calcium and magnesium carbonates occur
together in the mineral dolomite. Suppose you
heat a sample of the mineral to obtain the oxides,
CaO and MgO, and then treat the oxide sample
with hydrochloric acid.
CaO(s) + 2 HCl(aq) n CaCl2(aq) + H2O(ℓ)
MgO(s) + 2 HCl(aq) n MgCl2(aq) + H2O(ℓ)
If 7.695 g of the oxide sample requires 125 mL of
2.55 M HCl, what is the weight percent of each
oxide (CaO and MgO) in the sample?
133. Gold can be dissolved from gold-bearing rock by
treating the rock with sodium cyanide in the presence of oxygen.
4 Au(s) + 8 NaCN(aq) + O2(g) + 2 H2O(ℓ) n
4 NaAu(CN)2(aq) + 4 NaOH(aq)
(a) Name the oxidizing and reducing agents in
this reaction. What was oxidized, and what
was reduced?
(b) If you have exactly one metric ton (1 metric
ton = 1000 kg) of gold-bearing rock, what
volume of 0.075 M NaCN, in liters, do you
need to extract the gold if the rock is 0.019%
gold?
134. ▲ You mix 25.0 mL of 0.234 M FeCl3 with
42.5 mL of 0.453 M NaOH.
(a) What mass of Fe(OH)3 (in grams) will precipitate from this reaction mixture?
(b) One of the reactants (FeCl3 or NaOH) is
present in a stoichiometric excess. What is the
molar concentration of the excess reactant
remaining in solution after Fe(OH)3 has been
precipitated?
135. Atom Economy: One type of reaction used in the
chemical industry is a substitution, where one
atom or group is exchanged for another. In this
reaction, an alcohol, 1-butanol, is transformed
into 1-bromobutane by substituting Br for the
–OH group in the presence of sulfuric acid.
CH3CH2CH2CH2OH + NaBr + H2SO4 n CH3CH2CH2CH2Br + NaHSO4 + H2O
Chapter 4 / Stoichiometry: Quantitative Information about Chemical Reactions
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Calculate the % atom economy for the desired
product, CH3CH2CH2CH2Br.
136. Atom Economy: Ethylene oxide, C2H4O, is an
important industrial chemical [as it is the starting
place to make such important chemicals as ethylene glycol (antifreeze) and various polymers].
One way to make the compound is the chlorohydrin route.
C2H4 + Cl2 + Ca(OH)2 n C2H4O + CaCl2 + H2O
Another route is the modern catalytic reaction.
C2H4 + 1/2 O2 n C2H4O
(a) Calculate the % atom economy for the production of C2H4O in each of these reactions.
Which is the more efficient method?
(b) What is the percent yield of C2H4O if 867 g of
C2H4 is used to synthesize 762 g of the
product by the catalytic reaction?
In the Laboratory
137. Suppose you dilute 25.0 mL of a 0.110 M solution of
Na2CO3 to exactly 100.0 mL. You then take 10.0 mL
of this diluted solution and add it to a 250-mL volumetric flask. After filling the volumetric flask to the
mark with distilled water (indicating the volume of
the new solution is 250. mL), what is the concentration of the diluted Na2CO3 solution?
138. ▲ In some laboratory analyses, the preferred
technique is to dissolve a sample in an excess of
acid or base and then back-titrate the unreacted
acid or base with a standard base or acid. To
assess the purity of a sample of (NH4)2SO4 you
dissolve a 0.475-g sample of impure (NH4)2SO4 in
aqueous KOH.
(NH4)2SO4(aq) + 2 KOH(aq) n
2 NH3(aq) + K2SO4(aq) + 2 H2O(ℓ)
The NH3 liberated in the reaction is distilled from
the solution into a flask containing 50.0 mL of
0.100 M HCl. The ammonia reacts with the acid
to produce NH4Cl, but not all of the HCl is used
in this reaction. The amount of excess acid is
determined by titrating the solution with standardized NaOH. This titration consumes 11.1 mL
of 0.121 M NaOH. What is the weight percent of
(NH4)2SO4 in the 0.475-g sample?
139. Oyster beds in the oceans require chloride ions for
growth. The minimum concentration is 8 mg/L
(8 parts per million). To determine the concentration of chloride ion in a 50.0-mL sample of water,
you add a few drops of aqueous potassium chromate and then titrate the sample with 25.60 mL
of 0.001036 M silver nitrate. The silver nitrate
reacts with chloride ion. After the chloride ion is
completely precipitated, the silver nitrate reacts
with potassium chromate to give a red precipitate.
(a) Write a balanced net ionic equation for the
reaction of silver nitrate with chloride ions.
(b) Write a complete balanced equation and a net
ionic equation for the reaction of silver nitrate
with potassium chromate, indicating whether
each compound is water-soluble or not.
(c) What is the concentration of chloride ions in
the sample? Is it sufficient to promote oyster
growth?
140. ▲ A compound consisting of yttrium(III) ions,
barium(II) ions, both copper(II) and copper(III)
ions, and oxide ions is a superconducting material
at low temperatures (page 177). It has the formula
YBa2Cu3O7−x where x is a variable between 1 and
0. To find out the value of x, you dissolve
34.02 mg of the compound in 5 mL of 1.0 M
HCl. Bubbles of oxygen gas (O2) are observed as
the following reaction occurs:
YBa2Cu3O7−x(s) + 13 H+(aq) n Y3+(aq) + 2 Ba2+(aq) + 3 Cu2+(aq)
+ 1/4(1 − 2x) O2(g) + 13/2 H2O(ℓ)
You then boil the solution, cool it, and add 10 mL
of 0.70 M KI under argon. The following reaction
occurs:
2 Cu2+(aq) + 5 I−(aq) n 2 CuI(s) + I3−(aq)
When this reaction is complete, a titration of the
resulting solution with sodium thiosulfate
requires 1.542 × 10−4 mol S2O32−(aq).
I3−(aq) + 2 S2O32−(aq) n 3 I−(aq) + S4O62−(aq)
What is the value of x in YBa2Cu3O7−x?
141. You wish to determine the weight percent of
copper in a copper-containing alloy. After dissolving a 0.251-g sample of the alloy in acid, an excess
of KI is added, and the Cu2+ and I− ions undergo
the reaction
2 Cu2+(aq) + 5 I−(aq) n 2 CuI(s) + I3−(aq)
The liberated I3− is titrated with sodium thiosulfate according to the equation
I3−(aq) + 2 S2O32−(aq) n S4O62−(aq) + 3 I−(aq)
(a) Designate the oxidizing and reducing agents
in the two reactions above.
(b) If 26.32 mL of 0.101 M Na2S2O3 is required for
titration to the equivalence point, what is the
weight percent of Cu in the alloy?
142. ▲ A compound was isolated that can have either
of two possible formulas: (a) K[Fe(C2O4)2(H2O)2]
or (b) K3[Fe(C2O4)3]. To find which is correct, you
dissolve a weighed sample of the compound in
acid, forming oxalic acid, H2C2O4. You then titrate
Study Questions
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249
this oxalic acid with potassium permanganate,
KMnO4 (the source of the MnO4− ion). The balanced, net ionic equation for the titration is
5 H2C2O4(aq) + 2 MnO4−(aq) + 6 H3O+(aq) n 2 Mn2+(aq) + 10 CO2(g) + 14 H2O(ℓ)
Titration of 1.356 g of the compound requires
34.50 mL of 0.108 M KMnO4. Which is the
correct formula of the iron-containing compound: (a) or (b)?
143. ▲ Chromium(III) chloride forms many compounds with ammonia. To find the formula of
one of these compounds, you titrate the NH3 in
the compound with standardized acid.
Cr(NH3)xCl3(aq) + x HCl(aq) n
x NH4+(aq) + Cr3+(aq) + (x + 3) Cl−(aq)
Assume that 24.26 mL of 1.500 M HCl is used to
titrate 1.580 g of Cr(NH3)xCl3. What is the value
of x?
144. ▲ Thioridazine, C21H26N2S2, is a pharmaceutical
agent used to regulate dopamine. (Dopamine, a
neurotransmitter, affects brain processes that control
movement, emotional response, and ability to experience pleasure and pain.) A chemist can analyze a
sample of the pharmaceutical for the thioridazine
content by decomposing it to convert the sulfur in
the compound to sulfate ion. This is then trapped
as water-insoluble barium sulfate (see Figure 4.4).
SO4 (aq, from thioridazine) + BaCl2(aq) n
BaSO4(s) + 2 Cl−(aq)
2−
Suppose a 12-tablet sample of the drug yielded
0.301 g of BaSO4. What is the thioridazine
content, in milligrams, of each tablet?
145. ▲ A herbicide contains 2,4-D
(2,4-dichlorophenoxy­acetic acid), C8H6Cl2O3. A
1.236-g sample of the herbicide was decomposed
to liberate the chlorine as Cl− ion. This was precipitated as AgCl, with a mass of 0.1840 g. What
is the mass percent of 2,4-D in the sample?
OCH2CO2H
H
H
C
C
C
C
C
C
Cl
H
Cl
2,4-D (2,4-dichlorophenoxyacetic acid)
146. ▲ Sulfuric acid is listed in a catalog with a concentration of 95–98%. A bottle of the acid in the
stockroom states that 1.00 L has a mass of
1.84 kg. To determine the concentration of
250
sulfuric acid in the stockroom bottle, a student
dilutes 5.00 mL to 500. mL. She then takes four
10.00-mL samples and titrates each with standardized sodium hydroxide (c = 0.1760 M).
Sample
Volume NaOH (mL)
1
2
3
4
20.15
21.30
20.40
20.35
(a) What is the average concentration of the
diluted sulfuric acid sample?
(b) What is the mass percent of H2SO4 in the original bottle of the acid?
147. ▲ Anhydrous calcium chloride is a good drying
agent because it will rapidly pick up water.
Suppose you have stored some carefully dried
CaCl2 in a desiccator. Unfortunately, someone
did not close the top of the desiccator tightly,
and the CaCl2 became partially hydrated. A
150-g sample of this partially hydrated material
was dissolved in 80 g of hot water. When the
solution was cooled to 20 °C, 74.9 g of CaCl2 ⋅
6 H2O precipitated. Knowing the solubility of
calcium chloride in water at 20 °C is 74.5 g
CaCl2/100 g water, determine the water content
of the 150-g sample of partially hydrated
calcium chloride (in moles of water per mole of
CaCl2).
148. ▲ A 0.5510-g sample consisting of a mixture of
iron and iron(III) oxide was dissolved completely
in acid to give a solution containing iron(II) and
iron(III) ions. A reducing agent was added to
convert all of the iron to iron(II) ions, and the
solution was then titrated with a standardized
KMnO4 solution (0.04240 M); 37.50 mL of the
KMnO4 solution was required. Calculate the mass
percent of Fe and Fe2O3 in the 0.5510-g sample.
(Example 4.16 gives the equation for the reaction
of iron(II) ions and KMnO4.)
149. ▲ Phosphate in urine can be determined by spectrophotometry. After removing protein from the
sample, it is treated with a molybdenum compound to give, ultimately, a deep blue polymolybdate. The absorbance of the blue polymolybdate
can be measured at 650 nm and is directly related
to the urine phosphate concentration. A 24-hour
urine sample was collected from a patient; the
volume of urine was 1122 mL. The phosphate in a
1.00 mL portion of the urine sample was converted to the blue polymolybdate and diluted to
50.00 mL. A calibration curve was prepared using
phosphate-containing solutions. (Concentrations
are reported in grams of phosphorus (P) per liter
of solution.)
Chapter 4 / Stoichiometry: Quantitative Information about Chemical Reactions
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Solution (mass P/L)
Absorbance at 650 nm
in a 1.0-cm cell
1.00 × 10−6 g
0.230
g
0.436
3.00 × 10
−6
g
0.638
4.00 × 10
−6
g
0.848
Urine sample
0.518
12
Mass of product (g)
2.00 × 10
−6
152. ▲ A weighed sample of iron (Fe) is added to
liquid bromine (Br2) and allowed to react completely. The reaction produces a single product,
which can be isolated and weighed. The experiment was repeated a number of times with different masses of iron but with the same mass of
bromine (see graph below).
(a) What are the slope and intercept of the calibration curve?
(b) What is the mass of phosphorus per liter of
urine?
(c) What mass of phosphate did the patient
excrete in the one-day period?
150. ▲ A 4.000-g sample containing KCl and KClO4
was dissolved in sufficient water to give
250.00 mL of solution. A 50.00-mL portion of the
solution required 41.00 mL of 0.0750 M AgNO3
in a Mohr titration (page 234). Next, a 25.00-mL
portion of the original solution was treated with
V2(SO4)3 to reduce the perchlorate ions to chloride ions,
8 V3+(aq) + ClO4−(aq) + 12 H2O(ℓ) n
Cl−(aq) + 8 VO2+(aq) + 8 H3O+(aq)
and the resulting solution was titrated with
AgNO3. This titration required 38.12 mL of
0.0750 M AgNO3. What is the mass percent of KCl
and KClO4 in the mixture?
Summary and Conceptual Questions
The following questions may use concepts from this and
previous chapters.
151. Two beakers sit on a balance; the total mass is
167.170 g. One beaker contains a solution of KI;
the other contains a solution of Pb(NO3)2. When
the solution in one beaker is poured completely
into the other, the following reaction occurs:
Photos: © Charles D. Winters/Cengage
2 KI(aq) + Pb(NO3)2(aq) n 2 KNO3(aq) + PbI2(s)
Solutions of KI and
Pb(NO3)2 before reaction
Solutions after reaction
What is the total mass of the beakers and solutions after reaction? Explain completely.
10
8
6
4
2
0
0
1
2
3
4
Mass of Fe (g)
(a) What mass of Br2 is used when the reaction
consumes 2.0 g of Fe?
(b) What is the mole ratio of Br2 to Fe in the
reaction?
(c) What is the empirical formula of the product?
(d) Write the balanced chemical equation for the
reaction of iron and bromine.
(e) What is the name of the reaction product?
(f) Which statement or statements best describe
the experiments summarized by the graph?
(i)When 1.00 g of Fe is added to the Br2,
Fe is the limiting reagent.
(ii)When 3.50 g of Fe is added to the Br2,
there is an excess of Br2.
(iii)When 2.50 g of Fe is added to the Br2,
both reactants are used up completely.
(iv)When 2.00 g of Fe is added to the Br2,
10.8 g of product is formed. The percent
yield must therefore be 20.0%.
153. ▲ For the chemical reaction
2 H2(g) + O2(g) n 2 H2O(ℓ)
4.76 g of O2 was allowed to react with different
masses of H2. The data from the experiments are
in the following table.
Experiment
Mass
H2
Used
Mass
H 2O
Produced
Mass
H2 in
Excess
Mass
O2 in
Excess
1
0.200 g
1.79 g
0g
3.17 g
2
0.400 g
3.57 g
0g
1.59 g
3
0.600 g
5.36 g
0g
0g
4
0.800 g
5.36 g
0.200 g
0g
5
1.00 g
5.36 g
0.400 g
0g
Study Questions
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251
155. Let us explore a reaction with a limiting reactant.
Here, zinc metal is added to a flask containing
aqueous HCl, and H2 gas is a product.
(a) Prepare three graphs:
(i) Mass of H2O produced (y-axis) versus the
mass of H2 used (x-axis).
(ii) Mass of excess H2 versus the mass of H2
used.
(iii) Mass of excess O2 versus the mass of H2
used.
(b) In which experiment(s) was H2 the limiting
reactant? Explain.
(c) In which experiment(s) was O2 the limiting
reactant? Explain.
(d) In which experiment was the correct stoichiometric ratio of reagents used? Explain.
(e) Show that the law of conservation of mass is
followed in each experiment.
Zn(s) + 2 HCl(aq) n ZnCl2(aq) + H2(g)
© Charles D. Winters/Cengage
The three flasks each contain 0.100 mol of HCl.
Zinc is added to each flask in the following
quantities.
154. ▲ Aluminum and bromine react to form aluminum bromide according to the following balanced
chemical equation.
2 Al(s) + 3 Br2(ℓ) n Al2Br6(s)
Photos: © Charles D. Winters/Cengage
Flask 1:
7.00 g Zn
Before reaction
(a) Complete the following table for the reaction
of differing Al masses with 35.5 g of Br2.
Assume complete reaction of the limiting
reactant.
Experiment
1
1.00 g
2
2.00 g
3
4.00 g
4
6.00 g
Mass
Al2Br6
Produced
Mass
Al in
Excess
Mass
Br2 in
Excess
(b) In which experiment(s) was aluminum the
limiting reactant? Explain.
(c) In which experiments(s) was bromine the limiting reactant? Explain.
(d) Explain why the mass of aluminum bromide
produced does not continue to increase with
the mass of aluminum.
252
Flask 3:
1.31 g Zn
When the reactants are combined, the H2 inflates
the balloon attached to the flask. The results are
as follows:
Flask 1: Balloon inflates completely, but some Zn
remains when inflation ceases.
Flask 2: Balloon inflates completely. No Zn remains.
Flask 3: Balloon does not inflate completely. No Zn
remains.
After reaction
Mass
Al
Used
Flask 2:
3.27 g Zn
Explain these results. Perform calculations that
support your explanation.
156. Antacids are chemical compounds that can give
immediate relief from indigestion or heartburn
because they contain carbonate or hydroxide ions
that neutralize stomach acids. Some common
active ingredients include NaHCO3, KHCO3,
CaCO3, Mg(OH)2, and Al(OH)3. Although these
compounds give quick relief, they are not recommended for prolonged consumption. Calcium carbonate may contribute to the growth of kidney
stones, and calcium carbonate and aluminum
hydroxide may cause constipation. Magnesium
hydroxide, on the other hand, is a mild laxative
that can cause diarrhea. Antacids containing magnesium, therefore, are often combined with aluminum hydroxide since the aluminum counteracts
the laxative properties of the magnesium.
Chapter 4 / Stoichiometry: Quantitative Information about Chemical Reactions
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(a) Which of the compounds listed above
produce gas-forming reactions when combined with HCl?
(b) One tablet of Tums Regular Strength Antacid
contains 500. mg CaCO3.
(i)Write a balanced chemical equation for
the reaction of CaCO3 and stomach acid
(HCl).
(ii)What volume (in mL) of 0.500 M
HCl(aq) will react completely with one
tablet of Tums?
(c) The active ingredients in Rolaids® are CaCO3
and Mg(OH)2.
(i)Write a balanced chemical equation for
the reaction of Mg(OH)2 and HCl.
(ii)If 29.52 mL of 0.500 M HCl is required
to titrate one tablet of Rolaids® and the
tablet contains 550 mg of CaCO3, what
mass of Mg(OH)2 is present in one
­tablet?
(d) Maalox may be purchased in either a liquid or
solid form. One teaspoon of the liquid form
of Maalox contains a mixture of 200. mg of
Al(OH)3 and 200. mg of Mg(OH)2. What
volume of 0.500 M HCl(aq) will react completely with one teaspoon of Maalox?
(e) Which product neutralizes the greatest
amount of acid when taken in the quantities
presented above: one tablet of Tums or
Rolaids® or one teaspoon of Maalox?
157. Two students titrate different samples of the same
solution of HCl using 0.100 M NaOH solution
and phenolphthalein indicator (Figure 4.12). The
first student pipets 20.0 mL of the HCl solution
into a flask, adds 20 mL of distilled water and a
few drops of phenolphthalein solution, and
titrates until a lasting pink color appears. The
second student pipets 20.0 mL of the HCl solution into a flask, adds 60 mL of distilled water
and a few drops of phenolphthalein solution, and
titrates to the first lasting pink color. Each student
correctly calculates the molarity of an HCl solution. What will the second student’s result be?
(a) four times less than the first student’s result
(b) four times greater than the first student’s result
(c) two times less than the first student’s result
(d) two times greater than the first student’s result
(e) the same as the first student’s result
158. In most states, a person will receive a “driving
while intoxicated” (DWI) ticket if the blood
alcohol level (BAL) is 80 mg per deciliter (dL) of
blood or higher. Suppose a person is found to
have a BAL of 0.033 mol of ethanol (C2H5OH)
per liter of blood. Will the person receive a DWI
ticket?
159. Atom Economy: Benzene, C6H6, is a common
compound, and it can be oxidized to give maleic
anhydride, C4H2O3, which is used in turn to make
other important compounds.
H
H
H
C
C
C
C
O
C
C
H
H
+ 9⁄2 O2
H
H
C
C
C
O + 2 CO2 + 2 H2O
C
O
H
(a) What is the % atom economy for the synthesis
of maleic anhydride from benzene by this
reaction?
(b) If 972 g of maleic anhydride is produced from
exactly 1.00 kg of benzene, what is the percent
yield of the anhydride? What mass of the
by‑product CO2 is also produced?
160. Atom Economy: Maleic anhydride, C4H2O3, can
be produced by the oxidation of benzene (Study
Question 159). It can also be produced from the
oxidation of butene.
O
H
H2C
CH
CH2
CH3
+ 3 O2
H
C
C
C
O + 3 H2O
C
O
(a) What is the percent atom economy for the
synthesis of maleic anhydride from butene by
this reaction?
(b) If 1.02 kg of maleic anhydride is produced
from exactly 1.00 kg of butene, what is the
percent yield of the anhydride? What mass of
the by-product H2O is also produced?
Study Questions
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253
5
Principles of Chemical Reactivity:
Energy and Chemical Reactions
38
1
Sr
H
Strontium
Hydrogen
29
19
Cu
K
Potassium
Copper
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C hapt e r O ut li n e
5.1
Energy: Some Basic Principles
5.2
Specific Heat Capacity: Heating and Cooling
5.3
Energy and Changes of State
5.4
The First Law of Thermodynamics
5.5
Enthalpy Changes for Chemical Reactions
5.6
Calorimetry
5.7
Enthalpy Calculations
5.8
Product- or Reactant-Favored Reactions and Thermodynamics
The importance of energy is evident in everyday life—in heating and cooling homes,
in powering appliances, and in propelling vehicles, among other things. Most of
the energy used for these purposes is obtained by conducting chemical reactions,
largely by burning fossil fuels (natural gas, coal, and petroleum). Natural gas is used
for heating, coal and natural gas are burned to provide electrical power, and fuels
derived from petroleum are used in automobiles, trucks, and aircraft. In addition,
energy is required for living things: Chemical reactions in our bodies provide energy
for bodily functions, for movement, and to maintain body temperature.
Thermodynamics is the area of science that deals with energy and its relationship
to quantities such as heat and work. Determining the energy changes that occur when
chemical processes take place is an important factor in the study of chemical systems.
In this chapter, you will learn how to measure energy changes for chemical processes,
particularly those that result in heating and cooling, and how these can be used to
explore chemical reactions more thoroughly.
5.1 Energy: Some Basic Principles
World Energy Consumption In 2021,
• Describe the nature of energy transfers as heat.
• Understand the sign conventions of thermodynamics.
burning fossil fuels provided
about 84% of the total energy
used by people on our planet.
Nuclear power contributed about
5%. All other energy sources,
including renewable sources
(such as hydroelectric, solar,
wind, biomass, and geothermal)
provided about 12%.
Energy is defined as the capacity to do work and can be divided into two basic categories: kinetic energy (the energy associated with motion) and potential energy (the
energy that results from an object’s position, composition, or state). Chemists often
use the term thermal energy when referring to the kinetic energy of molecules.
Units of Energy The SI unit for
energy (the joule) is discussed on
page 37.
Goals for Section 5.1
• Recognize and use the language of thermodynamics: the system and its
surroundings; exothermic and endothermic reactions.
◀ The reaction of potassium and water. This reaction involves the transfer of energy between
the system and surroundings in the form of heat (thermal energy), work, and light.
© Charles D. Winters/Cengage
255
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Chemical energy is one type of potential energy. It is the energy associated with the
forces that hold atoms together as molecules or bind atoms and molecules together
as solids or liquids. In a chemical reaction, chemical energy (potential energy) is
converted to other forms of energy such as thermal energy or light (kinetic energy).
Although energy can be converted from one type into another, the total amount
of energy is conserved. This is formally stated in the law of conservation of energy:
Energy can neither be created nor destroyed. Or, stated differently, the total energy
of the universe is constant. To understand the importance of this law, some new
terminology must be introduced and the implications of a number of experiments
must be considered.
The Terminology of Thermodynamics
The definitions of terms (such
as energy, heat, and work)
used when discussing energy in
chemistry are more precise than
in everyday language.
Surroundings
System
© Charles D. Winters/Cengage
Systems and Surroundings
Surroundings
System
Figure 5.1 Systems and their
surroundings. Earth can be
considered a thermodynamic system,
with the rest of the universe as its
surroundings. A chemical reaction
(here the reaction of S and O2)
occurring in a laboratory is also
a system, with the flask and the
laboratory as its surroundings.
Thermal Equilibrium A general
feature of systems at equilibrium
is that there is no change on
a macroscopic level but that
processes still occur at the
particulate level. (Section 3.3,
page 145.)
In thermodynamics, the terms system and surroundings have precise and i­mportant
meanings. A system is defined as an object, or collection of objects, being s­ tudied
(Figure 5.1). The surroundings include everything outside the system that can
exchange energy and/or matter with the system. In the discussion that follows,
systems will need to be defined precisely. If a chemical reaction is carried out in
solution, for example, the system might be defined as the reactants, products, and
solvent. The surroundings would be the reaction flask and the air in the room and
anything else in contact with the flask with which it might exchange energy or matter. At the atomic level, the system could be a single atom or molecule, and the surroundings would be the atoms or molecules in its vicinity. This concept of a system
and its surroundings applies to nonchemical situations as well. To study the energy
balance on our planet, Earth might be defined as the system and outer space as
the surroundings. On a cosmic level, the solar system might be defined as the system being studied, and the rest of the galaxy would be the surroundings. The treatment of experimental data is dependent upon the choices made for the system and
surroundings.
Directionality and Extent of Transfer of Heat:
Thermal Equilibrium
Energy can be transferred between a system and its surroundings or between different parts of the system. One way that energy can be transferred is as heat; this
occurs when two objects at different temperatures are brought into contact. In
Figure 5.2, for example, the beaker of water and the piece of metal being heated
in a Bunsen burner flame have different temperatures. When the hot metal is
plunged into the cold water, energy is transferred as heat from the metal to the
water. The thermal energy (molecular motion) of the water molecules increases,
and the thermal energy of the metal atoms decreases. Eventually, the two objects
reach the same temperature, and the system has reached thermal equilibrium. The
distinguishing feature of thermal equilibrium is that, on the macroscopic scale,
no further temperature change occurs; both the metal and water are at the same
temperature.
Photos: © Charles D. Winters/Cengage
Figure 5.2 Energy transfer.
256
Energy transfer as heat occurs from the
hotter metal cylinder to the cooler water.
Eventually, the water and metal reach the
same temperature and are said to be in
thermal equilibrium.
Chapter 5 / Principles of Chemical Reactivity: Energy and Chemical Reactions
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A Closer Look
What Is Heat?
Two hundred years ago, scientists characterized heat as a real
substance called “caloric fluid.” The caloric hypothesis supposed that when a fuel burned and a pot of water was heated,
for example, caloric fluid was transferred from the fuel to the
water. Burning the fuel released caloric fluid, and the temperature of the water increased as the caloric fluid was absorbed.
However, the caloric-fluid concept was wrong. Experiments
by James Joule (1818–1889) and Benjamin Thompson
(1753–1814) that showed the interrelationship between
heat and other forms of energy, such as mechanical energy,
provided the key to disproving this idea. Even so, everyday language retains the influence of this early theory. For example,
people often speak of heat “flowing” as if it were a fluid.
From the discussion so far, you know one thing that “heat” is
not—but what is it? Heat is said to be a “process quantity.” It is
the process by which energy is transferred across the boundary of
a system owing to a difference in temperature between the two
sides of the boundary. In this process, the energy of the object at
the lower temperature increases, and the energy of the object at
the higher temperature decreases.
Heat is not the only process by which energy can be transferred. Another is work (as described in Section 5.4).
The idea of energy transfer by the processes of heat and work
is embodied in the definition of thermodynamics: the science of
heat and work.
Gregory_G/Shutterstock.com
A Rumford fireplace. Benjamin Thompson (1753–1814), also
known as Count Rumford, established his scientific reputation
through research on the explosive force of gunpowder. His
experience with explosives led to an interest in heat, and he
designed a classic experiment that showed the relationship between
work and heat. He is also known for developing the Rumford
fireplace, in which the back walls of the fire box are angled to
reflect heat into the room and the chimney is better designed to
carry away smoke. This design is still in use today.
This experiment illustrates two important principles:
1. Energy transfer as heat occurs spontaneously from an object at a higher temperature to an object at a lower temperature; the object whose temperature
increases gains thermal energy and the object whose temperature decreases
loses thermal energy.
2.
Transfer of energy as heat continues until both objects are at the same temperature
and thermal equilibrium is achieved.
For the specific case where energy is transferred only as heat within an isolated
system (that is, a system that cannot transfer either energy or matter with its surroundings), the quantity of energy lost as heat by the hotter object and the quantity
of energy gained as heat by the cooler object are numerically equal. This is required
by the law of conservation of energy.
When energy is transferred as heat between a system and its surroundings, the
directionality of this transfer is described as exothermic or endothermic (Figure 5.3).
•
In an exothermic process, energy is transferred as heat from a system to its surroundings. The energy of the system decreases and the energy of the surroundings increases. The energy transferred as heat is designated by the symbol q.
Because the system ends up with less energy than it started with, qsys < 0 (where
the subscript sys refers to the system).
•
An endothermic process is the opposite of an exothermic process. Energy is transferred as heat from the surroundings to the system, increasing the energy of the
system and decreasing the energy of the surroundings. For an endothermic process, qsys > 0 because the system ends up with more energy than it started with.
5.1 Energy: Some Basic Principles
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257
Photos: © Charles D. Winters/Cengage
Exothermic
qsys < 0
Endothermic
qsys > 0
System
System
Surroundings
Exothermic: energy transferred from system to surroundings
Surroundings
Endothermic: energy transferred from surroundings to system
Figure 5.3 Exothermic and endothermic processes. The symbol q represents the energy transferred as heat, and the subscript
sys refers to the system.
5.2 Specific Heat Capacity: Heating
and Cooling
Goal for Section 5.2
• Use specific heat capacity in calculations of energy transfers as heat involving
temperature changes.
Change of State A change of state
(such as melting or boiling) is a
change of an object’s physical
state (solid, liquid, or gas).
Heat capacity Heat capacity is
defined as the energy required
to change the temperature of
an object by 1 kelvin. Common
units for heat capacity are joules
per kelvin (J/K).
When an object is heated or cooled without a change in its physical state, the object’s
change in temperature is proportional to the quantity of thermal energy gained or lost by
the object. The energy gained or lost as heat (q, in joules) is described by Equation 5.1.
q = C × ∆T
(5.1)
The constant, C, is called the heat capacity. This constant, usually reported in units
of J/K, has a unique value for every object. The change in temperature, ∆T, is calculated as the final temperature minus the initial temperature.
∆T = Tfinal − Tinitial
(5.2)
For many calculations the specific heat capacity (c) is used. This is defined as
the energy transferred as heat that is required to raise the temperature of one gram
of a substance or a mixture by one kelvin. It is often reported in units of joules per
gram per kelvin (J/g ⋅ K). A few specific heat capacities are listed in Figure 5.4, and a
longer list is given in Appendix D (Table D.2).
The energy gained or lost as heat when a given mass of a substance is warmed
or cooled can be calculated using Equation 5.3.
q = c × m × ∆T
(5.3)
Here, q is the energy gained or lost as heat by a given mass of substance (m), c is the
specific heat capacity, and ∆T is the change in temperature.
Heat capacity can also be expressed on a per-mole basis. The amount of energy that
is transferred as heat in raising the temperature of one mole of a substance by one kelvin is the molar heat capacity. For water, the molar heat capacity is 75.4 J/mol ⋅ K. The
molar heat capacity of metals at room temperature is always near 25 J/mol ⋅ K.
Calculating a value of q using Equation 5.3 gives a result with an algebraic sign
that indicates the direction of energy transfer. This depends on whether the value of
∆T calculated using Equation 5.2 is positive or negative. A positive q indicates that the
energy of the system increases, while a negative q indicates that the energy of the system
decreases. For example, you can use the specific heat capacity of copper, 0.385 J/g ⋅ K, to
calculate the energy that must be transferred from the surroundings to a 10.0-g sample
of copper if the metal’s temperature is raised from 298 K (25 °C) to 598 K (325 °C).
258
Chapter 5 / Principles of Chemical Reactivity: Energy and Chemical Reactions
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Specific Heat Capacities of Some Elements and Compounds
H2O
© Charles D. Winters/Cengage
Cu
Fe
Al
Substances
Specific Heat
Capacity (J/g ? K)
Molar Heat
Capacity (J/mol ? K)
Al, aluminum
0.897
24.2
Fe, iron
0.449
25.1
Cu, copper
0.385
24.5
Au, gold
0.129
25.4
Water (liquid)
4.184
75.4
Water (ice)
2.06
37.1
Water (steam)
1.86
33.6
HOCH2CH2OH(),
ethylene glycol
(antifreeze)
2.39
14.8
All metals have
molar heat capacities
near 25 J/mol ⋅ K
Figure 5.4 Specific heat capacity. Metals have different values of specific heat capacity. However, their molar heat capacities are all
near 25 J/mol ⋅ K.
J
(10.0 g)(598 K − 298 K) = +1160 J
g∙K
Tfinal
Final temp.
Tinitial
Initial temp.
Notice that the answer has a positive sign. This indicates that the energy of the sample
of copper has increased by 1160 J, which is in accord with energy being transferred as
heat to the copper (the system) from the surroundings.
The relationship between energy, mass, and specific heat capacity has numerous implications. The high specific heat capacity of liquid water, 4.184 J/g ⋅ K, is a
major reason that large bodies of water have a profound influence on climate. In
spring, lakes warm up more slowly than the air. In autumn, the energy transferred
by a large lake as it cools moderates the drop in air temperature. The relevance of
specific heat capacity is also illustrated when food is wrapped in aluminum foil
(specific heat capacity 0.897 J/g ⋅ K) and heated in an oven. You can remove the foil
with your fingers after taking the food from the oven. The small mass of aluminum
foil used and its low specific heat capacity result in only a small quantity of energy
being transferred to your fingers (which have a larger mass and a higher specific
heat capacity).
Stockcreations/Shutterstock.com
q = 0.385
A practical example of knowing
about specific heat capacity. If
you are careful, it is possible to
remove the salmon from the grill
by grasping the edges of the
aluminum foil with unprotected
hands. Due to the small quantity
of aluminum and its low specific
heat capacity, only a small
quantity of energy is transferred.
Exam p le 5.1
Specific Heat Capacity
Problem How much energy must be transferred to raise the temperature of a cup of
coffee (250 mL) from 20.5 °C (293.7 K) to 95.6 °C (368.8 K)? Assume that water and coffee
have the same density (1.00 g/mL) and specific heat capacity (4.184 J/g ⋅ K).
What Do You Know? The energy required to warm a substance is related to
its specific heat capacity (c), the mass of the substance, and the temperature change
­(Equation 5.3). The mass of coffee, initial and final temperatures, and the value for c are
given in the problem.
5.2 Specific Heat Capacity: Heating and Cooling
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259
Strategy You can calculate the mass of coffee from the volume and density (mass =
volume × density) and the temperature change from the initial and final temperatures
(∆T = Tfinal − Tinitial). Use Equation 5.3 to solve for q.
Solution Mass of coffee = (250 mL)(1.00 g/mL) = 250 g
∆T = Tfinal − Tinitial = 368.8 K − 293.7 K = 75.1 K
q = c × m × ∆T
q = (4.184 J/g ⋅ K)(250 g)(75.1 K)
q = 79,000 J (or 79 kJ)
Think about Your Answer The positive value for the answer indicates that
energy has been transferred to the coffee. The thermal energy of the coffee is now higher.
Check Your Understanding
An experiment showed that 59.8 J was required to raise the temperature of 25.0 g of
­e thylene glycol (a compound used as antifreeze in automobile engines) by 1.00 K.
­Calculate the specific heat capacity of ethylene glycol from these data.
Quantitative Aspects of Energy Transferred as Heat
Specific heat capacity is a characteristic intensive property of a pure substance. It can
be determined experimentally by accurately measuring temperature changes that
occur when energy is transferred as heat from the substance to a known quantity of
water (whose specific heat capacity is known).
Suppose a 55.0-g piece of metal is heated in boiling water to 99.8 °C and then
dropped into cool water in an insulated beaker (Figure 5.5). Assume the beaker contains 225 g of water and its initial temperature (before the metal was dropped in)
was 21.0 °C. The final temperature of the metal and water is 23.1 °C. What is the
specific heat capacity of the metal? Here are the important aspects of this experiment.
•
The system in this experiment includes the metal and the water; the surroundings include the beaker and the environment.
•
Energy is transferred only within the system. (This means that energy is not transferred between the system and the surroundings. This assumption is good, but
not perfect; for a more accurate result, any energy transfer to the surroundings
must also be measured.)
Chemistry in Your Career
Randy Santos
Randy Santos
260
Randy Santos uses chemistry daily in his job at an
urgent care veterinary facility, where he is a ­veterinary
technician. There he works alongside veterinarians
to triage, stabilize, diagnose, and treat his animal
patients.
Santos draws on his chemistry education
(with degrees in veterinary nursing and health
­science) to understand the fundamental ­principles
behind medications and diagnostic testing.
Knowing how different drugs affect the body or
react with another substance or medication is
extremely important. “When mixing drugs in a
singular syringe, I’m always very observant of different reactions such as heat production, color
change/discolorations, or even precipitation.”
When administering anesthesia or sedatives, Santos must calculate the duration of drug action
based on how the body breaks down molecules.
Santos finds it rewarding to see his patients
recover and come back happy and healthy. He
adds, “I also really enjoy mastering my technical
skills and taking on the challenges the day brings.”
Chapter 5 / Principles of Chemical Reactivity: Energy and Chemical Reactions
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Hot metal (55.0 g iron)
99.8 °C
23.1 °C
Metal cools in
exothermic process.
∆T of metal is negative.
Immerse
hot metal
in water
qmetal is negative.
Water is warmed in
endothermic process.
21.0 °C
∆T of water is positive.
qwater is positive.
Cool water (225 g)
Figure 5.5 Transfer of energy as heat. When energy is transferred as heat from a hot metal
to cool water, the thermal energy of the metal decreases and that of the water increases. The
value of qmetal is thus negative and the value of qwater is positive.
•
Energy is transferred only as heat within the system.
•
The water and the metal end with the same temperature. (Tfinal is the same for both.)
•
The energy transferred as heat from the metal to the water, qmetal, has a negative
value because the temperature of the metal decreases. Conversely, qwater has a
positive value because its temperature increases.
•
The values of qmetal and qwater are numerically equal but have opposite signs.
Because of the law of conservation of energy, in an isolated system the sum of
the energy changes within the system must be zero. If energy is transferred only as
heat, then
q1 + q2 + q3 + . . . = 0 (5.4)
where the quantities q 1, q2, and so on represent the energies transferred as heat
for the individual parts of the system. For this specific problem, there are thermal
energy changes associated with the two components of the system, water and metal,
qwater and qmetal; thus
qwater + qmetal = 0
Each of these quantities is related individually to specific heat capacities, masses,
and changes of temperature, as defined by Equation 5.3. Thus
[cwater × mwater × (Tfinal − Tinitial, water)] + [cmetal × mmetal × (Tfinal − Tinitial, metal)] = 0
The specific heat capacity of the metal, cmetal , is the unknown in this example. Using
the specific heat capacity of water (4.184 J/g ⋅ K) and converting Celsius to kelvin
temperatures gives
Celsius to Kelvin Unit
Conversions Conversions are
not necessary when measuring
changes in temperature because
a change of 1° C is also a
change of 1 K. (See Problem
Solving Tip 5.1.)
[(4.184 J/g ⋅ K)(225 g)(296.3 K − 294.2 K)] + [(cmetal)(55.0 g)(296.3 K − 373.0 K)] = 0
cmetal = 0.47 J/g ⋅ K
Problem Solving Tip 5.1 Calculating DT
Virtually all calculations that
involve temperature in chemistry
require expressing temperature in
kelvins. In calculating ∆T, however,
you can use Celsius temperatures
because a kelvin and a Celsius
degree are the same size. That
is, the difference between two
temperatures is the same on both
scales. For example, the difference
between the boiling and freezing
points of water is
∆T, Celsius = 100 °C − 0 °C = 100 °C
∆T, kelvin = 373 K − 273 K = 100 K
5.2 Specific Heat Capacity: Heating and Cooling
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261
E xamp le 5.2
Using Specific Heat Capacity
Problem In an experiment like that shown in Figure 5.5, an 88.5-g piece of iron whose
temperature is 78.8 °C (352.0 K) is placed in a beaker containing 244 g of water at 18.8 °C
(292.0 K). When thermal equilibrium is reached, what is the final temperature? (Assume no
energy is transferred to warm the beaker and its surroundings.)
What Do You Know? Iron cools and the water warms until thermal equilibrium
is reached. The energies associated with the two changes are determined by the specific
heat capacities, masses, and temperature changes for each species. If iron and water are
defined as the system, the sum of these two energy quantities will be zero due to the law
of conservation of energy. The final temperature is the unknown in this problem. Masses
and initial temperatures are given; the specific heat capacities of iron and water can be
found in Appendix D or Figure 5.4.
Strategy The sum of the two energy quantities, qwater and qFe , is zero (qwater + qFe = 0).
Each energy quantity is defined using Equation 5.3; the value of ∆T in each is Tfinal − Tinitial.
You can use either kelvin or Celsius temperatures (Problem Solving Tip 5.1). Substitute the
given information into Equation 5.4 and solve.
Solution
qwater + qFe = 0
[cwater × mwater × (Tfinal − Tinitial, water)] + [cFe × mFe × (Tfinal − Tinitial, Fe)] = 0
[(4.184 J/g ⋅ K)(244 g)(Tfinal − 292.0 K)] + [(0.449 J/g ⋅ K)(88.5 g)(Tfinal − 352.0 K)] = 0
(1021 J/K)Tfinal − 298100 J + (39.74 J/K)Tfinal − 13990 J = 0
(1061 J/K)Tfinal − 312100 J = 0
(1061 J/K)Tfinal = 312100 J
Tfinal = 294 K (21 °C)
Think about Your Answer Be sure to notice that Tinitial for the metal and T­initial
for the water in this problem have different values. Also, the low specific heat capacity
and smaller quantity of iron result in the temperature of iron being reduced by about
60 degrees; in contrast, the temperature of the water has been raised by only a few
degrees. Finally, as expected, Tfinal (294 K) is between Tinitial, Fe and Tinitial, water.
Check Your Understanding
A 15.5-g piece of chromium, heated to 100.0 °C, is dropped into 55.5 g of water at
16.5 °C. The final temperature of the metal and the water is 18.9 °C. What is the specific
heat capacity of chromium? (Assume no energy is lost to the container or to the surrounding air.)
5.3 Energy and Changes of State
Goal for Section 5.3
• Use heat of fusion and heat of vaporization to calculate the energy transferred as
heat in changes of state.
A change of state refers to changes (such as melting or boiling) between the three
states of matter: solid, liquid, and gas. When a solid melts, its atoms, molecules, or
262
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ions move about vigorously enough to break free of the attractive forces holding
them in the rigid positions of the solid lattice. When a liquid boils, the particles
move much farther apart from one another, to distances at which attractive forces
are minimal. In both cases, energy must be furnished to overcome attractive forces
among the particles.
The energy transferred as heat required to convert a substance from a solid
to a liquid is called the heat of fusion. The energy transferred as heat to convert
a liquid to a vapor is called the heat of vaporization. Values for a few common
substances are given in Appendix D (Table D.3).
It is important to recognize that temperature is constant throughout a change
of state (Figure 5.6). During a change of state, the added energy is used to overcome
the forces holding one molecule to another, not to increase the temperature.
For water, the heat of fusion at 0 °C is 333 J/g, and the heat of vaporization at
100 °C is 2256 J/g. These values can be used to calculate the energy required for
a given mass of water to melt or evaporate, respectively. For example, the energy
required to convert 500. g of water from the liquid to gaseous state at 100 °C is
(2256 J/g)(500. g) = 1.13 × 106 J (= 1130 kJ)
Temperature Dependence
of the Heats of Fusion and
Vaporization Quantities such as
the heat of fusion and the heat
of vaporization of a substance
are dependent on temperature.
For example, the heat of
vaporization of water at 25 °C
is 2442 J/g. This value is slightly
larger than the value at 100 °C
(2256 J/g).
In contrast, to melt the same mass of ice to form liquid water at 0 °C requires only
167 kJ.
(333 J/g)(500. g) = 1.67 × 105 J (= 167 kJ)
Figure 5.7 gives a profile of the energy changes occurring as 500. g of ice at
−50 °C is converted to water vapor at 200 °C. This involves a series of steps:
(1) warming ice to 0 °C, (2) conversion to liquid water at 0 °C, (3) warming
liquid water to 100 °C, (4) evaporation at 100 °C, and (5) warming the water
vapor to 200 °C. Each step requires the input of energy. The energy transferred
as heat to raise the temperature of solid, liquid, and vapor can be calculated with
Equation 5.3, using the specific heat capacities of ice, liquid water, and water
vapor (which are different), and the energies for the changes of state can be calculated using heats of fusion and vaporization. These calculations are shown in
Example 5.3.
Iron,
2.0 kg
Photos: © Charles D. Winters
/Cengage
Ice, 2.0 kg
+ 500 kJ
0 °C
+ 500 kJ
0 °C
0 °C
0 °C
557 °C
State changes.
Temperature does NOT change.
Transferring 500 kJ of energy as heat to 2.0 kg of ice at 0 °C will cause
1.5 kg of ice to melt to water at 0 °C (and 0.5 kg of ice will remain).
No temperature change occurs.
Temperature changes.
State does NOT change.
In contrast, transferring 500 kJ of energy as heat to 2.0 kg of
iron at 0 °C will cause the temperature to increase to 557 °C
(and the metal to expand slightly but not melt).
Figure 5.6 Contrast between a change of state and an increase in temperature as a result of
adding energy.
5.3 Energy and Changes of State
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263
Figure 5.7 Energy transfer as
heat and the temperature change
as 500. g of water warms from
−50 °C to 200 °C (at 1 atm).
+200
Energy liberated
Temperature (°C)
+150
Boiling
+100
+50
0
−50
0
STEAM
(100 °C–200 °C)
LIQUID WATER (0 °C–100 °C)
Melting
Energy absorbed
ICE (−50 °C–0 °C)
200
400
600
800
1000
Heat (kJ)
1200
1400
1600
Exam pl e 5 .3
Energy and Changes of State
Problem Calculate the energy needed to convert 500. g of ice at −50.0 °C (223.2 K)
to steam at 200.0 °C (473.2 K). The heat of fusion of water at 0 °C is 333 J/g, and the heat
of vaporization at 100 °C is 2256 J/g. The specific heat capacities of ice, liquid water, and
water vapor are given in Appendix D.
What Do You Know? The overall process of converting ice at −50 °C to steam at
200 °C involves both temperature changes and changes of state (Figure 5.7); all require
input of energy as heat. Recall that melting occurs at 0 °C (273.2 K) and boiling at 100 °C
(373.2 K) at 1 atm pressure. You know the mass of the water and will need the specific heat
capacities of ice, liquid water, and steam from Appendix D. The heat of fusion of water
(333 J/g), and the heat of vaporization (2256 J/g) are given.
Strategy The problem is broken down into a series of steps:
Strategy Map
Step 1. Warm the ice from −50 °C to the melting point of ice, 0 °C.
Problem
Calculate energy required
to heat a mass of water from
−50.0 °C to steam at 200.0 °C.
Step 2. Melt the ice at 0 °C.
Step 3. Raise the temperature of the liquid water from 0 °C to the boiling point of water,
100 °C.
Step 4. Evaporate the water at 100 °C.
Step 5. Raise the temperature of the steam from 100 °C to 200 °C.
Data/Information
• Mass of water
• ∆T
• Heats of fusion and
vaporization of water
• Specific heat capacities
Step 6. Sum the energy transferred as heat in each step to find the total.
Use Equation 5.3 and the specific heat capacities of solid, liquid, and gaseous water to
calculate the energy transferred as heat associated with the temperature changes. Use the
heats of fusion and of vaporization to calculate the energy transferred as heat associated
with changes of state. The total energy transferred as heat is the sum of the energies of
the individual steps.
Solution
Step 1
Raise the temperature of ice from 250.0 °C to the melting point of ice, 0.0 °C.
Energy is being transferred as heat to raise the temperature of the ice, so use Equation 5.3
to calculate q from the specific heat capacity of ice, the mass, and the change in temperature to go from −50.0 °C (223.2 K) to the melting point of ice, 0.0 °C (273.2 K).
q1 = (2.06 J/g ⋅ K)(500. g)(273.2 K − 223.2 K) = 5.150 × 104 J
264
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Step 2
Melt the ice at 0.0 °C.
Energy is being transferred as heat to change the state from solid to liquid, so multiply
the mass by the heat of fusion at this temperature.
q2 = (500. g)(333 J/g) = 1.665 × 105 J
Step 3
Raise the temperature of the liquid water from 0 °C to the boiling point of
water, 100.0 °C.
Energy is being transferred as heat to raise the temperature of the liquid water, so use
Equation 5.3 to calculate q from the specific heat capacity of liquid water, the mass, and
the change in temperature to go from the melting point of ice, 0.0 °C (273.2 K), to the
boiling point of water, 100.0 °C (373.2 K).
q3 = (4.184 J/g ⋅ K)(500. g)(373.2 K − 273.2 K) = 2.092 × 105 J
Step 4
Evaporate the water at 100.0 °C.
Energy is being transferred as heat to change the state from liquid to gas, so multiply the
mass by the heat of vaporization at this temperature.
q4 = (500. g)(2256 J/g) = 1.128 × 106 J
Step 5
Raise the temperature of the steam from 100.0 °C to 200.0 °C.
Energy is being transferred as heat to raise the temperature of the steam. Use Equation 5.3
to calculate q from the specific heat capacity of the steam, the mass, and the change
in temperature to go from the boiling point of water, 100.0 °C (373.2 K), to the final
temperature, 200.0 °C (473.2 K).
q5 = (1.86 J/g ⋅ K)(500. g)(473.2 K − 373.2 K) = 9.300 × 104 J
Step 6
Sum the energy transferred as heat in each step to find the total.
qtotal = q1 + q2 + q3 + q4 + q5
qtotal = 1.65 × 106 J (or 1650 kJ)
Think about Your Answer The conversion of liquid water to steam is the largest
increment of energy by a considerable margin. (You may have noticed that when water is
heated at a steady rate on a stove it takes much less time to heat the water to boiling than
it takes to boil off the water.)
Check Your Understanding
Calculate the amount of energy necessary to raise the temperature of 1.00 L of ethanol (d = 0.7849 g/cm3) from 25.0 °C to its
boiling point (78.3 °C) and then to vaporize the liquid. (cethanol = 2.44 J/g ⋅ K; heat of vaporization at 78.3 °C = 38.56 kJ/mol.)
Exam p le 5.4
Change of State
Problem What is the minimum mass of ice at 0 °C that must be added to the contents
of a can of diet cola (340. mL) to cool the cola from 20.5 °C to 0.0 °C? Assume that the specific heat capacity and density of diet cola are the same as for water.
What Do You Know? The final temperature is 0 °C. Melting ice requires energy as
heat, and cooling the cola evolves energy as heat. The sum of the energy changes for the
two components in the system is zero; that is, the two energy changes (melting ice, cooling cola) will be the same magnitude but opposite in sign. (Assume there is no transfer of
energy between the surroundings and the system.) The density and specific heat capacity
of liquid water (Appendix D) are needed.
Strategy Assuming only energy changes within the system, qcola + qice = 0. The energy
evolved as the cola cools, qcola, is calculated using Equation 5.3. The initial temperature is
20.5 °C, and the final temperature is 0 °C. The mass of cola is calculated from the volume and
5.3 Energy and Changes of State
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265
density. The energy as heat required to melt the ice, qice, is equal to the heat of fusion at 0 °C
(333 J/g) multiplied by the mass of ice: qice = (333 J/g)(mice). The mass of ice is the unknown.
Solution The mass of cola is 340. g [(340. mL)(1.00 g/mL) = 340. g], and its temperature
changes from 293.7 K to 273.2 K. The heat of fusion of water is 333 J/g, and the mass of ice is
the unknown.
qcola + qice = 0
ccola × mcola × (Tfinal − Tinitial) + (heat of fusion of water)(mice) = 0
[(4.184 J/g ⋅ K)(340. g)(273.2 K − 293.7 K)] + [(333 J/g)(mice)] = 0
mice = 87.6 g
Think about Your Answer If more than 87.6 g of ice is added, the final temperature will still be 0 °C when thermal equilibrium is reached, but some ice will remain (see
the following Check Your Understanding problem). If less than 87.6 g of ice is added, the
final temperature will be greater than 0 °C. In this case, all the ice will melt, and the liquid
water formed by melting the ice will absorb additional energy to warm up to the final
temperature (an example is given in Study Question 87, page 296).
Check Your Understanding
To make a glass of iced tea, you pour 250 mL of tea, whose temperature is 18.2 °C, into a glass
containing five ice cubes. Each cube has a mass of 15 g. What quantity of ice will melt, and how
much ice will remain floating in the tea? Assume iced tea has a density of 1.0 g/mL and a specific heat capacity of 4.2 J/g ⋅ K, that energy is transferred only as heat within the system, ice is
at 0.0 °C, and no energy is transferred between the system and surroundings.
5.4 The First Law of Thermodynamics
Goals for Section 5.4
• Recognize how energy transferred as heat and work contributes to changes in the
internal energy of a system.
• Calculate the work done by a system when a gas expands against a constant pressure.
• Calculate changes in enthalpy and internal energy.
• Recognize state functions whose values are determined only by the state of the
system and not by the pathway by which the state was achieved.
Sublimation The change
of state in which a solid
transforms directly into a gas
is called sublimation. Different
materials undergo sublimation
at different combinations of
temperatures and pressures. At
standard atmospheric pressure,
dry ice undergoes sublimation
at −78.5 °C.
266
Recall that thermodynamics is the science of heat and work. Up to this point in the
chapter, heat has been discussed, but not work. Work (w) done by a system or on a
system also affects the energy in the system. If a system does work on its surroundings, energy must be expended by the system, and the system’s energy will decrease;
w is therefore negative, w < 0. Conversely, if work is done by the surroundings on a
system, the energy of the system will increase; w > 0.
A system doing work on its surroundings is illustrated in Figure 5.8. A small
quantity of dry ice, solid CO2, is sealed inside a plastic bag, and a weight (a book) is
placed on top of the bag. When energy is transferred as heat from the surroundings
to the dry ice, the dry ice changes directly from solid to gas at −78.5 °C in a process
called sublimation:
CO2(s, −78.5 °C) n CO2(g, −78.5 °C)
As sublimation proceeds, gaseous CO2 expands within the plastic bag, lifting the
book against the force of gravity. The system (the CO2 inside the bag) is expending
energy to do this work.
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Photos: © Charles D. Winters/Cengage
(a) Pieces of dry ice [CO2(s),−78.5 °C] are placed in a plastic bag. The dry
ice will sublime (change directly from a solid to a gas) upon the
input of energy.
(b) Energy is absorbed by CO2(s) when it sublimes, and the system
(the contents of the bag) does work on its surroundings by lifting the book
against the force of gravity.
Figure 5.8 Energy changes in a physical process (a phase change as solid CO2 changes to CO2 gas).
Even if the book were not on top of the plastic bag, work would have been
done by the expanding gas because the gas must push back the atmosphere when it
expands. Instead of raising a book, the expanding gas moves a part of the atmosphere.
Now think about this example in terms of thermodynamics, starting with the
identification of the system and the surroundings. The system is the CO2, initially a
solid and later a gas. The surroundings consist of objects that exchange energy with
the system. This includes the plastic bag, the book, the table-top, and the surrounding air. Sublimation requires energy, which is transferred as heat to the system (the
CO2) from the surroundings. At the same time, the system does work on the surroundings by lifting the book. An energy balance for the system will include both
quantities, energy transferred as heat and energy transferred as work.
This example can be generalized. For any system, you can identify energy transfers both as heat and as work between the system and surroundings. The change in
energy for a system is given explicitly by Equation 5.5,
Change in energy
content
Energy transferred as
work to or from the
system
∆U = q + w
(5.5)
Energy transferred as heat
to or from the system
which is a mathematical statement of the first law of thermodynamics: The energy
change for a system (∆U) is the sum of the energy transferred as heat (q) between
the system and its surroundings and the energy transferred as work (w) between the
system and its surroundings.
The equation defining the first law of thermodynamics is just a restatement
of the general principle of conservation of energy, which states that the energy of
the universe is constant. Because energy is conserved, all changes in energy of the
system must be accounted for. All energy transfers between a system and its surroundings occur by the processes of heat and work. Equation 5.5 thus states that
the change in the energy of the system is exactly equal to the sum of the energy
transfers (heat and/or work) from or to the surroundings.
The quantity U in Equation 5.5 has a formal name—internal energy. The internal
energy in a chemical system is the sum of the potential and kinetic energies inside the
system, that is, the energies of the atoms, molecules, or ions in the system. The potential
5.4 The First Law of Thermodynamics
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267
energy here is the energy associated with the attractive and repulsive forces between all
the nuclei and electrons in the system. It includes the energy associated with bonds in
molecules, forces between ions, and forces between molecules. The kinetic energy is the
energy of motion of the atoms, ions, and molecules in the system. Actual values of internal energy are rarely determined or needed. Instead, changes in internal energy (∆U) are
determined. In fact, Equation 5.5 provides a means for determining ∆U: Measure the
energy transferred as heat and work to or from the system.
The sign conventions for Equation 5.5 are important and are outlined in the
following table.
Sign Conventions for q and w of the System
Energy transferred as . . .
Sign Convention
Effect on Usystem
Heat to the system (endothermic)
q > 0 (+)
U increases
Heat from the system (exothermic)
q < 0 (−)
U decreases
Work done on the system
w > 0 (+)
U increases
Work done by the system
w < 0 (−)
U decreases
The work in the example involving the sublimation of CO2 (Figure 5.8) is of a
specific type, called P –V (pressure–volume) work. It is the work (w) associated with
a change in volume (∆V) that occurs against a resisting external pressure (P). For a
system in which the external pressure is constant, the value of P –V work can be calculated using Equation 5.6,
Work (at
constant pressure)
Change in volume
(5.6)
w = −P × ∆V
A Closer Look
Pressure
P–V Work
The example of a gas sealed in a cylinder with a movable piston can be used
to understand the work done by a system on its surroundings (or vice versa)
when the volume of a system changes.
If the gas in the cylinder is heated, it
expands, pushing the piston upward
until the internal gas pressure equals
the (constant) downward external pressure applied by the piston and the
atmosphere (see figure). Ideally, the piston
moves without friction, so that none of the
work done by an expanding gas is lost to
heating the cylinder walls.
The work required to move the piston
is calculated from a law of physics, w =
F × d, that is, work equals the magnitude
of the force (F ) applied times the distance
(d) over which the force is applied. Pressure
is defined as a force divided by the area
over which the force is applied: P = F /A,
where the force is a function of the
268
piston’s mass, the external air pressure,
and the Earth’s gravity. In this example,
the force is applied to a piston with an
area A. Substituting P × A for F in the
equation for work gives w = (P × A) × d.
The product of A × d is equal to the
change in the volume of the gas in the
cylinder, and, because ∆V = Vfinal − Vinitial,
this change in volume is positive. Finally,
because work done by a system on the
surroundings is defined as negative, this
means that
w = −P∆V.
Expanding the gas and moving the piston upward means the system has done
work on the surroundings.
This equation applies specifically to
expansion of a gas against a constant
pressure. If the pressure of a gas changes
as it is compressed or expanded, the
calculation of P–V work is more complicated, though possible.
A
d
V
F
Heat source
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To calculate this work in units of joules, the pressure is measured in pascals (1 Pa =
1 kg/m ⋅ s2) and the volume change is measured in cubic meters (m3).
In a constant-volume process, ΔV = 0. This means the energy transferred as
work will also be zero. Thus, the change in internal energy of the system under
constant-volume conditions is equal only to the energy transferred as heat (qv). The
subscript v indicates conditions of constant volume.
Calculating Work The SI unit
of pressure is the pascal
(1 Pa = 1 kg/m ⋅ s2), which
when multiplied by the volume
change in m3, gives work in
joules (1 J = 1 kg ⋅ m2/s2).
∆U = qv + wv
∆U = qv + 0 when wv = 0 because ∆V = 0
and so ∆U = qv
Exam p le 5.5
Energy and Work
Problem Nitrogen gas (1.50 L) is confined in a cylinder under constant atmospheric
pressure (1.01 × 105 pascals). The gas expands to a volume of 2.18 L when 882 J of energy
is transferred as heat from the surroundings to the gas. What is the change in the internal
energy of the gas?
What Do You Know? Energy as heat (882 J) is transferred at constant pressure
into the system; thus, qp = +882 J. The system does work on the surroundings when the
gas expands from 1.50 L to 2.18 L under a constant pressure of 1.01 × 105 pascals, thereby
transferring some energy back to the surroundings.
Strategy Calculate the work done by the system using wp = −P(∆V). The unit of work
is joules, provided that SI units are used for pressure and volume. The pressure is given
in SI units (pascals, Pa, where 1 Pa = 1 kg/(m ⋅ s2). To calculate work, the volume must
be converted to m3 (1 m3 = 1000 L). The change in internal energy of the gas is the sum
of the enthalpy change of the gas and the work done by the gas on the surroundings
(∆U = qp + wp).
Solution The change in volume of the gas in m3 is
(2.18 L − 1.50 L N2 gas)(1 m3/1000 L) = 6.8 × 10−4 m3
The work done by the system under conditions of constant pressure is
w = −P(∆V) = −(1.01 × 105 kg/m ⋅ s2)(6.8 × 10−4 m3)
= −68.7 kg ⋅ m2/s2 = −68.7 J
Finally, calculate the change in internal energy.
∆U = q + w = 882 J + (−68.7 J) = 813 J
Think about Your Answer In this example, the value of q is positive because
energy is transferred as heat to the system from the surroundings; this is an endothermic
process. The value of w is negative because the system does work on the surroundings as
the gas expands. The internal energy of the gas increases upon heating. However, the gas
does work on the surroundings as it expands against pressure, giving some of its energy
to the surroundings.
Check Your Understanding
Nitrogen gas (2.75 L) is confined in a cylinder under constant atmospheric pressure
(1.01 × 105 pascals). The volume of gas decreases to 2.10 L when 485 J of energy is transferred as heat to the surroundings. What is the change in internal energy of the gas?
5.4 The First Law of Thermodynamics
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269
Enthalpy
Most experiments in a chemical laboratory are carried out in beakers or flasks open to
the atmosphere, where the external pressure is constant. Similarly, chemical processes
that occur in living systems are open to the atmosphere. Because so many processes in
chemistry and biology are carried out under conditions of constant pressure, it is useful
to have a specific measure of the energy transferred as heat under this circumstance.
Under conditions of constant pressure,
∆U = qp + wp
where the subscript p indicates conditions of constant pressure. If the only type of
work that occurs is P –V work, then
∆U = qp − P∆V
Rearranging this gives
qp = ∆U + P∆V
The thermodynamic function called enthalpy, H, is defined as
H = U + PV
Energy Transferred as Heat
Processes at constant V:
∆U = qv
Processes at constant P:
∆H = qp
The change in enthalpy for a system at constant pressure is calculated from the
­following equation:
∆H = ∆U + P∆V
Thus,
∆H = qp
Enthalpy, Internal Energy, and Non-Expansion Work
The expression ∆H = qp was derived
under the assumption that only
­pressure–volume work occurs during
a chemical reaction. However, this is
not always the case. Other types of
work can occur in chemical reactions.
One example is the electrical work
done when a battery is discharged or
recharged. In a battery, an oxidation–
reduction reaction converts ­chemical
potential energy into electrical energy,
which in turn powers a device, such as a
cell phone. Chances are that the electrical
energy will not change the volume of either
the battery or the cell phone—at least, so
you hope—but work is still being done.
A thorough derivation of the enthalpy
change (∆H) takes into account other types of
work (wother), in addition to p
­ ressure–volume
work. In this derivation, ∆H is still defined as
∆U = qp + wp = qp − P∆V + wother
where the work at constant pressure (wp)
is the sum of pressure–volume work and
other work. Substituting this equation for
∆U into the equation for ∆H gives
∆H = qp − P∆V + wother + P∆V
which simplifies to
∆H = qp + wother
∆H = ∆U + P∆V
for a reaction at constant pressure. However, the internal energy change takes all
types of work into account.
270
This equation better describes the
change in enthalpy that occurs when a
battery discharges because batteries are
designed to produce electrical work. The
chemical components of a battery undergo
an enthalpy change as electrical current
is used. In an ideal system, all the energy
would be converted to electrical work. In a
real battery, some of the enthalpy change
is converted to heat. You may have noticed
that a battery can get warm (or even hot!)
when charging or discharging.
2p2play/Shutterstock.com
A Closer Look
For a system where the only type of work possible is P –V work (see ”A Closer Look:
Enthalpy, Internal Energy and Non-Expansion Work”), the change in enthalpy,
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∆H, is equal to the energy transferred as heat at constant pressure, qp. The directionality of energy transfer as heat (under conditions of constant pressure) is indicated
by the sign of ∆H.
•
Negative values of ∆H: energy is transferred as heat from the system to the surroundings (exothermic process).
•
Positive values of ∆H: energy is transferred as heat from the surroundings to the
system (endothermic process).
Enthalpy and Internal Energy
Differences The difference
between ∆H and ∆U is quite
small unless a large volume
change occurs. For example,
the difference between ∆H and
∆U for the conversion of ice to
liquid water, where there is only
a small change in volume, is
0.165 J/mol at 1 atm pressure.
For the conversion of liquid
water to water vapor at 373 K
(and 1 atm pressure) where there
is a large change in volume, the
difference is 3100 J/mol.
Under conditions of constant pressure and where the only type of work possible
is P –V work, ∆U (= qp − P∆V) and ∆H (= qp) differ by P∆V (the energy transferred
to or from the system as work). In many processes—such as the melting of ice—the
volume change, ∆V, is small, and hence the amount of energy transferred as work is
small. Under these circumstances, ∆U and ∆H have almost the same value. The amount
of energy transferred as work is significant, however, when the volume change is large,
such as when gases are formed or consumed. Thus, ∆U and ∆H have significantly different values for processes such as the evaporation or condensation of water, the sublimation of CO2, and chemical reactions in which the number of moles of gas changes.
State Functions
Internal energy and enthalpy share a significant characteristic—namely, changes in
these quantities depend only on the initial and final states. They do not depend
on the path taken going from the initial state to the final state. No matter how a
reaction proceeds from reactants to products, the values of ∆H and ∆U are always
the same. A quantity that has this property is called a state function.
Many commonly measured quantities, such as the pressure of a gas, the volume of a gas or liquid, and the temperature of a substance are state functions. For
example, if the final temperature of a substance is 75 °C and its initial temperature
was 25 °C, the change in temperature, ∆T, is calculated as
State Function A quantity that
depends only on the state of
a system, not on the path the
system takes to reach that state.
It does not matter if the substance was heated directly from 25 °C to 75 °C or if the
substance was heated from 25 °C to 95 °C and then cooled to 75 °C; the overall
change in temperature is still the same, 50 °C.
Not all quantities are state functions; some depend on the pathway taken to get
from the initial condition to the final condition. For instance, distance traveled is
not a state function (Figure 5.9). The travel distance from New York City to Denver
depends on the route taken. Nor is the elapsed time of travel between these two
locations a state function. In contrast, a change in altitude is a state function; in
going from New York City (at sea level) to Denver (1600 m above sea level), there is
an altitude change of 1600 m, regardless of the route followed.
Significantly, in the expansion of a gas, neither the energy transferred as heat nor
the energy transferred as work, individually, is a state function. However, their sum,
the change in internal energy, ∆U, is. The value of ∆U is fixed by Uinitial and Ufinal.
A transition between the initial and final states can be accomplished by different
routes having different values of q and w, but the sum of q and w for each path must
always give the same ∆U.
David Kotz
Tfinal − Tinitial = 75 °C − 25 °C = 50 °C.
Figure 5.9 State functions. ​
5.5 Enthalpy Changes for Chemical Reactions
Goal for Section 5.5
• Understand and use the enthalpy change for the conversion of reactants to products
in their standard states, ∆rH°.
There are many ways to climb
a mountain, but the change in
altitude from the base of the
mountain to its summit is the
same. The change in altitude is
a state function. The distance
traveled to reach the summit
is not.
5.5 Enthalpy Changes for Chemical Reactions
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271
Enthalpy changes accompany chemical reactions. For example, the standard
reaction enthalpy, ∆rH°, for the decomposition of water vapor to hydrogen and
oxygen at 25 °C is +241.8 kJ/mol-rxn.
Moles of Reaction, mol-rxn This
concept was also described
in one of the methods shown
for solving limiting reactant
problems on page 200.
Notation for Thermodynamic
Parameters NIST (National
Institute for Standards and
Technology) and IUPAC
(International Union of Pure
and Applied Chemistry) specify
that descriptors of functions
such as ∆H should be written
as a subscript, between the
∆ and the thermodynamic
function. Common subscripts
include “r” for reaction, “f” for
formation, “c” for combustion,
“fus” for fusion, and "vap" for
vaporization.
Fractional Stoichiometric
Coefficients ​When writing
balanced equations to define
thermodynamic quantities,
chemists often use fractional
stoichiometric coefficients. For
example, to define ∆rH for the
decomposition or formation of
1 mol of H2O, the coefficient for
O2 must be ½.
H2O(g) n H2(g) + 1⁄2 O2(g) ∆rH° = +241.8 kJ/mol-rxn
The positive sign of ∆rH° indicates that the decomposition is an endothermic process.
There are two important things to know about ∆rH°.
1. The designation of ∆rH° as a standard enthalpy change (where the superscript °
indicates standard conditions) means that the pure, unmixed reactants in their
standard states have formed pure, unmixed products in their standard states.
The standard state of an element or a compound is defined as the most stable
form of the substance in the physical state that exists at a pressure of 1 bar
(1 × 105 Pa, a pressure slightly lower than standard atmospheric pressure at sea
level) and at a specified temperature. [Most sources report standard reaction
enthalpies at 25 °C (298 K).]
2. The “per mol-rxn” designation in the units for ∆ rH° means that this is the
enthalpy change for a mole of reaction (where rxn is an abbreviation for
­reaction). One mole of reaction takes place when a chemical reaction occurs
in exactly the amounts specified by the coefficients of the balanced ­chemical
­equation. For example, for the reaction H2O(g) n H2(g) + ½ O2(g), a mole of
reaction has occurred when 1 mol of water vapor has been converted ­completely
to 1 mol of H2(g) and ½ mol of O2(g).
Now consider the opposite reaction, the combination of hydrogen and oxygen
to form 1 mol of water. The magnitude of the enthalpy change for this reaction is
the same as that for the decomposition reaction, but the sign of ∆rH° is reversed.
The exothermic formation of 1 mol of water vapor from 1 mol of H2 and ½ mol of
O2 transfers 241.8 kJ to the surroundings (Figure 5.10).
H2(g) + 1⁄2 O2(g) n H2O(g) ∆rH° = −241.8 kJ/mol-rxn
The value of ∆rH° depends on the chemical equation used. The equation for
the formation of water can be written without a fractional coefficient for O2.
2 H2(g) + O2(g) n 2 H2O(g) ∆rH° = −483.6 kJ/mol-rxn
Figure 5.10 The exothermic
combustion of hydrogen. The
reaction involves the transfer of
energy between the system and
surroundings in the form of heat,
work, and light.
When the balloon breaks,
the candle flame ignites
the hydrogen.
Photos: © Charles D. Winters/Cengage
A lighted candle is
brought up to a balloon
filled with hydrogen gas.
1/ O (g)
2 2
272
+
H2(g)
∆r H° = −241.8 kJ/mol-rxn
H2O(g)
Chapter 5 / Principles of Chemical Reactivity: Energy and Chemical Reactions
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The value of ∆rH° for 1 mol of this reaction, the formation of 2 mol of water, is twice the
value for the formation of 1 mol of water.
It is important to identify the states of reactants and products in a reaction
because the magnitude of ∆rH° depends on whether they are solids, liquids, or
gases. For the formation of 1 mol of liquid water from the elements, the enthalpy
change is −285.8 kJ.
H2(g) + 1⁄2 O2(g) n H2O() ∆rH° = −285.8 kJ/mol-rxn
Notice that this value is not the same as ∆rH° for the formation of 1 mol of water
vapor from hydrogen and oxygen. The difference between the two values is equal to the
enthalpy change for the condensation of 1 mol of water vapor to 1 mol of liquid water
(−44.0 kJ).
These examples illustrate several general features of enthalpy changes for chemical
reactions.
•
Enthalpy changes are specific to the reaction being carried out. The identities
of reactants and products, and their states (s, , g), are important, as are the
amounts of reactants and products.
•
The enthalpy change depends on the number of moles of reaction, that is, the
number of times the reaction as written is carried out.
•
∆rH° has a negative value for an exothermic reaction; it has a positive value for
an endothermic reaction.
•
Values of ∆rH° are numerically the same, but opposite in sign, for chemical
reactions that are the reverse of each other.
Changes in Chemical Energy in a
Chemical Reaction
• If a reaction is exothermic, the
potential energy of the reactants is greater than that of the
products (PEreact > PEprod).
• If a reaction is endothermic,
the potential energy of the
reactants is less than that of the
products (PEreact < PEprod).
Standard reaction enthalpies can be used to calculate the energy transferred
as heat under conditions of constant pressure for any given mass of a reactant or
­product. Suppose you want to know the energy transferred to the surroundings as
heat if 454 g of butane, C4H10, is burned (at constant pressure), given the equation
for the exothermic combustion and the enthalpy change for the reaction.
2 C4H10(g) + 13 O2(g) n 8 CO2 (g) + 10 H2O() ∆rH° = −5755 kJ/mol-rxn
Two steps are needed. First, find the amount of propane present in the sample:
 1 mol C 4H10 
454 g C 4H10 
 7.811 mol C 4H10
 58.12 g C 4H10 
Second, use ∆rH° to determine ∆H° for this amount of propane:
 1 mol-rxn   5755 kJ 
H° 7.811 mol C 4H10 

 22,500 kJ
 2 mol C 4H10   1 mol-rxn 
Exam p le 5.6
Calculating the Enthalpy Change for a Reaction
Problem Sucrose (table sugar, C12H22O11) can be oxidized to CO2 and H2O, and the
enthalpy change for the reaction can be measured.
C12H22O11(s) + 12 O2(g) n 12 CO2(g) + 11 H2O(𝓵) ∆rH° = −5645 kJ/mol-rxn
What is the enthalpy change when 5.00 g of sugar is burned under conditions of constant
pressure?
5.5 Enthalpy Changes for Chemical Reactions
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273
What Do You Know? The balanced equation for the combustion and the value of
∆rH° are given. Also, the mass of sugar is given.
Strategy First determine the amount (mol) of sucrose in 5.00 g, and then use this with
the value given for the enthalpy change for the oxidation of 1 mol of sucrose.
Solution
5.00 g sucrose 1 mol sucrose
1.461 102 mol sucrose
342.3 g sucrose
 1 mol-rxn   5645 kJ 
H° 1.461 102 mol sucrose 
 1 mol sucrose   1 mol-rxn 
∆H° = −82.5 kJ
Just a Spoonful of Sugar
A (level) teaspoonful of sugar
(about 3.5 g) supplies about
15 Calories (dietary Calories;
the conversion is 4.184 kJ =
1 Cal). A single spoonful of
sugar doesn’t have a large
caloric content. But will you use
just one level teaspoonful?
Think about Your Answer The calculated value is negative, as expected for a
combustion reaction. The magnitude of ∆H° agrees with the fact that the mass of sucrose
used, 5.00 g, is significantly less than the mass of one mole of sucrose (342.3 g).
Check Your Understanding
The combustion of ethane, C2H6, has an enthalpy change of −2857.3 kJ for the reaction as
written below. Calculate ∆H° for the combustion of 15.0 g of C2H6.
2 C2H6(g) + 7 O2(g) n 4 CO2(g) + 6 H2O(g) ∆rH° = −2857.3 kJ/mol-rxn
5.6 Calorimetry
Goal for Section 5.6
• Describe how to measure and calculate the quantity of energy transferred as heat in a
reaction by calorimetry.
The energy evolved or absorbed as heat in a chemical or physical process can be measured by calorimetry. The apparatus used in this kind of experiment is a calorimeter.
Thermometer
Cardboard or
Styrofoam lid
Nested
Styrofoam cups
Reaction
occurs in
solution.
Figure 5.11 A coffee-cup
calorimeter.​ A chemical reaction
produces a change in temperature
of the solution in the calorimeter.
The Styrofoam container is fairly
effective in preventing the transfer
of energy as heat between the
solution and its surroundings.
Because the cup is open to the
atmosphere, this is a constantpressure measurement.
274
Constant-Pressure Calorimetry, Measuring ∆H
A constant-pressure calorimeter can be used to measure the quantity of energy
transferred as heat under constant-pressure conditions, that is, the enthalpy change
for a chemical reaction.
The constant-pressure calorimeter used in general chemistry laboratories is often
a coffee-cup calorimeter. This inexpensive device consists of two nested ­Styrofoam
coffee cups with a loose-fitting lid and a thermometer (Figure 5.11) or thermocouple. Styrofoam, a fairly good insulator, minimizes energy transfer as heat between
the system and the surroundings. The reaction occurs in solution in the cup. If the
reaction is exothermic, it releases energy as heat to the solution, and the temperature of the solution rises. If the reaction is endothermic, energy is absorbed as heat
from the solution, and the temperature of the solution decreases. The change in
temperature of the solution is measured. Knowing the mass and specific heat capacity of the solution and the temperature change, the enthalpy change for the reaction
can be calculated.
In this type of calorimetry experiment, it is convenient to define the chemicals and the solution as the system. The surroundings are the cup and everything
beyond the cup. Assuming that there is no energy transfer to the cup or beyond,
the energy is transferred only as heat within the system. Two energy changes occur
within the system. One is the energy released or gained as heat by the chemical
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reaction (qr) itself. This corresponds to either releasing potential energy stored in
the reactants or absorbing energy and converting it to potential energy stored in the
products, respectively. This energy is labeled as qr. The other energy change is the
corresponding energy gained or lost as heat by the solution (qsolution). Based on
the law of conservation of energy,
qr + qsolution = 0
The value of qsolution can be calculated from the specific heat capacity, mass, and
change in temperature of the solution. The quantity of energy evolved or absorbed
as heat for the reaction (qr) is the unknown in the equation.
The accuracy of a calorimetry experiment depends on the accuracy of the measured quantities (temperature, mass, specific heat capacity). In addition, it depends
on how closely the assumption is followed that there is no energy transfer beyond
the solution. A coffee-cup calorimeter is an inexpensive apparatus, but the results
obtained with it are not highly accurate, largely because this assumption is poorly
met. However, the calorimeters used in research laboratories more effectively limit
the energy transfer between system and surroundings. In addition, it is also possible
to correct for the minimal energy transfer that does occur between the system and
the surroundings.
Exam p le 5.7
Using a Coffee-Cup Calorimeter
Problem You place 0.0500 g of magnesium chips in a coffee-cup calorimeter and then
add 100.0 mL of 1.00 M HCl. The reaction that occurs is
Mg(s) + 2 HCl(aq) n H2(g) + MgCl2(aq)
The temperature of the solution increases from 22.21 °C (295.36 K) to 24.46 °C (297.61 K).
What is the enthalpy change for the reaction per mole of Mg? Assume that the specific
heat capacity of the solution is 4.20 J/g · K, the density of the HCl solution is 1.00 g/mL,
and that no energy as heat is lost to the surroundings.
Strategy Map
Problem
Calculate DrH per mol for
reaction of Mg with HCl.
Data/Information
• Mass of Mg and HCl solution
• ∆T
• Specific heat capacity
What Do You Know? You know the mass of magnesium as well as the volume and
molarity of the HCl. You know that energy is evolved as heat in this reaction because the
temperature of the solution rises. The sum of the energy evolved as heat in the reaction,
qr, and the energy absorbed as heat by the solution, qsolution, will be zero, that is, qr + qsolution
= 0. The value of qsolution can be calculated from data given; qr is the unknown.
Strategy Solving the problem has four steps.
Step 1. Calculate qsolution from the values of the mass, specific heat capacity, and ∆T
using Equation 5.3.
Step 2. Calculate qr, assuming no energy transfer as heat occurs beyond the solution,
that is, qr + qsolution = 0.
Step 3. Calculate the amount of magnesium.
Step 4. Use the value of qr and the amount of Mg to calculate the enthalpy change per
mole of Mg.
Solution
Step 1
Calculate qsolution.
Use Equation 5.3 to calculate qsolution. The mass of the solution in this equation is the mass
of 100.0 mL of the HCl solution (because its density is 1.00 g/mL, this mass is 100. g) plus
the mass of Mg.
5.6 Calorimetry
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275
qsolution = (100.0 g HCl solution + 0.0500 g Mg)(4.20 J/g · K)(297.61 K − 295.36 K)
= 945.5 J
Step 2
Calculate qr.
qr + qsolution = 0
qr + 945.5 J = 0
qr = −945.5 J
Step 3
Calculate the amount of Mg.
0.0500 g Mg Step 4
1 mol Mg
0.002057 mol Mg
24.31 g Mg
Calculate the enthalpy change per mole of Mg.
The value of qr (Step 2) resulted from the reaction of 0.002057 mol Mg. The enthalpy
change per mole of Mg is determined by dividing qr by this amount of Mg.
∆rH = (−945.5 J/0.002057 mol Mg)
= −4.60 × 105 J/mol Mg (= −4.60 × 102 kJ/mol Mg)
Think about Your Answer The calculation gives the correct sign of qr and ∆rH.
The negative sign indicates that this is an exothermic reaction. The balanced equation
states that one mole of magnesium is involved in one mole of reaction. The calculated
enthalpy change per mole of reaction, ∆rH, is therefore −460. kJ/mol-rxn.
Check Your Understanding
Assume 200. mL of 0.400 M HCl is mixed with 200. mL of 0.400 M NaOH in a coffee-cup calorimeter. The temperature of the
solutions before mixing was 25.10 °C; after mixing and allowing the reaction to occur, the temperature is 27.78 °C. What is the
enthalpy change when one mole of acid is neutralized? (Assume that the densities of all solutions are 1.00 g/mL and their specific
heat capacities are 4.20 J/g · K.)
Constant-Volume Calorimetry, Measuring ∆U
Calorimetry, DU, and DH The two
types of calorimetry (constant
volume and constant pressure)
highlight the differences between
enthalpy and internal energy.
The energy transferred as heat
at constant pressure, qp, is, by
definition, ∆H, whereas the
energy transferred as heat at
constant volume, qv , is ∆U.
276
Constant-volume calorimetry is often used to evaluate the energy released by the
combustion of fuels and the caloric value of foods. A weighed sample of a combustible solid or liquid is placed inside a bomb, often a cylinder about the size of
a large fruit juice can with thick steel walls and ends (Figure 5.12). The bomb is
placed in a water-filled container with well-insulated walls. After filling the bomb
with pure oxygen, the sample is ignited, usually by an electric spark. The heat
generated by the combustion reaction warms the system, which is the bomb, its
contents, and the water. Assessment of energy transfers as heat within the system
shows that
qr + qbomb + qwater = 0
where qr is the energy released as heat by the reaction, qbomb is the energy involved in
heating the calorimeter bomb, and qwater is the energy involved in heating the water
in the calorimeter. Because the volume does not change in a constant-volume calorimeter, energy transfer as work does not occur. Therefore, the energy transferred as
heat at constant volume (qv) is equal to the change in internal energy, ∆U.
Chapter 5 / Principles of Chemical Reactivity: Energy and Chemical Reactions
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Thermometer
Water
Stirrer
Ignition
wires
Charles D. Winters/Science Source
Insulated
outside
container
(a)
Steel
bomb
Sample
dish
Steel
container
The sample burns in
pure oxygen, warming
the bomb.
The heat generated warms
the water and ΔT is measured
by the thermometer.
(b)
Figure 5.12 Constant-volume calorimeter. (a) A combustible sample is placed inside a sealed metal container or “bomb.” (b) The
bomb is filled with oxygen and submerged inside an insulated, water-filled container. The sample is then burned. Energy transferred as
heat from the reaction warms the bomb and the water surrounding it. By measuring the increase in temperature, the energy transferred
as heat in the reaction can be determined.
Exam p le 5.8
Constant-Volume Calorimetry
Problem Octane, C8H18, a primary constituent of gasoline, burns in air:
C8H18(𝓵) + 25⁄2 O2(g) n 8 CO2(g) + 9 H2O(𝓵)
Strategy Map
Problem
Calculate DU per mol for
combustion of 1.15 g octane.
Data/Information
• Mass of octane
• Mass of water in calorimeter
• cwater
• Cbomb
• ∆T
A 1.15-g sample of octane is burned in a constant-volume calorimeter similar to that
shown in Figure 5.12. The calorimeter is in an insulated container with 1.20 kg of water.
The temperature of the water and the bomb rises from 25.00 °C (298.15 K) to 34.43 °C
(307.58 K). The heat required to raise the bomb’s temperature (its heat capacity), Cbomb ,
is 837 J/K. (a) What is the heat of combustion per gram of octane? (b) What is the heat of
combustion per mole of octane?
What Do You Know? There are energy changes for the three components of this
system: the energy evolved in the reaction, qr; the energy absorbed by the water, qwater;
and the energy absorbed by the calorimeter, qbomb. You know the following: the molecular
formula of octane, masses of the sample and the calorimeter water, Tinitial, Tfinal, Cbomb, and
cwater . You can assume no energy loss to the surroundings.
Strategy
Step 1. Calculate qwater, the energy transferred as heat to the water.
Step 2. Calculate qbomb, the energy transferred as heat to the bomb.
Step 3. Calculate qr, the energy transferred as heat in the reaction. The sum of all
energies transferred as heat in the system is qr + qbomb + qwater = 0.
5.6 Calorimetry
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277
Step 4. Calculate the heat of combustion per gram of octane by dividing qr by the mass
of octane combusted.
Step 5. Calculate the heat of combustion per mole of octane by multiplying the heat of
combustion per gram by the molar mass of octane.
Solution
Step 1
Calculate qwater, the energy transferred as heat to the water.
Use Equation 5.3 with the specific heat capacity, mass, and change in temperature for the
water to calculate qwater.
qwater = cwater × mwater × ∆T = (4.184 J/g · K)(1.20 × 103 g)(307.58 K − 298.15 K)
= +4.735 × 104 J
Step 2
Calculate, qbomb, the energy transferred as heat to the bomb.
Notice that it is the heat capacity of the bomb provided, not the specific heat capacity.
You can obtain qbomb by multiplying the heat capacity of the bomb by the change in temperature (Equation 5.1).
qbomb = (Cbomb)(∆T) = (837 J/K)(307.58 K − 298.15 K) = 7.893 × 103 J
Step 3
Calculate, qr, the energy transferred as heat in the reaction.
qr + qwater + qbomb = 0
qr + 4.735 × 104 J + 7.893 × 103 J = 0
qr = −5.524 × 104 J (or −55.24 kJ)
Step 4
Calculate the heat of combustion per gram of octane.
Divide the value of qr obtained by the mass of octane used.
Heat of combustion per gram = −55.24 kJ/1.15 g octane = −48.03 kJ/g = −48.0 kJ/g
Step 5
Calculate the heat of combustion per mole of octane.
Use the molar mass of octane to convert from the heat of combustion per gram of octane
to the heat of combustion per mole of octane.
Heat of combustion per mole of octane = (−48.03 kJ/g)(114.2 g/mol)
= −5.49 × 103 kJ/mol
Think about Your Answer Because the volume does not change, no energy
transfer in the form of work occurs. The change of internal energy, ∆rU, for the combustion
of C8H18(𝓵) is −5.49 × 103 kJ/mol.
Check Your Understanding
A 1.22-g sample of ordinary table sugar (sucrose, C12H22O11) is burned in a bomb calorimeter. The temperature of 1.50 × 103 g of
water in the calorimeter rises from 25.00 °C to 27.84 °C. The heat capacity of the bomb is 837 J/K, and the specific heat capacity of
the water is 4.184 J/g · K. Calculate (a) the heat evolved per gram of sucrose and (b) the heat evolved per mole of sucrose.
5.7 Enthalpy Calculations
Goals for Section 5.7
• Apply Hess’s law to find the enthalpy change for a reaction.
• Know how to draw and interpret energy level diagrams.
• Use standard molar enthalpies of formation, ∆f H°, to calculate the standard
enthalpy change for a reaction, ∆r H°.
Chemists and chemical engineers often want to know if a reaction is expected to
be exothermic or endothermic and to what extent. You may be able to perform a
278
Chapter 5 / Principles of Chemical Reactivity: Energy and Chemical Reactions
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calorimetry experiment to determine the enthalpy change for the reaction, or this
information may already be available on the Internet or in reference books. Even if
it is not, you can often calculate or estimate the enthalpy change using data about
other reactions for which the enthalpy changes are known. This section outlines
how to use these data.
Hess’s Law
The enthalpy change for a reaction can be measured by calorimetry for many,
but not all, chemical processes. Consider, for example, the oxidation of carbon
(as graphite) to form carbon monoxide.
C(s) + 1/2 O2(g) n CO(g)
The primary product of the reaction is CO2 and not CO, even if a deficiency of
oxygen is used. As soon as CO is formed, it will react with O2 to form CO2. It is not
possible to measure the change in enthalpy for this reaction by calorimetry because
the reaction cannot be carried out in a way that allows CO to be the sole product.
The enthalpy change for the reaction forming CO(g) from C(s) and O2(g) can
be determined indirectly, however, from enthalpy changes for other reactions for
which values of ∆rH° can be measured. The calculation is based on Hess’s law,
which states that if a reaction is the sum of two or more other reactions, ∆rH° for
the overall process is the sum of the ∆rH° values of those reactions.
The enthalpy change for the oxidation of C(s) to CO(g) at 25 °C can be determined indirectly from thermochemical data obtained from two reactions that can
be studied by calorimetry. These reactions are the oxidation of CO(g) and the oxidation of C(s), both of which form CO2(g) as the sole product.
Equation 1:
CO(g) + 1⁄2 O2(g) n CO2(g)
∆rH°1 = −283.0 kJ/mol-rxn
Equation 2:
C(s) + O2(g) n CO2(g)
∆rH°2 = −393.5 kJ/mol-rxn
These equations can be manipulated to give equations that add together to yield
the desired net equation. To have CO(g) appear as a product in the net equation,
Equation 1 is reversed. The sign of the standard enthalpy change is also reversed
(Section 5.5). Equation 2 contains C(s) on the correct side of the equation and in
the correct stoichiometric amount; it is left unchanged. Adding these two equations
gives the equation for the oxidation of C(s) to CO(g).
Equation 1′:
CO2(g) n CO(g) + 1⁄2 O2(g)
∆rH°1′ = +283.0 kJ/mol-rxn
Equation 2:
C(s) + O2(g) n CO2(g)
∆rH°2 = −393.5 kJ/mol-rxn
Equation 3:
C(s) + 1⁄2 O2(g) n CO(g)
∆rH°3 = −110.5 kJ/mol-rxn
The enthalpy change for the overall reaction (∆rH°3) is equal to the sum of the
enthalpy changes for reactions 1′ and 2.
∆rH°3 = ∆rH°1′ + ∆rH°2
= +283.0 kJ/mol-rxn + (−393.5 kJ/mol-rxn)
= −110.5 kJ/mol-rxn
Hess’s law also applies to physical processes. The enthalpy change for the reaction of H2(g) and O2(g) to form 1 mol of H2O vapor is different from the enthalpy
change to form 1 mol of liquid H2O. The difference is the negative of the enthalpy of
vaporization of water, ∆rH°2 (= −∆vapH°) as shown in the following example using
values that are accurate at 25 °C.
Equation 1:
H2(g) + 1⁄2 O2(g) n H2O(g)
∆rH°1 = −241.8 kJ/mol-rxn
Equation 2:
H2O(g) n H2O(𝓵)
∆rH°2 = −44.0 kJ/mol-rxn
Equation 3:
H2(g) + 1⁄2 O2(g) n H2O()
∆rH°3 = −285.8 kJ/mol-rxn
5.7 Enthalpy Calculations
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279
Exam pl e 5 .9
Using Hess’s Law
Problem Suppose you want to know the enthalpy change for the formation of
­methane, CH4, from solid carbon (as graphite) and hydrogen gas:
C(s) + 2 H2(g) n CH4(g) ∆rH° = ?
The enthalpy change for this reaction cannot be measured in the laboratory because the
reaction is very slow. However, the enthalpy changes can be measured for the combustion of carbon, hydrogen, and methane.
Equation 1:
C(s) + O2(g) n CO2(g)
∆rH°1 = −393.5 kJ/mol-rxn
Equation 2:
H2(g) + 1⁄2 O2(g) n H2O()
∆rH°2 = −285.8 kJ/mol-rxn
Equation 3:
CH4(g) + 2 O2(g) n CO2(g) + 2 H2O()
∆rH°3 = −890.3 kJ/mol-rxn
Use this information to calculate ∆rH° for the formation of methane from its elements.
What Do You Know? This is a Hess’s law problem. You need to adjust the three equations so they can be added together to give the desired equation, C(s) + 2 H2(g) n CH4(g).
When an adjustment in an equation is made, you also need to adjust the enthalpy change.
Strategy Map
Strategy
Problem
Hess’s Law: Calculate ∆rH° for
targeted reaction from ∆rH°
values for other reactions.
Step 1. Arrange the equations with known ∆rH° values so the reactants and products of
the desired equation appear on the correct sides. You may need to reverse some of the
equations to do this. Remember that if you reverse a chemical equation, you must also
reverse the sign of its ∆rH°.
Step 2. Adjust the amounts of substances in the equations with known ∆rH° values so
they match the amounts in the desired equation. You may need to multiply some of the
equations by a factor to do this. Remember that if you multiply a chemical equation by a
factor, you must also multiply its ∆rH° by that factor.
Data/Information
Three reactions with known
∆rH° values
Step 3. Ensure that other substances in the equations with known ∆rH° values cancel
when the equations are added.
Step 4. Add the equations and their ∆rH° values to obtain the result.
Solution
Step 1
Arrange the equations with known DrH° values so that the reactants and products of the desired equation appear on the correct sides.
The desired equation has C(s) and H2(g) on the left and CH4(g) on the right. In Equations
1 and 2, C(s) and H2(g) already appear as reactants, so these two equations are left in
the direction given. To have CH4(g) appear as a product in the overall reaction, reverse
­Equation 3, which changes the sign of its ΔrH.
Equation 3′:
CO2(g) + 2 H2O() n CH4(g) + 2 O2(g)
∆rH°3′ = −∆rH°3 = +890.3 kJ/mol-rxn
Step 2
Adjust the amounts of substances in the equations with known DrH° values so
they match the amounts in the desired equation.
The desired equation has 1 mol C(s), 2 mol H2(g), and 1 mol CH4(g). Equation 1 uses 1
mol C(s), and Equation 3´ produces 1 mol CH4(g), so these two equations are left alone.
­Equation 2, however, only uses 1 mol H2(g), rather than the 2 mol needed. Therefore,
­multiply the stoichiometric coefficients in Equation 2 by 2 and multiply its ΔrH° by 2.
Equation 2′:
2 H2(g) + O2(g) n 2 H2O()
∆rH°2′ = 2 ∆rH°2 = 2(−285.8 kJ/mol-rxn) = −571.6 kJ/mol-rxn
280
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Step 3
Ensure that other substances in the equations with known DrH° values cancel
when the equations are added.
The desired equation does not include any CO2(g), H2O(), or O2(g). Equation 1 has 1 mol
CO2(g) on the right side and Equation 3´ has 1 mol CO2(g) on the left side, so they cancel.
Likewise, Equation 2´ has 2 mol H2O() on the right side, and Equation 3´ has 2 mol H2O()
on the left, canceling these. Finally, Equation 1 and Equation 2´ each have 1 mol O2(g) on
the left, giving a total of 2 mol O2(g) on the left, which cancel the 2 mol O2(g) on the right
side of Equation 3´.
Step 4
Add the equations and their DrH° values to obtain the result.
You now have three equations that, when added together, give the equation for the formation of methane from carbon and hydrogen. The sum of their ∆rH° values is ∆rH° for the
desired equation.
Equation 1:
C(s) + O2(g) n CO2(g)
∆rH°1 = −393.5 kJ/mol-rxn
Equation 2′:
2 H2(g) + O2(g) n 2 H2O() ∆rH°2′ = 2 ∆rH°2 = −571.6 kJ/mol-rxn
Equation 3′:
CO2(g) + 2 H2O() n CH4(g) + 2 O2(g)
∆rH°3′ = −∆rH°3 = +890.3 kJ/mol-rxn
Net Equation:
C(s) + 2 H2(g) n CH4(g)
∆rH°net = ∆rH°1 + ∆rH°2′ + ∆rH°3′
∆rH°net = (−393.5 kJ/mol-rxn) + (−571.6 kJ/mol-rxn)
+ (+890.3 kJ/mol-rxn)
= −74.8 kJ/mol-rxn
Thus, for the formation of 1 mol of CH4(g) from the elements, ∆rH° = −74.8 kJ/mol-rxn.
Think about Your Answer Notice that the enthalpy change for the formation of
the compound from its elements is exothermic, as it is for the great majority of compounds.
Check Your Understanding
Use Hess’s law to calculate the enthalpy change for the formation of CS2() from C(s) and S(s) [C(s) + 2 S(s) n CS2()] from the
following enthalpy values.
C(s) + O2(g) n CO2(g)
∆rH°1 = −393.5 kJ/mol-rxn
S(s) + O2(g) n SO2(g)
∆rH°2 = −296.8 kJ/mol-rxn
CS2() + 3 O2(g) n CO2(g) + 2 SO2(g)
∆rH°3 = −1103.9 kJ/mol-rxn
Energy Level Diagrams
When using Hess’s law, it is often helpful to represent enthalpy data schematically
in an energy level diagram. In such drawings, the various substances being s­ tudied—
the reactants and products in a chemical reaction, for example—are placed on an
arbitrary energy scale. The relative enthalpy of each is given by its position on the
vertical axis, and numerical differences in enthalpy between them are shown by
vertical arrows. Such diagrams can give you a visual perspective of the magnitude
and direction of enthalpy changes and show how the enthalpies of the substances
are related.
Energy level diagrams that summarize two examples of Hess’s law discussed
earlier are shown in Figure 5.13. In Figure 5.13a, the elements C(s) and O2(g) are
at the highest enthalpy. The reaction of carbon and oxygen to form CO2(g) lowers
the enthalpy by 393.5 kJ. This can occur either in a single step, shown on the left in
Figure 5.13a, or in two steps via initial formation of CO(g), as shown on the right.
Similarly, in Figure 5.13b, the mixture of H2(g) and O2(g) is at the highest enthalpy.
5.7 Enthalpy Calculations
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281
H2(g) + 1 O2(g)
C(s) + O2(g)
2
∆rH°3 =
−∆rH°1 + ∆rH°2
= −110.5 kJ
∆rH°1 =
−241.8 kJ
CO(g) + 1 O2(g)
Energy
2
Energy
Figure 5.13 Energy level
diagrams. (a) Relating enthalpy
changes in the formation of
CO2(g). (b) Relating enthalpy
changes in the formation of
H2O(). Enthalpy changes
associated with changes between
energy levels are given alongside
the vertical arrows.
∆rH°2 =
−393.5 kJ
∆rH°3 =
∆rH°1 + ∆rH°2
= −285.8 kJ
∆rH°1 =
−283.0 kJ
H2O(g)
∆rH°2 =
−44.0 kJ
H2O(ℓ)
CO2(g)
(a) The formation of CO2 can occur in a single step
or in a succession of steps. ∆rH° for the overall
process is −393.5 kJ, no matter which path is
followed.
(b) The formation of H2O(ℓ) can occur in a single
step or in a succession of steps. ∆rH° for the
overall process is −285.8 kJ, no matter which
path is followed.
Both liquid and gaseous water have lower enthalpies, with the difference between
the two being the enthalpy of vaporization.
The enthalpy change for a reaction is a state function; that is, the enthalpy
change from reactants to products does not depend on the path taken. Energy diagrams illustrate this point. Chemists often want to know the enthalpy change for
one step of a reaction. If the overall enthalpy change and the enthalpy changes for
all the steps but one are known, then the unknown change can be calculated.
Standard Enthalpies of Formation
∆f H ° Values Consult the National
Institute for Standards and
Technology website (https://
webbook.nist.gov/chemistry)
for an extensive compilation of
enthalpies of formation.
Units for Enthalpy of Formation ​
The units for values of ∆f H° are
usually given simply as k J/mol
rather than as kJ/mol-rxn.
However, because an enthalpy
of formation is defined as the
change in enthalpy for the
formation of 1 mol of compound,
it is understood that “per mol”
also means “per mol-rxn.”
Calorimetry and the application of Hess’s law have provided a great many ∆rH°
values for chemical reactions. The table in Appendix L, for example, lists standard
molar enthalpies of formation, Df H°. The standard molar enthalpy of formation
is the enthalpy change for the formation of 1 mol of a compound directly from its
component elements in their standard states.
Several examples of standard molar enthalpies of formation may be helpful to
illustrate this definition.
∆f H° for NaCl(s): At 25 °C and a pressure of 1 bar, Na is a solid, and Cl2 is
a gas. The standard enthalpy of formation of NaCl(s) is defined as the enthalpy
change that occurs when 1 mol of NaCl(s) is formed from 1 mol of Na(s) and
1
⁄2 mol of Cl2(g).
Na(s) + 1⁄2 Cl2(g) n NaCl(s) ∆f H° 5 −411.12 kJ/mol
Notice that a fraction is required as the coefficient for the chlorine gas in this equation because the definition of ∆f H° specifies the formation of one mole of NaCl(s).
∆f H° for NaCl(aq): The standard enthalpy of formation for an aqueous solution
of a compound refers to the enthalpy change for the formation of a 1 mol/L solution
of the compound starting with the elements. It is thus the enthalpy of formation of
NaCl(s) plus the enthalpy change that occurs when the substance dissolves in water
(13.85 kJ/mol).
Na(s) + 1⁄2 Cl2(g) n NaCl(aq) ∆f H° 5 −407.27 kJ/mol
∆f H° for C2H5OH(): At 25 °C and 1 bar, the standard states of the elements are
C(s, graphite), H2(g), and O2(g). The standard enthalpy of formation of C2H5OH()
is defined as the enthalpy change that occurs when 1 mol of C2H5OH() is formed
from 2 mol of C(s), 3 mol of H2(g), and ½ mol of O2(g).
282
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2 C(s) + 3 H2(g) + 1⁄2 O2(g) n C2H5OH() ∆f H° 5 −277.0 kJ/mol
Notice that the reaction defining the enthalpy of formation for liquid ethanol
is not a reaction that a chemist can carry out in the laboratory. This illustrates an
important point: The enthalpy of formation of a compound does not necessarily
correspond to a reaction that can be carried out.
Appendix L lists values of ∆f H° for some common substances at 25 °C, and a
review of these values leads to some important observations.
•
The standard enthalpy of formation for an element in its standard state is zero.
•
Most ∆f H° values are negative, indicating that the formation of most compounds from their elements is exothermic. Very few values are positive, and
these represent compounds that are unstable with respect to decomposition to
the elements. (One example is NO(g) with ∆f H° 5 +90.29 kJ/mol.)
•
Values of ∆f H° can often be used to compare the stabilities of related compounds. Consider the values of ∆f H° for the hydrogen halides. Hydrogen
­fluoride is the most stable of these compounds with respect to decomposition
to the elements, whereas HI is the least stable (as indicated by ∆f H° of HF being
the most negative value and that of HI being the most positive).
Df H° Values of Hydrogen Halides
Compound
∆f H° (kJ/mol)
HF(g)
−273.3
HCl(g)
−92.31
HBr(g)
−35.29
HI(g)
+25.36
Enthalpy Change for a Reaction
Using standard molar enthalpies of formation and Equation 5.7, it is possible to
calculate the enthalpy change for a reaction under standard conditions.
∆rH° 5 ∑n∆f H°(products) − ∑n∆f H°(reactants)
(5.7)
In this equation, the symbol ∑ (the Greek capital letter sigma) means “take the
sum.” To find ∆rH°, add up the molar enthalpies of formation of the products, each
multiplied by its stoichiometric coefficient n, and subtract from this the sum of the
molar enthalpies of formation of the reactants, each multiplied by its stoichiometric
coefficient. This equation is a consequence of the definition of ∆f H° and Hess’s law
(see “A Closer Look: Hess’s Law and Equation 5.7,” page 284).
Suppose you want to know how much energy is required to decompose 1 mol
of calcium carbonate (limestone) to calcium oxide (lime) and carbon dioxide under
standard conditions:
Stoichiometric Coefficients In
Equation 5.7 a stoichiometric
coefficient, n, represents
the number of moles of the
substance per mole of reaction.
D 5 Final 2 Initial Equation
5.7 is another example of the
convention that a change (∆) is
always calculated by subtracting
the value for the initial state (the
reactants) from the value for the
final state (the products).
CaCO3(s) n CaO(s) + CO2(g) ∆rH° 5 ?
You would use the following enthalpies of formation (from Appendix L):
Compound
∆f H° (kJ/mol)
CaCO3(s)
−1207.6
CaO(s)
−635.1
CO2(g)
−393.5
and then use Equation 5.7 to find the standard enthalpy change for the reaction, ∆rH°.
  1 mol CaO   635.1 kJ   1 mol CO2   393.5 kJ  
r H°  
 



  1 mol-rxn   mol CaO   1 mol-rxn   1 mol CO2  
  1 mol CaCO3   1207.6 kJ  



  1 mol-rxn   1 mol CaCO3  
179.0 kJ/mol-rxn
The decomposition of limestone to lime and CO2 is endothermic. That is, energy
(179.0 kJ) must be supplied to decompose 1 mol of CaCO3(s) to CaO(s) and CO2(g).
5.7 Enthalpy Calculations
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283
A Closer Look
Hess’s Law and Equation 5.7
The decomposition of calcium ­carbonate can be used to show
that Equation 5.7 is an application of Hess's law.
CaCO3(s) n CaO(s) + CO2(g)
∆rH° = ?
Because enthalpy is a state function, the change in enthalpy
for this reaction is independent of the route from reactants
to products (see page 271). Imagine an alternate route from
reactant to products that involves first converting the reactant
(CaCO3) to elements in their standard states, then recombining
these elements to give the reaction products. Notice that the
enthalpy changes for these processes are the enthalpies of formation
of the reactants and products in the equation for the decomposition
of calcium carbonate:
That is, the change in enthalpy for the reaction is equal to the
enthalpies of formation of products (CO 2 and CaO) minus the
enthalpy of formation of the reactant (CaCO3), which is exactly,
what you do when using Equation 5.7. The relationship among
these enthalpy quantities is illustrated in the energy level
diagram.
Energy level diagram for the decomposition of CaCO3(s)
3
O (g)
2 2
Ca(s) + C(s) +
C(s) + O2(g) n CO2(g) ∆f H°[CO2(g)] = ∆r H°2
Ca(s) + 1⁄2 O2(g) n CaO(s)
∆f H°[CaO(s)] = ∆r H°3
CaCO3(s) n CaO(s) + CO2(g)
∆r H°net
Energy, kJ
CaCO3(s) n Ca(s) + C(s) + 3⁄2 O2(g) −∆f H°[CaCO3(s)] = ∆r H°1
∆rH°1 =
−∆f H°[CaCO3(s)]
= +1207.6 kJ
∆r H°net = ∆r H°1 + ∆rH°2 + ∆rH°3
∆rH°2 + ∆rH°3 =
(−635.1 kJ) + (−393.5 kJ)
CaO(s) + CO2(g)
∆r H° = ∆f H°[CaO(s)] + ∆f H°[CO2(g)] − ∆f H°[CaCO3(s)]
∆rH°net = ∆rH°1 + ∆rH°2 + ∆r H°3
= + 179.0 kJ
CaCO3(s)
Exampl e 5 .1 0
Using Enthalpies of Formation
Problem Nitroglycerin, C3H5(NO3)3 , is a powerful explosive that forms four different
gases when detonated:
2 C3H5(NO3)3(𝓵) n 3 N2(g) + 1⁄2 O2(g) + 6 CO2(g) + 5 H2O(g)
Strategy Map
Calculate the enthalpy change that occurs when 10.0 g of nitroglycerin is detonated. The
standard enthalpy of formation of nitroglycerin, ∆f H°, is −364 kJ/mol. Use Appendix L to
find other ∆f H° values that are needed.
Problem
Calculate DrH° for reaction of a
given mass of compound.
What Do You Know? From Appendix L, ∆f H°[CO2(g)] = −393.5 kJ/mol, ∆f H°[H2O(g)] =
−241.8 kJ/mol, and ∆f H° = 0 kJ/mol for N2(g) and O2(g). The mass and ∆f H° for nitroglycerin
are also given.
Strategy
Data/Information
• Mass and ∆fH° of compound
• ∆fH° values for products
• Balanced equation
Step 1. Substitute the enthalpy of formation values for products and reactants into
Equation 5.7 to determine the enthalpy change for 1 mol of reaction. This represents the
enthalpy change for the detonation of 2 mol of nitroglycerin.
Step 2. Determine the amount (mol) represented by 10.0 g of nitroglycerin.
Step 3. Calculate the enthalpy change using the amount nitroglycerin and the enthalpy
change for 1 mol of reaction.
284
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Solution
Step 1
Calculate DrH°.
Use Equation 5.7, the ∆f H° values, and the coefficients of the balanced equation to
calculate ∆rH°, the enthalpy change when a mole of reaction occurs.
 6 mol CO2 
 5 mol H2O 
r H° 
H°[CO2(g)] 
H°[H2O(g)]
 1 mol-rxn  f
 1 mol-rxn  f
 2 mol C3H5(NO3)3 

 f H°[C3H5(NO3)3()]

1 mol-rxn
 6 mol CO2   393.5 KJ 
 5 mol H2O   241.8 KJ 
r H° 

  1 mol-rxn   1 mol H O 
 1 mol-rxn   1 mol CO2 


2 

 2 mol C3H5(NO3 )3  
364 KJ


 2842.0 KJ/mol-rxn

1 mol-rxn

  1 mol C3H5(NO3 )3 
Step 2
Determine the amount of compound.
The problem asks for the enthalpy change using 10.0 g nitroglycerin. Determine the
amount of nitroglycerin in 10.0 g.
 1 mol nitroglycerin 
10.0 g nitroglycerin 
 0.04403 mol nitroglycerin
 227.1 g nitroglycerin 
Step 3
Calculate the enthalpy change when this amount of compound reacts.
Convert the amount of nitroglycerin to mol-rxn using the coefficient of nitroglycerin in the
balanced equation and then multiply by ∆rH°.

  2842.0 kJ 
1 mol-rxn
H° 0.04403 mol nitroglycerin 


 2 mol nitroglycerin   1 mol-rxn 
= −62.6 kJ
Think about Your Answer The large negative value of ∆rH° is in accord with the
fact that this reaction is highly exothermic.
Check Your Understanding
Calculate the standard enthalpy of combustion for benzene, C6H6.
C6H6(𝓵) + 15/2 O2(g) n 6 CO2(g) + 3 H2O(𝓵) ∆rH° = ?
The enthalpy of formation of benzene is known [∆f H°[C6H6(𝓵)] = +49.0 kJ/mol], and other values needed can be found in Appendix L.
5.8 Product- or Reactant-Favored
Reactions and Thermodynamics
An extensive study of thermodynamics will ultimately provide answers to four
questions.
1. How do you measure and calculate the energy changes associated with physical
changes and chemical reactions?
5.8 Product- or Reactant-Favored Reactions and Thermodynamics
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285
© Charles D. Winters/Cengage
2. What is the relationship between energy changes, heat, and work?
3. How can you determine whether a chemical reaction is product-favored or
reactant-favored at equilibrium?
4. How can you determine whether a chemical reaction or physical process will
occur spontaneously, that is, without outside intervention?
Figure 5.14 The productfavored oxidation of iron. Iron
powder, sprayed into a Bunsen
burner flame, is rapidly oxidized.
The reaction is exothermic and is
product-favored.
The first two questions were addressed earlier in this chapter, while the other
two will be considered later in the book. Nonetheless, the stage has been set to
connect all of these issues. First, in Chapter 3 you learned that chemical reactions
proceed toward equilibrium, and spontaneous changes occur in a way that allows a
system to approach equilibrium. Reactions in which reactants are largely ­converted
to products when equilibrium is reached are called product-favored at equilibrium.
­Reactions in which only small amounts of products are present at equilibrium are
called reactant-favored at equilibrium (pages 146–147).
Look back at the many chemical reactions that you have seen. All combustion reactions are exothermic. For example, the oxidation of iron (Figure 5.14) is
exothermic.
4 Fe(s) + 3 O2(g) n 2 Fe2O3(s)
 2 mol Fe2O3   825.5 kJ 
r H° 2 f H°[Fe2O3(s)] 
1651.0 kJ/mol-rxn
 1 mol-rxn   1 mol Fe2O3 
The reaction has a negative value for ∆ rH°, and it is also product-favored at
equilibrium.
Conversely, the decomposition of calcium carbonate is endothermic.
CaCO3(s) n CaO(s) + CO2(g) ∆rH° = +179.0 kJ/mol-rxn
The decomposition of CaCO3 proceeds to an equilibrium that favors the reactants;
that is, it is reactant-favored at equilibrium.
Are all exothermic reactions product-favored at equilibrium and all endothermic reactions reactant-favored at equilibrium? From these examples, you might
formulate that idea as a hypothesis that can be tested by experiment and by examination of other examples. You would find that in most cases, product-favored reactions have negative values of ∆rH°, and reactant-favored reactions have positive
values of ∆rH°. But this is not always true; there are exceptions.
Clearly, a further discussion of thermodynamics must be tied to the concept of
equilibrium. This relationship, and the complete discussion of the third and fourth
questions, will be presented in Chapter 18.
Applying Chemical Principles
5.1 Gunpowder
Gunpowder has been used in fireworks, explosives, and firearms
for over one thousand years. Until the late 1800s, gunpowder
was a mixture of saltpeter (KNO3), charcoal (largely C), and
sulfur. Today, this mixture is known as black powder. A simplified version of the reaction occurring when it explodes is the
following:
2 KNO3(s) + 3 C(s) + S(s) n K2S(s) + N2(g) + 3 CO2(g)
Although black powder was used for hundreds of years, it has
some disadvantages as a propellant: it produces a large quantity
of white smoke, and the residues from the reaction are corrosive.
286
Modern firearms use smokeless powders. These ­powders are
primarily composed of nitrocellulose (also known as g­ uncotton)
or a mixture of nitrocellulose and n
­ itroglycerin. Nitrocellulose is the product of the reaction of cotton (­cellulose, with
an empirical formula of C6H10O5) with nitric acid. The fully
nitrated product has the empirical formula C 6H7(NO3)3O2.
Decomposition of nitrocellulose and nitroglycerin releases
more energy than the comparable mass of black powder. Just
as important, this is a better propellant because all of the
products are gaseous.
Chapter 5 / Principles of Chemical Reactivity: Energy and Chemical Reactions
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(b) Determine the enthalpy change that occurs when 1.00 g of
black powder decomposes according to the stoichiometry
of the balanced equation. (Even though black powder is
a mixture, assume that 1 mol of black powder consists of
exactly 2 mol of KNO3, 3 mol of C, and 1 mol of S.)
2. The enthalpy of reaction of guncotton depends on the degree
of nitration of the cellulose. When 0.725 g of a particular
sample of guncotton is decomposed in a bomb c­ alorimeter,
the temperature of the system increases by 1.32 K.
­Assuming the bomb has a heat capacity of 691 J/K and the
calorimeter contains 1.200 kg of water, what is the energy of
reaction per gram of this guncotton?
3. The decomposition of nitroglycerin (C3H5N3O9) produces
carbon dioxide, nitrogen, water, and oxygen gases.
(a) Write a balanced chemical equation for the decomposition
of nitroglycerin.
(b) If the decomposition of 1.00 g nitroglycerin releases
6.23 kJ/g of energy in the form of heat, what is the standard
molar enthalpy of formation of nitroglycerin?
Black gunpowder. Black gunpowder has been known for over
1000 years. This photo shows one of the disadvantages of
black powder: the great amount of smoke produced.
Questions
Reference
1. The standard enthalpies of formation of KNO3(s) and K2S(s)
are −494.6 kJ/mol and −376.6 kJ/mol, respectively.
(a) Determine the standard enthalpy change for the reaction
of black powder based on the given balanced equation.
J. Kelly, Gunpowder, Alchemy, Bombards, and P
­ yrotechnics:
The History of the Explosive That Changed the World,
New York: Basic Books, 2004.
In recent years, an increasing amount of evidence shows that
Earth’s climate is changing. It is warming in many areas,
and one specific reason is the use of fossil fuels, such as
gasoline, for transportation. In addition, fossil fuel supplies
are declining as humankind has ever greater energy needs.
One way to move away from a reliance on fossil fuels is to
use renewable fuels from biological sources. One such fuel
is ethanol (C2H6O), the majority of which is produced from
corn. In the United States, ethanol is added to most gasoline. Why?
Even though burning ethanol produces CO2, some ­consider
ethanol to be a carbon-neutral fuel because as the corn
plant grew, it removed CO2 from the air. Finally, ethanol and
­ethanol-gasoline mixtures burn more cleanly than pure gasoline, leading to less air pollution.
Most ethanol-containing fuels currently used in the United
States are a mixture of 10% ethanol and 90% gasoline (E10).
A small fraction of fuel is sold as E85—a blend of gasoline
with 51–85% ethanol. However, this can only be used in
vehicles with engines designed for fuels with a high ethanol
content (flexible fuel engines). In 2020 there were about
3900 E85 stations in the United States, and in 2018 over
21 ­million vehicles were equipped to use it.
Is a goal of replacing gasoline completely with ethanol reasonable? This is a lofty goal, given that present gasoline consumption in the United States is about 140 billion gallons annually.
The problem is that even if all of the corn grown in the United
States is converted to ethanol, the supply will still be inadequate.
It is clear that there must be more emphasis on ways to derive
Jeffrey Isaac Greenberg 16+/Alamy Stock Photo
5.2 The Fuel Controversy—Alcohol and Gasoline
Ethanol available at a service station. E85 fuel is a blend of
gasoline with 51–85% ethanol. Be aware that you can only use
E85 in vehicles designed for the fuel. In an ordinary vehicle,
the ethanol leads to deterioration of seals in the engine and fuel
system.
ethanol from other sources, such as cellulose from cornstalks
and various grasses.
Beyond this, there are other problems associated with
ethanol. One is that it cannot be distributed through a
pipeline system as gasoline can. Any water in the pipeline
would dissolve in the ethanol, which causes the fuel value
to decline.
Applying Chemical Principles
Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
287
Questions
For the purposes of this analysis, octane (C8H18) is used as a
substitute for the complex mixture of hydrocarbons in gasoline. Data needed for this question (in addition to the data in
Appendix L) are:
∆fH° [C8H18()] = −250.1 kJ/mol
Density of ethanol = 0.785 g/mL
Density of octane = 0.699 g/mL
1. Calculate ∆rH° for the combustion of ethanol and octane producing carbon dioxide gas and liquid water, and compare the
values per mole and per gram. Which provides more energy
per mole? Which provides more energy per gram?
2. Compare the energy produced per liter of the two fuels.
Which produces more energy for a given volume (something
useful to know when filling your gas tank)?
3. What mass of CO2, a greenhouse gas, is produced per liter of
fuel (assuming complete combustion)?
4. Now compare the fuels on an energy-equivalent basis. What
volume of ethanol would have to be burned to get the same
energy as 1.00 L of octane? When you burn enough ethanol
to have the same energy as a liter of octane, which fuel produces more CO2?
Think–Pair–Share
1. You are tasked with determining the specific heat capacity
(in J/g ? K) for an unknown metal. The following equipment
and supplies are available: a 25.0-g piece of the metal, a
200-mL insulated container, a graduated cylinder, water,
ice, and a thermometer that can measure temperature
changes accurately to ±0.02 °C.
(a) Outline the steps for an experimental procedure to determine the specific heat capacity of the metal using the
available equipment and supplies.
(b) Identify possible sources of error in your experimental procedure that might cause an inaccurate result. (Assume the
graduated cylinder and thermometer are both accurate
and precise, and that human error is not a source of error.)
(c) Suggest some ideas for how to correct for some of these
error sources.
2. For introductory laboratories, resealable plastic bags are a
convenient way to conduct experiments involving gas-forming
reactions. Energy as heat or work can transfer in or out of
the plastic bag, but reactants and products remain trapped.
In an experiment, nitric acid reacts with a small amount of
copper in a sealed plastic bag according to the following balanced chemical equation:
Cu(s) + 4 HNO3(aq) n Cu(NO3)2(aq) + 2 NO2(g) + 2 H2O()
As the reaction proceeds, the contents of the bag become
warm, and the bag inflates with a brown gas (NO2).
(a) Define the system and the surroundings for this
experiment.
(b) Does energy as heat (q) flow into the system or out of the
system? What is the sign of q?
(c) Is work (w) done in this experiment? If so, is work done by
the system on the surroundings, or by the surroundings on
the system? What is the sign of w?
3. The following table shows the enthalpy of combustion for
some fuels in units of kJ/mol, kJ/L, and kJ/g. The enthalpy of
combustion per volume assumes a temperature of 25 °C and
atmospheric pressure (1 atm). Note that although gasoline
is a mixture of many hydrocarbons, it is often represented
as octane.
288
Fuel
∆H°
(kJ/mol)
∆H°
(kJ/L)
∆H°
(kJ/g)
Hydrogen,
H2(g)
−285.8
−11.7
−141.8
Methane,
CH4(g)
−890.2
−36.4
−55.5
Ethane,
C2H6(g)
−1560.6
−63.8
−51.9
Ethanol,
C2H6O()
−1376.5
−23,590
−29.9
Octane,
C8H18()
−5490
−33,814
−48.1
(a) When determining which fuel provides the most energy
in a car engine, is it best to compare the enthalpy change
in a combustion reaction by moles, volume, or mass?
Explain your reasoning? Based on your decision, which fuel
provides the most energy (at the given temperature and
pressure).
(b) Gasoline sold in the United States is often a blend of 10%,
15%, or even up to 85% ethanol by volume. Identify at least
one advantage and one disadvantage of using gasoline
blended with ethanol versus ethanol-free gasoline?
(c) Why are the enthalpies of combustion per liter of hydrogen, methane, and ethane much lower than those of ethanol and octane?
(d) The enthalpy of combustion of hydrogen per gram is
nearly three times that of a hydrocarbon. And unlike a
hydrocarbon, which produces the greenhouse gas CO 2
upon combustion, the combustion of hydrogen produces only water. Unfortunately, there are multiple issues
in replacing gasoline with hydrogen as a fuel. What are
some of the problems that must be overcome if hydrogen
is to compete with, or possibly replace, gasoline as a fuel
in vehicles?
Chapter 5 / Principles of Chemical Reactivity: Energy and Chemical Reactions
Copyright 2023 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook a