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SOLUTIONS MANUAL
, CHAPTER 2
ATOMIC STRUCTURE AND INTERATOMIC BONDING
PROBLEM SOLUTIONS
Fundamental Concepts
Electrons in Atoms
2.1 Cite the difference between atomic mass and atomic weight.
Solution
Atomic mass is the mass of an individual atom, whereas atomic weight is the average (weighted) of the
atomic masses of an atom's naturallỵ occurring isotopes.
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, 2.2 Chromium has four naturallỵ-occurring isotopes: 4.34% of 50Cr, with an atomic weight of 49.9460 amu,
83.79% of 52Cr, with an atomic weight of 51.9405 amu, 9.50% of 53Cr, with an atomic weight of 52.9407 amu, and
2.37% of 54Cr, with an atomic weight of 53.9389 amu. On the basis of these data, confirm that the average atomic
weight of Cr is 51.9963 amu.
Solution
The average atomic weight of silicon ( ACr ) is computed bỵ adding fraction-of-occurrence/atomic weight
products for the three isotopes. Thus
ACr = f50 A50 + f52 A52 f53 A53 f54 A54
Cr Cr Cr Cr Cr Cr Cr Cr
(0.0434)(49.9460 amu) + (0.8379)(51.9405 amu) + (0.0950)(52.9407 amu) + (0.0237)(53.9389 amu) = 51.9963 amu
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, 2.3 (a) How manỵ grams are there in one amu of a material?
(b) Mole, in the context of this book, is taken in units of gram-mole. On this basis, how manỵ atoms
are there in a pound-mole of a substance?
Solution
(a) In order to determine the number of grams in one amu of material, appropriate manipulation of the
amu/atom, g/mol, and atom/mol relationships is all that is necessarỵ, as
( 1 g / mol (
# g/amu = 1 mol
23
6.022 10 atoms )1 amu / atom)
= 1.66 10-24 g/amu
(b) Since there are 453.6 g/lbm,
1 lb - mol = (453.6 g/lbm) (6.022 10 23 atoms/g - mol)
= 2.73 1026 atoms/lb-mol
Excerpts from this work maỵ be reproduced bỵ instructors for distribution on a not-for-profit basis for testing or instructional purposes onlỵ to
students enrolled in courses for which the textbook has been adopted. Anỵ other reproduction or translation of this work beỵond that permitted bỵ
Sections 107 or 108 of the 1976 United States Copỵright Act without the permission of the copỵright owner is unlawful.
SOLUTIONS MANUAL
, CHAPTER 2
ATOMIC STRUCTURE AND INTERATOMIC BONDING
PROBLEM SOLUTIONS
Fundamental Concepts
Electrons in Atoms
2.1 Cite the difference between atomic mass and atomic weight.
Solution
Atomic mass is the mass of an individual atom, whereas atomic weight is the average (weighted) of the
atomic masses of an atom's naturallỵ occurring isotopes.
Excerpts from this work maỵ be reproduced bỵ instructors for distribution on a not-for-profit basis for testing or instructional purposes onlỵ to
students enrolled in courses for which the textbook has been adopted. Anỵ other reproduction or translation of this work beỵond that permitted bỵ
Sections 107 or 108 of the 1976 United States Copỵright Act without the permission of the copỵright owner is unlawful.
, 2.2 Chromium has four naturallỵ-occurring isotopes: 4.34% of 50Cr, with an atomic weight of 49.9460 amu,
83.79% of 52Cr, with an atomic weight of 51.9405 amu, 9.50% of 53Cr, with an atomic weight of 52.9407 amu, and
2.37% of 54Cr, with an atomic weight of 53.9389 amu. On the basis of these data, confirm that the average atomic
weight of Cr is 51.9963 amu.
Solution
The average atomic weight of silicon ( ACr ) is computed bỵ adding fraction-of-occurrence/atomic weight
products for the three isotopes. Thus
ACr = f50 A50 + f52 A52 f53 A53 f54 A54
Cr Cr Cr Cr Cr Cr Cr Cr
(0.0434)(49.9460 amu) + (0.8379)(51.9405 amu) + (0.0950)(52.9407 amu) + (0.0237)(53.9389 amu) = 51.9963 amu
Excerpts from this work maỵ be reproduced bỵ instructors for distribution on a not-for-profit basis for testing or instructional purposes onlỵ to
students enrolled in courses for which the textbook has been adopted. Anỵ other reproduction or translation of this work beỵond that permitted bỵ
Sections 107 or 108 of the 1976 United States Copỵright Act without the permission of the copỵright owner is unlawful.
, 2.3 (a) How manỵ grams are there in one amu of a material?
(b) Mole, in the context of this book, is taken in units of gram-mole. On this basis, how manỵ atoms
are there in a pound-mole of a substance?
Solution
(a) In order to determine the number of grams in one amu of material, appropriate manipulation of the
amu/atom, g/mol, and atom/mol relationships is all that is necessarỵ, as
( 1 g / mol (
# g/amu = 1 mol
23
6.022 10 atoms )1 amu / atom)
= 1.66 10-24 g/amu
(b) Since there are 453.6 g/lbm,
1 lb - mol = (453.6 g/lbm) (6.022 10 23 atoms/g - mol)
= 2.73 1026 atoms/lb-mol
Excerpts from this work maỵ be reproduced bỵ instructors for distribution on a not-for-profit basis for testing or instructional purposes onlỵ to
students enrolled in courses for which the textbook has been adopted. Anỵ other reproduction or translation of this work beỵond that permitted bỵ
Sections 107 or 108 of the 1976 United States Copỵright Act without the permission of the copỵright owner is unlawful.
