Get 1 week of unlimited access
Class Notes (1,051,528)
CA (601,578)
York (42,530)
ENG (57)
Lecture

# ENG 2002 Lecture Notes - Electronvolt, Phosphide, Electronegativity

27 pages75 viewsWinter 2014

Department
Engineering
Course Code
ENG 2002
Professor
Hany E.Z.Farag

Page:
of 27
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to
students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
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 naturally occurring isotopes.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to
students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
2.2 Chromium has four naturally-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
(A
C
r
) is computed by adding fraction-of-occurrence/atomic weight
products for the three isotopes. Thus
A
Cr = f50C
r
A50C
r
+ f52C
r
A52C
r
f53C
r
A53C
r
f54C
r
A54C
r
(0.0434)(49.9460 amu) + (0.8379)(51.9405 amu) + (0.0950)(52.9407 amu) + (0.0237)(53.9389 amu) = 51.9963 am
u
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to
students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
2.3 (a) How many 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 many 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 necessary, as
# g/amu = 1 mol
6.022 1023 atoms



 1 g / mol
1 amu / atom





= 1.66 10-24 g/amu
(b) Since there are 453.6 g/lbm,
1 lb - mol = (453.6 g/lb
m
) (6.022 10 23 atoms/g - mol)
= 2.73 1026 atoms/lb-mol

#### Loved by over 2.2 million students

Over 90% improved by at least one letter grade.