Assessment 1: Bonding
1. How many grams are there in one amu of a material? (5 pts)
a.
b.
c.
d.
2.25 x 10^-24 g/amu
1.49 x 10^-25 g/amn
1.66 x 10^-24 g/amu
1.89 x 10^-26 g/amu
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

 1 g / mol 
1 mol
# g/amu = 


23
6.022  10 atoms 1 amu / atom
= 1.66  10-24 g/amu

2. 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? (5 pts)
a.
b.
c.
d.
2.73 x 10^26 atoms/lb-mol
2.34 x 10^24 atoms/lb-mol
2.51 x 10^24 atoms/lb-mol
2.85 x 10^25 atoms/lb-mol
Solution
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

3. Match the electronic structure to the element
Fe ____
Al_____
Cu_____
O _____
A. 1s22s22p4
B. 1s22s22p63s23p63d64s2
C. 1s22s22p63s23p1
D. 1s22s22p63s23p63d104s1
Fe – B (5 pts) Al – C (5 pts) Cu – D (5 pts) O – A (5 pts)
4. A gold O-ring is used to form a gastight seal in a high-vacuum chamber. The ring is
formed from a 80-mm length of 1.5-mm-diameter wire. Calculate the number of gold
atoms in the O-ring.
a. 4.16 x 10^21
b. 4.16 x 10^25
c. 8.33 x 10^21
d. 8.33 x 10^25
5. Calculate the force of attraction between a K+ and an O2- ion the centers of which are
separated by a distance of 1.5 nm.
a. 3.55 x 10^-10 N
b. 4.01 x 10^-10 N
c. 1.34 x 10^-10 N
d. 2.05 x 10^-10 N
Solution
The attractive force between two ions FA is just the derivative with respect to the
interatomic separation of the attractive energy expression, Equation 2.8 in the text, which
is just
FA =
dEA
dr
 A 
d 
A
 r 
=
=
dr
r2
The constant A in this expression is defined in footnote 3. Since the valences of the K+ and O2
ions (Z1 and Z2) are +1 and -2, respectively, Z1 = 1 and Z2 = 2, then
FA =
=

(Z1e) (Z 2 e)
40r 2
(1)(2)(1.602  1019 C) 2

(4)()
(8.85  1012 F/m) (1.5  10 9 m) 2
= 2.05  10-10 N
Copper is added to silver to make the metal stronger and more durable Sterling silver contains
92.5 wt% silver and 7.5 wt% copper. A small sterling silver spoon has a mass of 100 grams.
Calculate the number of copper and silver atoms in the spoon. (5pts)
Sterling silver: 92.5 wt % Ag + 7.5 wt % Cu Mass of spoon = 100 gr
Mass of silver = mAg = 92.5 gr ⇒ number of silver atoms =
6.02×1023 atoms⁄mol
×
gr
107.9 ⁄mol
92.5 g
= 5.2 × 1023 atoms (Ag)
Mass of copper = mCu = 7.5 gr ⇒ number of copper atoms =
6.02×1023 atoms⁄mol
gr
63.55 ⁄mol
× 7.5 g
= 7.1 × 1022 atoms (Cu)