Answers to Isotope and Average
Atomic Mass Worksheet
1.
127 (0.80) = 101.6
126 (0.17) =
21.42
128 (0.03) = + 3.84
126.86 amu
2.
197 (0.50) = 98.5
198 (0.50) = + 99
197.5 amu
3.
55 (0.15) =
56 (0.85)
8.25
= + 47.6
55.85 amu
4.
1 (0.99)
=
.99
2 (0.008)
=
.016
3 (0.002) = + .006
1.012 amu
5.
14 (0.95) =
15 (0.03)
=
13.3
.45
16 (0.02) = + .32
14.07 amu
6.
a. Find the missing percentage:
100%  (92.23% + 4.67%) = 3.10%
b. Multiply each mass number by its
percentage and set it equal to the
average atomic mass:
28(.9223) + 29(.0467) + x(.0310) = 28.1087
25.8244 + 1.3543 + .0310x = 28.1087
27.1787 + .0310x = 28.1087
.0310x = 0.93
x = 30
c. The isotope is Si-30.
7.
a. Find the missing percentage:
100%  (.35% + 2.25% + 11.6% + 57% + 17.3%) = 11.5%
b. Multiply each mass number by its
percentage and set it equal to the
average atomic mass:
78(.0035) + 80(.0225) + 82(.116) + 84(.57) + 86(.173) + x(.115) = 83.888
.273 + + 1.8 + 9.512 + 47.88 + 14.878 + .115x = 83.888
74.343 + .115x = 83.888
.115x = 9.545
x = 83
c. The isotope is Kr-83.
8.
C-12 Abundance = x
C-13 Abundance = 1  x
12x + 13(1  x) = 12.011
12x + 13  13x = 12.011
13  x = 12.011
 x =  .989
C-12 = 98.90%
x = .989
C-13 = 1.10%
1  x = .011
9.
Cl-35 Abundance = x
Cl-37 Abundance = 1  x
35x + 37(1  x) = 35.453
35x + 37  37x = 35.453
37  2x = 35.453
2 x =  1.547
Cl-35 = 77.35%
x = .7735
Cl-37 = 22.65%
1  x = .2265
10.
Br-79 Abundance = x
Br-81 Abundance = 1  x
79x + 81(1  x) = 79.904
79x + 81  81x = 79.904
81  2x = 79.904
2 x =  1.096
Br-79 = 54.80%
x = .548
Br-81 = 45.20%
1  x = .452