Foundations of Matter Worksheet #4
Average atomic mass(2 points, min 6, max 8)
1. What is the atomic mass of element Y if Y consists of 57.25% of atoms with a mass of 120.90 amu
and 42.75% of atoms with a mass of 122.90 amu?
AAM = [ (57.25% * 120.90amu) + (42.75% * 122.90amu)]/100%
AAM = [6922%amu + 5254%amu] / 100 = 12176%amu / 100%
= 121.76 amu
2. If element X consists of 60.4% of atoms with a mass of 68.9 amu each and 39.6% of atoms with a
mass of 70.9 amu each, what is the atomic mass of element X?
AAM = [ (60.4% * 68.9amu) + (39.6% * 70.9amu)]/100%
AAM = [4160%amu + 2.81x103 %amu] / 100 = 6970 % amu / 100%
= 69.7 amu
3. Calculate the average atomic mass of oxygen. The naturally occurring element consists of 99.759%
with a mass of 15.99491 amu, 0.037% with a mass of 16.99914 amu, and 0.204% with a mass of
17.99916 amu.
AAM = [ (99.759% * 15.99491amu) + (0.037% * 16.99914amu) + (0.204% *
17.99916)]/100%
AAM = [1595.6%amu + 0.63%amu + 3.67%amu] / 100 = 1599.9% amu / 100%
= 15.999 amu
4. The element boron consists of two isotopes: 10B, which has a mass of 10.01 amu and 11B, which has
a mass of 11.01 amu. The atomic mass of boron is 10.81 amu. What is the percent abundance of
each of the two isotopes?
%10B = x
%11B = 100% - x
10.81amu = [(x * 10.01amu) + (100% - x)* 11.01amu] / 100%
1081%amu = (x * 10.01amu) + (100% - x)* 11.01amu
1081%amu = (10.01amu)x + 1101%amu - (11.01amu)x
1081%amu = (-1.00amu)x + 1101amu%
-20.%amu = (-1.00amu)x
20.% = x, 20.% 10K , 80.% 11B
5. There are two naturally occurring isotopes of rubidium: 85Rb, which has a mass of 84.91 amu and
87
Rb, which has a mass of 86.92 amu. The atomic mass of rubidium is 85.47 amu. What is the
percent abundance of each of the isotopes?
%85Rb = x
%87Rb = 100% - x
85.47amu = [(x * 84.91amu) + (100% - x)* 86.92amu] / 100%
8547%amu = (x * 84.91amu) + (100% - x)* 86.92amu
8547%amu = (84.91amu)x + 8692%amu - (86.92amu)x
8547%amu = (-2.01amu)x + 8692amu%
-145%amu = (-2.01amu)x
72.1% = x, 72.1% 85Rb , 27.9% 87Rb
6. Naturally occurring bromine consists of two isotopes: 79Br, which has a mass of 78 amu and 81Br,
which has a mass of 80.916 amu. The atomic mass of bromine is 79.904 amu. What is the percent
abundance of each of the isotopes?
%79Br = x
%81Br = 100% - x
79.904amu = [(x * 78amu) + (100% - x)* 80.916amu] / 100%
7990.4%amu = (x * 78amu) + (100% - x)* 80.916amu
7990.4%amu = (78amu)x + 8091.6%amu - (80.916amu)x
7990.4%amu = (-3amu)x + 8091.6amu%
-101%amu = (-3amu)x
30% = x, 30% 79Br , 70% 81Br
25
Half-Life Problems(2 points, min 6, max 8)
7. How much of a 100.0 g sample of Au-198 is left over after 8.10 days if its half-life is 2.70 days?
8.10 days / 2.70 days = 3 half-life = 1/8 remains
1/8*100.0g = 12.50g left over
8. A 50.0 g sample of N-16 decays to 12.5 g in 14.4 seconds. What is its half-life?
12.5g/50.0g = ¼ remains = 2 half-life
14.4 sec = x sec
2
1
14.4 sec = 2 x sec
x = 7.2 sec
9. The half-life of K-42 is 12.4 hours. How much of a 750 g sample is left after 62.0 hours?
62.0hrs / 12.4hrs = 5 half-life = 1/32 remains
1/32 * 750g = 23g left over
10. What is the half-life of Tc-99 if a 500 g sample decays to 62.5 g in 639,000 years?
62.5g / 500 g = 1/8 remains = 3 half-life
639,000 years/3 = 213,000 years
11. The half-life of Th-232 is 1.4x1010 years. If there are 25.0 g of the sample left after 2.8 x 1010 years,
how many grams were in the original sample?
2.8x1010 yrs / 1.4x1010 yrs = 2nd half-life = ¼ remains
1 = 25.0g
4
x
X = 25.0g * 4= 100.g
12. There are 5.0 g of I-131 after 40.35 days. How many grams were in the original sample if its half-life
is 8.07 days?
40.35 days / 8.07 days = 5th half-life = 1/32 remains
1 = 5.0g
32
x
X = 5.0g * 32 = 160 g
Identify the following as alpha, beta, gamma, or neutron radiation. (½ point, min 3)
1
a. 0 n _neutron_
0
b. -1 e __beta_
4

0
c. 2 He _alpha_
d. 0
__gamma_
13. Nuclear decay with no mass and no charge _gamma radiation_
14. An electron _beta radiation_
15. Least penetrating nuclear decay _alpha radiation_
16. Most damaging nuclear decay to the human body __gamma radiation_
17. Nuclear decay that can be stopped by skin or paper _alpha radiation_
18. Nuclear decay that can be stopped by aluminum __beta radiation_
Complete the following nuclear equations(2 points, min 6, max 8)
19. Write a nuclear equation for the alpha decay of
20. Write a nuclear equation for the beta decay of
231
91
Md 
Fr 
223
87
149
21. Write a nuclear equation for the alpha decay of 62 Sm
22. Write a nuclear equation for the beta decay of
165
61
249
146
24. Write a nuclear equation for the alpha decay of 62 Sm
198
25. Write a nuclear equation for the beta decay of 85 At
e +

4
2

223
88
227
87
Es
Ra
He +
145
60
Nd
0
1
e + 165
62 Sm


He +
0
1
Pm 
23. Write a nuclear equation for the alpha decay of 101 Md
4
2
4
2
4
2
0
1
He +
He +
e +
198
86
245
99
142
60
Es
Nd
Rn