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

  1. 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%

= 69.7amu

  1. 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

  1. 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

  1. There are two naturally occurring isotopes of rubidium: 85Rb, which has a mass of 84.91 amu and 87Rb, 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

  1. 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

Half-Life Problems(2 points, min 6, max 8)

  1. 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

  1. 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

  1. 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

  1. 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 = 3half-life

639,000 years/3 = 213,000 years

  1. 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?

2.8x1010yrs / 1.4x1010yrs = 2nd half-life = ¼ remains

1 = 25.0g

4 x

X = 25.0g * 4= 100.g

  1. 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)

a._neutron_ b. __beta_ c. _alpha_d. __gamma_

  1. Nuclear decay with no mass and no charge _gamma radiation_
  2. An electron _beta radiation_
  3. Least penetrating nuclear decay_alpha radiation_
  4. Most damaging nuclear decay to the human body__gamma radiation_
  5. Nuclear decay that can be stopped by skin or paper _alpha radiation_
  6. Nuclear decay that can be stopped by aluminum__beta radiation_

Complete the following nuclear equations(2 points, min 6, max 8)

  1. Write a nuclear equation for the alpha decay of
  2. Write a nuclear equation for the beta decay of
  3. Write a nuclear equation for the alpha decay of
  4. Write a nuclear equation for the beta decay of
  5. Write a nuclear equation for the alpha decay of
  6. Write a nuclear equation for the alpha decay of
  7. Write a nuclear equation for the beta decay of