1. What is average atomic mass?
Average atomic mass is the weighted average of the atomic masses of the naturally occurring isotopes of an element

2. Calculating Average Atomic Mass
EXAMPLE
Boron has two isotopes:
B-10 (mass amu) 19.8% abundance
B-11 (mass amu) 80.2% abundance
Calculate the average atomic mass.
(.198) (10.013) + (.802) ( ) =
1.98 amu amu = amu

3. Calculating Average Atomic Mass
Calculate the average atomic mass of Mg.
Isotope amu (78.99%)
Isotope amu (10.00%)
Isotope 3 – amu (11.01%)
(23.985)(.7899)+(24.986)(.1000)+(25.982)(.1101)
18.95 amu amu amu = amu

4. Average Atomic Mass
Helium has two naturally occurring isotopes, He-3 and He-4. The atomic mass of helium is amu. Which isotope is more abundant in nature?
He-4 is more abundant in nature because the atomic mass is closer to the mass of He-4 than to the mass of He-3.

5. Calculating % Abundance
Set one of the isotope’s relative abundance equal to X.
The relative abundances must add up to 1; therefore, the other isotope’s relative abundance would be equal to 1-X.
Remember, the relative mass is equal to the relative abundance times the mass number. Remember, the relative masses must add up to the average atomic mass.

6. Calculating % Abundance
Setup an algebraic equation where the relative masses of each isotope is set equal to the average atomic mass for the element.  
(X • mass number) + [(1-x) • mass number] = avg. atomic mass
*When solving for X, keep 4 decimal places. This will allow the % abundance to have 2 decimal places.

7. Calculating % Abundance
X is the relative abundance, so multiply this by 100 to make it % abundance.
To obtain the other isotope’s % abundance, the 2 % abundances must add up to 100.

8. Calculating % Abundance
Determine the % abundance for each isotope of antimony. Antimony exists as Sb-121 and Sb-123.
Potassium exists as K-39 and K-41. Determine the % abundance for each isotope of potassium.

9. Isotopic Pennies – number of pre and post 1982
a. Let X be the number of pre-1982 pennies
b. Let 10-X be the number of post-1982 pennies
c. (X)(3.1g) + (10-X)(2.5g) = mass of 10 pennies
pre post-82
EXAMPLE (Mass of a sample of pennies is 31.0g)
(X)(3.1g) + [(10-X)(2.5g)] = 31.0 g
3.1X X = 31.0g
.6X + 25 = 31.0g
.6X = 6.0g X = 6.0g/.6
X = 10 pre-82 pennies
10-X = 0 pre-82 pennies

10. Isotopic Penny Lab- Average Atomic Mass
Calculate percent of pre-82 and post-82 pennies
# of pre-82 pennies x 100% # post-82 pennies x 100%
Calculate the average atomic mass of coinium
(% pre-82)(3.1g) + (% post-82)(2.5) = average atomic mass