Irvington High School  AP Chemistry
Mr. Markic
Name______________________________
Number __ Date __/__/__
3  Mass Relationships in Chemical Reactions
A t o m i c
M a s s
1. The atomic masses of 35Cl17 (75.53 percent) and 37Cl17 (24.47 percent) are 34.968 amu and 36.95
amu, respectively. Calculate the average atomic mass of chlorine. The percentages in parentheses
denote the relative abundances.
(34.968 amu)(0.7553)  (36.956 amu)(0.2447)  35.45 amu
2. The atomic masses of 6Li3 and 7Li3 are 6.0151 amu and 7.0160 amu, respectively. Calculate the
natural abundances of these two isotopes. The average atomic mass of Li is 6.941 amu.
Strategy: Each isotope contributes to the average atomic mass based on its relative abundance. Multiplying the
mass of an isotope by its fractional abundance (not percent) will give the contribution to the average atomic mass of
that particular isotope.
It would seem that there are two unknowns in this problem, the fractional abundance of 6Li and the fractional
abundance of 7Li. However, these two quantities are not independent of each other; they are related by the fact that
they must sum to 1. Start by letting x be the fractional abundance of 6Li. Since the sum of the two abundance’s must
be 1, we can write
Abundance 7Li  (1  x)
Solution:
Average atomic mass of Li  6.941 amu  x(6.0151 amu)  (1  x)(7.0160 amu)
6.941  1.0009x  7.0160
1.0009x  0.075
x  0.075
x  0.075 corresponds to a natural abundance of 6Li of 7.5 percent. The natural abundance of 7Li is
(1  x)  0.925 or 92.5 percent.
3. What is the mass in grams of 13.2 amu?
 6.022  1023 amu 
The unit factor required is 



1g


? g  13.2 amu 
1g
6.022  10
23
 2.19  1023 g
amu
4. How many amu are there in 8.4 g?
 6.022  1023 amu 
The unit factor required is 



1g


? amu  8.4 g 
6.022  1023 amu
= 5.1  1024 amu
1g
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