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Question
Calculate the % relative abundance of B-10 and B-11, if its average atomic mass is 10.804 amu.
Solution
The average atomic mass of Boron = 10.804 amu.
% relative abundance of B – 10 =?
% relative abundance of B – 11 =?
Let the fraction of relative abundance of B – 10 = x
Let the fraction of relative abundance of B – 11 = y
x + y = 1
y = 1 – x
Relative abundance = x (10) + (1 – x) (11) = 10.804 amu
⇒ 10x + 11 – 11x = 10.804 amu
⇒ 11 – x = 10.804 amu
⇒ -x = 10.804 – 11
⇒ -x = -0.196
⇒ x = 0.196
x = % abundance of B – 10 = 0.196 × 100 = 19.6 %
y = % abundance of B – 11 = 100 – 19.6 = 80.4 %
Percentage abundance of B – 10 = 19.6 %
Percentage abundance of B – 11 = 80.4 %.
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