Question

Calculate the % relative abundance of B – 10 and B – 11, if its average atomic mass is 10.804 amu.

Answer 1

% of relative abundance can be calculated by the formula. Average atomic mass of the element

= Atomic mass of 1st isotope × abundance of 1st isotope + Atomic mass of 2nd isotope × abundance of 2nd isotope 

∴ Average atomic mass of Boron 

= Atomic mass of B – 0 × abundance of B -10 + Atomic mass of B – 11 × abundance of B – 11

Let the abundance of B – 10 be ‘x’ and B – 11 be (1 – x)

So, 10.804 = 10 × x + 11 (1 – x)

10.804 = 10x + 11 – 11x

x = 11 – 10.804 

x = 0.196

1 -x = 1 – 0.196 = 0.804 

Therefore % abundance of B – 10 is 19.6% and B – 11 is 80.4%

[OR] 

Let the % of the isotope B – 10 = x 

Then the % of the isotope B – 11 = 100 – x

1100 – x = 1080.4 

x = 19.6

% abundance of B – 10 = 19.6% 

% abundance of B – 11 = 80.4%