Question

The element boron occurs in nature as two isotopes having atomic masses 10u and 11u. What are the percentage abundances of these isotopes in a sample of boron having average atomic mass of 10.8u.

Answer 1

Calculating the percentage abundance of different isotopes of Boron:

Let the % of B -10 isotope = x

Now, the % of B-11 isotope = (100-x)

The formula for calculation of average atomic mass $=\frac{{\displaystyle \sum _{n=1}^{\infty }}\left(Isotopicmass\times percentageabundance\right)}{100}$

Therefore, the average atomic mass of boron $=\frac{10\times \mathrm{x}+11(100-\mathrm{x})}{100}$

Now,

$10.8=\frac{10\mathrm{x}+1100-11\mathrm{x}}{100}$

$1080=10\mathrm{x}+1100-11\mathrm{x}$

$11\mathrm{x}-10\mathrm{x}=1100-1080$

$\mathrm{x}=20$

Percentage of B-isotope = 20%

Percentage of B- 11 Isotope = (100 - 20) = 80%