Question

Naturally occurring boron consists of two isotopes whose atomic weights are $10.01$ and $11.01$. The atomic weight of natural boron is $10.81$. Calculate the percentage of each isotope in natural boron.

Answer 1

Let the percentage of boron isotope with atomic weights 10.01 be x% and that of atomic weight 11.01 be (100-x)%. Average atomic mass of boron is 10.81.

$\text{Average atomic mass}$ = $\frac{\text{Atomic weight of first isotope}\times x+\text{atomic weight of second isotope}\times (100-x)}{100}$

$10.81=\frac{10.01x+11.01\times (100-x)}{100}$

$x=20$

$(100-x)=80$

Hence, natural abundance of isotope with atomic weight 10.01 is 20%

natural abundance of isotope with atomic weight 11.01 is 80%