Question

A sample of an element $\mathrm{E}$ contains two isotopes ${}_8{}^{16}\mathrm{E}$ and ${}_8{}^{18}\mathrm{E}$. If the average atomic mass of this sample of the element is 16.2 u, calculate the percentage of the two isotopes in this sample.

Answer 1

Given:

Isotope ${}_8{}^{16}\mathrm{E}$: Atomic mass$=16\mathrm{u}$

Isotope ${}_8{}^{18}\mathrm{E}$: Atomic mass$=18\mathrm{u}$

Average atomic mass$=16.2\mathrm{u}$

Formula used:

Average atomic mass$=\frac{\%\mathrm{abundance}\mathrm{of}\mathrm{isotope}-1}{100}\times \mathrm{At}.\mathrm{mass}\mathrm{of}\mathrm{isotope}-1+\frac{\%\mathrm{abundance}\mathrm{of}\mathrm{isotope}-2}{100}\times \mathrm{At}.\mathrm{mass}\mathrm{of}\mathrm{isotope}-2$

Calculate the percentage of isotope ${}_8{}^{16}\mathrm{E}$:

Let the percentage abundance of isotope ${}_8{}^{16}\mathrm{E}$ be $\mathrm{x}$.

Then, the percentage abundance of isotope ${}_8{}^{18}\mathrm{E}$ will be $100-\mathrm{x}$.

Thus,

Average atomic mass$=\frac{\%\mathrm{abundance}\mathrm{of}\mathrm{isotope}-1}{100}\times \mathrm{At}.\mathrm{mass}\mathrm{of}\mathrm{isotope}-1+\frac{\%\mathrm{abundance}\mathrm{of}\mathrm{isotope}-2}{100}\times \mathrm{At}.\mathrm{mass}\mathrm{of}\mathrm{isotope}-2$

Substitute the values,

$$\begin{gathered} 16.2=\frac{x}{100}\times 16+\frac{100-\mathrm{x}}{100}\times 18 \\ 16.2=\frac{16\mathrm{x}}{100}+\frac{1800-18\mathrm{x}}{100} \\ 16.2=\frac{1800-2\mathrm{x}}{100} \\ 1620=1800-2\mathrm{x} \\ 2\mathrm{x}=180 \\ \mathrm{x}=90 \end{gathered}$$

Thus,

The percentage abundance of isotope ${}_8{}^{16}\mathrm{E}$$=\mathrm{x}=90\%$.

Calculate the percentage of isotope ${}_8{}^{18}\mathrm{E}$:

The percentage abundance of isotope ${}_8{}^{18}\mathrm{E}$$=100-\mathrm{x}$

Thus,

The percentage abundance of isotope ${}_8{}^{18}\mathrm{E}$$=100-90$

The percentage abundance of isotope ${}_8{}^{18}\mathrm{E}$$=10\%$.

Thus, the percentage of the two isotopes ${}_8{}^{16}\mathrm{E}$ and ${}_8{}^{18}\mathrm{E}$ in this sample is $90\%$ and $10\%$ respectively.