The average atomic mass of a sample of an element $\text{'}\mathrm{X}\text{'}$ is $16.2\mathrm{u}$. What is the percentage of each isotope ${}_8{}^{16}\mathrm{X}$ and ${}_8{}^{18}\mathrm{X}$ in the sample?

From the given data,

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

Let the percentage of the isotope ${}_8{}^{16}\mathrm{X}$ be $\mathrm{A}\%$ and the percentage of the isotope ${}_8{}^{18}\mathrm{X}$ be $(100-\mathrm{A})\%$.

Therefore,

$$\begin{gathered} 16\times \frac{\mathrm{A}}{100}+18\times \frac{100-\mathrm{A}}{100}=16.2 \\ 1800-2\mathrm{A}=1620 \\ \Rightarrow \mathrm{A}=90 \\ \%\mathrm{of}{}_8{}^{16}\mathrm{X}=\mathrm{A}\%=90\% \\ \%\mathrm{of}{}_8{}^{18}\mathrm{X}=(100-\mathrm{A})\%=10\% \end{gathered}$$

Hence, the percentage of the isotopes ${}_8{}^{16}\mathrm{X}$ and ${}_8{}^{18}\mathrm{X}$ in the sample are 90% and 10% respectively.