Question.
The average atomic mass of a sample of an element $\mathrm{X}$ is $16.2 \mathrm{u}$. What are the percentages of isotopes ${ }_{8}^{16} \mathrm{X}$ and ${ }_{8}^{18} \mathrm{X}$ in the sample?
The average atomic mass of a sample of an element $\mathrm{X}$ is $16.2 \mathrm{u}$. What are the percentages of isotopes ${ }_{8}^{16} \mathrm{X}$ and ${ }_{8}^{18} \mathrm{X}$ in the sample?
Solution:
Suppose percentage of isotope ${ }_{8}^{16} \mathrm{X}=\mathrm{x}$
Then percentage of isotope ${ }_{8}^{18} \mathrm{X}=(100-\mathrm{x})$
$\therefore$ Average atomic mass $=\frac{16 \times x+18(100-x)}{100}$
$=\frac{1800-2 x}{100}=18-0.02 x$
$\therefore 18-0.02 \mathrm{x}=16.2$ (Given)
or $0.02 \mathrm{x}=1.8$ or $\mathrm{x}=\frac{1.8}{0.02}=90$
Hence, percentage of ${ }_{8}^{16} \mathrm{X}=90 \%$
and $\quad$ percentage of ${ }_{8}^{18} \mathrm{X}=100-90=10 \%$
Suppose percentage of isotope ${ }_{8}^{16} \mathrm{X}=\mathrm{x}$
Then percentage of isotope ${ }_{8}^{18} \mathrm{X}=(100-\mathrm{x})$
$\therefore$ Average atomic mass $=\frac{16 \times x+18(100-x)}{100}$
$=\frac{1800-2 x}{100}=18-0.02 x$
$\therefore 18-0.02 \mathrm{x}=16.2$ (Given)
or $0.02 \mathrm{x}=1.8$ or $\mathrm{x}=\frac{1.8}{0.02}=90$
Hence, percentage of ${ }_{8}^{16} \mathrm{X}=90 \%$
and $\quad$ percentage of ${ }_{8}^{18} \mathrm{X}=100-90=10 \%$