Question

The average atomic mass of a sample of an element $X$ is 16.2u. What are the percentages of isotopes $^{16}X_8$ and $^{18}X_8$ in the sample?

Answer 1

It is given that the average atomic mass of the sample of element X is 16.2 u.

Let the percentage of isotope $^{18}X_8$ be $y$ %. Thus, the percentage of isotope $^{16}X_8$ will be $(100 - y)$ %.

Therefore,

$18 \times \frac{y}{100}$ + $16 \times \frac{(100 -y) }{ 100}$ = $16.2$

$\frac{18y}{100}$ + $\frac{16(100-y)}{100 }$= $16.2$

$18y$ + $\frac{1600 -16y}{100}$ = $16.2$

$18y$ + $1600 - 16y$ = $1620$
$2y$+ $1600$ = $1620$
$2y$ = $1620 - 1600$
$y$= $10$

Therefore, the percentage of isotope $^{18}X_8$ is 10%.
And, the percentage of isotope $^{18}X_8$ is (100 - 10) % = 90%.