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%.