Question

An oleum sample was labbled as 101.8 oleum. What is the percentage of free $S{O}_3$ in this sample?

Answer 1

$101.8\%$ oleum means that ${100}_g$ of this oleum when reacts with water, $101.8g{H}_{2}S{O}_4$ is formed by following reaction.

$H_{2}S{O}_4+S{O}_3+H_{2}O\to 2H_{2}S{O}_4$

$\Rightarrow$ mole of free $S{O}_3$ = moles of $H_{2}O$ added in the sample

$\because$ moles of $H_{2}O$ added = $\frac{101.8-100}{18}=\frac{1.8}{18}=0.1$

$\therefore$ mass of $S{O}_3$ in 100g oleum = $0.1\times 80=8g$

$\therefore \%$ of free $S{O}_3=\frac{8}{10}\times 100\%=8\%$