A container with 100 gram oleum is labeled as 109% by mass. % of ${\mathrm{SO}}_3$ remaining in container is ____?

$\text{ Oleum is a mixture of }{\mathrm{H}}_2{\mathrm{SO}}_4\text{ and }{\mathrm{SO}}_3$

Let mass of oleum be 100 gram

$\text{ Let mass of }S{O}_3={{}^{\text{'}}x}^{\text{'}}g$

$\text{ Then mass of }{\mathrm{H}}_2{\mathrm{SO}}_4=(100-x)g$

$n_{{\mathrm{SO}}_3}=\frac{\text{ Weight of }{\mathrm{SO}}_3}{\text{ gram molecular weight of }{\mathrm{SO}}_3}=\frac{x}{80}$

$$\begin{gathered} {\mathrm{SO}}_3+{\mathrm{H}}_2\mathrm{O}\to {\mathrm{H}}_2{\mathrm{SO}}_4 \\ n_{{\mathrm{SO}}_3}=n_{{\mathrm{H}}_2{\mathrm{SO}}_4} \end{gathered}$$

$\therefore \text{ mass of }{\mathrm{H}}_2{\mathrm{SO}}_4\text{ formed }=\frac{98x}{80}$

$\text{ mass of }{\mathrm{H}}_2{\mathrm{SO}}_4\text{ already present + mass of }{\mathrm{H}}_2{\mathrm{SO}}_4\text{ formed }=\text{ Total mass of oleum }$

$(100-x)+\frac{98x}{80}=109$

$$\begin{gathered} \frac{98x}{80}-x=109-100 \\ x=40 \mathrm{g} \end{gathered}$$

$\therefore 40\%\text{ of }S{O}_3\text{ is remaining }$