An organic compound is heated with ${\mathrm{Na}}_2{\mathrm{O}}_2$ and then treated with conc. ${\mathrm{H}}_2{\mathrm{SO}}_4$ followed by addition of ${\mathrm{BaCl}}_2$.  0.233 grams of ${\mathrm{BaSO}}_4$ is formed from 0.32 gram of organic compound. Percentage of sulphur in the compound is

Molar mass of BaSO₄ is 233 g/mol and mass of BaSO₄ is 0.233 g. 32 g of sulphur present in one mole of BaSO₄ (or 233 g of BaSO₄). Mass of sulphur present in 0.233 g of BaSO₄

$$\begin{gathered} \mathrm{Mass} \mathrm{of} \mathrm{sulphur}=\frac{32 \mathrm{g}}{233 \mathrm{g}}\times 0.233 \mathrm{g}=0.032 \mathrm{g} \\ \% \mathrm{of} \mathrm{sulphur}=\frac{\mathrm{Mass} \mathrm{of} \mathrm{sulphur}}{\mathrm{Mass} \mathrm{of} \mathrm{organic} \mathrm{compound}}\times 100 \\ \% \mathrm{of} \mathrm{sulphur}=\frac{0.032 \mathrm{g}}{0.32 \mathrm{g}}\times 100=10\% \end{gathered}$$