Given, Mass of organic compound = $0.2175 \text{ g}$ Mass of barium sulfate ($\mathrm{BaSO}_{4}$) = $0.5825 \text{ g}$ Molar mass of $\mathrm{BaSO}_{4} = 137.33 + 32 + (4 \times 16) = 233.33 \text{ g/mol}$ We know that $233.33 \text{ g}$ of $\mathrm{BaSO}_{4}$ contains $32 \text{ g}$ of sulfur. So, $0.5825 \text{ g}$ of $\mathrm{BaSO}_{4}$ contains: $$ \begin{aligned} \text{Mass of S} &= \frac{32}{233.33} \times 0.5825 \text{ g} \\ &= 0.07987 \text{ g} \end{aligned} $$ Percentage of sulfur in the organic compound: $$ \begin{aligned} \% S &= \frac{\text{Mass of S}}{\text{Mass of organic compound}} \times 100 \\ &= \frac{0.07987}{0.2175} \times 100 \\ &= 36.72 \% \end{aligned} $$ Thus, the percentage of sulfur is $36.72 \%$.