Percentage of Sulphur Estimation by Carius Method

To determine the percentage of sulphur in the organic compound using the Carius method, we use the mass of the precipitate ($\text{BaSO}_4$) formed and the molar masses provided.

Given Data:

• Mass of the organic substance ($m$) = $0.2 \text{ g}$
• Mass of $\text{BaSO}_4$ formed ($m_{\text{BaSO}_4}$) = $0.6 \text{ g}$
• Molar mass of sulphur ($\text{S}$) = $32 \text{ g mol}^{-1}$
• Molar mass of barium sulphate ($\text{BaSO}_4$) = $231 \text{ g mol}^{-1}$

Formula: $\text{Percentage of Sulphur} = \left( \frac{\text{Molar mass of S}}{\text{Molar mass of BaSO}_4} \right) \times \left( \frac{\text{Mass of BaSO}_4}{\text{Mass of organic substance}} \right) \times 100$

Calculation: $\text{Percentage of Sulphur} = \left( \frac{32}{231} \right) \times \left( \frac{0.6}{0.2} \right) \times 100$

$\text{Percentage of Sulphur} = \frac{32}{231} \times 3 \times 100$

$\text{Percentage of Sulphur} = \frac{96}{231} \times 100 = \frac{32}{77} \times 100$

$\text{Percentage of Sulphur} \approx 41.56\%$

Rounding off to the nearest integer gives $42\%$ (or $41.56\%$).