In Carius method, 0.75 g of an organic compound gave 1.2 g of barium sulphate, find percentage of sulphur (molar mass 32 g $$mol^{-1}$$ ) Molar mass of barium sulphate is 233 g $$mol^{-1}$$

This problem uses the Carius method for estimating sulphur in an organic compound. In this method, the organic compound is heated with fuming nitric acid in a sealed tube, which oxidises sulphur to sulphate ions. These sulphate ions are then precipitated as barium sulphate ($$BaSO_4$$) by adding barium chloride solution.

Mass of organic compound = 0.75 g

Mass of $$BaSO_4$$ obtained = 1.2 g

Molar mass of S = 32 g/mol

Molar mass of $$BaSO_4$$ = 233 g/mol

Using the formula: moles = mass / molar mass

$$\text{Moles of } BaSO_4 = \frac{1.2}{233} = 0.005150 \text{ mol}$$

Since each molecule of $$BaSO_4$$ contains exactly one sulphur atom, the stoichiometric relationship is:

$$\text{Moles of S} = \text{Moles of } BaSO_4 = 0.005150 \text{ mol}$$

$$\text{Mass of S} = \text{Moles of S} \times \text{Molar mass of S} = 0.005150 \times 32 = 0.16481 \text{ g}$$

$$\% S = \frac{\text{Mass of S}}{\text{Mass of compound}} \times 100 = \frac{0.16481}{0.75} \times 100$$

$$= 21.97\%$$

The correct answer is Option 4: 21.97%.