During " S " estimation, 160 mg of an organic compound gives 466 mg of barium sulphate. The percentage of Sulphur in the given compound is _______ %. (Given molar mass in $$gmol^{-1}$$ of Ba : 137, S : 32, O : 16)

In the estimation of sulphur, the organic compound is converted to $$BaSO_4$$.

The formula for percentage of sulphur is $$\% S = \frac{32}{233} \times \frac{\text{Mass of } BaSO_4}{\text{Mass of compound}} \times 100$$.

Here the molar mass of $$BaSO_4$$ is $$137 + 32 + 4(16) = 233$$ g/mol.

For a compound mass of 160 mg and a $$BaSO_4$$ precipitate of 466 mg, substitution into the formula gives $$\% S = \frac{32}{233} \times \frac{466}{160} \times 100$$.

This can be rewritten as $$= \frac{32 \times 466}{233 \times 160} \times 100$$ and then as $$= \frac{14912}{37280} \times 100$$.

Since $$\frac{466}{233} = 2$$, the calculation simplifies to $$\% S = \frac{32 \times 2}{160} \times 100 = \frac{64}{160} \times 100 = 40\%$$.

The answer is 40%.