In a Carius estimation, 0.28 g of a sulphur-containing organic compound gave 0.466 g of BaSO4. Using Ba = 137, S = 32, O = 16, the percentage of sulphur is

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In a Carius estimation, 0.28 g of a sulphur-containing organic compound gave 0.466 g of $BaSO X_{4}$ . Using Ba = 137, S = 32, O = 16, the percentage of sulphur is

Molar mass of

BaSO X_{4}

= 137 + 32 + 64 = 233 and each mole holds 32 g of S. Mass of S =

\frac{32}{233} \times 0.466 = 0.06400

g. %S =

\frac{0.06400}{0.28} \times 100 = 22.86%