Question

In the estimation of sulphur by Carius method, 0.468 g of an organic sulphur compound afforded 0.668 g of barium sulphate. Find out the percentage of sulphur in the given compound.

Answer 1

Total mass of organic compound = 0.468 g [Given]
Mass of barium sulphate formed = 0.668 g [Given]
1 mol of BaSO4 = 233 g of $BaS{O}_4=32$ g of sulphur
Thus, 0.668 g of $BaS{O}_4$ contains $\frac{32\times 0.668}{233}g$ of sulphur = 0.0917 g of sulphur
Therefore, percentage of sulphur $=\frac{0.0197}{0.468}\times 100=19.59\%$
Hence, the percentage of sulphur in the given compound is 19.59 %.