# 12.35.    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.

Given data,

Total mass of the OC = 0.468 g
Mass of barium sulphate formed = 0.668 g

We know that,

1 mol of BaSO4 = 233 g of $BaSO_{4}$ = 32 g of sulphur
Thus, 0.668 g of $BaSO_{4}$ contains = $\frac{32\times 0.668}{233}$ g of sulphur = 0.0917 g of sulphur

Therefore, the percentage(%) of sulphur
$= \frac{0.0197}{0.468}\times 100$
= 19.59 %

Hence, the percentage of sulphur in the given compound is 19.59 %.

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