# In the estimation of sulphur by Carius method, 0.468 g of an organic sulphur compound afforded 0.668 g of barium sulphate.

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.

Solution:

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 BaSO4 = 32 g of sulphur

Thus, $0.668 \mathrm{~g}$ of $\mathrm{BaSO}_{4}$ contains $\frac{32 \times 0.668}{233} \mathrm{~g}$ of sulphur $=0.0917 \mathrm{~g}$ 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 $19.59 \%$.