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 

Mass of barium sulphate formed = 0.668 g 

1 mol of BaSO₄ = 233 g of BaSO₄ contains 32 g of Sulphur 

Thus, 0.668 g of BaSO₄ solution contains 

= \(\frac{32\times0.668}{233}\) g of Sulphur 

= 0.0917 g of Sulphur 

Therefore, Percentage of Sulphur, 

= \(\frac{0.0197}{0.468}\) × 100 

= 19.59% 

Hence, the percentage of Sulphur in the given compound is 19.59%