In an estimation of sulphur by Carius method 0.2 g of the substance gave 0.6 g of BaSO$$_4$$. The percentage of sulphur in the substance is _______%. (Given molar mass in g mol$$^{-1}$$ S: 32, BaSO$$_4$$: 231)

In Carius method, the Sulphur $$(S)$$ is converted into Barium Sulphate $$(BaSO_{4})$$.

Given:

• Mass of substance $$= 0.2 g$$
• Mass of $$BaSO_{4}$$ formed $$= 0.6 g$$
• Molar mass of $$BaSO_{4}$$ $$=231$$
• Atomic mass of $$S = 32$$

Mass of sulphur in $$0.6 g$$ of $$BaSO_{4}$$​:

$$Mass\ of\ S=\frac{32}{231}\times\ 0.6=0.0831g$$

Percentage of Sulphur:

$$\ \%\ S=\frac{0.0831g}{0.2}\times\ 100$$

$$\ \%\ S=41.56\ \%\ $$

The percentage of sulphur in the substance is 41.56%