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

In the sulphur by Carius method, 0.468 g of an organic sulphur compoun

1.	In the sulphur by Carius method, 0.468 g of an organic sulphur compound afforded 0.668 g of barium sulphate. Find the percentage of sulphur in the given compound.
Answer» Solution :Here, the MASS of the SUBSTANCE taken = 0.468 g Mass of `BaSO_(4)` FORMED = 0.668 g Now 1 mole of `BaSO_(4) -= 1g` atom of S or `(137 + 32 + 4 xx 16) = 233 g` of `BaSO_(4) -= 32 g` of S Applying the relation, Percentage of SULPHUR `= (32)/(233) xx ("Mass of " BaSO_(4) " formed")/("Mass of substance taken") xx 100` `= (32)/(233) xx (0.668)/(0.468) xx 100 = 19.60`.