Here, mass of the substance taken = 0.157 g
Mas of `BaSO_(4)` ppt. formed = 0.4813 g
Now, 1 mole of `BaSO_(4) -= 1 g` atom of S
or (137 + 32 + 64) = 233 g of `BaSO_(4) -= 32 g` of S
i.e., 233 g of `BaSO_(4)` contain sulphur = 32 g
`:.` 0.4813 g of `BaSO_(4)` will contain sulphur `= (32)/(233) xx 0.4813 g`
`:.` % age of sulphur in the compound `= (32)/(233) xx (0.4813)/(0.157) xx 100 = 42.10`