Question

In a Carius determination, $0.234g$ of an organic substance gave $0.334g$ of barium sulphate. Calculate the percentage of sulphur in the given compound. $[Ba=137,S=32,O=16]$

Answer 1

Mass of Organic Substance = 0.234 g

Mass of Barium Sulphate = 0.334 g

Molecular Mass of $BaS{O}_4$ = 233 $g{mol}^{-1}$

233 g of $BaS{O}_4$ contain 32 g of Sulphur

0.334 g of barium sulphate would contain $\frac{32}{233}\times 0.334$ = 0.0458 g

% of Sulphur = $\frac{0.0458}{0.234}\times 100$ = $19.6$%