Question

$0.1254$g of an organic compound gave $0.1292$g of barium sulphate in the estimation of sulphur by Carius method. Find out the percentage of sulphur in the given organic compound.

(Atomic mass $Ba=137;S=32;O=16$).

• A. 12.02
• B. 10.23
• C. 14.15
• D. 45.32

Answer 1

The correct option is C 14.15

Total mass of organic compound = 0.1254 g

Mass of barium sulphate formed = 0.1292 g

1 mole of barium sulphate = 233 g so 233 g of $BaS{O}_4$ contain 1 mole of sulphur weight 32 g

thus 0.1292 g barium sulphate contains

$\frac{0.1292\times 32}{233}=0.01774$ g of sulphur

therefore percentage of sulphur is $\frac{0.01774\times 100}{0.1254}=14.15$