Question

$0.32$ g of an aorganic compound gave $0.233$ g of $BaS{O}_4$. Determine the percentage of sulphur in the compound.
(At. mass of $Ba=137,S=32,O=16$)

• A. $1.0$
• B. $10.0$
• C. $23.5$
• D. $32.4$

Answer 1

The correct option is C $23.5$

Molar mass of $BaS{O}_4$ = 0.233

Suppose the mass of org. com. Is = m grams

Let the mass of barium sulphate formed = 0.233

32 g of S is present in 1 mol of $BaS{O}_4$

Therefore, 233 g of $BaS{O}_4$ contain 32g of $S$

= _0.233g of $BaS{O}_4$ contain (32x 0.233x100)/32 g of sulphur

% of $S$ = ( 32x 0.233x100)/(32) = 23.3%