Question

In the sulphur estimation, $0.471g$ of an organic compound gave $1.44g$ of barium sulphate. The percentage of sulphur in the compound is ___ $\%$. (Nearest integer)
(Atomic Mass of $Ba=137u$ )

Answer 1

Answer: 42

Molecular mass of $BaS{O}_4=233g$
$\because 233BaS{O}_4$ contain $\to 32g$ sulphur
$\therefore 1.44gBaS{O}_4$ contain $\to \frac{32}{233}\times 1.44g$ sulphur
given : $0.471g$ of organic compound
$\%$ of $S=\frac{32\times 1.44}{233\times 0.471}\times 100$
$=41.98\%\approx 42\%$
OR

$\Rightarrow n_s=n_{BaS{O}_4}=\frac{1.44}{233}$
$\Rightarrow W_s=\frac{1.44}{233}\times 32g$
therefore $\%S=\frac{W_s}{W_{\text{o.C. }}}\times 100$
$=\frac{1.44\times 32}{233\times 0.471}\times 100$
$=\frac{46.08}{109.743}\times 100$
$=41.98\simeq 42$