Question

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

Answer 1

Molecular mass of $BaSO _{4}=233\, g$
$\because 233 \,BaSO _{4}$ contain $\rightarrow 32\ g$ sulphur
$\therefore 1.44\, g\, BaSO _{4}$ contain $\rightarrow \frac{32}{233} \times 1.44\, g$ sulphur
given : $0.471\, g$ of organic compound
$\%$ of $S =\frac{32 \times 1.44}{233 \times 0.471} \times 100$
$=41.98 \% \approx 42 \%$
OR

$\Rightarrow n _{ s }= n _{ BaSO _{4}}=\frac{1.44}{233}$
$\Rightarrow W _{ s }=\frac{1.44}{233} \times 32\, g$
therefore $\% S =\frac{ W _{ s }}{ W _{\text {o.C. }}} \times 100$
$=\frac{1.44 \times 32}{233 \times 0.471} \times 100$
$=\frac{46.08}{109.743} \times 100$
$=41.98 \simeq 42$