Mass of pure organic compound = 0.20 g

Mass of barium sulphate obtained = 0.40 g

% of sulphur in the compound = ?

Molar mass of $$BaS{O_4} = 134\,g\,mo{l^{ - 1}} + 32\,g\,mo{l^{ - 1}} + (16 \times 4)\,g\,mo{l^{ - 1}} = 233\,g\,mo{l^{ - 1}}$$

Moles of $$BaS{O_4}:$$

Moles $$ = {{mass} \over {molar\,mass}}$$

$$ = {{0.40\,g} \over {233\,g\,mo{l^{ - 1}}}} = 0.001717\,mol$$

Moles of sulphur:

From the formula $$BaS{O_4}$$, 1 mole of $$BaS{O_4}$$ contains 1 mole of sulphur (s). Therefore, the moles of sulphur in the sample is equal to the moles of $$BaS{O_4}$$.

Moles of S = 0.001717 mol

Mass of sulphur :

Mass = moles $\times$ molar mass

Moles of S = 0.001717 mol, substitute this value as

Mass of S = 0.001717 mol $\times$ 32 g mol$^{-1}$.

= 0.054944 g

Percentage of sulphur in the organic compound:

The formula is

Percentage of $$S = {{mass\,of\,S} \over {mass\,of\,organic\,compound}} \times 100$$

Substituting the values,

Percentage of $$S = {{0.054944\,g} \over {0.20\,g}} \times 100$$

$$ = 0.27472 \times 100$$

$$ = 27.472\% $$

$$ = 274.72 \times {10^{ - 1}}\% $$

$$ = 275 \times {10^{ - 1}}\% $$