Mole ConceptHard

Question

In Carius method, 0.75 g of an organic compound gave 1.2 g of barium sulphate, find percentage of sulphur (molar mass $32\text{ }g{\text{ }mol}^{- 1}$ ). Molar mass of barium sulphate is $233\text{ }g{\text{ }mol}^{- 1}$.

Options

A.$4.55\%$
B.$10.30\%$
C.$21.97\%$
D.$16.48\%$

Solution

$\frac{n_{{BaSO}_{4}} \times 32}{{\text{ }W}_{\text{(unknown comp.)~}}} \times 100$

$$= \frac{1.2 \times 32}{233} \times \frac{100}{0.75} = 21.97\% $$

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