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

Question

Accepted Answer

D. 1. **Calculate the moles of BaSO$_4$ formed.** Given
$$
n(\mathrm{BaSO}_4) = \frac{1.2}{233}\ \mathrm{mol}
$$

2. **Determine the moles (and mass) of sulphur.** Since there's 1 S atom per formula unit,
$$
n(\mathrm{S}) = \frac{1.2}{233}\ \mathrm{mol} \quad \Rightarrow \quad m(\mathrm{S}) = \frac{1.2}{233}	imes 32 = 0.1648\ \mathrm{g}
$$

3. **Calculate the percentage of sulphur** in the 0.75 g compound sample:
$$
\%\mathrm{S} = \frac{0.1648}{0.75}	imes 100 = 21.97\%
$$

**Hence, the answer is option D.**

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

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