Percentage of Sulphur (Carius Method)

From the reaction sequence given in Q-17, the molecular formula of compound $$M$$ is already known (see the previous answer) and it contains exactly one sulphur atom per molecule. Its molar mass is

$$M_M = 379\;\text{g mol}^{-1}$$

In the Carius method the entire sulphur present in the sample is oxidised to sulphate ion and is quantitatively precipitated as $$\mathrm{BaSO_4}$$. For every mole of sulphur, one mole of $$\mathrm{BaSO_4}$$ is obtained:

$$\text{S (in sample)} \;\longrightarrow\; \mathrm{SO_4^{2-}} \;\longrightarrow\; \mathrm{BaSO_4}$$

1. Moles of $$M$$ taken:

$$n_M = \frac{3.79}{379} = 0.010\;\text{mol}$$

2. Since each molecule of $$M$$ contains one sulphur atom, the moles of sulphur are also 0.010 mol.

3. Hence, moles of $$\mathrm{BaSO_4}$$ formed:

$$n_{\mathrm{BaSO_4}} = 0.010\;\text{mol}$$

4. Mass of $$\mathrm{BaSO_4}$$ precipitated (molar mass $$= 137+32+64 = 233\;\text{g mol}^{-1}$$):

$$m_{\mathrm{BaSO_4}} = n_{\mathrm{BaSO_4}}\times 233 = 0.010 \times 233 = 2.33\;\text{g}$$

Therefore, from 3.79 g of compound $$M$$ the mass of $$\mathrm{BaSO_4}$$ obtained is

2.33 g.