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

We have an organic compound whose mass is given as $$0.471\ \text{g}$$. During sulphur estimation by the classical Carius method, all the sulphur present in the compound is converted into barium sulfate, $$\text{BaSO}_{4}$$. The experiment yields $$1.44\ \text{g}$$ of $$\text{BaSO}_{4}$$.

First, we need to know how much sulphur is present in this $$\text{BaSO}_{4}$$. For that we recall the molar mass (also called molecular weight) of barium sulfate:

$$\text{BaSO}_{4}: \quad M = M_{\text{Ba}} + M_{\text{S}} + 4\,M_{\text{O}}$$

Substituting the atomic masses, we get

$$M = 137\ \text{u} + 32\ \text{u} + 4 \times 16\ \text{u} = 137 + 32 + 64 = 233\ \text{u}$$

This means that in every $$233\ \text{g}$$ of $$\text{BaSO}_{4}$$, there are exactly $$32\ \text{g}$$ of sulphur. Hence the mass ratio of sulphur to barium sulfate is

$$\frac{m_{\text{S}}}{m_{\text{BaSO}_{4}}} = \frac{32}{233}$$

Using this ratio, the actual mass of sulphur in our sample of $$\text{BaSO}_{4}$$ is

$$m_{\text{S}} = 1.44\ \text{g} \times \frac{32}{233}$$

Carrying out the multiplication step by step,

$$1.44 \times 32 = 46.08$$

and now dividing by $$233$$,

$$m_{\text{S}} = \frac{46.08}{233}\ \text{g} \approx 0.198\ \text{g}$$

So, $$0.198\ \text{g}$$ of sulphur is present in the original $$0.471\ \text{g}$$ sample of the organic compound.

The percentage of sulphur is obtained from the general formula

$$\%\text{S} = \frac{\text{mass of S in the sample}}{\text{mass of the organic compound}} \times 100$$

Substituting the values just found,

$$\%\text{S} = \frac{0.198}{0.471} \times 100$$

Performing the division first,

$$\frac{0.198}{0.471} \approx 0.420$$

and then multiplying by $$100$$,

$$\%\text{S} \approx 42.0\%$$

Rounding to the nearest integer as required, we obtain $$42\%$$.

So, the answer is $$42\%$$.