JEE Chemistry 2025 Question with Solution

Given: Mass of the organic compound = $0.20 \, \text{g}$, mass of barium sulphate formed = $0.40 \, \text{g}$.

Find: Percentage of sulphur in the compound.

All the sulphur in the organic compound is converted into $\text{BaSO}_4$. Since $1$ mole of $\text{BaSO}_4$ contains $1$ mole of sulphur, we use the mass relation between sulphur and barium sulphate.

Molar mass of $\text{BaSO}_4$ is

$$137 + 32 + 64 = 233 \, \text{g mol}^{-1}$$

Thus, $233 \, \text{g}$ of $\text{BaSO}_4$ contain $32 \, \text{g}$ of sulphur.

So, mass of sulphur in $0.40 \, \text{g}$ of $\text{BaSO}_4$ is

$$\text{Mass of S} = \frac{32}{233} \times 0.40 = 0.0549 \, \text{g}$$

Now percentage of sulphur in the compound is

$$\%\,\text{S} = \frac{0.0549}{0.20} \times 100 = 27.45 \approx 27.5$$

Therefore, the percentage of sulphur in the compound is $27.5\%$.

Given: Mass of sample = $0.20 \, \text{g}$, mass of $\text{BaSO}_4$ formed = $0.40 \, \text{g}$.

Find: Percentage of sulphur in the compound.

First calculate the moles of $\text{BaSO}_4$ formed:

$$\text{Moles of BaSO}_4 = \frac{0.40}{233} = 0.00172 \, \text{mol}$$

Because the stoichiometry between sulphur and $\text{BaSO}_4$ is $1:1$, moles of sulphur are also $0.00172 \, \text{mol}$.

Mass of sulphur is

$$\text{Mass of S} = 0.00172 \times 32 = 0.05504 \, \text{g}$$

Now calculate the percentage of sulphur:

$$\%\,\text{S} = \frac{0.05504}{0.20} \times 100 = 27.5$$

Therefore, the required numerical value is $27.5$.