Question

In the sulphur estimation, 0.20 g of a pure organic compound gave 0.40 g of barium sulphate. The percentage of sulphur in the compound is

Hint:

To find the percentage of an element in a compound, divide the mass of the element by the mass of the sample and multiply by 100.

Answer 1

Correct Answer: 27.5

Given that the molar mass of sulphur (S) is 32 g/mol and that of barium sulphate (BaSO\(_4\)) is 233 g/mol, the mass of sulphur in the compound can be calculated from the mass of barium sulphate produced:
The moles of barium sulphate formed: \[ \text{Moles of BaSO}_4 = \frac{0.40 \, \text{g}}{233 \, \text{g/mol}} = 0.00172 \, \text{mol} \]
The moles of sulphur in the compound are equal to the moles of BaSO\(_4\) because of the 1:1 stoichiometry of BaSO\(_4\) and sulphur.
The mass of sulphur is: \[ \text{Mass of S} = 0.00172 \, \text{mol} \times 32 \, \text{g/mol} = 0.05504 \, \text{g} \]
The percentage of sulphur in the compound is: \[ % \text{S} = \frac{0.05504 \, \text{g}}{0.20 \, \text{g}} \times 100 = 27.5% \] Thus, the correct percentage of sulphur is \( \boxed{27.5%} \).