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

Utilizing the molar mass of sulfur (S) at 32 g/mol and barium sulfate (BaSO\(_4\)) at 233 g/mol, the mass of sulfur present in the compound is determined from the mass of barium sulfate generated. The calculation proceeds as follows: The moles of barium sulfate formed are calculated: \[ \text{Moles of BaSO}_4 = \frac{0.40 \, \text{g}}{233 \, \text{g/mol}} = 0.00172 \, \text{mol} \] Due to the 1:1 stoichiometry between BaSO\(_4\) and sulfur, the moles of sulfur in the compound are equivalent to the moles of BaSO\(_4\). The mass of sulfur is then determined: \[ \text{Mass of S} = 0.00172 \, \text{mol} \times 32 \, \text{g/mol} = 0.05504 \, \text{g} \] The percentage of sulfur in the compound is computed: \[ % \text{S} = \frac{0.05504 \, \text{g}}{0.20 \, \text{g}} \times 100 = 27.5% \] Consequently, the accurate percentage of sulfur is \(27.5\%\).