Question:
During "S" estimation, 160 mg of an organic compound gives 466 mg of barium sulphate. The percentage of Sulphur in the given compound is %.
(Given molar mass in g mol\(^{-1}\) of Ba: 137, S: 32, O: 16)
During "S" estimation, 160 mg of an organic compound gives 466 mg of barium sulphate. The percentage of Sulphur in the given compound is %.
(Given molar mass in g mol\(^{-1}\) of Ba: 137, S: 32, O: 16)
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In stoichiometry problems, always ensure to use the correct molar mass to calculate moles and then find the percentage of the element based on its molecular composition.
Updated On: Mar 17, 2025
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Solution and Explanation
We are given that 160 mg of an organic compound gives 466 mg of barium sulphate (BaSO4). First, calculate the moles of BaSO4 using its molar mass: \[ \text{Millimoles of BaSO4} = \frac{466}{233} = 2 \, \text{mol} \] Next, calculate the amount of Sulphur (S) in the sample. Since the molar mass of BaSO4 consists of 32 g of Sulphur (S) per 233 g of BaSO4, the percentage of Sulphur is: \[ \%S = \frac{466}{233} \times 32 \times \frac{100}{160} = 40\% \] Thus, the percentage of Sulphur in the given compound is 40%.
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