Question

Compound (X) has molar mass \(76\). \(2 \times 10^{-3}\) mole of X gives \(0.4813\,g\) \(BaSO_4\) as precipitate. What is the % of S in compound (X)? Mention answer in \(10^{-1}\) form.

• A. 215
• B. 325
• C. 435
• D. 515

Answer 1

The Correct Option is C

To find the percentage of sulfur (\(S\)) in compound \(X\), we need to determine how much \(S\) is present in the given amount of compound, and then relate that to the molar mass of \(X\).

We are provided with the following information:

• Molar mass of compound \(X\) is \(76 \, \text{g/mol}\)
• \(2 \times 10^{-3} \, \text{mol}\) of \(X\) yields a \(BaSO_4\) precipitate weighing \(0.4813 \, \text{g}\)

First, calculate the amount of \(S\) in the \(BaSO_4\) precipitate:

• The molar mass of \(BaSO_4\) is approximately \(233 \, \text{g/mol}\) (Barium \(137\), Sulfur \(32\), Oxygen \(16 \times 4\)).

To find the moles of \(BaSO_4\):

\[ \text{Moles of } BaSO_4 = \frac{0.4813 \, \text{g}}{233 \, \text{g/mol}} = 0.002065 \, \text{mol} \]

Since 1 mole of \(BaSO_4\) contains 1 mole of \(S\), the moles of \(S\) are also \(0.002065 \, \text{mol}\).

Then, calculate the mass of \(S\) in these moles:

\[ \text{Mass of } S = 0.002065 \, \text{mol} \times 32 \, \text{g/mol} = 0.06608 \, \text{g} \]

To find the percentage of sulfur in compound \(X\):

\[ \text{Percentage of } S = \left( \frac{\text{Mass of } S}{\text{Mass of sample}} \right) \times 100 \]

\[ \text{Mass of sample of } X = 2 \times 10^{-3} \, \text{mol} \times 76 \, \text{g/mol} = 0.152 \, \text{g} \]

\[ \text{Percentage of } S = \left( \frac{0.06608}{0.152} \right) \times 100 = 43.45\% \]

Thus, the percentage of sulfur \(S\) in compound \(X\) is approximately \(43.5\%\) which in \(10^{-1}\) form is \(435\).