In a Carius determination, 0.15g of an organic substance gave 0.34g of barium sulfate. Calculate the percentage of sulphur in the given compound. [Ba=137,S=32,O=16]

Options

A. 19.6%
B. 31.1%
C. 18.4%
D. 10.9%

Explanation

To solve the problem of calculating the percentage of sulfur in the given organic compound based on the Carius determination, we can follow these steps:

### Step 1: Identify the mass of the organic substance and barium sulfate
- Mass of organic substance = 0.15 g
- Mass of barium sulfate (BaSO₄) = 0.34 g

### Step 2: Calculate the molar mass of barium sulfate (BaSO₄)
- The molar mass of barium sulfate can be calculated using the atomic masses:
- Ba = 137 g/mol
- S = 32 g/mol
- O = 16 g/mol (there are 4 oxygen atoms in sulfate)

Molar mass of BaSO₄ = Ba + S + 4(O)
= 137 + 32 + 4(16)
= 137 + 32 + 64
= 233 g/mol

### Step 3: Determine the mass of sulfur in the barium sulfate
- From the molar mass, we know that in 233 g of BaSO₄, there are 32 g of sulfur.
- To find the mass of sulfur in 0.34 g of BaSO₄, we can set up a proportion:

\[ \text{Mass of S} = \left(\frac{32 \text{ g S}}{233 \text{ g BaSO₄}}\right) \times 0.34 \text{ g BaSO₄} \]

Calculating this gives:

\[ \text{Mass of S} = \frac{32}{233} \times 0.34 \approx 0.0467 \text{ g} \]

### Step 4: Calculate the percentage of sulfur in the organic compound
- The percentage of sulfur in the organic compound can be calculated using the formula:

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

Substituting the values gives:

\[ \text{Percentage of S} = \left(\frac{0.0467 \text{ g}}{0.15 \text{ g}}\right) \times 100 \approx 31.13\% \]

### Final Answer
The percentage of sulfur in the given organic compound is approximately **31.1%**.

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