Question

0.48 g of an organic compound on complete combustion produced 0.22 g of CO₂ . The percentage of C in the given organic compound is :

• A. 25
• B. 50
• C. 12.5
• D. 87.5

Answer 1

The Correct Option is C

The combustion of an organic compound produces carbon dioxide (CO₂) and water (H₂O). The mass of carbon in the compound can be calculated from the mass of CO₂ produced.

Molecular weight of CO₂: The molecular weight of CO₂ is 44 g/mol.

Mass of CO₂ produced: Given as 0.22 g.

Moles of CO₂ produced:

\(\text{Moles of CO₂} = \frac{\text{Mass of CO₂}}{\text{Molar mass of CO₂}} = \frac{0.22 \, \text{g}}{44 \, \text{g/mol}} = 0.005\ mol\)

Carbon in CO₂: Each mole of CO₂ contains 1 mole of carbon. Therefore, moles of carbon = moles of CO₂ = 0.005 mol.

Mass of carbon in the compound:

\(\text{Mass of carbon} = \text{moles of carbon} \times \text{molar mass of carbon} = 0.005 \, \text{mol} \times 12 \, \text{g/mol} = 0.06 \, \text{g}\)

Percentage of carbon in the organic compound:

\(\text{Percentage of C} = \left( \frac{\text{Mass of carbon}}{\text{Mass of compound}} \right) \times 100 = \left( \frac{0.06 \, \text{g}}{0.48 \, \text{g}} \right) \times 100 = 12.5\%\)

Thus, the percentage of carbon in the given organic compound is 12.5%, making option (C) the correct answer.

Answer 2

To calculate the percentage of C in the compound, we need to determine the moles of CO₂ produced.

Moles of CO₂ = Mass of CO₂ / Molar mass of CO₂

Moles of CO₂ = 0.22 g / 44 g/mol = 0.005 moles of CO₂

Each mole of CO₂ corresponds to 1 mole of carbon atoms, so the mass of carbon in the sample is:

Mass of carbon = 0.005 moles × 12 g/mol = 0.06 g

Now, calculate the percentage of carbon:

Percentage of C=(Mass of CTotal mass of compound)×100=(0.06 g0.48 g)×100=12.5%

\[\text{Percentage of C} = \left(\frac{\text{Mass of C}}{\text{Total mass of compound}}\right) \times 100 = \left(\frac{0.06 \, \text{g}}{0.48 \, \text{g}}\right) \times 100 = 12.5\%\]