JEE MAIN - Chemistry (2025 - 3rd April Morning Shift - No. 23)
0.5 g of an organic compound on combustion gave 1.46 g of $\mathrm{CO}_2$ and 0.9 g of $\mathrm{H}_2 \mathrm{O}$. The percentage of carbon in the compound is _______________. (Nearest integer)
[Given : Molar mass (in $\left.\mathrm{g} \mathrm{mol}^{-1}\right) \mathrm{C}: 12, \mathrm{H}: 1, \mathrm{O}: 16$ ]
Explanation
To determine the percentage of carbon in the organic compound, follow these steps:
Calculate Moles of CO₂:
Given: 1.46 g of CO₂ is produced.
Molar mass of CO₂ = 44 g/mol.
Moles of CO₂ = $\frac{1.46 \text{ g}}{44 \text{ g/mol}}$.
Determine Moles of Carbon Atoms:
Each molecule of CO₂ contains one atom of carbon.
Thus, moles of carbon = moles of CO₂.
Calculate Mass of Carbon:
Molar mass of carbon (C) = 12 g/mol.
Mass of carbon = Moles of carbon × 12 g/mol.
Mass of carbon = $\frac{1.46}{44} \times 12$.
Calculate Percentage of Carbon:
Total mass of the organic compound = 0.5 g.
Percentage of carbon = $\left(\frac{\text{Mass of carbon}}{0.5 \text{ g}}\right) \times 100\%$.
Hence, Percentage of carbon = $\frac{1.46}{44} \times \frac{12}{0.5} \times 100$.
Percentage of carbon = 79.63%.
Therefore, the percentage of carbon in the compound is approximately 80%.
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