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 $\left.\mathrm{g} \mathrm{mol}^{-1}\right) \mathrm{C}: 12, \mathrm{H}: 1, \mathrm{O}: 16$ ]

<p>To determine the percentage of carbon in the organic compound, follow these steps:</p> <p><p><strong>Calculate Moles of CO₂:</strong></p> <p><p>Given: 1.46 g of CO₂ is produced.</p></p> <p><p>Molar mass of CO₂ = 44 g/mol.</p></p> <p><p>Moles of CO₂ = $\frac{1.46 \text{ g}}{44 \text{ g/mol}}$.</p></p></p> <p><p><strong>Determine Moles of Carbon Atoms:</strong></p> <p><p>Each molecule of CO₂ contains one atom of carbon.</p></p> <p><p>Thus, moles of carbon = moles of CO₂.</p></p></p> <p><p><strong>Calculate Mass of Carbon:</strong></p> <p><p>Molar mass of carbon (C) = 12 g/mol.</p></p> <p><p>Mass of carbon = Moles of carbon × 12 g/mol.</p></p> <p><p>Mass of carbon = $\frac{1.46}{44} \times 12$.</p></p></p> <p><p><strong>Calculate Percentage of Carbon:</strong></p> <p><p>Total mass of the organic compound = 0.5 g.</p></p> <p><p>Percentage of carbon = $\left(\frac{\text{Mass of carbon}}{0.5 \text{ g}}\right) \times 100\%$.</p></p> <p><p>Hence, Percentage of carbon = $\frac{1.46}{44} \times \frac{12}{0.5} \times 100$.</p></p> <p><p>Percentage of carbon = 79.63%.</p></p></p> <p>Therefore, the percentage of carbon in the compound is approximately 80%.</p>