An organic compound weighing 500 mg , produced 220 mg of $\mathrm{CO}_2$, on complete combustion. The percentage composition of carbon in the compound is _________ $\%$. (nearest integer)

(Given molar mass in $\mathrm{g} \mathrm{mol}^{-1}$ of $\mathrm{C}: 12, \mathrm{O}: 16$ )

<p><p><strong>Convert the mass of $\mathrm{CO}_2$ to moles:</strong></p> <p>The molar mass of $\mathrm{CO}_2$ is 44 g/mol. Thus, the number of moles of $\mathrm{CO}_2$ produced is calculated as follows:</p> <p>$ \mathrm{n}_{\mathrm{CO}_2} = \frac{220 \times 10^{-3} \, \text{g}}{44 \, \text{g/mol}} = 5 \times 10^{-3} \text{ moles} $</p></p> <p><p><strong>Calculate the mass of carbon in $\mathrm{CO}_2$:</strong></p> <p>Since each mole of $\mathrm{CO}_2$ contains one mole of carbon, the moles of carbon are the same: $5 \times 10^{-3}$ moles. The molar mass of carbon is 12 g/mol, so the mass of carbon is:</p> <p>$ \mathrm{m}_{\mathrm{C}} = 5 \times 10^{-3} \, \text{moles} \times 12 \, \text{g/mol} = 60 \times 10^{-3} \, \text{g} = 60 \, \text{mg} $</p></p> <p><p><strong>Calculate the percentage composition of carbon:</strong></p> <p>To find the percentage by mass of carbon in the compound, use the formula:</p> <p>$ \% \, \text{Carbon} = \left( \frac{60 \, \text{mg}}{500 \, \text{mg}} \right) \times 100 = 12\% $</p></p> <p>Thus, the percentage composition of carbon in the compound is approximately 12%.</p>