An organic compound weighing 500 mg, produced 220 mg of $$CO_2$$ on complete combustion. The percentage composition of carbon in the compound is .......... %. (nearest integer)
(Given molar mass in g $$mol^{-1}$$ of C : 12, O : 16)

Step 1: Find the mass of carbon present in $$220 \text{ mg}$$ of $$CO_2$$.
Molar mass: $$CO_2$$ has $$12 \text{ g}$$ of C in $$44 \text{ g}$$ of $$CO_2$$.
Hence the mass fraction of C in $$CO_2$$ is $$\frac{12}{44} = \frac{3}{11}$$.

Mass of C obtained from combustion = $$220 \text{ mg} \times \frac{3}{11}$$.
Calculate: $$\frac{220}{11} = 20$$, so $$20 \times 3 = 60 \text{ mg}$$.

Step 2: Calculate percentage of carbon in the original compound.
Original sample mass = $$500 \text{ mg}$$.
Percentage of C = $$\frac{60 \text{ mg}}{500 \text{ mg}} \times 100$$.

Simplify: $$\frac{60}{500} = 0.12$$, therefore percentage of C = $$0.12 \times 100 = 12\%$$.

Nearest integer value = $$12$$.