0.5 g of an organic compound on combustion gave 1.46 g of $$CO_2$$ and 0.9 g of $$H_2O$$. The percentage of carbon in the compound is _________. (Nearest integer)
[Given : Molar mass (in g mol$$^{-1}$$) C : 12, H : 1, O : 16]

The percentage of an element in an unknown organic compound can be obtained from its combustion data by first calculating the mass of that element present in the measured amounts of $$CO_2$$ and $$H_2O$$, then relating that mass to the original mass of the compound.

Mass of organic compound taken: $$0.5 \text{ g}$$.

Step 1: Find mass of carbon present
• Every mole of $$CO_2$$ contains one mole of carbon.
• Molar mass of $$CO_2$$ is $$44 \text{ g mol}^{-1}$$, and the mass of carbon in each mole is $$12 \text{ g}$$.

Therefore the mass of carbon obtained from $$1.46 \text{ g}$$ of $$CO_2$$ is
$$\text{mass of C}=1.46\; \frac{12}{44}$$

Simplifying,
$$\text{mass of C}=1.46 \times 0.2727 = 0.398 \text{ g (approximately)}$$.

Step 2: Calculate percentage of carbon
Percentage of carbon $$= \frac{\text{mass of C in product}}{\text{mass of original sample}}\times 100$$

$$\%\,\text{C}= \frac{0.398}{0.5}\times 100 = 79.6$$

Rounding to the nearest integer, the percentage of carbon is $$\mathbf{80}$$.

Hence, the required percentage of carbon in the organic compound is 80 %.