On complete combustion of 0.492 g of an organic compound containing C, H and O, 0.7938 g of $$CO_2$$ and 0.4428 g of $$H_2O$$ was produced. The % composition of oxygen in the compound is _____ (Nearest Integer)

The mass of the organic compound was 0.492 g and it contains C, H, and O. When combusted, it produced 0.7938 g of $$CO_2$$ and 0.4428 g of $$H_2O$$.

Since each mole of $$CO_2$$ (44 g) contains 1 mole of C (12 g), we calculate the mass of carbon as follows:

$$\text{Mass of C} = \frac{12}{44} \times 0.7938 = \frac{12 \times 0.7938}{44} = \frac{9.5256}{44} = 0.2163 \text{ g}$$

Similarly, every mole of $$H_2O$$ (18 g) contains 2 moles of H (2 g), so the mass of hydrogen is given by:

$$\text{Mass of H} = \frac{2}{18} \times 0.4428 = \frac{0.8856}{18} = 0.0492 \text{ g}$$

From the above masses, we find the mass of oxygen in the compound by subtracting the masses of C and H from the total mass:

$$\text{Mass of O} = 0.492 - 0.2163 - 0.0492 = 0.2265 \text{ g}$$

This gives the percentage of oxygen as:

$$\% \text{ of O} = \frac{0.2265}{0.492} \times 100 = 46.04\%$$

Therefore, the percentage composition of oxygen in the compound is 46% (nearest integer).