The complete combustion of 0.492 g of an organic compound containing 'C', 'H' and 'O' gives 0.793 g of CO$$_2$$ and 0.442 g of H$$_2$$O. The percentage of oxygen composition in the organic compound is ______ (nearest integer)

We need to find the percentage of oxygen in an organic compound containing C, H, and O. The mass of carbon is determined from CO$$2$$: Mass of CO$$2$$ = 0.793 g, its molar mass is 44 g/mol, and each mole contains 12 g of carbon, giving $$\text{Mass of C} = \frac{12}{44} \times 0.793 = \frac{9.516}{44} = 0.2163 \text{ g}$$. The mass of hydrogen is found from H$$2$$O: Mass of H$$2$$O = 0.442 g, molar mass 18 g/mol, each mole contains 2 g of hydrogen, hence $$\text{Mass of H} = \frac{2}{18} \times 0.442 = \frac{0.884}{18} = 0.04911 \text{ g}$$.

Subtracting these from the total mass of the compound (0.492 g) gives the mass of oxygen as $$\text{Mass of O} = 0.492 - 0.2163 - 0.04911 = 0.2266 \text{ g}$$. The percentage of oxygen is then $$\% \text{ O} = \frac{0.2266}{0.492} \times 100 = 46.06\%$$. Rounding to the nearest integer yields 46, which is the required percentage of oxygen.