On complete combustion of 0 . 492g of an organic compound...

Organic compound $+ {O}_{2} ⟶ {CO}_{2}\left(g\right) + {H}_{2}O$

$0 . 7938$ gram $0 . 4428$ gram

Weight of the organic compound=$0 . 492$ gram

Number of moles of ${CO}_{2}$=Number of moles of $C$

$= \dfrac{0 . 7938}{44}Weightofcarbon = \left[\dfrac{0 . 7938}{44}\right]12 = 0 . 216g$

Weightofhydrogen = \left[\dfrac{0 . 4428}{18}\right]2 = 0 . 0491$g</p><p>Weightofoxygen = \left[0 . 492 - 0 . 216 - 0 . 0491\right] = 0 . 2269g$

$$ of $O = \dfrac{0 . 2249}{0 . 492} \times 100 = 46 . 11 \approx 46$