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Q.
$44g$ of a sample on complete combustion gives $88$ $gm$ $CO_2$ and $36$ $gm$ of $H_2O$. The molecular formula of the compound may be:
Some Basic Concepts of Chemistry
Solution:
$CO_{2} = \frac{88}{44} = 2$ mole of $CO_{2} = 2$ mole of $C$
$H_{2}O = \frac{36}{18} = 2$ mole of $H_{2}O = 4$ mole of $H$
Mass of $C$ + Mass of $H$ + Mass of $O = 44$
$\Rightarrow 24 + 4 + x = 44; x = 16$
$\therefore $ mole of $O = 1$ and molecular formula is $C_{2}H_{4}O$.