Question

$44g$ 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:

• A. $C_4H_6$
• B. $C_2H_6O$
• C. $C_2H_4O$
• D. $C_3H_6O$

Answer 1

Answer: (C)

$C_2H_4O$

Answer 2

$CO_{2} = \frac{88}{44} = 2$ 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 $C_{2}H_{4}O$.