Question

$44gm$ of a sample on complete combustion gives $88gmC{O}_2$ and $36gm$ of $H_{2}O$. The molecular formula of the compound may be:

• A. $C_4{H}_6$
• B. $C_2{H}_{6}O$
• C. $C_2{H}_{4}O$
• D. $C_3{H}_{6}O$

Answer 1

The correct option is C $C_2{H}_{4}O$

$C{O}_2=\frac{88}{44}=2$ moles of $C{O}_2=2$ mole of $C$
$H_{2}O=\frac{36}{18}=2$ moles of $H_{2}O=2$ 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 $C_2{H}_{4}O$.