Question

0.30 g of an organic compound C, H and Oxygen on combustion yields 0.44g $C{O}_2$ and 0.18 g $H_{2}O$. If one mole of compound weighs 60, then molecular formula of the compound is

• A. CH₂O
• B. C₃H₈O
• C. C₄H₈O
• D. C₂H₄O₂

Answer 1

The correct option is D

C₂H₄O₂

Percentage of $C=\frac{12}{44}\times \frac{0.44}{0.30}\times 100=40=\frac{40}{12}=3.33=1$

Percentage of $H=\frac{2}{18}\times \frac{0.18}{0.30}\times 100=6.66=2$

Percentage of $O=100$ $=(\%C+\%H)$ $=100–(40+6.66)$ $=53.34\%$
$=\frac{53.34}{16}=3.33=1$

$E.F=C{H}_{2}O$

= M.F = n $\times$ E.F

$n=\frac{60}{30}=2$
$M.F.=2\times C{H}_{2}O=C_2{H}_4{O}_2$