An organic compound has $\mathrm{C}$ and $\mathrm{H}$ percentage in the ratio $6: 1$ and $\mathrm{C}$ and…
ratio $3: 4$ the compound is
- $\mathrm{HCHO}$
- $\mathrm{CH}_{3} \mathrm{OH}$
- $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}$
- $(\mathrm{COOH})_{2}$
Solution
$($ Total $=15)$
$\%$ of $\mathrm{C}=\frac{6}{15} \times 100=40$
$40 / 12=3.33=1$
$\%$ of $\mathrm{H}=\frac{1}{15} \times 100=6.6$
$6.6 / 1=6.6=2$
$\%$ of $\mathrm{O}=\frac{8}{15} \times 100=53.3$
$53.3 / 16=3.3=1$
simple ratio: $\mathrm{CH}_{2} \mathrm{O}$
$\therefore$ The compound is $\mathrm{HCHO}$ ~
Asked in: JEE-TOPICTESTS-CHEMISTRY