Question

The ratio of mass percent of $C$ and $H$ of an organic compound $\left(C_X{H}_Y{O}_Z\right)$ is $6:1$. If one molecule of the above compound $\left(C_X{H}_Y{O}_Z\right)$ contain half as much oxygen as required to burn one molecule of compound $C_X{H}_Y$ completely to $C{O}_2$ and $H_{2}O$. The empirical formula of compound $C_X{H}_Y{O}_Z$ is :

• A. $C_3{H}_4{O}_2$
• B. $C_2{H}_4{O}_3$
• C. $C_3{H}_6{O}_3$
• D. $C_2{H}_{4}O$

Answer 1

Answer: (B)

$C_2{H}_4{O}_3$

$C_x{H}_y{O}_z$

$C_x{H}_y+n{O}_2\to xC{O}_2+\frac{y}{2}{H}_{2}O$

$n=\frac{4x+y}{4}$ and amount of $O$ required, $m=n\ast 2$

Given that $z=m/2=n$

and mass percent ratio of $C$:$H$ $::6:1=$ $12:2$

therefore for each $C$ two $H$ are present in the molecule.

For $x=1$,$y=2$ $\therefore z=\frac{4\times 1+2}{4}$

$z=1.5$

Emperical formula is $C_2{H}_4{O}_3$

Option $B$ is correct.