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 $\left(C_X{H}_Y{O}_Z\right)$ contains 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. $C_3{H}_6{O}_3$
• B. $C_2{H}_4{O}^{}$
• C. $C_3{H}_4{O}_2$
• D. $C_2{H}_4{O}_3$

Answer 1

The correct option is D $C_2{H}_4{O}_3$

$\frac{12X}{Y}=\frac{6}{1}$ (ratio of masses)

$2X=Y$ for $C_X{H}_Y{O}_Z$
Equation for combustion of $C_X{H}_Y$

$C_X{H}_Y+(X+\frac{Y}{4})O_2\to XC{O}_2+\frac{Y}{2}{H}_{2}O$
Number of oxygen atoms in $C_X{H}_Y{O}_Z$=Z
Number of oxygen atoms required for combustion of $C_X{H}_Y$= $2(X+\frac{Y}{4})$

$\frac{1}{2}(2X+\frac{Y}{2})=Z$

$\Rightarrow (X+\frac{Y}{4})=Z$

$\Rightarrow (X+\frac{2X}{4})=Z$

$\Rightarrow Z=\frac{3X}{2}$
$X:2X:\frac{3X}{2}$
$2X:4X:3X$
$2:4:3$
Hence, $C_2{H}_4{O}_3$