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

Answer: (D)

Element	Relative mass	Relative mole	Simplest whole number ratio
C	6	$\frac{6}{12}=0.5$	1
H	1	$\frac{1}{1}=1$	2

So, $X=1,Y=2$

Equation for combustion of $C_x{H}_Y$

$C_X{H}_Y+\left(X+\frac{Y}{4}\right)O_2⟶XC{O}_2+\frac{Y}{2}{H}_{2}O$

Oxygen atoms required $=2\left(X+\frac{Y}{4}\right)$

As per information,

$2\left(X+\frac{Y}{4}\right)=2Z$

$\Rightarrow \left(1+\frac{2}{4}\right)=Z$

$\Rightarrow Z=1.5$

Molecule can be written

$C_x{H}_Y{O}_z$

$C_1{H}_2{O}_{3/2}$

$\Rightarrow C_2{H}_4{O}_3$