Question

The ratio of mass percent of $C\text{and}H$ 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 $CO_{2}$ and $H_{2}O$ . The empirical formula of compound $C_{X}H_{Y}O_{Z}$ is

• A. $C_{2}H_{4}O_{3}$
• B. $C_{3}H_{6}O_{3}$
• C. $C_{2}H_{4}O$
• D. $C_{3}H_{4}O_{2}$

Answer 1

Answer: (A)

Ratio of mass % of C and H in $C_{X}H_{Y}O_{Z}$ is 6 : 1.
Therefore, Ratio of mole % of C and H in $C_{X}H_{Y}O_{Z}$ will be 1 : 2.
Therefore x : y = 1 : 2, which is possible in options 1, 2 and 3.
Now oxygen required to burn $C_{X}H_{Y}$
$C_{X}H_{Y}+\left(x + \frac{y}{4}\right)O_{2} \rightarrow xCO_{2}+\frac{y}{2}H_{2}O$
Now z is half of oxygen atoms required to burn $C_{X}H_{Y}$
So $Z=\frac{\left(2 X + \frac{Y}{2}\right)}{2}=\left(X + \frac{Y}{4}\right)$
Now putting values of x and y from the given options
Option A, $x=2,y=4$
$Z=\left(2 + \frac{4}{4}\right)=3$
Option B, $x=3,y=6$
$Z=\left(3 + \frac{6}{4}\right)=4.5$
Therefore correct option is 1 $\left(C_{2} H_{4} O_{3}\right)$