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. $C_2{H}_4{O}_3$
• 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 $C_X{H}_Y$

$C_X{H}_Y+\left(x+\frac{y}{4}\right)O_2\to xC{O}_2+\frac{y}{2}{H}_{2}O$

Now z is half of oxygen atoms required to burn $C_X{H}_Y$

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

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 $\left(C_2{H}_4{O}_3\right)$