Question

The ratio of mass per cent 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 ${CO}_2$ and $H_{2}O$ .The empirical formula of compound $C_x{H}_y{O}_z$ is :

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

Answer 1

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

According to the question, the reaction involved is :

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

Atoms of oxygen involved in the reaction in the reactant and the product side is,

$2n=2x+$$\frac{y}{2}$

$\Rightarrow n=\frac{4x+y}{4}........\left(i\right)$

Given, compound $C_x{H}_y{O}_z$ contains half as much oxygen as required to burn one molecule of compound $C_x{H}_y$.

$\therefore z=n$ (since, 2n atoms of O are req. to burn one molecule of compound $C_x{H}_y$).

Now, the mass ratio of C and H is $6:1$ or $12:2.$ Hence, ratio of their no. of moles will be $1:2$. (no. of moles = $\frac{mass}{Mol.mass}$)

$i.e\text{For every single C atom, there are 2 H atoms.}$

For $x=1$ and $y=2,$

$z=\frac{4x+y}{4}$ (from equation (i))

$z=\frac{4\times 1+2}{4}$ = $\frac{6}{4}=1.5$

Thus, in the simplest whole no. ratio, $x=2,y=4$ and $z=3$ and the required empirical formula will be $C_2{H}_4{O}_3$.

Option C is correct.