### JEE Main On 08 April 2018 Question 7

##### Question: The ratio of mass percent of C and H of an organic compound $ (C _{X}H _{Y}O _{Z}) $ is 6 : 1. If one molecule of the above compound $ (C _{X}H _{Y}O _{Z}) $ contains half as much oxygen as required to burn one molecule of compound $ C _{X}H _{Y} $ completely to $ CO _2 $ and $ H _2O $ . The empirical formula of compound $ C _{X}H _{Y}O _{Z} $ is: [JEE Main Online 08-04-2018]

#### Options:

A) $ C _3H _4O _2 $

B) $ C _2H _4O _3 $

C) $ C _3H _6O _3 $

D) $ C _2H _4O $

## Show Answer

#### Answer:

Correct Answer: B

#### Solution:

Mole ratio of $ \text{C:H}\Rightarrow \frac{6}{12}:\frac{1}{1}\Rightarrow 1:2 $

$ {C_X}{H_y}\text{+}( \text{x+}\frac{y}{4} ){O_2}\xrightarrow{{}}xC{O_2}\text{+}\frac{y}{2}{H_2}O $

According to questions $ \text{2z=2}( x+\frac{y}{4} )……..(1) $

$ \frac{x}{y}=\frac{1}{2}……….(2) $ From (1) & (2) $ $

$ C _{x}H _{y}O _{z}\equiv C _{x}H _{2x}{O _{\frac{3x}{2}}}\equiv C _2H _4O _3 $