The ratio of mass percent of C and H of an organic compound $\left(\mathrm{C}_{\mathrm{X}} \mathrm{H}_{\mathrm{Y}} \mathrm{O}_{\mathrm{Z}}\right)$ is $6: 1$. If one molecule of the above compound ( $\mathrm{C}_{\mathrm{X}} \mathrm{H}_{\mathrm{Y}} \mathrm{O}_{\mathrm{Z}}$ ) contains half as much oxygen as required to burn one molecule of compound $\mathrm{C}_{\mathrm{X}} \mathrm{H}_{\mathrm{Y}}$ completely to $\mathrm{CO}_{2}$ and $\mathrm{H}_{2} \mathrm{O}$. The empirical formula of compound $\mathrm{C}_{\mathrm{X}} \mathrm{H}_{\mathrm{Y}} \mathrm{O}_{\mathrm{Z}}$ is :

(A) $\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{3}$

(B) $\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}_{3}$

(C) $\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}$

(D) $\mathrm{C}_{3} \mathrm{H}_{4} \mathrm{O}_{2}$

JEE Main · 2018 · 08 Apr 2018 · Offline · Some Basic Concepts of Chemistry

The ratio of mass percent of C and H of an organic compound

(C_{X} H_{Y} O_{Z})

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_{2} O

. The empirical formula of compound

C_{X} H_{Y} O_{Z}

is :

Answer: (A)

A C:H mass ratio of

6 : 1

means a mole ratio of

\frac{6}{12} : \frac{1}{1} = 1 : 2

, so

x : y = 1 : 2

. Burning

C X_{x} H X_{y}

completely needs

(x + \frac{y}{4})

molecules of

O X_{2}

, that is

2 (x + \frac{y}{4})

oxygen atoms, and the compound holds half of this, so

z = x + \frac{y}{4}

. Taking the smallest whole-number fit,

x = 2

and

y = 4

give

z = 2 + 1 = 3

, so the empirical formula is

C X_{2} H X_{4} O X_{3}

JEE Main · 2018 · 08 Apr 2018 · Offline · Single correct option