Some Basic Concepts of Chemistry

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_{2}O. The empirical formula of compound C_{X}H_{Y}O_{Z} is

C

_{2}H_{4}O_{3}C

_{3}H_{6}O_{3}C

_{2}H_{4}OC

_{3}H_{4}O_{2}

Answer

A.

C_{2}H_{4}O_{3}

The ratio of mass % of C and H in C_{x}H_{y}O_{z} is 6: 1.

Therefore,

The 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}

${C}_{x}{H}_{y}+\left(x+\frac{y}{4}\right){O}_{2}\stackrel{}{\to}xC{O}_{2}+\frac{y}{2}{H}_{2}O$

Now z is half of the oxygen atoms required to burn C_{x}H_{y}

$\therefore z=\frac{(2x+{\displaystyle \frac{y}{2}})}{2}=\left(x+\frac{y}{4}\right)$

Now putting values of x and y from the given options:

Option (1), x = 2, y = 4

$z=\left(2+\frac{4}{4}\right)=3\phantom{\rule{0ex}{0ex}}Option2,x=3,y=6\phantom{\rule{0ex}{0ex}}z=\left(3+\frac{6}{4}\right)=4.5\phantom{\rule{0ex}{0ex}}Thereforecorrenct\left({C}_{2}{H}_{4}{O}_{3}\right)$