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 (${{C}_{X}}{{H}_{Y}}{{O}_{Z}}$) contain half as much oxygen is required to burn one molecule of compound ${{C}_{X}}{{H}_{Y}}$ completely to $C{{O}_{2}}$ and ${{H}_{2}}O$. The empirical formula of the compound ${{C}_{X}}{{H}_{Y}}{{O}_{Z}}$is:
(A) ${{C}_{3}}{{H}_{4}}{{O}_{2}}$
(B) ${{C}_{2}}{{H}_{4}}{{O}_{3}}$
(C) ${{C}_{3}}{{H}_{6}}{{O}_{3}}$
(D) ${{C}_{2}}{{H}_{4}}O$

Answer 1

Hint: Divide the question into two parts and solve accordingly. From the ratio of the mass percentage of carbon and hydrogen find the ratio of mass. The atomic mass of carbon is 12 and the atomic mass of hydrogen is 1. Now from the second part calculate the ratio of oxygen.

Complete step by step solution:
So, in this question, two information are given, from the first part,
The formula of given organic compound = ${{C}_{X}}{{H}_{Y}}{{O}_{Z}}$
The ratio of the mass percentage of C and H is 6:1,
We know that the atomic mass of carbon = $12$
The atomic mass of hydrogen = $1$
For calculating the mass ratio,
If x atoms of carbon are present and y atoms of hydrogen are present, then the equation will be:
$12\text{ x }x\text{ = 1 x }y$
We can say carbon is six times the hydrogen, the above equation will be:
$(12\text{ x }x)\text{ = (1 x }y)6$
$12x=6y$
$y=2x$
So from this, the number of hydrogen atoms is twice the number of carbon atoms.
Now, from the second part, to burn one molecule of ${{C}_{x}}{{H}_{y}}$, the equation will be:
${{C}_{x}}{{H}_{2x}}+{{O}_{2}}\to xC{{O}_{2}}+\dfrac{2x}{2}{{H}_{2}}O$
${{C}_{x}}{{H}_{2x}}+{{O}_{2}}\to xC{{O}_{2}}+x{{H}_{2}}O$
So, we can calculate the number of oxygen atom involved in this reaction:
Oxygen = $2x+x=3x$
From the question, $Z=\dfrac{3x}{2}$
So, we have, $X=x,\text{ }Y=2x,\text{ }and\text{ }Z=\dfrac{3x}{2}$
So, from all the options, (b)- ${{C}_{2}}{{H}_{4}}{{O}_{3}}$fits in the above equation.

Therefore, the correct answer is an option (B)- ${{C}_{2}}{{H}_{4}}{{O}_{3}}$.

Note: It must be noted that the ratio of mass percentage is given not the ratio of the mass of carbon and oxygen. The equation of burning the compound must be written correctly.