Question

Given,$0.30g$ of an organic compound containing $C,H$ and $O$ on combustion yield $0.44g$ $C{O_2}$ and $0.18g{\text{ }}{H_2}O.$ If one mol of compound weights $60,$ then molecule formula of the compound is:
(A) $C{H_2}O$
(B) ${C_3}{H_8}O$
(C) ${C_4}{H_6}O$
(D) ${C_2}{H_4}{O_2}$

Answer 1

Hint:
Empirical formula is the formula which impulses the smallest whole number ratio of the constituent atom within a molecule. It is the simplest positive integer ratio of the atoms present in the compound.

Complete step by step answer:
As the empirical formula gives the simplest ratio of the number of the different atoms present in the molecule whereas the molecular formula of the molecule provides the actual number of each different atom present in the molecule. The molecular formula is the multiple of the empirical formula.
Now, let's calculate the molecular formula of the given compound step by step.
Step$1$:- Calculating the % of $C,H\& O$ for the given organic compound:-
(i) $\% $ of $C = \;\left( {\dfrac{{12}}{{44}}} \right) \times \left( {\dfrac{{0.44}}{{0.3}}} \right) \times 100 = 40$
Here $12$ is the atomic weight of carbon.
(ii) $\% $ of $H = \left( {\dfrac{2}{{18}}} \right) \times \left( {\dfrac{{0.18}}{{0.3}}} \right) \times 100 = 6.66$
(III) $\% $ of $O$ $ = (100 - 40 - 6.666) = 53.34$
We calculated % of $'O'$ by subtraction formula.
Step $2$ :- Let’s calculates empirical formula by generating ratios
=--
So, empirical formula is :- $C{H_2}$
Step$3$:-
Given $1$ mole of the compound $ = 60g$
$n = \left\{ {\dfrac{{Weight}}{{Molecule\,weight}}} \right\}\;\;\;\;\;Molecular\;weight \Rightarrow 60$
$n \times \left( {Equivalent\;weight} \right) = 60$
\[n = \dfrac{{60}}{{30}} \Rightarrow 2\]
Step$4$:-
So, molecular formula will be:-
$n \times \left\{ {Empirical\;formula} \right\}$
$2 \times \left\{ {C{H_2}O} \right\}$
$ \Rightarrow {C_2}{H_4}{O_2}$
Hence, option D is the correct answer.

Additional information
${C_2}{H_4}{O_2}$ Acetic acid
The above is the structure and the molecular formula of the acetic acid. The molecular formulas are associated with gram molecular masses and the empirical formula is calculated by only knowing the percent composition of a compound.

Note: The basic difference in writing the molecular formula and the empirical formula is that the molecular formula uses subscript as the actual number of atoms present in the molecule whereas the empirical formula uses the subscript which is the smallest integer.