Question

An organic compound (mol. wt. $=44$) (X) contains $54.54\%$ C and $9.09\%$ of H. With $PC{{l}_{5}}$, (X) gives a compound of molecular weight $99$. On oxidation it gives an acid of molecular weight $60$. What is (X)?

Answer 1

First of all, find out the number of moles of each element and then find out the unit factor to know the empirical formula of the compound X. The number of moles can be found from the percentages and their molecular weights.

Complete step by step solution:
Given that,
An organic compound X has a molecular weight of $44$.
The compound has $54.54\%$ of carbon, $9.09\%$ of hydrogen.
With $PC{{l}_{5}}$, the compound X yields a compound having molecular weight $99$ and whereas on oxidation, it gives an acid of molecular weight $60$.
So, we see that on oxidation the compound X gives an acid. Therefore, we can infer that it must have either an aldehyde group or an alcohol group or a ketone group. In other words, we can say that oxygen is present in the compound X.
So, the percentage of oxygen present in the compound will be
$100-(54.54+9.09)=100-63.63=36.37\%$
Now, we have to find out the empirical formula of the compound.
So, to find out the empirical formula we have to find out the number of moles and then the unit factor of each element present in the compound.

Element	$\%$ by weight	$X_i$	$\dfrac{X_i}{{smallest}{X_i}}$
C	54.54	$\dfrac{54.54}{12} =4.545$	$\dfrac{4.54}{2.27} =2$
H	9.09	$\dfrac{9.09}{1} =9.09$	$\dfrac{9.09}{2.27} =4$
O	36.37	$\dfrac{36.37}{16} =2.27$	$\dfrac{2.27}{2.27} =1$

So, in the compound X there will be two carbon atoms, four hydrogen atoms and one oxygen atom.
So, the formula may be ${{C}_{2}}{{H}_{4}}O$.
The molecular weight will be $12\times 2+1\times 4+16\times 1=44g$.
Now, the number of the functional group will be $=\dfrac{6-4}{2}=1$.
So, the compound will have one $-CHO$ group and the compound is an aldehyde.

Hence, the compound X is $C{{H}_{3}}-CHO$ and on oxidation it gives $C{{H}_{3}}-COOH$ whose molecular mass is $60$.

Note: It is important to note that the empirical formula does not provide the actual number of atoms of each element present in one molecule; instead it gives us the simplest formula of a compound from which we can guess what could be the compound.