Question

A compound on analysis gave the following percentage composition by weight : Hydrogen $= 9.09\%$, Oxygen $= 36.36\%$, Carbon $54.55\%$. It’s vapour density is $44$. If the molecular formula of the compound is ${C_x}{H_y}{O_z}$, then find the value of $x,y$ and $z$.
A.) $x = 4,y = 8,z = 2$
B.) $x = 3,y = 6,z = 3$
C.) $x = 5,y = 10,z = 4$
D.) None of the above.

Answer 1

To solve this type of question, we need to first make a table consisting of the number of moles and least ratio for each element and find the empirical formula. From empirical formula we will find empirical formula weight. Then it will give a multiplying factor for the empirical formula which can be used to get the required molecular formula.

Complete step by step answer:
Step $1$.) First of all we will make a table in which the first column will consist of all elements given in question.
Second column will be the number of moles which is divided by dividing the given mass($W$) by the molecular mass($M$) of each element. That is,
Number of moles $=$ $\dfrac{W}{M}$
Third column will be of least ratio that is obtained by dividing the number of moles of each compound with the number of moles of that compound which has the lowest number of moles as shown in table.
Thus table can be given as:
Elements
Number of moles
Least ratio
For C
$\dfrac{{54.55}}{{12}} = 4.54$
$\dfrac{{4.54}}{{2.27}} = 2$
For H
$\dfrac{{9.09}}{1} = 9.09$
$\dfrac{{9.09}}{{0.27}} = 4$
For O
$\dfrac{{36.36}}{{16}} = 2.27$
$\dfrac{{2.27}}{{2.27}} = 1$

Step $2$.) In this step, we will write an empirical formula which can be given from the least ratio of each compound. For above table, the empirical formula can be written as:
Empirical formula $= {C_2}{H_4}O$
As carbon has least ration $2$, hydrogen has least ratio $4$ and oxygen has least ratio $1$.
Step $3$.) In this step we will calculate empirical formula weight which can be written from the above given empirical formula. In it we will multiply each atom’s molecular weight by its total number of atoms in empirical formula.
Empirical weight $= 12 \times 2 + 1 \times 4 + 16$
It is because carbon has molecular weight $12$ and there are $2$ carbons in empirical formula, Hydrogen has molecular weight $1$ and there are $4$ hydrogen in empirical formula and oxygen has molecular weight $16$ and there is only one oxygen.
Step $4$.) In this step we will find molecular weight from vapour density . As we are given in question that vapour density is $44$.
Now, relation between molecular formula and viscosity density is :
Molecular weight $= 2 \times$ Vapour density
Thus we can say that,
Molecular formula $= 2 \times 44$
Molecular formula $= 88$
Step 5.) In this step, we will calculate molecular formula which is given by :
Molecular formula $= n \times$ (Empirical formula) -(1)
Here, $n = \dfrac{M}{{E.F.W.}}$
Where $M$is molecular weight and $E.F.W.$ is empirical formula weight
$n = \dfrac{{88}}{{44}} \\\ n = 2 \\\$
Now, from equation $- (1)$
Molecular formula $= 2 \times ({C_2}{H_4}O)$
$= {C_4}{H_8}{O_2}$
Hence, the required molecular formula is given by ${C_4}{H_8}{O_2}$.

Note:
In such a type of question when the molecular weight is not given then we need to find it using vapour density. We can find molecular weight from vapour density. Molecular weight can be given by multiplying vapour density by $2$.