Question

A compound contains 4.07% hydrogen,24.27% carbon and 71.65% chlorine. Its molar mass is 98.96g. What are its empirical and molecular formulas?

Answer 1

Hint: The answer to this question is based on the basic definitions of empirical formula and molecular formula where firstly the percentage is to be converted into grams and then the given molecular mass is divided by empirical formula mass which gives the required answer.

Complete answer:
In the lower classes of chemistry, we have studied about the molecular formula and also the empirical formula calculation and let us now see this in detail and calculate both the quantities with the given data.
To start with, let us first concert the percentiles into grams. If we consider that these percentiles of individual atoms in the total composition considered to be hundred then,
Total amount of hydrogen present in 100g of compound is 4.07g
Total amount of carbon present will be24.27g
Total amount of chlorine present will be 71.65g.
Now, we shall calculate the number of moles of each atom present in the required compound.
Thus,
\[n=\dfrac{Wt.}{Atomic.wt.}\]
For hydrogen, \[{{n}_{H}}=\dfrac{4.07}{1.008}=4.04moles\]
For carbon, \[{{n}_{C}}=\dfrac{24.27}{12}=2.02moles\] and
For chlorine, \[{{n}_{Cl}}=\dfrac{71.65}{35.5}=2.02moles\]
Now, according to the standard rule of empirical formula, we have to divide the above values with the one smallest value among those, thus the ratio will be
\[{{n}_{H}}:{{n}_{C}}:{{n}_{Cl}}=4.04:2.02:2.02\]
Now, by dividing this with smallest value among those that is 2.02 we get,
\[{{n}_{H}}:{{n}_{C}}:{{n}_{Cl}}=2:1:1\]
Now, empirical formula will be ${{H}_{2}}CCl$
Now, to find the molecular formula, we have to calculate the empirical formula mass that is,
\[\left( 1.008\times 2 \right)+\left( 12.01\times 1 \right)+\left( 35.5\times 1 \right)=49.48g\]
Now, according to the rule, we have to divide the molecular formula mass by empirical formula mass and this will be,
\[n=\dfrac{98.96}{49.48}=2\]
Thus, the value of n is 2.
Now, we have to multiply this value of n to the empirical formula found and that is $({{H}_{2}}CCl)\times 2={{H}_{4}}{{C}_{2}}C{{l}_{2}}$
This molecular formula can be written accordingly as ${{C}_{2}}{{H}_{4}}C{{l}_{2}}$

Note:
Note that empirical formula and molecular formula are different and do not be confused because empirical formula shows the simplest whole number ratio of the atoms in a compound and molecular formula shows number of each type of atom in a molecule.