Question

A compound contains 4.07% hydrogen, 24.27% carbon and 71.65% chlorine. Its molar mass is 98g. What are its empirical and molecular formulas? (Atomic mass of C=12, H=1, Cl=35)

Answer 1

First we calculate the empirical formula in 2 steps: (1) Divide the given percentage with the elements atomic weight; (2) Divide the values obtained with the smallest value obtained. This ratio of all atoms gives the empirical formula. Now we divide the molecular weight by the empirical weight which gives us the multiple which when multiplied with the empirical weight gives us the molecular formula.

Complete step by step solution:
-First of all let us see what is an empirical formula and a molecular formula.
A molecular formula is made of the chemical symbols of the constituent elements which are followed by their numeric subscripts which tell us about the number of atoms of each element present in the molecule. While an empirical formula basically expresses the ratio between the numbers of atoms of different elements present in a molecule of some compound.
In short the empirical formulas represent the simplest whole integer ratio of the constituent atoms of any compound and a multiple of a compound’s empirical formula gives us its molecular formula.
-We will now start with the determination of the empirical formula of the compound. This is done in 2 steps:
(1)First we will divide the given percentages of the atoms with their molecular masses to calculate the number of moles.
For Hydrogen: Given percentage = 4.07% and its molar mass = 1
So, for H: $\dfrac{{4.07}}{1} = 4.07$
For Carbon: Given percentage = 24.27% and its molar mass = 12
So, for C: $\dfrac{{24.27}}{{12}} = 2.02$
For Chlorine: Given percentage = 71.65% and its molar mass = 35.5
So, for Cl: $\dfrac{{71.65}}{{35.5}} = 2.01$
(2)Now, in a compound the constituents are present in a simple whole number ratio. So to get the whole number we will divide the moles calculated in the above step by the smallest number of moles.
We can see that the moles of Cl are smallest. So, we will divide the moles of C, H and Cl by the moles of Cl.
We will now divide all the values with the lowest value obtained to get a whole number.
For Hydrogen: $\dfrac{{4.07}}{{2.01}} = 2$
For Carbon: $\dfrac{{2.02}}{{2.01}} = 1$
For Chlorine: $\dfrac{{2.01}}{{2.01}} = 1$
Hence the empirical formula of the compound will be: $C{H_2}Cl$
-We will now calculate the weight of the empirical formula. E.Wt. = $\left( {12 \times 1} \right) + \left( {1 \times 2} \right) + \left( {35.5 \times 1} \right)$
$\therefore 49.5g$
-The molar mass of the compound in given in the question to be: M.Wt. = 98 g
The molecular formula shows the actual number of atoms of each of the elements in a compound.
The relation between the molecular formula (MF) and the empirical formula (EF) is:
$M.F. = \dfrac{{M.Wt.}}{{E.Wt.}} \times E.F$
Putting values in the above formula we get: = $\dfrac{{98}}{{49.5}} \times C{H_2}Cl$
$\implies 2 \times C{H_2}Cl$
$\therefore {C_2}{H_4}C{l_2}$

Hence, the empirical formula is $C{H_2}Cl$ and the molecular formula is ${C_2}{H_4}C{l_2}$.

Additional information:
In the case of simpler molecules, the empirical formula and molecular formula can be the same. For example: in case of water or Sodium chloride.

Note: Both empirical and molecular formulas indicate the ratios in which different elements are present in the compound. But they do not give any information about the structure or bonding between them. For that we need to understand concepts of chemical bonding.