Question

The chemical analysis of a compound reveals that it contained $ \text{Na =14}\text{.31 }%\text{ } $ , $ \text{S = 9}\text{.97 }%\text{ } $ , $ \text{H = 6}\text{.22 }%\text{ } $ and $ \text{O = 69}\text{.5 }%\text{ } $ . The molecular mass of the compound is found to be 322 amu. If all the hydrogen is present in the compound in the form of water of crystallization, find the molecular weight of the compound.

Answer 1

Hint: The empirical formula is the simplest way to represent the molecular composition of any compound or it is the simplest whole number ratio of the atoms present in a compound. The molecular formula is a multiple of the empirical formula. We shall calculate the mass/atomic mass ratio for each compound and then divide each of them by the smallest ratio to get a simple integer.

Complete Step by Step Answer
Let us first find out the empirical formula of the compound from the percentage of the elements present. It contained $ \text{Na =14}\text{.31 }%\text{ } $ , $ \text{S = 9}\text{.97 }%\text{ } $ , $ \text{H = 6}\text{.22 }%\text{ } $ and $ \text{O = 69}\text{.5 }%\text{ } $ , therefore the empirical formula is as follows:

Element	Mass %	Atomic mass	Mass %/atomic mass	Simplest mass ratio
Sodium	$ \text{14}\text{.31} $	23	$ \dfrac{14.31}{23}=0.622 $	$ \dfrac{0.622}{0.311}=2 $
Sulphur	$ \text{9}\text{.97} $	32	$ \dfrac{9.97}{32}=0.311 $	$ \dfrac{0.311}{0.311}=1 $
Hydrogen	$ \text{6}\text{.22} $	1	$ \text{6}\text{.22} $ $ $	$ \dfrac{6.22}{0.311}=20 $
Oxygen	$ \text{69}\text{.5} $	16	$ \dfrac{69.5}{16}=4.34 $	$ \dfrac{4.34}{0.311}=14 $

As obtained from the table, the empirical formula of the compound is: $ \text{N}{{\text{a}}_{\text{2}}}\text{S}{{\text{O}}_{\text{14}}}{{\text{H}}_{\text{20}}} $
Let us calculate the empirical mass of the compound, which is equal to: $ \left[ \left( 23\times 2 \right)+32+\left( 1\times 20 \right)+\left( 14\times 16 \right) \right]=322 $
The given molecular mass of the compound is also 322 g. therefore as the empirical mass of the compound and the molecular mass of the compound are both equal therefore the empirical formula and the molecular formula of the compound are same.
Now, we need to find out the number of water molecules present in the compound in the form of water of crystallization. In each water molecule, there are “two hydrogen atoms” and “one oxygen atom”. Therefore for 20 hydrogen atoms to form the water molecules, 10 atoms of oxygen would be required to form 10 water of molecules and the remaining 4 oxygen atoms would be the part of the main compound.
Therefore the molecular formula of the compound with the water of crystallization is: $ \text{N}{{\text{a}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\text{. 10}\left( {{\text{H}}_{\text{2}}}\text{O} \right) $ , sodium sulphate decahydrate. The common name for this salt is Glauber’s salt.

Note
The biggest advantage of using the empirical formula is that it provides the ratio of the elements in the simplest form and it becomes to determine the molecular formula of the compound in that way, as the molecular formula is the multiple of the empirical formula.