Question

Calculate the mass percentage of different elements present in sodium sulphate $\left( {{\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}} \right)$.

Answer 1

The mass percentage is one of the ways which is generally used for representing the concentration or molarity of the present element in any compound or the mixture of components.

Complete step by step answer:
We know that the mass percentage of any element can be determined with the help of their molar mass and the molar mass of the given compound.
First we have to calculate the molar mass of sodium sulphate i.e. $\left( {{\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}} \right)$
= 23 x 2 + 32 + 16 x 4
= 142
Therefore the molar mass of the compound is 142 g/mol.
Now, The molar mass of sodium is 23 g/mol.
The molar mass of oxygen is 16 g/mol.
The molar mass of sulphur is 32 g/mol.
In the formula of sodium sulphate the number of sodium atoms are two, the number of Sulphur atoms is one and the number of oxygen atoms are four.
The mass percentage of each element can be calculated by using the formula given below.
${\rm{Mass}}\;{\rm{Percentage}}\;{\rm{of}}\;{\rm{element}} = \dfrac{{{\rm{Mass}}\;{\rm{of}}\;{\rm{element}}}}{{{\rm{Mass}}\;{\rm{of}}\;{\rm{compound}}}} \times 100\%$
Substitute the respective values in the above equation.
For Sodium:
${\rm{Mass}}\;{\rm{Percentage}}\;{\rm{of}}\;{\rm{element}} = \dfrac{{{\rm{23}} \times {\rm{2}}}}{{{\rm{142}}}} \times 100$
= 32.39 %
= 32.4 %
For Sulphur:
${\rm{Mass}}\;{\rm{Percentage}}\;{\rm{of}}\;{\rm{element}} = \dfrac{{{\rm{32}}}}{{{\rm{142}}}} \times 100$
= 22.5 %
For oxygen:
${\rm{Mass}}\;{\rm{Percentage}}\;{\rm{of}}\;{\rm{element}} = \dfrac{{16 \times 4}}{{{\rm{142}}}} \times 100$
= 45 %

Hence, the mass percent of sodium is $32.4\%$, the mass percentage of Sulphur is $22.5\%$, and the mass percentage of oxygen is $45\%$.

Note: The mass percentage of element can be evaluated by the fraction of element’s mass to the compound’s mass. The overall mass value is multiplied by $100\%$.