Question

What is the molar mass of Freon, and what is the molecular formula \[\left( {9.92\% C,58.6\% Cl,31.8\% F} \right)\] ?

Answer 1

Hint: Given are the percentages of compositions of elements in the freon. From these percentages and molar mass, the moles of each atom will be calculated. By dividing each component moles with the smallest mole gives the mole ratio. Empirical formulas can be written from mole ratio.

Complete answer:
Freon is a pollutant that can be released from the refrigerators and aeroplanes. It has a molecular composition of carbon, chlorine and fluorine.
Given percentages of composition are \[9.92\% C\] , \[58.6\% Cl\] , and \[31.8\% F\]
The molar mass of carbon is \[12\] amu. Thus, the number of moles of carbon will be \[\dfrac{{9.92}}{{12}} = 0.826\]
The molar mass of chlorine is \[35.5\] amu. Thus, the number of moles of carbon will be \[\dfrac{{58.6}}{{35.5}} = 1.653\]
The molar mass of fluorine is \[19\] amu. Thus, the number of moles of carbon will be \[\dfrac{{31.8}}{{19}} = 1.68\]
In the obtained moles of carbon, chlorine and fluorine. The number of moles of carbon were small.
Divide the moles of each atom with moles of carbon to obtain the mole ratio.
\[C:Cl:F = \dfrac{{0.826}}{{0.826}}:\dfrac{{1.653}}{{0.826}}:\dfrac{{1.68}}{{0.826}}\]
The ratio will be \[C:Cl:F = 1:2:2\]
Thus, the empirical formula of freon is \[CC{l_2}{F_2}\]
The empirical formula and molecular formula are the same for freon. The molecular mass will be \[12 + 2 \times 35.5 + 2 \times 19 = 121gmo{l^{ - 1}}\]
The molecular formula of freon is \[CC{l_2}{F_2}\] and the molecular mass of freon is \[121gmo{l^{ - 1}}\]

Note:
Empirical formula is a shortest representation of molecular formula which is obtained from the mole ratio of atoms. For some compounds both empirical formula and molecular formula are the same. Freon has the same empirical and molecular formula. Molar mass is also known as molecular mass.