Question

On,analysis a substance was found to have the following percentage composition. K=31.84,Cl=28.98 and O=39.18. Calculate its molecular formula if its molecular mass is 122.5 ?

Answer 1

Dear Student


% of K = 31.84

% of Cl = 28.98

% of O = 39.18

Let us consider, if the mass of the substance is 100 g, then-

Mass of K = 31.84 g

Mass of Cl = 28.98 g

Mass of O = 39.18 g


So, No. of Moles = $\frac{Mass}{Molar\hspace{0.17em}Mass}$

Thus, Moles of K = $\frac{31.84g}{39g/mol}$ = 0.816 mol

where, molar mass of K = 29 g/mol


​Moles of Cl = $\frac{28.98g}{35.4g/mol}$ = 0.818
where, molar mass of Cl = 35.4 g/mol


​Moles of O = $\frac{39.18g}{16g/mol}$ = 2.44 mol

where, molar mass of O = 16



Now, taking the Simple Whole Ratio, we get,

K = $\frac{0.816mol}{0.816mol}$ = 1

Cl = $\frac{0.818mol}{0.816mol}$ = 1.002 $\approx$ 1

O = $\frac{2.44mol}{0.816mol}$ = 2.99 $\approx$3

So, the Empirical formula = KClO₃

Thus, Weight of Emprical Formula = (1 x 39) + (1 x 35) + (3 x 16) = 39 + 35 + 48 = 122 g/mol

The molecular mass given in the question = 122.5 g/mol

​

Now, Molecular formula = $\frac{MolecularMass}{Empiricalformulamass}$= $\frac{122.5g/mol}{122g/mol}$ = 1.004

Therefore, we can consider it as one

So the Molecular formula of the substance will be KClO₃



Regards