A compound contains 4.07% hydrogen, 24.27% carbon and 71.65% chlorine by mass. Its molar mass is 98.96 g mol^(-1).

Given : 

Percentage of H, C and Cl = 4.07% , 24.27% and 71.65% by mass respectively.

To find : 

Empirical formula and molecular formula

Calculation : 

Step I :

Check whether the sum of all the percentages is 100.

4.07 + 24.27 + 71.65 = 99.99 ≈ 100

Therefore,

No need to consider presence of oxygen atom in the molecule.

Step II : 

Conversion of mass percent to grams. Since we are having mass percent, it is convenient to use 100 g of the compound as the starting material. Thus, in the 100 g sample of the above compound, 4.07 g hydrogen 24.27 g carbon and 71.65 g chlorine are present.

Step III : 

Convert into number of moles of each element. Divide the masses obtained above by respective atomic masses of various elements.

Moles of hydrogen = \(\frac{4.07g}{1.008g}\) = 4.04 mol

Moles of carbon = \(\frac{24.27g}{12.000g}\) = 2.0225 mol

Moles of chlorine = \(\frac{71.65g}{35.453g}\) = 2.021 mol

Steps IV :

Divide the mole values obtained above by the smallest value among them.

\(\frac{4.04}{2.021}\) = 2,

\(\frac{2.0225}{2.021}\) = 1 and

\(\frac{2.021}{2.021}\) = 1

Hence, 

The ratio of number of moles of 2 : 1 : 1 for H : C : Cl.

In case the ratios are not whole numbers, then they may be converted into whole number by multiplying by the suitable coefficient.

Step V : 

Write empirical formula by mentioning the numbers after writing the symbols of respective elements. CH₂Cl is thus, the empirical formula of the above compound.

Step VI : 

Writing molecular formula 

a. Determine empirical formula mass :

 Add the atomic masses of various atoms present in the empirical formula.

For CH₂Cl, 

Empirical formula mass = 12.000 + 2 × 1.008 + 35.453 = 49.469 g mol^-1

b. Divide molar mass by empirical formula mass :

r = \(\frac{Molar\,mass}{Empirical\,formula\,mass}\) 

= \(\frac{98.96g\,mol^{-1}}{49.469g\,mol^{-1}}\) 

= 2

c. Multiply empirical formula by r obtained above to get the molecular formula : 

Molecular formula = r × empirical formula

∴ Molecular formula is 2 × CH₂Cl 

i.e. C₂H₄Cl₂.

∴ The empirical formula of the compound is CH₂Cl and the molecular formula of the compound is C₂H₄Cl₂.

[Note : The question is modified to include the determination of molecular formula of the compound.]