A compound contains 4.07% hydrogen, 24.27% carbon and 71.65% chlorine. Its molar mass is 98.96 g. What are its empirical and molecular formulas ?

Explanation

Step 1 : Conversion of mass per cent to grams : Since we are having mass per cent, 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 is present, 24.27 g carbon is present and 71.65 g chlorine is present.
Step 2 : Convert into number moles of each element : Divide the masses obtained above by respective atomic masses of various elements.
Moles of hydrogen = `(4.07g)/(1.008gmol^(-1))=4.04`
Moles of carbon = `(24.27g)/(12.01gmol^(-1))=2.021`
Moles of chlorine = `(71.65g)/(35.453mol^(-1))=2.021`
Step 3 : Divide the mole value obtained above by the smallest number : Since 2.021 is smallest value, division by it gives a ratio 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 4 : These numbers indicate the relative number of atoms of the elements. Write empirical formula by mentioning the number after writing the symbols of respective elements : `CH_(2)Cl` is, thus, empirical formula of the above compound.
Step 5 : Writing molecular formul :
(a) Determine empirical formula mass. Add the atomic masses of various atoms present in the empirical formula.
For `CH_(2)Cl`, empirical formula mass is `12.01+2xx1.008+35.453=49.48g`
(b) Divide molar mass by empirical formula mass
`("Molar mass of the compound")/("Empirical formula mass")=(98.96g)/(49.48g)=2=(n)`
( c) Multiply empirical formula by n obtained above to get the molecular formula
Empirical formula = `CH_(2)Cl`, n = 2. Hence molecular formula is `C_(2)H_(4)Cl_(2)`.