A compound contains `4.07% H,. 24.27% C`, and `71.65% Cl`. If its molar mass is `98.96`, the molecular formula will be

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, 4.07g hydrogen, 24.27g carbon and 71.65g chlorine are present.
Step 2. Convert into number moles of each element
Divide the masses obtained above by respective atomic masses of various elements. This gives the number of moles of constituent elements in the compound

Moles of hydrogen `=(4.07 g) /( 1.008 g) =4.04`
Moles of carbon `=(24.27g) /(12.01g)=2.021`
Moles of clorine` =(71.65 g) /(35.453 g) =2.021`
Step 3. Divide each of the mole values obtained above by the smallest number amongst them
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. Write down the empirical formula by mentioning the numbers after writing the symbols of respective elements
`CH_(2)Cl` is, thus, the empirical formula of the above compound.
Step 5. Writing molecular formula
(a) Determine empirical formula mass by adding 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`
`=49.48 g`
(b) Divide Molar mass by empirical formula mass

`("molaar mass ")/("Empirical formula mass ") =(98.96 g) /(49.48 g) `
(c) Multiply empirical formula by n obtained above to get the molecular formula
Empirical formula `=CH_(2) Cl,n=2 ` Hence molecular fromula is `C_(2) H_(4) Cl_(2)`