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

Explanation

Step-1 : Conversion of mass per cent 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, 4.07g 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)/(1.008g) = 4.04`
Moles of Carbon `=(24.27g)/(12.01g) = 2.021`
Moles of Chlorine `=(71.65 g)/(35.453g)=2.021`
Step-3 : Divide the mole value obtained above by the smallest number :
Since 2.021 is the smallest value, division by it gives a ratio of `2:1:1` for `H:C:Cl`
In case the ratios are not whole number, then, they may be converted into whole number by multiplying by the suitable coessicient.
Step-4 : Write empirical formula by mentioning the numbers after writing the symbols of respective elements :
`CH_(2)Cl` is the empirical formula of the above compound.
Step-5 : Writing molecular formula :
(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 + 2 xx 1.008 + 35.453=49.48g`
(b) Divide molar mass by empirical formula mass : `("Molar mass")/("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)`