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

Since we are having mass percent, it is converient to use ,100 g of the compound as the starting material. Thus, in the 100 g. Sample of the above compound, 4.07 g hy-drogen, 24.27 g carbon, and 71.65 g chlo-rine are present.
Stap 2. Convertion into number of moles of each element:
Divide the masses obtained above by re-spective atomic masses of various elements.
Moles of hydrogen= `(4.07 g)/(1.008 g "mol"^(-1))=4.04`
Moles of carbon = `(24.27 g)/(1.2.0lg"mol"^(-1))=2.021`
Moles of chlorine = `(71.65 g)/(35.453 g "mol"^(-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 num-ber by multiplying by the suitable coefficient.
Step 4. These mumbers indicate the rela-tive number of atoms of the elements. Write empirical formula by mentioning the numbers after writing the symbols of re-spective elements :
`CH_(2)Cl` is thus, 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 =49.48 g
(b) Divide Molar mass by empirical formula mass
= `("Molar mass of the compound ")/("Empirical formula mass " )=(98.96 g)/(49.48 g)=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)`