A compound contains 4.07 % hydrogen, 24.27% carbon and 71.65 % chlorine by mass. Its molar mass is 98.96 g. What is its empirical formula ? Atomic masses of hydrogen, carbon and chlorine are 1.008, 12.000 and 35.453 u, respectively

Explanation

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 is 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 ` = (4.07 g)/(1.008 g) = 4.04`
Moles of carbons =` (24.27g)/(12.01) = 2.0225`
Moles of chlorine ` = (71.65g)/(35.453 g) = 2.021`
Steps IV : Divide the mole values obtained above by the smallest value among them. Since 2.021 is smallest value, division by it gives a ratio of 2:1:1 for H:C:Cl. In case the ratio 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. `CH_2Cl ` 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 `CH_2Cl` , empirical formula mass is 12.01 + 2 x 1.008 + 35.453 = 49.48 g
b. Divide molar mass by empirical formula mass
` therefore ("molar mass")/(empirical formula mass") = (98.96g)/(49.48g)`
` therefore r = 2`
c. multiply empirical formula by r obtained above to get the molecular formula.
Molecular formula = r x empirical formula
molecular formula is 2` xx CH_2Cl i.e. C_2H_4Cl_2` .