Question

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

Answer 1

                    Hydrogen          Carbon             Chlorine
       mass      4.07                 24.27                 71.65
       RAM         1                      12                     35.5
no. of moles  4.07/1               24.27/12            71.65/35.5
                      4.07                    2.02                2.02
divide by the   4.07/2.02          2.02/2.02         2.02
 least no            2                           1                 1

Empirical formula therefore is H2CCl

the molecular formula is (EF)n=MF


(1+1+12+35.5)=49.5......98.96/49.5=2  n=2

(H2CCl)2=H4C2Cl2

therefore the molecular formula is C2H4Cl2 (dichloroethane)

Answer 2

In 100g of sample of the compound, 4.07g of hydrogen, 24.27g of carbon and 71.65g of chlorine are present. Moles of hydrogen= 4.07g/ 1g = 4.0  Moles of carbon= 24.27g/ 12g = 2.0  Moles of chlorine= 71.65g/ 35g= 2.0  Since 2.0 is the smallest value, so by dividing each of the mole values obtained by this smallest value we will get a ratio of 2:1:1 for H:C:Cl.   Thus, the empirical formula of the compound is CH2Cl.  For CH2Cl, empirical formula mass= 12+ (2×1) +35 = 49g.  Molar mass/ empirical formula = 98.96g/ 49g = 2=n  Therefore, Empirical formula = CH2Cl  n=2  Hence, molecular formula= C2H4Cl2.