Question

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?

Answer 1

Hey there!


Let total mass of the molecule = 100g


Mass of C in the molecule = 24.27g


Mass of H in the molecule = 4.07g


Mass of Cl in the molecule =71.65 g


Number of moles of C in molecule = 24.27/ molar mass of carbon = 24.27 /12 = 2.02

Number of moles of H in molecule = 4.07/ molar mass of hydrogen = 4.07 /1 = 4.07

Number of moles of Cl in molecule = 71.65/ molar mass of nitrogen = 71.65 /35.45 =2.02

 

Ratio of the number of moles of C, H & Cl

i.e.

C:H:Cl = 2.02 : 4.07 : 2.02 = 1:2:1

So the empirical formula will be..

CH2Cl

Empirical formula mass = 12.01+2+35.45 = 49.5 g

 

Now

n = Molecular mass/ empirical formula mass = 98.96/49.5 = 2

 

Molecular formula = n x empirical formula = 2(CH2Cl) = C2H4Cl2


Hope it Helped!☺️

Answer 2

heya......


                   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)


tysm.#Gozmit