Question

Q4. 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 formula?

Answer 1

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

Answer 2

Hey mate!

Here's your answer!!

Number of Moles of H
= 4.07/1 = 4 (almost)
Number of Moles of C
= 24.27/12 = 2(almost)
Number of Moles of Cl
= 71/35.5 = 2​(almost)

Ratio
C:H:Cl = 2:4:2 = 1:2:1

Empirical Formula = CH2Cl

Empirical Formula Mass =12+2+35 = 49

N = 98.96/46 = 2(almost)

Molecular formula
= n x empirical formula
= C(2)H(5)Cl(2)

$hope \: it \: helps \: you.$
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