A compound contains 4.07% hydrogen, 24.27% Carbon and 71.65% Chlorine. Its molar mass is 90.96g. What are it's empirical and molecular formula?

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Solution

Hydrogen

Carbon

Chlorine

Mass

4.07

24.27

71.65

RAM

1

12

35.5

No. of moles

4.07/1= 4.07

24.27/12 = 2.02

71.65/35.5 = 2.02

Divided by the least number

4.07/2.02 = 2

2.02/2.02= 1

2.02/2.02= 1

Emprical formula therefore is H2CCL
The molecular formula is (EF)n = MF [EF = emprical formula , MF= molecular formula ]
(1+1+12+35.5) = 49.5
= 90.96/49.5
= 1.8 = 2
Then the molecular formula will be, (H2CCL) =H4C2CL2
Therefore the molecular formula is C2H4CL2 then it can be namedd as Dichloroethane

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	Hydrogen	Carbon	Chlorine
Mass	4.07	24.27	71.65
RAM	1	12	35.5
No. of moles	4.07/1= 4.07	24.27/12 = 2.02	71.65/35.5 = 2.02
Divided by the least number	4.07/2.02 = 2	2.02/2.02= 1	2.02/2.02= 1