C=82.76% H=17.24% Molecular formula ???

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Explanation:

Element  %composition  Mol. mass  No. of C atoms  Simplest       Rounding

                                                                                           ratio             off  

C                     82.76                   12            6.89             1× 2                 2              

H                    17.24                      1              17.24              2.5× 2           5

Number of carbon atoms present =  

12

82.76

=6.89

Number of hydrogen atoms present =  

1

17.24

=17.24

simplest ratio =  

6.89

6.89

=1  

simplest ratio =  

6.89

17.24

=2.5

∴ Empirical fromula is C  

2

H  

5

Empirical formula mass= 2×12+5 = 29

∴ V.D. = 29

Mol. mass = 2 × V.D.

=58

n=  

Empiricalformulamass

Molecularmass

=  

29

58

=2

∴ Molecular formula = C  

4

H  

10

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