Question

A compound has the following percentage composition by mass: [3] Carbon – 54.55%, Hydrogen – 9.09% and Oxygen – 36.26%. Its vapour density is 176. Find the Empirical and Molecular formula of the compound [C=12, H = 1, O = 16]​

Answer 1

Answer:

C4H8O2

Explanation:

​Let amount of organic = 100 gm

amount of carbon = 54.55 g

amount of hydrogen = 9.09 g

Amount of oxygen = 36.26 g

number of mole of carbon = 34.55/12  

 =4.54583

number of mole of hydrogen = 9.09 /1.018

=9

number of mole of oxygen =  36.26/16

=2.26

4.54833:9:2.26

2:4:1

organic compounds Empirical formula=C2H40

Molecular weight = 24+4+16=44  

(for empirical)

Molecular weight = 2× vapour density  

                           =2×44=88 g

To molecular formula be = C4H8O2

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