an organic compound has c=40.687%, h=5.085% and o=54.228%. the vapour density of compound is 59. calculate the molecular formula of the compound

Given,

C = 40.687%, H = 5.085% and O = 54.228%.

Therefore, in 100 g of compound,

C = 40.687 g

H = 5.085 g

O = 54.228 g

Moles of C = 40.687/12 = 3.4

Moles of H = 5.085/1 = 5

Moles of O = 54.228/16 = 3.4

Ratio of moles of C, H, and O = 3.4 : 5 : 3.4

Dividing by 3.4

C : H : O = 1: 1.5 :1

Multiplying by 2 to get whole number ratio

C : H : O = 2 : 3: 2

Therefore, empirical formula of the compound = C₂H₃O₂

Empirical formula mass = 12 x 2 + 1 x 3 + 2 x 16 = 59 u

Vapour density = 59

Now,

 Molecular mass = 2 x Vapour density

  = 2 x 59 u

And

n =  Molecular mass / vapour density

 = 2 x 59 / 59

  =2

Hence, the molecular formula =

  = C₄H₆O₄