Question

an organic compound contains c=40.687,h=5.085% ,o=54.288% calculate molecular formula if vapour density of compound is 59 ​

Answer 1

Answer:

C4H6O4

Explanation:

Given composition is

C=40.687,H=5.085% ,O=54.288%

So in 100g of compound we will have c=40.687 g ,h=5.085 g ,o=54.288 g

Calculation of moles:-

Carbon-  40.687/12 = 3.39 or 3.4

Hydrogen-  5.085/1 = 5

Oxygen- 54.288/16 = 3.4

Ratio of moles of C, H, O is 3.4:5:3.4

Dividing by 3.4 we get 1:1.5:1

Multiplying by 2 we get 2:3:2

So Empirical Formula: C2H3O2

Now E.F mass is 12*2 + 3*1 + 16*2 which is 59 u

And molecular mass= 2*Vapour density = 2*59

n= M.F.M/E.F.M ie. 2*59/59 = 2

Molecular formula is n x E.F which is C4H6O4

Hope it helps.