Question

A organic compound who's molecular weight is 74 contain 48.56 percentage of carbon 8.11 percentage of hydrogen 43.240 of oxygen.determine in molecular mass?

Answer 1

Given,
The organic compound contains
48.56 % of carbon
8.11 % of hydrogen and
43.24 % of oxygen
Consider the compound as a whole is 100 g
Therefore,
Number of moles of C = 48.56 g / 12 = 4.05
Number of moles of H = 8.11 g / 1 = 8.11
Number of moles of O = 43.24 / 16 = 2.70

We divide by the smallest number
For C = 4.05 / 2.70 = 1.5
For H = 8.11 / 2.70 = 3
For O = 2.70 / 2.70 = 1
Thus formula = C_1.5H₃O
So as to get whole numbers. we can multiply by 2

Hence,
C₃H₆O₂ is the empirical formula
as in the question, molecular weight = (12x3)+(1x6)+(16x2) = 74 g/mol