0.30 g of an organic compound containing C, H and Oxygen on combustion yields 0.44 g `CO_(2)`, and 0.18 g of H:0. If one mole of compound weighs 60, then molecular formula of the compound is

Options

A. `C_(4)H_(6)O `
B. `CH_(2)O`
C. `C_(2)H_(4)O_(2)`
D. `C_(3)H_(8)O`

Answer

C

Explanation

Given 0.30 g orgnaic compoud : molecular mass 60
% of carbon ` = (12)/(100) xx (0.44)/(0.30) xx 100 =39.9 =~40%`
% of Hydrogen `= (2)/(100) xx (0.18)/(0.30) xx 100 = 6.6%`
% of oxygen `= 100 - (40+6.6) = 53.4 %`
Empirical formula :

Empirical formula ` = CH_(2)O` empirical mass = 30
Molecular formula ` = n xx ["empirical formua"]`
`N = ("Molecular mass of compound")/("Empirical formula mass ") = ( 60)/(30) = 2 `
Molecular formlua ` = 2 xx [CH_(2)O] = C_(2)H_(4)O_(2)`