`0.30g` of an organic compound containing `C, H`, and `O` an combustion yields `0.44g` of `CO_(2)` and `0.18g` of `H_(2) O`. If its molecular mass is `60 mu` the molecular mass is formula will be

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 0.30g organic compound , molecular mass 60.
% of carbon `= 12/44 x 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", "Element Percentage", "At. mass", "Relative Number of atoms", "Simplest ratio of atoms"),("Carbon", 40%, 12, 40/12 = 3.33, 3.33/3.33 =1 =2),("Hydrogen", 6.6%, 1, 6.6/1 = 6.6, 6.6/3.33 = 1.98), ("Oxygen", 53.4%, 16, 53.4/16 = 3.33 ,3.33/3.33=1):}`
Empirical formula= `CH_(2)O` empirical Mass=30
Molecular formula =n `xx` [empirical formula]
`N = ("Molecular mass of compound")/("Empirical formula mass") = 60/30 = 2`
Molecular formula `= 2 xx [CH_(2)O] = C_(2)H_(4)O_(2)`