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Question:

The ratio of mass percent of C and H of an organic compound (CXHYOZ) is 6:1. If one molecule of the above compound (CXHYOZ) contains half as much oxygen as required to burn one molecule of compound CXHY completely to CO2 and H2O. The empirical formula of compound CXHYOZ is:

C2H4O3

C3H6O3

C3H4O2

C2H4O

Solution:

CxHyOz
CxHy + nO2 → xCO2 + y/2H2O
n = (4x + y)/4
and amount of O required, m = n * 2
Given that z = m/2 = n
and mass percent ratio of C:H :: 6:1 = 12:2
therefore for each C two H are present in the molecule.
For x = 1, y = 2 ∴ z = (4*1 + 2)/4 = 1.5
Empirical formula is C2H4O3
Option B is correct.