JEE Main | 2018General Organic ChemistryEasy

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 :

Options

A.

C2H4O

B.

C3H4O2

C.

C2H4O3

D.

C3H6O3

Solution

12xy=61
2x=y for CxHyOz
CxHy(g) + x+y4 O2(g) → xCO2(g) + y2H2O(l)
no. of oxygen atom in CxHyOz = z
no. of oxygen atom required for CxHy
combustion is x+y4×2=2x+y2
So z = 122x+y2
z = x +y4
z = x + 2x4=3x2
x : 2x : 3x2
2x : 4x : 3x
2 : 4 : 3
Hence C2H4O3

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