NTA Abhyas JEE Main2020ChemistryHydrocarbonsPractice
The ratio of mass percent of C and H of an organic compound C X H Y O Z is 6 : 1. If one molecule of the above compound C X H Y O Z contains half as much oxygen as required to burn one molecule of compound C X H Y completely to C O 2 and H 2 O . The empirical formula of compound C X H Y O Z is
Options
- AC 2 H 4 O 3
- BC 3 H 6 O 3
- CC 2 H 4 O
- DC 3 H 4 O 2
Correct answer
A. C 2 H 4 O 3
Step-by-step solution
Ratio of mass % of C and H in C X H Y O Z is 6 : 1. Therefore, Ratio of mole % of C and H in C X H Y O Z will be 1 : 2. Therefore x : y = 1 : 2, which is possible in options 1, 2 and 3. Now oxygen required to burn C X H Y C X H Y + x + y 4 O 2 → x C O 2 + y 2 H 2 O Now z is half of oxygen atoms required to burn C X H Y So Z = 2 X + Y 2 2 = X + Y 4 Now putting values of x and y from the given options Option A, x = 2 , y = 4 Z = 2 + 4 4 = 3 Option B, x = 3 , y = 6 Z = 3 + 6 4 = 4.5 Therefore correct option is 1 C 2 H