⇦ | ⇨ |
An organic compound containing C, H, N gave the following analysis: C=40%; H=13.33%; N=46.67%. Its empirical formula would be
Options
(a) C₂H₇N₂
(b) CH₅N
(c) CH₄N
(d) C₂H₇N
Correct Answer:
CH₄N
Explanation:
At wt of C = 12
Rel Number for C = 40/12 = 3.66
Ratio for C = 3.66/3.33 = 1.09
At wt of H = 1
Rel Number for H = 13.33/1 = 13.33
Ratio for H = 13.33/3.33 = 4
At wt of N = 14
Rel Number for N = 46.67/14 = 3.33
Ratio for N = 3.33/3.33 = 1
Hence empirical formula is CH₄N
Related Questions: - The boiling point of 0.2 mol kg⁻¹ solution of X in water is greater than equimolal
- An organic compound ‘X’ having molecular formula C₅H₁₀O yields phenyl hydrozone
- How much energy is released when 6 moles of octane is burnt in air
- How enzymes increases the rate of reaction
- Number of molecules in one litre of water is close to
Question Type: Memory
(964)
Difficulty Level: Easy
(1008)
Topics: Basic Concepts of Chemistry
(94)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The boiling point of 0.2 mol kg⁻¹ solution of X in water is greater than equimolal
- An organic compound ‘X’ having molecular formula C₅H₁₀O yields phenyl hydrozone
- How much energy is released when 6 moles of octane is burnt in air
- How enzymes increases the rate of reaction
- Number of molecules in one litre of water is close to
Question Type: Memory (964)
Difficulty Level: Easy (1008)
Topics: Basic Concepts of Chemistry (94)
Subject: Chemistry (2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends
Leave a Reply