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An organic compound containing C, H and O gave the following analysis:
C=40%; H=6.66%; Its empirical formula would be
Options
(a) C₃H₆O
(b) CHO
(c) CH₂O
(d) CH₄O
Correct Answer:
CH₂O
Explanation:
At wt of C = 12
Rel Number for C = 40/12 = 3.22
Ratio for C = 3.66/3.33 = 1
At wt of H = 1
Rel Number for H = 6.66/1 = 6.66
Ratio for H = 6.66/3.33 = 2
At wt of O = 16
% of O = 100 – (40+6.66) = 53.34%
Rel Number for O = 53.34/16 = 3.33
Ratio for O = 3.33/3.33 = 1
Hence empirical formula is CH₂O
Related Questions: - The ratio of the difference in energy between first and second Bohr orbit
- CrO₃ dissolves in aqueous NaOH to give
- Which of the following is obdained when acetone is treated with bleaching powder
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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 ratio of the difference in energy between first and second Bohr orbit
- CrO₃ dissolves in aqueous NaOH to give
- Which of the following is obdained when acetone is treated with bleaching powder
- Lanthanide contraction is due to increase in
- Which of the following elements are analogous to the lanthanides
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
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