| ⇦ |
| ⇨ |
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: - Oxidation number of Fe in Fe₃O₄ is
- The number of acidic protons in H₃PO₃ are
- Which statements is wrong about pH and H⁺
- Which of the following amino acid is optically inactive
- Impurities present in the ore react to form a fusible substance known as
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
- Oxidation number of Fe in Fe₃O₄ is
- The number of acidic protons in H₃PO₃ are
- Which statements is wrong about pH and H⁺
- Which of the following amino acid is optically inactive
- Impurities present in the ore react to form a fusible substance known as
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