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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
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- Tertiary butyl alcohol gives tertiary butyl chloride on treatment with
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Question Type: Memory (964)
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Subject: Chemistry (2512)
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