An organic compound containing c.h.n.gave following analysis c=40%,h=13.33%
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The question is incomplete though i know the whole question nitrogen=46.67%
atomic wt of C = 12
Relative C = 40/12 = 3.66
Ratio for C = 3.66/3.33 = 1.09
At wt of H = 1
Relative H = 13.33/1 = 13.33
Ratio for H = 13.33/3.33 = 4
At wt of N = 14
Relative N = 46.67/14 = 3.33
Ratio for N = 3.33/3.33 = 1
Hence empirical formula is CH₄N
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