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Question

A certain alkaloid has 70.8%C,6.2%H and 4.1% of N and rest O. The empirical formula is


A
C20H21NO4
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B
C20H20NO4
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C
C21H20NO3
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D
C20H10NO3
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Solution

The correct option is A C20H21NO4
% of O = 100(%C+%H+%N)=100(70.8+6.2+4.1)=18.9
Element%%comp/at.massSimplest ratioEmpirical formulaC70.870.812=5.95.90.2928=20.1520H6.26.21=6.26.200.2928=21.1721N4.14.114=0.29280.29280.2928=11O18.918.916=1.1811.1810.2928=4.0334
Therefore empirical formula is : C20H21NO4

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