An organic compound contains 20.0 % carbon 6.66% hydrogen 47. 33% nitrogen and the rest was oxygen its molecular mass is 60 gram mole -1 the molecular formula of compound is
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I'm using the format Element -- Percentage Composition -- Atomic mass -- Atomic ratio.
First,
Carbon -- 40% -- 12 -- 40/12 = 3.33
Hydrogen -- 6.66% -- 1 -- 6.66/1 = 6.66
Oxygen -- 53.34% -- 16 -- 53.34/16 = 3.33
The least atomic ratio from these is 3.33
So, to get the simplest ratio, divide each by 3.33
We get C:H:O = 1:2:1
So, the empirical formula is CH₂O
Empirical formula Weight (EFW) = 12 + 2 + 16 = 30
Now, given that, molecular mass is 60 (MM = 60).
Let the molecular formula be (CH₂O)n where n is a positive integer.
n is given my MM / EFW
So, n = 60/30 = 2
So, Molecular formula = (CH₂O)2 = C₂H₄O₂
which can be written as CH₃COOH (acetic acid)
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