An organic compound on analysis gave H = 6.48 % and O = 51.42 % and the rest is
carbon. Determine its empirical formula and molecular formula given that the vapour
density of the compound is 60.
[H = 1, C = 12, O = 16]
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Answer:
% of Carbon =100−(51.42+6.8)=42.1
Element % Ratio Simplest Ratio(nearest whole number)
Carbon 41.78/12=3.48 3.48/3.21=1
Hydrogen 6.8/1=6.8 6.8/3.2=2
Oxygen 51.42/16=3.21 3.21/3.21=1
Hence, empirical formula is CH
2
O .
Explanation:
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