An organic compound containing C, H and O gave the following analysis:
C=40%; H=6.66%; Its empirical formula would be

Options

(a) C₃H₆O
(b) CHO
(c) CH₂O
(d) CH₄O

Correct Answer:

CH₂O

Explanation:

At wt of C = 12
Rel Number for C = 40/12 = 3.22
Ratio for C = 3.66/3.33 = 1
At wt of H = 1
Rel Number for H = 6.66/1 = 6.66
Ratio for H = 6.66/3.33 = 2
At wt of O = 16
% of O = 100 – (40+6.66) = 53.34%
Rel Number for O = 53.34/16 = 3.33
Ratio for O = 3.33/3.33 = 1
Hence empirical formula is CH₂O