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A dibasic acid containing C, H and O was found to contain C = 26.7% and H = 2.2%. The vapour density of its dimethyl ester was found to be 73. The molecular formula of the acid is

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${CH}_{2} O_{2}$

$C_{3} H_{3} O_{6}$

$C_{3} H_{3} O_{4}$

$C_{2} H_{2} O_{4}$

answer is B.

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Detailed Solution

$C = 26.7 % \frac{26.7}{12} = \frac{2.22}{2.2} = 1$

$H = 2.2 % \frac{2.2}{1} = \frac{2.2}{2.2} = 1$

$O = 71 % \frac{71}{16} = \frac{4.44}{2.2} = 2$

$E.F = {CHO}_{2}$

$E:F.wt = 12 + 1 + 32$

$= 45$

$M.Wt = 2 \times 73$

$= 146$

$n = \frac{M.F}{E.F} = \frac{146}{45} = 3.2 = 3$

$M.F = {(E.F)}_{n}$

$C_{3} H_{3} O_{6}$

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