A dibasic organic acid gave the following results on analysis : `0.2496 gm` of acid gave `0.3168 gm CO_2` and `0.0864 gm H_2 O` Further, `0.1092 gm` of acid was exactly neutralised by `21 ml` of `N//10 NaOH`. Calculate the molecular formula of the acid and write down its structural formula.

Explanation

Percentahe of `C = (12)/(44) xx (Wt. of CO_2 xx 100)/(Wt. "of compound")`
=`(12 xx 0.3168 xx 100)/(44 xx 0.2496) = 34.61 %`
Percentage of `H = (2)/(18) xx (Wt. of H_2 O xx 100)/(Wt. "of compound")`
=`(2 xx 0.0864 xx 100)/(18 xx 0.2496) = 3.84 %`
Percentahe of `O = [100 - (34.61 + 2.84)]`
Calculation of `EF` :
`{:(,"Simpleast ration","Whole number ration"),(C=34.61/12=2.88,2.88/2.88=1,3),(H=3.84/1=3.84,3.84/2.88=1.3,4),(O=61.54/16=3.84,3.84/2.88=1.3,4):}`
`EF` of acid `= C_3 H_4 O_4, EFW = 36 + 4 + 64 = 104`
Calculation of `Ew` of acid.
Equivalent of acid = Equivalent of base
`(0.1092)/(Ew) = 21 xx (1)/(10) xx 10^(-3) eq.`
`Ew = (0.1092 xx 10)/(21 xx 10^-3) = 52`
`Mw` of acid = `Ew xx basicity = 52 xx 2 = 104`
Structure of dibasic acid :
= `104 - 90 =14`
`[Wt. of 2 (COOH) = 45 xx 2 = 90]`
`(Wt. of RCH = 14)` It should be `(CH_2 -)` group.
Therefore, structure of acid is :
.