percentage composition of elements
(i) percentage of carbon `=34.6` (given)
(ii). Percentage of hydrogen `=3.85` (given)
(iii). Percentage of oxygen
=100-`("percentage of " C+ " percentage of H")`
`=100-(34.6+3.85)=61.55`
(b).
(c). Calculation of the molecular formula of the acid: Empirical formula `(C_3H_3O_4)` mass of the acid
`=3xx12+4xx1+4xx16=104` (given)
but molecular mass of the acid `=104` (given)
`n=("Molecular formula mass")/("Empirical formula mass")=(104)/(104)=1`
Molecular formula of the acid
`=nxx`(Empirical formula)`=1xx(C_3H_4O_4)=C_3H_4O_4`
(d). Calculation of the basicity of the acid:
mEq. of `NaOH=7.33xx0.01` mEq. of acid `=0.0733`
`=0.0733`
Equivalelnt of acid `=0.0733xx10^-3`
equivalent of acid `=(weight)/(Ew)`
Ew of acid `=("weight")/("Equivalentof acid")=(3.812xx10^-3gm)/(0.0733xx10^-3)`
`=approx52gm`
basicity `(n)=("molecular mass" )/("equivalent")=(104)/(52)=2`
(e). Structure of the acid:
The acidic character of an organic acid is due to the presence of a corboxylic group `(-COOH)` as the functional group Since the basicity of the acid is 2, the given acid contains two `(-COOH)` groups. OUt of the molecular formula `(C_3H_4O_4),2xxC))H=C_2H_2O_4` is accounted for by the two carboxylic groups. rest of molecule of the acid `=(C_3H_4O_4-C_2H_2O_4)`
`=CH_2`
Thus, the structure of the acid is `CH_2(COOH)_2`, i.e., malonic acid.
