A dibasic organic acid gave the following results: C = 34.62%, H = 3.84%, 0.1075g of this acid consumes 20mL of 0.1N NaOH for complete neutralisation. Find out the molecular formula of the acid.

A dibasic organic acid gave the following results: C = 34.62%, H = 3.8

1.	A dibasic organic acid gave the following results: C = 34.62%, H = 3.84%, 0.1075g of this acid consumes 20mL of 0.1N NaOH for complete neutralisation. Find out the molecular formula of the acid.
Answer» Solution :Calculation of empirical formula: `{:("Element","Percentage","At.mass","Relative number of atoms","Simplest ratio of atoms"),("Carbon",34.62,12,(34.62)/(12)=2.88,(2.88)/(2.88)=1xx3=3),("Hydrogen",3.84,1,(3.84)/(1)=3.84,(3.84)/(2.88)=1.33xx3 =4),("Oxygen",overset(61.54)("by difference"),16,(61.54)/(16)=3.84,(3.84)/(2.88)=1.33 xx 3 =4):}` Empirical formula of the acid `= C_(3)H_(4)O_(4)` Empirical formula mass `= (3 xx 12) +(4 xx1) +(4 xx 16) = 104` Calculation of molecular mass: `20mL 0.1N NaOH -= 0.1075g` acid `20 xx 0.1 ML N NaOH -= 0.1075g` acid So, `1000 mL 1N NaOH -= (0.1075)/(20 xx 0.1) xx 1000g` acid `-= 53.75g` acid Eq. mass of the acid `= 53.75` MOL. mass of the acid = Eq. mass `xx` Basicity `= 53.75 xx 2 = 107.50` `n = ("Mol. mass")/("Emp. mass") = (107.50)/(104.0) ~~1` Molecular formula `= C_(3)H_(4)O_(4)`