On analyisis, 0.2 gm of a monobasic acid gave 0.505 gm of `CO_2` and 0.0864 gm of `H_2O.0.305` gm of this acid required `25ml` of `(N)/(10)NaOH` for complete neutralisation. Find the molecular formula of the acid.

Explanation

Calculation of percentage compression
(i). Percentage of carbon `=(12)/(44)xx("mass of "CO_2" produced")/("mass of substance of taken")xx100`
`=(12)/(44)xx(0.505)/(0.2)xx100=68.86%`
(ii). Percentage of hydrogen
`=(2)/(18)xx("mass of "H_2O" produced")/("mass of substance taken")xx100`
(iii) percentage of oxygen `=100-`("Percentage of "C`+`" percetage of H")
`=100-(68.86+4.8)`
`=26.34%`
(b).
(c) Calculation of molecular mass:
mEq. of `NaOH=25xx(1)/(10)=2.5` mEq. of acid `=2.5`
Equivelent of acid`=2.5xx10^-3`
equivalength of acid `=`
`Ew=("weight")/("equivalent of acid")=(0.305)/(2.5xx10^-3)`
molecular formula:
(d). calculation of molecular formula
Empirical formula `(C_7H_6O_2)` mass of the compound `=(7xx12+6xx1+16xx2)=122`
`n=("molecular mass")/("empirical formula mass")=(122)/(122)=1` Thus, molecular formula of the compound `=nxx`("Empirical formula")`=1xx(C_7H_6O_2)=C_7H_6O_2`