02g of a monobasic organic acid containing C, H and O on combustion gave 0.505g of CO_(2) and 0.0892 g of H_(2)O. 0.183g of this acid required 15cc of N/10 NaOH for exact neutralisation. Find the molecular formula of the acid.

02g of a monobasic organic acid containing C, H and O on combustion ga

1.	02g of a monobasic organic acid containing C, H and O on combustion gave 0.505g of CO_(2) and 0.0892 g of H_(2)O. 0.183g of this acid required 15cc of N/10 NaOH for exact neutralisation. Find the molecular formula of the acid.
Answer» Solution :Moles of `C=1 xx` moles of `CO_(2)= (0.505)/(44)=0.0115` Wt. of `C= 0.0115 xx 12= 0.1380g` Moles of `H=2 xx` moles of `H_(2)O=2 xx (0.0892)/(18)=0.0099` wt of `H= 0.0099 xx1= 0.0099g` Wt of O = wt of the compound -(wt. of C + wt. of H) `=0.2 -(0.1380 + 0.0099)= 0.0521g` Moles of `O= (0.0521)/(16)=0.0033` `THEREFORE` moles of `C: H: O= 0.0115: 0.0099: 0.0033` `=115: 99: 33` `=3.5:3:1=7:6:2` Empirical formula is `C_(7)H_(6)O_(2)` Now, since the acid is monobasic, mol wt. = eq. wt `therefore` EQUIVALENT of the acid `=(0.183)/("eq.wt")=(0.183)/("mol wt")` `therefore` m.e. of the acid `=(0.183)/("mol. wt") xx 1000` `=(183)/("mol.wt")` m.e. of the BASE `=(1)/(10) xx 15=1.5` `because` m.e. of the acid= m.e. of the base `(183)/("mol wt")=1.5 therefore` mol. wt.=122 Since the empirical formula weight, i.e., `84+6+32+122`, is equal to the molecular formula weight, the molecular formula of the acid is the same as the empirical formula. Hence the formula is `C_(7)H_(6)O_(2)`.