0.45g of a dibasic organic acid upon combustion produced 0.44g of CO_(2) and 0.09g of H_(2)O. 0.76g of its silver salt when ignited gave 0.54g of pure silver. What is the fomula of the acid? (Ag=108)

0.45g of a dibasic organic acid upon combustion produced 0.44g of CO_(

1.	0.45g of a dibasic organic acid upon combustion produced 0.44g of CO_(2) and 0.09g of H_(2)O. 0.76g of its silver salt when ignited gave 0.54g of pure silver. What is the fomula of the acid? (Ag=108)
Answer» Solution :Moles of C in `CO_(2)=1 xx` moles of `CO_(2)` `=1 xx (0.44)/(44)= 0.01` Weight of `C= 0.01 xx 12= 0.12g` Moles of H in `H_(2)O=2 xx` moles of `H_(2)O` `2 xx (0.09)/(18)=0.01` Weight of `H= 0.01 XX1= 0.01g` Weight of O = wt. of acid - (wt. of C + wt. of H) `=0.45-(0.12+ 0.01)g` = 0.32g Mole of `O= (0.32)/(16)= 0.02` Mole of `C: H: O= 0.01 : 0.01: 0.02` `=1: 1: 2` Empirical formula is `CHO_(2)` Now 0.76g of Ag salt of the dibasic acid gives 0.54 g of pure silver, i.e., `{:("dibasic acid",rarr Ag " salt"rarr,Ag),(,0.76g,0.54g),(,"(contains2 Ag atoms)",):}` We know that the Ag salt of the dibasic acid is formed by the replacement of 2H atoms of the acid by 2Ag atoms. Applying POAC for the Ag atoms, moles of Ag atom in Ag salt= moles of Ag in the product `2XX` moles of Ag salt= moles of Ag in the product `2 xx (0.76)/("mol. wt. of Ag salt") = (0.54)/(108)` Molecular weight of the salt = 304 `therefore` molecular weight of the acid = mol. wt of salt `-2 xx` at wt of Ag + 2`xx` at. wt. of H `=304 -216+2=90` Hence, `("molecular formula weight")/(" empirical formula weight")= (90)/(45)=2` `therefore` molecular formula is `(CHO_(2))_(2)` i..e., `(COOH)_(2)`