An organic compound contains C, H and O 0.30g of this compound on combustion yielded 0.44g of CO_(2) and 0.18g of H_(2)O. If the weight of 1 mole of the compound is 60, show that the molecular formula is C_(2)H_(4)O_(2)

An organic compound contains C, H and O 0.30g of this compound on comb

1.	An organic compound contains C, H and O 0.30g of this compound on combustion yielded 0.44g of CO_(2) and 0.18g of H_(2)O. If the weight of 1 mole of the compound is 60, show that the molecular formula is C_(2)H_(4)O_(2)
Answer» SOLUTION :Moles of C in `CO_(2)=1 xx` moles of `CO_(2)= (0.44)/(44)=0.01` Weight of `C+ 0.01 xx12g = 0.12g` Moles of H in `H_(2)O=2 xx` moles of `H_(2)O= (2 xx 0.18)/(18)=0.02` Weight of `H= 0.02 xx 1= 0.02g` Weight of O= weight of compound -(wt. of C + wt. of H) `=0.30 -(0.12+ 0.02)g` =0.16g `THEREFORE` mole of `O= (0.16)/(16)=- 0.01` `therefore` mole of `C: H: O = 0.01 : 0.02 : 0.01 =1:2:1` `therefore` EMPIRICAL formula = `CH_(2)O` Now, since the weight of 1 mole is the molecular weight in grams, the molecular weight of the compound is 60. `therefore ("molecular formula weight")/("empirical formula weight") = (60)/(30)=2` Thus, the molecular formula is `(CH_(2)O)_(2)` i..e, `C_(2)H_(4)O_(2)`