`0.30g` of an organic compound containing `C, H`, and `O` an combustion yields `0.44g` of `CO_(2)` and `0.18g` of `H_(2) O`. If its molecular mass is `60 mu` the molecular mass is formula will be

Explanation

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)`