Let `n_(C), n_(H)` and `n_(0)` be the number of moles of elements peresent.
Step 1 :
`n_(C) = (40.92g)/(12.01g mol^(-1)) = 3.407 "mol C"`
`n_(H) = (4.58g)/(4.008g "mol"^(-1)) = 4.54 "mol H"`
`n_(O) = (54.58g)/(16.00 "g mol"^(-1)) = 3.406 "mol O"`
Hence, we write the formula `C_(3.407) H_(4.54) O_(3.406)` which gives the identity and the ratious of atoms present However, since chemical formula are always written with whole numbers, we cannot have `3.407 C` atoms, `4.54 H` atoms, and `3.406 O` atoms.
Step 2 : We turn some of the subsciprs into the whole numbers by dividing all the subsrips by the smallest subscript `(3.406)`.
Step2 : We turn some of the subsrips into whole numbers by dividing all the subsrips by the smallest subsript `(3.406)`,
`C: (3.407)/(3.406) = 1 , H : (4.54)/(3.406) = 1.33, O, (3.406)/(3.406) = 1`
The gives us `CH_(1.33) O` as the formula for asorcbie acid. Next we turn 1.33, the subsciprt for `H`, into an interger (whole number) by the trail-and-corror procceduce (where we multiply the subcripts by small interges until whole numbers are found with minimum possible appocimation). In the present case, the multiplication of the subsciprits by 3 is needed :
`1.33xx1 = 1.33`
`1.33xx2 = 2.66`
`1.33xx3 = 3.99 = 4`
Thus, multipying all the subscripts by 3, we obtain `C_(3) H_(4) O_(3)` as the empirical formula for ascorbic acid.