Simplest or empirical formula: Ascorbie acid (vitamin `C`) a white crystalline solid, that is present in fruits and vegetables, curves scury and may help prevent the common cold. It is composed of `40.92%C, 4.58% H`, and `54.50% O` by mass. Determine its emprical formula.
Strategy: Because one mole of atoms of any given element is `6.022xx10^(23)` atoms, the ratio of the numbers of atoms the ratio of the numbers of atoms in any sample fo a compound is the same as the ratio fo motes of atoms is that compound. Thus, the energy is to, find the relative number of moles of each element in the compound and then use the numbers to establish the mole ratio of the elements which in turn give the subsrips in the empral formula. This calculation si carried out as follow:
Step 1: Becuase the sum of all the percentages is `100%` it is conventient to solve this type of problem by considerinf exactly `100g` of the substance, which means `100.00g` of ascoric acid contains `40.92g` of `C, 4.58g` of `H`, and `54.50g` of `O`
Next we need to calculate the number of moles of atoms of each element in the compound.
Step 2: We then obtain teh simplest whole number ratio between these number that gives that ratio of atoms in the sample and hence, the simplest or empirical formula for the compound.

Explanation

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.