Answer
The empirical formula is $C_{3}H_{4}O_{4}$.
Work Step by Step
Strategy: 1- Use percentage composition to find the mass of each element in the compound. For this assume you have 100 g compound. If for example you are given 21% as the percentage composition of an element X in the compound, this is $\frac{21}{100}\times100g$ = 21 g of element X since you assumed a mass 100 g for the compound. Thus the mass of the element equals the numerical value of percentage composition. 2- Convert the masses to moles, using conversion factor 3- Divide each mole number by the smallest one in order to find the smallest integers. 4- If you do not get integers, than find a whole number to multiply the results of step 3 to get integers.
The compound contains these percentages by mass of each element: 34.6% C, 3.9% H, and 61.5% O. Assuming 100 g compound we get the masses of each element: 34.6 g C, 3.9 g H, 61.5 g O.
Now we convert masses to moles:
Moles of C: 34.6 g$\times\frac{1 mol(C)}{12.0 g (C)}$ = 2.88 mol C
Moles of H: 3.9 g$\times\frac{1 mol(H)}{1.00 g (H)}$ = 3.90 mol H
Moles of oxygen: 61.5g$\times\frac{1 mol (O)}{16.0 g (O) }$ = 3.84 mol O
Divide the mole number by the smallest one.
For C: $\frac{2.88mol}{2.88mol}$ = 1.00
For H: $\frac{3.90mol}{2.88mol}$ = 1.35
For O: $\frac{3.84mol}{2.88mol}$ = 1.33
The numbers we found for H and O are not integers.
If we multiply by 2 we will get 1.35x2= 2.70 for H and 1.33x2= 2.66 for O that still are not integers (even if we round them).
If we multiply by 3 we get 1.35x3= 4.05 for H and 1.33x3= 3.99 for O. If we round these numbers, to consider experimental error, than we get the integer 4.
Therefore we found that we have to multiply all by 3 and we get:
for C: 1.0x3 = 3.0
for H : 1.35 x 3 = 4.0
for O: 1.33 x 3 = 4.0
So the empirical formula is $C_{3}H_{4}O_{4}$.
