Answer
The empirical formula is $C_{3}H_{4}O_{2}$.
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: 50.0% C, 5.6% H, and 44.4% O. Assuming 100 g compound we get the masses of each element: 50.0 g C, 5.6 g H, 44.4 g O.
Now we convert masses to moles:
Moles of C: 50.0 g$\times\frac{1 mol(C)}{12.0 g (C)}$ = 4.17 mol C
Moles of H: 5.6 g$\times\frac{1 mol(H)}{1.00 g (H)}$ = 5.60 mol H
Moles of oxygen: 44.4g$\times\frac{1 mol (O)}{16.0 g (O) }$ = 2.78 mol O
Divide the mole number by the smallest one.
For C: $\frac{4.17mol}{2.78mol}$ = 1.50
For H: $\frac{5.60mol}{2.78mol}$ = 2.02
For O: $\frac{2.78mol}{2.78mol}$ = 1.00
We round the last digit, that is subject to experimental error, and we get $C_{1.5}H_{2}O_{1}$.
We see that subscript for C is not integers, but that one can be turned into integer by multiplying by 2. Remember we have to multiply by 2 the subscript for H and O as well in order to get ratio of elements in the compound.
So we have for C: 1.5x2 = 3
for H : 2 x 2 = 4
for O: 1 x 2 = 2
So the empirical formula is $C_{3}H_{4}O_{2}$.