Answer
The empirical formula is $C_{3}H_{3}O$.
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: 65.4% C, 5.5% H, and 29.1% O. Assuming 100 g compound we get the masses of each element: 65.4 g C, 5.5 g H, 29.1 g O.
Now we convert masses to moles:
Moles of C: 65.4 g$\times\frac{1 mol(C)}{12.0 g (C)}$ = 5.45 mol C
Moles of H: 5.5 g$\times\frac{1 mol(H)}{1.00 g (H)}$ = 5.50 mol H
Moles of oxygen: 29.1g$\times\frac{1 mol (O)}{16.0 g (O) }$ = 1.82 mol O
Divide the mole number by the smallest one.
For C: $\frac{5.45mol}{1.82mol}$ = 3.00
For H: $\frac{5.00mol}{1.82mol}$ = 3.02
For O: $\frac{1.82mol}{1.82mol}$ = 1.00
We round the last digit, that is subject to experimental error, and we get the ratio 3:3:1.
So the empirical formula is $C_{3}H_{3}O$.
