Example 1: A 14.0 g sample of a compound that contains only carbon and hydrogen is completely burned. The resulting products are 20.49 g of CO2 and 12.59 g of H2O. If the molar mass of the compound is known to be 30 g/mol, then what is the empirical and molecular formula of the compound?
Step 1: Calculate the number of moles of CO2 that are formed. This is the same as the number of moles of C in the original compound.
CO2 molar mass = C + 2O = (12.01) + 2(16.00) = 44.01 g/mol
(20.49 g CO2)(1mol/44.01 g) = 0.4656 mol
CO2 therefore 0.4656 mol C in the original compound.
Step 2: Calculate the number of moles of H2O that is formed. Since there are 2 H atoms per H2O molecule, this is 1/2 the amount of H in the original compound.
H2O molar mass = 2H + O = 2(1.01) + 16.00 = 18.02 g/mol
(12.59 g H2O)(1 mol/18.02 g) = 0.6987 mol
H2O therefore twice as much H (1.397 mol) in the original compound.
Step 3: Substitute the moles into a trial formula C0.4656H1.397
Step 4: Divide by the smallest trial subscript, so that you can
simplfy to a whole number ratio 
Therefore, CH3 is the empirical formula.
Step 5: Calculate a trial molar mass for the empirical formula
CH3 molar mass = 1C + 3H = (12.01) + 3(1.01) = 15.04 g/mol.
Step 6: Divide the actual molar mass of the compound by the trial molar mass, to find how many formula units there are in the real compound
(30 g/mol)/(15.04 g/mol) = 1.995 which is almost 2
So there are 2(CH3) units in the actual molecular formula, C2H6
Example 2: 31.0 g of rust (an iron oxide) are heated together with charcoal. A pellet of iron with a mass of 21.6 g is formed. What is the empirical formula for the rust?
You need to realize in the problem that the oxygen is not included in the final mass, since it formed carbon dioxide gas with the charcoal, and escaped into the air. However, you can figure out how much oxygen there was using the law of conservation of mass.
Step 1: Calculate the mass of oxygen that was in the original compound. Since the original compound contains both iron and oxygen, the difference in mass at the end is the mass of oxygen that is reacted: 31.0 g - 21.6 g = 9.4 g
Calculate the moles of oxygen atoms that were present.
(9.4 g O atoms)(1 mol/16.00 g) = 0.59 mol O atoms
Step 2: Calculate the moles of iron atoms that were present.
(21.6 g Fe atoms)(1 mol/55.85 g) = 0.387 mol Fe atoms
Step 3: Substitute these numbers into a trial formula: Fe0.387O0.59
Step 4: Divide by the smallest trial subscript, so that you can
simplfy to a whole number ratio
You can't have a fractional ratio (1.52) in a chemical formula. However, if you doubled this you'd have Fe2O3.04, which probably means that the formula rounds off to Fe2O3.