Determine the empirical formula of an oxide of iron,which has 69.9% iron and 30.1% dioxygen by mass.
Iron:% by mass= 69.9Atomic mass= 55.85Moles of element( relative no. of moles)= 69.9/55.85 = 1.25Simplest molar ratio = 1.25/1.25 = 1Simplest whole no. molar ratio= 2Oxygen:% by mass = 30.1Atomic mass = 16.00Moles of element =30.1/16.00 = 1.88Simplest molar ratio = 1.88/1.25 = 1.5Simplest whole no. molar ratio = 3Fe2O3
Empirical formula of a compound may be defined as the formula which gives the simplest whole number ratio of the atoms of various elements present in one molecule of the compound.Form the available data Percentage of iron = 69.9Percentage of oxygen= 30.1 Total percentage of iron & oxygen= 69.9+30.1= 100%Step 1 calculation of simplest whole number ratios of the elements\tElementPercentageAtomic massAtomic ratioSimplest ratioSimplest whole no ratioFe69.955.8469.9/55.84=1.251.25= 12O30.11630.1/16 = 1.881.88=1.53\tStep 2 writing the empirical formula of the compoundThe empirical formula of the compound = Fe2\xa0O3