NCERT Chmistry (XI) 1.3 Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% dioxygen by mass.

Method - 2 

The easiest way to do this is to assume that you have 100g of compound. Then you will have 69.9g of iron and 30.1g of oxygen.

You will need to find how many moles there are of each. Divide these masses by the molar mass of each element (for oxygen, just use the molar mass of O, not O2, as the calculation is easier with atoms than dioxygen):

69.9g / 55.845g/mol = 1.25mol Fe
30.1g / 15.999g/mol = 1.88mol O

Next, find the ratio of these 2 molar amounts. If we have 1 mole of Fe, we will have 1.88 / 1.25 = 1.50 mol O. Make everything whole numbers now: 1.50 is half of 3, so multiply both by 2. We get 2 moles of Fe to every 3 moles of O, so the empirical formula is Fe2O3.

Method - 3 

Divide through by atomic mass
Fe = 69.9/55.845 = 1.25
O = 30.1*15.99 = 1.88
Again divide by smallest number;
Fe = 1.0
O = 1.5

Bring to whole integers by multiplying by 2

Fe = 2
O = 3

Empirical formula = Fe2O3

Method - 4

Iron oxide has Fe = 69.9%

= 30.1%

Empirical formula can be calculated with help of following table:

Element	% mass	Atomic mass	Relative no. of moles (% mass/ atomic mass)	Simple ratio (Relative no. of moles/ smallest relative no. of moles)	Simplest whole no. ratio
Fe	69.9	56	69.9/ 56 = 1.25	1.25/ 1.25= 1	1 × 2 = 2
O	30.1	16	30.1/ 16 = 1.89	1.89/ 1.25= 1.5	1.5 × 2= 3

Therefore, empirical formula is Fe₂O_3.

_{Method - 5}