Because the compound is an oxide it contains only nitrogen and oxygen.
The % oxygen in the compound is 100-25.9% =74.1%
Remember in empirical formula we can treat the % as having 100g of the compound, there would be 25,9 g nitrogen and 74.1 g oxygen.
Steps in empirical formula
1. Find the number of moles of each element
Nitrogen 25.9/14.0 = 1.85; Oxygen 74.1/16.0 =4.63
2. Dividing by the smallest # of moles will give us the ration of moles of element in the compound
N : O
1.85/1.85 4.63/1.85
1 :2.5
Empirical formula is N2O5
Molecular formula when obtained gives the molecular weight or molar mass of the compound.
Because we are dealing with a gas and the conditions given 1.00 atm and 273.15 represents STP,
we should know that one mole of any gas at STP occupies 22.4 liters and the mass of one mole of a gas is the same as its molar /molecular mass.
From the information given, we can find the molar/molecular mass
(22.4 L /1 mol gas) x (54g /11.2 L) = 108 g/mol.
Empirical formula is related to molecular formula as follows
n x{empirical formula wt } = molecular weight
Empirical formula weight = 2(14.0) + 5(16.0) = 108g
n=1 which means that the empirical formula and the molecular formula is the same N2O5
