The atomic mass of nitrogen is 14.01 g/mol and that of oxygen is 16.00 g/mol. So, we have 1.286 g N x (1 mol N / 14.01 g N) = 0.0917 mol N and 2.204 g O x (1 mol O / 16.00 g O) = 0.1378 mol O.
Dividing each mole value by the smallest mole value (0.0917), we get a mole ratio of N:O = 1:1.5. Since we can’t have half an atom in a formula, we multiply both numbers by 2 to get a whole-number ratio: N:O = 2:3.
So, the empirical formula of the compound is N₂O₃.
To find the molecular formula, we need to know the molar mass of the compound. You mentioned that it is 152.0 g/mol. The molar mass of N₂O₃ is (2 x 14.01) + (3 x 16.00) = 76 g/mol.
Dividing the molar mass of the compound by the molar mass of its empirical formula gives us a factor that tells us how many empirical formula units are in one molecule of the compound: 152.0 g/mol ÷ 76 g/mol = 2.
Multiplying each subscript in the empirical formula by this factor gives us the molecular formula: N₄O₆.