1. Determine the Empirical Formula
* Convert grams to moles:
- Moles of Nitrogen (N): 1.52 g / 14.01 g/mol = 0.108 mol
- Moles of Oxygen (O): 3.47 g / 16.00 g/mol = 0.217 mol
* Find the simplest whole-number ratio: Divide both mole values by the smaller one (0.108 mol):
- N: 0.108 mol / 0.108 mol = 1
- O: 0.217 mol / 0.108 mol ≈ 2
* The empirical formula is NO₂
2. Determine the Molecular Formula
* Calculate the empirical formula mass:
- NO₂: (14.01 g/mol) + 2(16.00 g/mol) = 46.01 g/mol
* Find the whole-number multiple:
- Divide the given molar mass range (90-95 g/mol) by the empirical formula mass:
- 90 g/mol / 46.01 g/mol ≈ 1.96
- 95 g/mol / 46.01 g/mol ≈ 2.06
- The whole-number multiple is likely 2.
* Multiply the empirical formula subscripts by 2:
- The molecular formula is N₂O₄
3. Determine the Accurate Mass
* Calculate the molar mass of N₂O₄:
- 2(14.01 g/mol) + 4(16.00 g/mol) = 92.02 g/mol
Therefore, the molecular formula of the compound is N₂O₄ and its accurate mass is 92.02 g/mol.
