To determine the empirical formula, we need to find the ratio of the elements in the compound based on the given percentages. Let's assume we have 100 g of the compound to make calculations easier. 1. Calculate the mass of each element in the 100 g sample: - Carbon (C): 26.7% of 100 g = 26.7 g - Hydrogen (H): 2.20% of 100 g = 2.20 g - Oxygen (O): 71.1% of 100 g = 71.1 g 2. Convert the masses of each element to moles: - Moles of C = 26.7 g / 12.01 g/mol (atomic weight of C) ≈ 2.22 mol - Moles of H = 2.20 g / 1.01 g/mol (atomic weight of H) ≈ 2.18 mol - Moles of O = 71.1 g / 16.00 g/mol (atomic weight of O) ≈ 4.44 mol 3. Find the simplest ratio of moles by dividing by the smallest number of moles (in this case, 2.18 mol): - C: 2.22 mol / 2.18 mol ≈ 1 - H: 2.18 mol / 2.18 mol = 1 - O: 4.44 mol / 2.18 mol ≈ 2 Therefore, the empirical formula is CH₂O. 4. Calculate the empirical formula weight: - Empirical formula weight = 12.01 g/mol (C) + 1.01 g/mol (H) + 16.00 g/mol (O) = 29.02 g/mol 5. Find the molecular formula by comparing the empirical formula weight to the given molecular weight: - Molecular weight / Empirical weight = 90.04 g/mol / 29.02 g/mol ≈ 3 6. Multiply the subscripts in the empirical formula by 3 to find the molecular formula: - Molecular formula = C₃H₆O₃ Therefore, the empirical formula is CH₂O and the molecular formula is C₃H₆O₃.