1. Empirical Formula
* Assume a 100g sample: This makes the percentages directly translate to grams.
* Convert grams to moles: Divide each element's mass by its molar mass:
* C: 19.53 g / 12.01 g/mol = 1.626 mol
* H: 2.44 g / 1.01 g/mol = 2.416 mol
* O: 13.02 g / 16.00 g/mol = 0.814 mol
* Br: 65.01 g / 79.90 g/mol = 0.813 mol
* Find the simplest whole-number ratio: Divide each mole value by the smallest mole value (0.813 mol):
* C: 1.626 mol / 0.813 mol ≈ 2
* H: 2.416 mol / 0.813 mol ≈ 3
* O: 0.814 mol / 0.813 mol ≈ 1
* Br: 0.813 mol / 0.813 mol ≈ 1
* Empirical formula: The empirical formula is C₂H₃OBr
2. Molecular Formula
* Calculate the empirical formula mass: (2 * 12.01) + (3 * 1.01) + 16.00 + 79.90 = 138.95 g/mol
* Determine the ratio between the molecular mass and empirical formula mass: 245.8 g/mol / 138.95 g/mol ≈ 1.77
* Multiply the subscripts in the empirical formula by the ratio: Since the ratio is close to 1.77, we can round it to 2.
* C₂H₃OBr * 2 = C₄H₆O₂Br₂
Therefore, the empirical formula is C₂H₃OBr, and the molecular formula is C₄H₆O₂Br₂.
