- For problem 1: Assume a 100 g sample, so you have 63.52 g Fe and 36.48 g S. Convert to moles: Fe = 63.52 g / 55.85 g/mol = 1.137 mol; S = 36.48 g / 32.06 g/mol = 1.138 mol. Divide by the smallest number of moles (1.137): Fe = 1.137/1.137 = 1.000; S = 1.138/1.137 ≈ 1.001. The ratio is approximately 1:1, so the empirical formula is FeS.
- For problem 2: Assume a 100 g sample, so you have 26.56 g K, 36.41 g Cr, and the remainder is oxygen: 100 - 26.56 - 36.41 = 37.03 g O. Convert to moles: K = 26.56 g / 39.10 g/mol = 0.6793 mol; Cr = 36.41 g / 52.00 g/mol = 0.7002 mol; O = 37.03 g / 16.00 g/mol = 2.314 mol. Divide by the smallest number of moles (0.6793): K = 0.6793/0.6793 = 1.000; Cr = 0.7002/0.6793 ≈ 1.031; O = 2.314/0.6793 ≈ 3.406. Multiply by 2 to get whole numbers: K = 2.000, Cr = 2.062 ≈ 2, O = 6.812 ≈ 7. The empirical formula is K₂Cr₂O₇.
- For problem 3: You are given masses in a 60.00 g sample: 38.43 g Pb, 17.83 g C, 3.74 g H. Convert to moles: Pb = 38.43 g / 207.20 g/mol = 0.1855 mol; C = 17.83 g / 12.01 g/mol = 1.485 mol; H = 3.74 g / 1.01 g/mol = 3.703 mol. Divide by the smallest number of moles (0.1855): Pb = 0.1855/0.1855 = 1.000; C = 1.485/0.1855 ≈ 8.005; H = 3.703/0.1855 ≈ 19.96 ≈ 20.0. The ratio is 1:8:20, so the empirical formula is PbC₈H₂₀.
Parent Tip: Review the logic above to help your child master the concept of empirical and molecular formula worksheet.
