This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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Alright m — let's do this. Problem 3.44 a) Empirical Formula Step 1: Calculate the mass of each element in 100 g of the compound. Given 30.45 mass % N, the remaining percentage is oxygen. Mass of N = 30.45 g Mass of O = 100.00 g - 30.45 g = 69.55 g Step 2: Convert the mass of each element to moles. Atomic mass of N = 14.01 g/mol Atomic mass of O = 16.00 g/mol Moles of N = 30.45 g14.01 g/mol = 2.1734 mol Moles of O = 69.55 g16.00 g/mol = 4.3469 mol Step 3: Determine the simplest whole-number mole ratio. Divide each mole value by the smallest number of moles (2.1734 mol). Ratio for N = 2.1734 mol2.1734 mol = 1 Ratio for O = 4.3469 mol2.1734 mol ≈ 2.000 The mole ratio of N:O is 1:2. The empirical formula is NO_2. b) Molecular Formula Step 4: Calculate the empirical formula mass. Empirical formula mass of NO_2 = (1 × 14.01 g/mol) + (2 × 16.00 g/mol) = 14.01 g/mol + 32.00 g/mol = 46.01 g/mol Step 5: Determine the multiplier (n) and the molecular formula. Given molar mass = 90 ± 5 g/mol. We will use 90 g/mol for the calculation. n = Molar massEmpirical formula mass = 90 g/mol46.01 g/mol ≈ 1.956 ≈ 2 Multiply the subscripts in the empirical formula by n=2. Molecular formula = (NO_2)_2 = N_2O_4. The molar mass of N_2O_4 is 2(14.01) + 4(16.00) = 28.02 + 64.00 = 92.02 g/mol, which is within the 90 ± 5 g/mol range. The molecular formula is N_2O_4. Problem 3.45 a) Empirical Formula Step 1: Calculate the mass of each element in 100 g of the compound. Given 79.1 mass % Cl, the remaining percentage is silicon. Mass of Cl = 79.1 g Mass of Si = 100.0 g - 79.1 g = 20.9 g Step 2: Convert the mass of each element to moles. Atomic mass of Si = 28.09 g/mol Atomic mass of Cl = 35.45 g/mol Moles of Si = 20.9 g28.09 g/mol = 0.7440 mol Moles of Cl = 79.1 g35.45 g/mol = 2.2313 mol Step 3: Determine the simplest whole-number mole ratio. Divide each mole value by the smallest number of moles (0.7440 mol). Ratio for Si = 0.7440 mol0.7440 mol = 1 Ratio for Cl = 2.2313 mol0.7440 mol ≈ 2.999 ≈ 3 The mole ratio of Si:Cl is 1:3. The empirical formula is SiCl_3. b) Molecular Formula Step 4: Calculate the empirical formula mass. Empirical formula mass of SiCl_3 = (1 × 28.09 g/mol) + (3 × 35.45 g/mol) = 28.09 g/mol + 106.35 g/mol = 134.44 g/mol Step 5: Determine the multiplier (n) and the molecular formula. Given molar mass = 269 g/mol. n = Molar massEmpirical formula mass = 269 g/mol134.44 g/mol ≈ 1.999 ≈ 2 Multiply the subscripts in the empirical formula by n=2. Molecular formula = (SiCl_3)_2 = Si_2Cl_6. The molar mass of Si_2Cl_6 is 2(28.09) + 6(35.45) = 56.18 + 212.70 = 268.88 g/mol, which is very close to 269 g/mol. The molecular formula is Si_2Cl_6. Drop the next question!