Empirical and Molecular Formulas - PowerPoint PPT Presentation

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Empirical and Molecular Formulas

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  1. Empirical and Molecular Formulas Part 2: Calculating the empirical formula

  2. When you complete this presentation, you will be able to • determine the empirical formula of a compound from the percent composition of that compound • determine the molecular formula of a compound from the empirical formula and molar mass of the compound Objectives:

  3. Chemical formulas that have ... • the lowest whole number ratio of atoms in the compound are empiricalformulas. • the total number of atoms in the compound are molecularformulas. Introduction

  4. If we know the chemical formula of a compound, then we can find the percent composition of each of the atomsin the compound. • We use the molar mass of the compound and the average atomic masses of the atoms in the compound. atomic mass of atoms percent composition = ✕100% molar mass of compound Introduction

  5. If we know the percent composition of a compound, then we can find the empiricalformula of the compound. • First, we assume that we have 100 g of the compound and find the mass of each atom in the compound. • Second, we find the number of molsof each atom. • Third, we find the lowest whole number ratio of mols. Finding the Empirical Formula

  6. For example, we have a compound with 52.2% C, 13.1% H, and 34.7% O. Step 1: Assume 100 g of compound  mC = 52.2 g, mH = 13.1 g, and mO = 34.7g Step 2: Find the number of moles of each atom 52.2 g mH mC = 4.35 mol 34.7 g mO 13.1g = = = nO = nC = nH = MH 1.01 g/mol 16.0 g/mol 12.0 g/mol MO MC = 13.0 mol = 2.17 mol Finding the Empirical Formula

  7. For example, we have a compound with 52.2% C, 13.1% H, and 34.7% O. Step 1: Assume 100 g of compound  mC = 52.2 g, mH = 13.1 g, and mO = 34.7g Step 2: Find the number of moles of each atom nC = 4.35 mol, nH = 13.0 mol, and nO = 2.17 mol Step 3: Find the lowest whole number ratio of mols • We find the proper ratios by dividing the lowest number of mols into the higher number of mols. • We will be dividing nO into nC and nO into nH. Finding the Empirical Formula

  8. For example, we have a compound with 52.2% C, 13.1% H, and 34.7% O. Step 1: Assume 100 g of compound  mC = 52.2 g, mH = 13.1 g, and mO = 34.7g Step 2: Find the number of moles of each atom nC = 4.35 mol, nH = 13.0 mol, and nO = 2.17 mol Step 3: Find the lowest whole number ratio of mols 2 nC 4.35 mol 6 13.0 mol nH = = = = 2.17 mol 2.17 mol nO 1 1 nO Finding the Empirical Formula

  9. For example, we have a compound with 52.2% C, 13.1% H, and 34.7% O. Step 1: Assume 100 g of compound  mC = 52.2 g, mH = 13.1 g, and mO = 34.7g Step 2: Find the number of moles of each atom nC = 4.35 mol, nH = 13.0 mol, and nO = 2.17 mol Step 3: Find the lowest whole number ratio of mols This gives the empirical formula: C2H6O nC 2 nH 6 = = nO 1 nO 1 Finding the Empirical Formula

  10. For example, we have a compound with 44.9% K, 18.4% S, and 36.7% O. Step 1: Assume 100 g of compound  mK = 44.9 g, mS = 18.4 g, and mO = 36.7g Step 2: Find the number of moles of each atom 44.9 g mS mK = 1.15 mol 36.7 g mO 18.4 g = = = nO = nK = nS= 32.1 g/mol Mk MS 16.0 g/mol 39.1 g/mol MO = 0.573 mol = 2.29 mol Finding the Empirical Formula

  11. For example, we have a compound with 44.9% K, 18.4% S, and 36.7% O. Step 1: Assume 100 g of compound  mK = 44.9 g, mS = 18.4 g, and mO = 36.7g Step 2: Find the number of moles of each atom nK = 1.15 mol, nS = 0.573 mol, and nO = 2.29 mol Step 3: Find the lowest whole number ratio of mols • We find the proper ratios by dividing the lowest number of mols into the higher number of mols. • We will be dividing nS into nK and nS into nO. Finding the Empirical Formula

  12. For example, we have a compound with 44.9% K, 18.4% S, and 36.7% O. Step 1: Assume 100 g of compound  mK = 44.9 g, mS = 18.4 g, and mO = 36.7g Step 2: Find the number of moles of each atom nK = 1.15 mol, nS = 0.573 mol, and nO = 2.29 mol Step 3: Find the lowest whole number ratio of mols 2 nK 1.15 mol 4 2.29 mol nO = = = = 0.573 mol 0.573 mol nS 1 1 nS Finding the Empirical Formula

  13. For example, we have a compound with 44.9% K, 18.4% S, and 36.7% O. Step 1: Assume 100 g of compound  mK = 44.9 g, mS = 18.4 g, and mO = 36.7g Step 2: Find the number of moles of each atom nK = 1.15 mol, nS = 0.573 mol, and nO = 2.29 mol Step 3: Find the lowest whole number ratio of mols This gives the empirical formula: K2SO4 nK 2 nO 4 = = nS 1 nS 1 Finding the Empirical Formula

  14. There is a three step process to finding the empirical mass of a compound when we know the percent composition. • First, we assume that we have 100 g of the compound and find the mass of each atom in the compound. • Second, we find the number of mols of each atom. • Third, we find the lowest whole number ratio of mols. Summary