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

2. Objectives: 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

3. Introduction Chemical formulas that have... the lowest whole number ratio of atoms in the compound are empirical formulas. the total number of atoms in the compound are molecular formulas.

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

5. Finding the Empirical Formula If we know the percent composition of a compound, then we can find the empirical formula 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 mols of each atom. Third, we find the lowest whole number ratio of mols.

6. Finding the Empirical Formula For example, we have a compound with 52.2% C, 13.1% H, and 34.7% O. Step 1: Assume 100 g of compound  m C = 52.2 g, m H = 13.1 g, and m O = 34.7g Step 2: Find the number of moles of each atom n C = mCmC MCMC = 52.2 g 12.0 g/mol = 4.35 mol n H = mHmH MHMH = 13.1g 1.01 g/mol = 13.0 mol n O = mOmO MOMO = 34.7 g 16.0 g/mol = 2.17 mol

7. Finding the Empirical Formula For example, we have a compound with 52.2% C, 13.1% H, and 34.7% O. Step 1: Assume 100 g of compound  m C = 52.2 g, m H = 13.1 g, and m O = 34.7g Step 2: Find the number of moles of each atom  n C = 4.35 mol, n H = 13.0 mol, and n O = 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 n O into n C and n O into n H.

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

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

10. Finding the Empirical Formula For example, we have a compound with 44.9% K, 18.4% S, and 36.7% O. Step 1: Assume 100 g of compound  m K = 44.9 g, m S = 18.4 g, and m O = 36.7g Step 2: Find the number of moles of each atom n K = mKmK MkMk = 44.9 g 39.1 g/mol = 1.15 mol n S = mSmS MSMS = 18.4 g 32.1 g/mol = 0.573 mol n O = mOmO MOMO = 36.7 g 16.0 g/mol = 2.29 mol

11. Finding the Empirical Formula For example, we have a compound with 44.9% K, 18.4% S, and 36.7% O. Step 1: Assume 100 g of compound  m K = 44.9 g, m S = 18.4 g, and m O = 36.7g Step 2: Find the number of moles of each atom  n K = 1.15 mol, n S = 0.573 mol, and n O = 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 n S into n K and n S into n O.

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

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

14. Summary 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.