1. Simplest (Empirical) and Molecular Formulas

2. Molecular Formula - shows the actual number of atoms Example: C 6 H 12 O 6 Simplest Formula - shows the ratio between atoms Example: CH 2 O

3. Given that a compound contains 12.7% C, 2.1% H and 85.2% Br, calculate its simplest (empirical) formula Step 1: Assume you have 100 g of the substance therefore, mass of C = 12.7 gmH = 2.1 gmBr = 85.2g MolarMass=12.01g/molMH = 1.01 g/molMBr = 79.90 Step 2: Calculate the number of moles of each using n= m/M nC =nH =nBr = = 1.06 mol = 2.1 mol = 1.07 mol 12.7g 12.01 2.1g 1.01 85.2g 79.90

4. Step 3: Ratio Calculation Divide by the smallest number of moles to figure out the ratio between the atoms. C=H =Br = C = 1H = 1.98Br = 1.01 Therefore the simplest formula is CH 2 Br 1.06 mol 2.1 mol 1.06 mol 1.07 mol 1.06 mol

5. Calculate the molecular formula for the compound in the previous example if its molar mass is 190 g/mol Step 1. Calculate the molar mass for the empirical formula, CH 2 Br. M CH2Br = 12.01 + 2(1.01) + 79.90 = 93.93 g/mol

6. Step 2. Divide the molar mass by the simplest (empirical) formula molar mass. = = 2 Step 3. Multiply this number by the empirical formula. 2 x CH 2 Br Therefore, the molecular formula is C 2 H 4 Br 2 Molar mass Simplest formula molar mass 190 g/mol 93.93 g/mol

7. Example 2: What is the empirical formula of a compound that contains 69.9 g Fe, and 30.1 g O? Step 1: Assume the mass is 100 g. Step 2: Calculate the # of moles of each element using n=m/M mFe = 69.9gmO = 30.1 g Mfe = 55.85 g/molMO = 16.00 g/mol nFe = 1.25 molnO = 1.88 mol

8. Step 3: Find the ratio by dividing by the smallest # of moles. Fe = O = Fe = 1 x 2O = 1.5 x 2 = 2 = 3 In this case, multiply by a factor (2)to get a whole number ratio. Step 4: Use the simplest whole number ratio: Fe 2 O 3 1.25 mol 1.88 mol 1.25 mol

9. Try these: p. 209, 211, 218: #11, 14, 15, 16, 18, 19, 20 p. 214 # 2, 6 p. 230 #12, 17, 18a Answers: page 231