2. Empirical Formula Calculations Problem: A compound is found to contain the following % by mass: 69.58% Ba 6.090% C 24.32% O What is the empirical formula? Solution: Step 1: Imagine that you have 100 g of the substance.  % will become mass in grams. Step 2: Calculate the # of moles  Mass ÷ Atomic Mass Step 3: Express moles as the simplest ratio  Divide through by the lowest number. Step 4: Write the simplest formula.

3. Empirical Formula Calculations Step 1: 69.58 g Ba 6.090 g C 24.32 g O Step 2: Ba: 69.58 g  137.33 g/mol = 0.5066 mol Ba C: 6.090 g  12.01 g/mol = 0.5070 mol C O: 24.32 g  16.00 g/mol = 1.520 mol O Step 3: Step 4: Empirical Formula is BaCO 3 Mol ratio mol 1.520 ÷ 0.5066 = 3.0000.5070 ÷ 0.5066 = 1.0010.5066 ÷ 0.5066 = 1 1.5200.50700.5066 OCBa

4. Example Problem: A compound is found to contain the following % by mass: 29.1 % Na 40.5 % S 30.4 % O What is the empirical formula?

5. Step 1: Assume 100 g 29.1 g Na 40.5 g S 30.4 g O Step 2: Calculate moles Na: 29.1 g  22.99 g/mol = 1.266 mol Na S: 40.5 g  32.06 g/mol = 1.263 mol S O: 30.4 g  16.00 g/mol = 1.90 mol O Step 3: Find Mole Ratio Step 4: Empirical Formula is Na 2 S 2 O 3 Example 1.901.2631.266 mol ratio mol 1.90 ÷ 1.263 = 1.501.263 ÷ 1.263 = 11.266 ÷ 1.263 = 1.00 OSNa **Multiply by the lowest whole number that gives whole number answers**

6. Molecular Formula Calculations Determine the molecular formula for the compound whose empirical formula is CH 2 O Molar mass of the compound must be given in the problem. Determine molar mass of the empirical formula. For CH 2 O it is 30 g/mol (12+2+16). Divide the molar mass of the compound by the molar mass of the empirical formula to get a factor: 150 g/mol  30 g/mol = 5 Multiply each subscript in the empirical formula by this factor. C 5 H 10 O 5 is the molecular formula.