1. Determining Empirical Formula from Mass % Data To convert the mass % composition obtained from a combustion analysis into an empirical formula, we must convert the mass % of each type of atom into the relative number of atoms. –To do this, assume that we have 100g of sample –The mass % will then be in grams 

2. Determining Molecular Formulas Once we know the empirical formula, we need one more piece of information to determine the molecular formula: The Molar Mass Say we know the empirical formula of a compound is C 3 H 4 O 3. –All we know about this compound at this point is the ratio of the 3 elements. –We don’t know the exact number of each type of atom in the molecule. –Is the Molecular Formula C 6 H 8 O 6, C 12 H 16 O 12 or C 18 H 24 O 18 ? 

3. G: Molarity Hands down one of the most important concepts you need to master is you are going to stay in the sciences. Period. The Molar Concentration, c, of a solute in solution is the number of moles of solute divided by the volume of the solution (in liters). –Also referred to as Molarity

4. Molarity  The symbol M is used to denote the molarity of the solution 1M NaCl = 1 mole NaCl per liter of H 2 O

5. G4: Dilutions Frequently in the laboratory, you will need to make dilutions from a stock solution. This involves taking a volume from the stock and bringing it to a new volume with solvent. In order to perform these dilutions, we can use the following equation: c 1 V 1 = c 2 V 2 Where: c 1 = Stock concentration V 1 = Volume removed from stock c 2 = Target conc of new sol’n V 2 = Volume of new solution 

6. Law of Conservation of Matter “Matter can neither be created nor destroyed” – Antoine Lavoisier, 1774 If a complete chemical reaction has occurred, all of the reactant atoms must be present in the product(s)

7. Law of Conservation of Matter a) b) Stoichiometry coefficients are necessary to balance the equation so that the Law of Conservation of Matter is not violated 6 molecules of Cl 2 react with 1 molecule of P 4 3 molecules of Cl 2 react with 2 molecules of Fe

8. Example of Using Stoichiometric Coefficients

9. Balancing Chemical Reactions Let’s look at Oxide Formation Metals/Nonmetals may react with oxygen to form an oxide with the formula M x O y Example 1: Iron reacts with oxygen to give Iron (III) Oxide Fe (s) + O 2 (g) → Fe 2 O 3 (s)

10. How do we solve it? Step 1: Look at the product. There are 3 atoms of oxygen in the product, but we start with an even number of oxygen atoms. –Let’s convert the # of oxygens in the product to an even number Fe (s) + O 2 (g) → Fe 2 O 3 (s) Result: Fe (s) + O 2 (g) → 2Fe 2 O 3 (s)

11. How do we Solve It? Then, balance the reactant side and make sure the number/type of atoms on each side balance. Fe (s) + O 2 (g) → 2Fe 2 O 3 (s) Balanced Equation: 4Fe (s) + O 2 (g) → 2Fe 2 O 3 (s)

12. How do we Solve It? Example 2: Sulfur and oxygen react to form sulfur dioxide. S (s) + O 2 (g) → SO 2 (g) Step 1: Look at the reaction. We lucked out! Balanced Equation: S (s) + O 2 (g) → SO 2 (g)

13. How do we Solve It? Example 3: Phosphorus (P 4 ) reacts with oxygen to give tetraphosphorus decaoxide. P 4 (s) + O 2 (g) → P 4 O 10 (s) Step 1: Look at the reaction. The phosphorus atoms are balanced, so let’s balance the oxygens. Balanced Equation: P 4 (s) + 5O 2 (g) → P 4 O 10 (g)

14. How do we Solve It? Example 4: Combustion of Octane (C 8 H 18 ). C 8 H 18 (l) + O 2 (g) → CO 2 (g) + H 2 O (g) Step 1: Look at the reaction. Then: –Balance the Carbons C 8 H 18 (l) + O 2 (g) → 8CO 2 (g) + H 2 O (g)

15. How do we Solve It? Step 2: Balance the Hydrogens C 8 H 18 (l) + O 2 (g) → 8CO 2 (g) + 9H 2 O (g) C 8 H 18 (l) + O 2 (g) → 8CO 2 (g) + H 2 O (g) Step 3: Balance the Oxygens –Problem! Odd number of oxygen atoms –Solution: Double EVERY coefficient (even those with a value of ‘1’)

16. How do we Solve It? Step 3 (cont’d): Balance the Oxygens 2C 8 H 18 (l) + 25O 2 (g) → 16CO 2 (g) + 18H 2 O (g) C 8 H 18 (l) + 12.5O 2 (g) → 8CO 2 (g) + 9H 2 O (g) Step 4: Make sure everything checks out

17. Review of Balancing Equations