1. It’s time to learn about...

2. Stoichiometry

3. Stoichiometry : Mole Ratios to Determining Grams of Product At the conclusion of our time together, you should be able to: 1. Review the conversion of particles or grams to moles 2. Determine mole ratios from a balanced chemical equation 3. Determine the amount of product produced when given the amount of reactants

4. Review the Molar Mass of Compounds The molar mass (MM) of a compound is determined by adding up all the atomic masses for the molecule (or compound) ◦ Ex. Molar mass of CaCl 2 ◦ Avg. Atomic mass of Calcium = 40.08g ◦ Avg. Atomic mass of Chlorine = 35.45g ◦ Molar Mass of calcium chloride = 40.08 g/mol Ca + (2 X 35.45) g/mol Cl  110.98 g/mol CaCl 2 17 Cl 35.45 Cl 20 Ca 40.08a

5. Practice Calculate the Molar Mass of calcium phosphate ◦ Formula = ◦ Masses elements: ◦ Molar Mass = 310.18 g Ca 3 (PO 4 ) 2

6. Calculations Moles molar mass Avogadro’s number Grams Moles Particles Everything must go through Moles!!!

7. Flowchart Atoms or Molecules Moles Mass (grams) Divide by 6.02 X 10 23 Multiply by 6.02 X 10 23 Multiply by atomic/molar mass from periodic table Divide by atomic/molar mass from periodic table

8. Chocolate Chip Cookies!! 1 cup butter 2 eggs 1/2 cup white sugar 1 teaspoon salt 1 cup packed brown sugar 1 teaspoon vanilla extract 2 1/2 cups all-purpose flour 1 teaspoon baking soda 2 cups semisweet chocolate chips Makes 3 dozen How many eggs are needed to make 3 dozen cookies? How much butter is needed for the amount of chocolate chips used? How much brown sugar would I need if I had 1½ cups white sugar?

9. Cookies and Chemistry…Huh!?!? Just like chocolate chip cookies have recipes, chemists have recipes as well Just like chocolate chip cookies have recipes, chemists have recipes as well Instead of calling them recipes, we call them reaction equations Instead of calling them recipes, we call them reaction equations Furthermore, instead of using cups and teaspoons, we use moles Furthermore, instead of using cups and teaspoons, we use moles Lastly, instead of eggs, butter, sugar, etc. we use chemical compounds as ingredients Lastly, instead of eggs, butter, sugar, etc. we use chemical compounds as ingredients

10. Chemistry Recipes Looking at a reaction tells us how much of something you need to react with something else to get a product (like the cookie recipe) Be sure you have a balanced reaction before you start!  Example: 2 Na + Cl 2  2 NaCl  This reaction tells us that by mixing 2 moles of sodium with 1 mole of chlorine we will get 2 moles of sodium chloride  What if we wanted 4 moles of NaCl?  10 moles?  50 moles?

11. Practice Write the balanced reaction for hydrogen gas reacting with oxygen gas. 2 H 2 + O 2  2 H 2 O ◦ How many moles of reactants are needed? ◦ What if we wanted 4 moles of water? ◦ What if we had 3 moles of oxygen, how much hydrogen would we need to react, and how much water would we get? ◦ What if we had 50 moles of hydrogen, how much oxygen would we need, and how much water would be produced?

12. Mole Ratios These mole ratios can be used to calculate the moles of one chemical from the given amount of a different chemical Example: How many moles of chlorine are needed to react with 5 moles of sodium (without any sodium left over)? Na + Cl 2  NaCl 2 Na + Cl 2  2 NaCl

13. 5 mol Na Question Answer 2 Na + Cl 2  2 NaCl # moles of Cl 2 = x 1 mol Cl 2 2 mol Na = 2.5 mol Cl 2

14. Mole-Mole Conversions How many moles of sodium chloride will be produced if you react 2.6 moles of chlorine gas with an excess (more than you need) of sodium metal? 2 Na + Cl 2  2 NaCl 2.6 mol Cl 2 x # moles of NaCl = 2 mol NaCl 1 mol Cl 2 = 5.2 mol NaCl

15. Mass-Mole We can also start with mass and convert to moles of product or another reactant We use molar mass and the mole ratio to get to moles of the compound of interest ◦ Calculate the number of moles of the combustion of ethane (C 2 H 6 ) needed to produce 10.0 g of water ◦ 2 C 2 H 6 + 7 O 2  4 CO 2 + 6 H 2 0

16. Mass-Mole 2 C 2 H 6 + 7 O 2  4 CO 2 + 6 H 2 0 = 0.185 mol C 2 H 2 10.0 g H 2 O x 1 mol H 2 O 18.02 g H 2 O x 2 mol C 2 H 2 6 mol H 2 O

17. Mass-Mass Conversions Most often we are given a starting mass and want to find out the mass of a product we will get (called theoretical yield) or how much of another reactant we need to completely react with it (no leftover ingredients!) Now we must go from grams to moles, to mole ratios, and back to grams of the compound we are interested in.

18. Mass-Mass Conversion Ex. Calculate how many grams of ammonia are produced when you react 2.00 g of nitrogen with excess hydrogen. N 2 + 3 H 2  2 NH 3 2.00 g N 2 1 mol N 2 2 mol NH 3 17.06 g NH 3 28.02 g N 2 1 mol N 2 1 mol NH 3 = 2.4 g NH 3

19. Practice Mass-Mole-Mass How many grams of calcium nitride are produced when 2.00 g of calcium reacts with an excess of nitrogen? Ca + N 2  Ca 3 N 2 3 Ca + N 2  Ca 3 N 2

20. Mass-Mole-Mass = 2.47 g Ca 3 N 2 2.00 g Ca x 1 mol Ca 40.08 g Ca x 1 mol Ca 3 N 2 3 mol Ca x 148.26 g Ca 3 N 2 1 mol Ca 3 N 2

21. Stoichiometry : Mole Ratios to Determining Grams of Product Let’s see if you can: 1. Review the conversion of particles or grams to moles 2. Determine mole ratios from a balanced chemical equation 3. Determine the amount of product produced when given the amount of reactants

22. Mass-Mole-Molecules: Determine the number of molecules in 73 g of water 73 g H 2 O # H 2 O molecules = x 1 mol H 2 O 18.02 g H 2 O = 2.4 x 10 24 molecules H 2 O x 6.02x10 23 molecules 1 mol H 2 O

23. Try this one: Calculate the mass in grams of iodine required to react completely with 0.50 moles of aluminum. Al + I 2  AlI 3 2 Al + 3 I 2  2 AlI 3 = 190 g I 2 0.50 mol Al x 3 mol I 2 2 mol Al x 253.80 g I 2 1 mol I 2

24. Try this one: Calculate the mass in grams of iodine required to react completely with 0.50 g of aluminum. Al + I 2  AlI 3 2 Al + 3 I 2  2 AlI 3 = 7.1 g I 2 x 3 mol I 2 2 mol Al x 253.80 g I 2 1 mol I 2 0.50 g Al x 1 mol Al 26.98 g Al