1. Stoichiometry Chapter 12

2. Chocolate Chip Cookies!! 1 cup butter ;1/2 cup white sugar 1 cup packed brown sugar 1 teaspoon vanilla extract 2 eggs ; 2 1/2 cups all-purpose flour 1 teaspoon baking soda ; 1 teaspoon salt 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 many eggs would we need to make 9 dozen cookies? How much brown sugar would I need if I had 1 ½ cups white sugar?

3. Cookies and Chemistry  Just like chocolate chip cookies have recipes, chemists have recipes as well.  Instead of calling them recipes, we call them reaction equations.  Furthermore, instead of using cups and teaspoons, we use moles.  Lastly, instead of eggs, butter, sugar, etc. we use chemical compounds as ingredients.  Thus, a recipe is much like a balanced equation.

4. Stoichiometry  Greek for “measuring elements”  The calculations of quantities in chemical reactions based on a balanced equation.  We can interpret balanced chemical equations 4 ways

5. 1.First, we can think in terms of Particles: Examples: 2H 2 + O 2 → 2H 2 O Two molecules of hydrogen and one molecule of oxygen form two molecules of water. 2 Al 2 O 3  Al + 3O 2 2 formula units of Al 2 O 3 decompose to 4 atoms of Al and 3 molecules of O 2

6. Lets Look at a Grand scale: Example: 2H 2 + O 2 → 2H 2 O 2 molecules of hydrogen and 1 molecule of oxygen form 2 molecules of water. 2 dozen molecules of hydrogen and 1 dozen molecules of oxygen form 2 dozen molecules of water. 2 x (6.02 x 10 23 ) molecules of hydrogen and 1 x (6.02 x 10 23 ) molecules of oxygen form 2 x (6.02 x 10 23 ) molecules of water. 2 moles of hydrogen and 1 moles of oxygen form 2 moles of water.

7.  2. In terms of Moles : The coefficients tell us how many moles of each substance  2Al 2 O 3  Al + 3O 2  2Na + 2H 2 O  2NaOH + H 2  Remember: A balanced equation is a Molar Ratio

8.  3. In terms of Mass : The Law of Conservation of Mass applies  We can check using moles  2H 2 + O 2  2H 2 O  2 moles H 2 2.02 g H 2 1 mole H 2 = 4.04 g H 2 1 mole O 2 32.00 g O 2 1 mole O 2 = 32.00 g O 2 2 moles H 2 O 18.02 g H 2 O 1 moles H 2 O = 36.04 g H 2 O 4.04 g H 2 + 32.00g O 2 = 36.04 g H 2 O Mass Conserved

9.  4. In terms of Volume : 2H 2 + O 2  2H 2 O  At STP, 1 mol of any gas = 22.4 L  (2 x 22.4 L H 2 ) + (1 x 22.4 L O 2 )  (2 x 22.4 L H 2 O)  NOTE: mass and atoms are always conserved- however, molecules, formula units, moles, and volumes will not necessarily be conserved

10. 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

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

12. A mole ratio is a conversion factor derived from the coefficients of a balanced chemical equation interpreted in terms of moles. Writing and Using Mole Ratios

13.  The three mole ratios derived from the balanced equation above are: N 2 (g) + 3H 2 (g)  2NH 3 (g) 2 mol NH 3 1 mol N 2 3 mol H 2 2 mol NH 3 Look again at the balanced equation for the production of ammonia.

14. KNOWN moles of nitrogen = 0.60 mol N 2 UNKNOWN moles of ammonia = ? mol NH 3 Analyze : List the known and the unknown. N 2 + 3H 2  2NH 3 The conversion is mol N 2  mol NH 3. 1 mol N 2 combines with 3 mol H 2 to produce 2 mol NH 3. To determine the number of moles of NH 3, the given quantity of N 2 is multiplied by the form of the mole ratio from the balanced equation that allows the given unit to cancel. How many moles of NH3 are produced when 0.60 mol of nitrogen reacts with hydrogen?

15. Write the mole ratio that will allow you to convert from moles N 2 to moles NH 3. Calculate Solve for the unknown. 1 mol N 2 2 mol NH 3

16. Multiply the given quantity of N 2 by the mole ratio in order to find the moles of NH 3. Calculate Solve for the unknown. 0.60 mol N 2  = 1.2 mol NH 3 2 mol NH 3 1 mol N 2 Remember that the mole ratio must have N 2 on the bottom so that the mol N 2 in the mole ratio will cancel with mol N 2 in the known.

17. The ratio of 1.2 mol NH 3 to 0.60 mol N 2 is 2:1, as predicted by the balanced equation. Evaluate Does the result make sense?

18. MOLE – MOLE CALCULATIONS  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)?  2 Na + Cl 2  2 NaCl 5 moles Na 1 mol Cl 2 2 mol Na = 2.5 moles Cl 2

19. 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?

20. Mole-Mass Conversion Most of the time in chemistry, the amounts are given in grams instead of moles We still go through moles and use the mole ratio, but now we also use molar mass to get to grams Example: How many grams of chlorine are required to react completely with 5.00 moles of sodium to produce sodium chloride? 2 Na + Cl 2  2 NaCl 5.00 moles Na 1 mol Cl 2 70.90g Cl 2 2 mol Na 1 mol Cl 2 = 177g Cl 2

21. TRY: Calculate the mass in grams of Iodine required to react completely with 0.50 moles of aluminum

22. Mass-Mole Conversion 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 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 10.0 g H 2 O 1 mol H 2 O 2 mol C 2 H 6 18.0 g H 2 O 6 mol H 2 0 = 0.185 mol C 2 H 6

23. TRY: Calculate how many moles of Oxygen are required to make 10.0g of aluminum oxide.

24. Mass-Mass Calculations  Steps to follow: 1. Correctly balance the equation. 2. Convert the given amount into moles. 3. Set up mole ratios. 4. Use mole ratios to calculate moles of desired chemical. 5. Convert moles back into final unit  Example: Calculate how many grams of ammonia are produced when you react 2.00g of nitrogen with excess hydrogen.  N 2 + 3 H 2  2 NH 3 2.00g N 2 1 mol N 2 2 mol NH 3 17.06g NH 3 28.02g N 2 1 mol N 2 1 mol NH 3 = 2.4 g NH 3

25.  TRY : How many grams of calcium nitride are produced when 2.00 g of calcium reacts with an excess of nitrogen?

26. Volume to Volume Calculations  How many liters of CH 4 at STP are required to completely react with 17.5 L of O 2 ?  CH 4 + 2O 2  CO 2 + 2H 2 O 17.5 L O 2 1 mol O 2 1 mol CH 4 22.4 L CH 4 22.4 L O 2 2 mol O 2 1 mol CH 4 = 8.75 L CH 4

28. Limiting Reactant- Chocolate Chip Cookies!! 1 cup butter, 1/2 cup white sugar 1 cup packed brown sugar, 1 teaspoon vanilla extract 2 eggs, 2 1/2 cups all-purpose flour 1 teaspoon baking soda, 1 teaspoon salt 2 cups semisweet chocolate chips Makes 3 dozen If we had the specified amount of all ingredients listed, could we make 4 dozen cookies? What if we had 6 eggs and twice as much of everything else, could we make 9 dozen cookies? What if we only had one egg, could we make 3 dozen cookies?

29. Limiting Reactant  The limiting reagent is the reactant you run out of first.  The excess reagent is the one you have left over.  The limiting reagent determines how much product you can make  The chemical that makes the least amount of product is the “limiting reagent”.  You can recognize limiting reagent problems because you be given 2 amounts of chemical..

30. Limiting Reactant: Example  10.0g of aluminum reacts with 35.0 grams of chlorine gas to produce aluminum chloride. Which reactant is limiting, which is in excess, and how much product is produced? 2 Al + 3 Cl 2  2 AlCl 3  Start with Al:  Now Cl 2 : 10.0 g Al 1 mol Al 2 mol AlCl 3 133.5 g AlCl 3 27.0 g Al 2 mol Al 1 mol AlCl 3 = 49.4g AlCl 3 35.0g Cl 2 1 mol Cl 2 2 mol AlCl 3 133.5 g AlCl 3 71.0 g Cl 2 3 mol Cl 2 1 mol AlCl 3 = 43.9g AlCl 3

31. Finding Excess Reagent  15.0 g of potassium reacts with 15.0 g of iodine. Calculate which reactant is limiting and how much product is made.  2 K + I 2  2 KI  We found that Iodine is the limiting reactant, and 19.6 g of potassium iodide are produced. 15.0 g I 2 1 mol I 2 2 mol K 39.1 g K 254 g I 2 1 mol I 2 1 mol K = 4.62 g K USED! 15.0 g K – 4.62 g K = 10.38 g K EXCESS Given amount of excess reactant Amount of excess reactant actually used

32. Percent Yield 32 Stoichiometry allows us to calculate the amounts of reactants required or the amounts of products generated from a chemical reaction. Chemical reactions frequently do not proceed to completion. Hence the amount of product recovered is often less than would be predicted from stoichiometric calculations. In these situations it is helpful to calculate a percent yield. The Theoretical Yield is defined as the amount of product(s) calculated using Stoichiometry calculations The Actual Yield is the amount of product that an actually be recovered when the reaction is done in a lab. The Percent Yield is calculated as follows Actual Yield Theoretical Yield X 100

33. Percent Yield 22.50 g Cu 34.14 g Cu 33 Iron reacts with copper sulfate in a single replacement reaction as follows Fe (s) + CuSO 4 (aq)  FeSO 4 (aq) + Cu (s) 30.00 grams of iron metal were added to excess copper sulfate dissolved in a water solution. 22.50 grams of copper were recovered. Calculate the theoretical yield of copper in this experiment. 1. First calculate the theoretical yield (30.00 g Fe) (1 mol Fe) (1 mol Cu ) (63.55 g Cu) (55.85 g Fe) (1 mol Fe ) (1 mol Cu) 2. Divide the actual yield by the theoretical yield and multiply by 100 = 34.14 g Cu X 100 = 65.91%

34. Percent Yield 34 Iron reacts with copper sulfate in a single replacement reaction as follows Fe (s) + CuSO 4 (aq)  FeSO 4 (aq) + Cu (s) 30.00 grams of iron metal were added to excess copper sulfate dissolved in a water solution. 22.50 grams of copper were recovered. Calculate the theoretical yield of copper in this experiment.

35. Percent Yield 22.50 g Cu 34.14 g Cu 35 Iron reacts with copper sulfate in a single replacement reaction as follows Fe (s) + CuSO 4 (aq)  FeSO 4 (aq) + Cu (s) 30.00 grams of iron metal were added to excess copper sulfate dissolved in a water solution. 22.50 grams of copper were recovered. Calculate the theoretical yield of copper in this experiment. 1. First calculate the theoretical yield (30.00 g Fe) (1 mol Fe) (1 mol Cu ) (63.55 g Cu) (55.85 g Fe) (1 mol Fe ) (1 mol Cu) 2. Divide the actual yield by the theoretical yield and multiply by 100 = 34.14 g Cu X 100 = 65.91%