2. “Stoichiometry” Mr. Mole

3. u First… –A bit of review

4. The Arithmetic of Equations u OBJECTIVES: Interpret balanced chemical equations in terms of: a) moles b) representative particles c) mass

5. Remember Avocado, I mean, AVOGADRO?

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. Remember calculating Molar Mass?

9. Molar Mass Molar mass is determined by adding the atomic masses for the atoms, molecules or compounds you are working with Molar mass is determined by adding the atomic masses for the atoms, molecules or compounds you are working with Ex. Molar mass of CaCl 2 Ex. Molar mass of CaCl 2 Avg. Atomic mass of Calcium = 40g Avg. Atomic mass of Calcium = 40g Avg. Atomic mass of Chlorine = 35.5g Avg. Atomic mass of Chlorine = 35.5g Molar Mass of calcium chloride = 40 g/mol Ca + (2 X 35.5) g/mol Cl  111 g/mol CaCl 2 Molar Mass of calcium chloride = 40 g/mol Ca + (2 X 35.5) g/mol Cl  111 g/mol CaCl 2 20 Ca 40a 17 Cl 35.5 Cl

10. Practice Calculate the Molar Mass of calcium phosphate Calculate the Molar Mass of calcium phosphate Formula = Formula = Masses elements: Masses elements: Ca: 3 Ca’s X 40 = Ca: 3 Ca’s X 40 = P: 2 P’s X 31 = P: 2 P’s X 31 = O: 8 O’s X 16 = O: 8 O’s X 16 = Molar Mass = Molar Mass = Ca 3 (PO 4 ) 2 120 g 62 g 128 g 120g Ca + 62g P + 128g O = 310 g/mol

11. Let’s make some Cookies! u When baking cookies, a recipe is usually used, telling the exact amount of each ingredient. If you need more, you can double or triple the amount u Thus, a recipe is much like a balanced equation.

12. 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?

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

14. Practice u 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 produced? 2 mol H 2 1 mol O 2 4 mol H 2 2 mol O 2 6 mol H 2, 6 mol H 2 O 25 mol O 2, 50 mol H 2 O

16. Stoichiometry is… u Greek for “measuring elements” Pronounced “stoy kee ahm uh tree” u Defined as: calculations of the quantities in chemical reactions, based on a balanced equation. u There are 4 ways to interpret a balanced chemical equation

17. #1. In terms of Particles u An Element is made of atoms u A Molecular compound (made of only nonmetals) is made up of molecules (Don’t forget the diatomic elements) u Ionic Compounds (made of a metal and nonmetal parts) are made of formula units

18. Example: 2H 2 + O 2 → 2H 2 O  Two molecules of hydrogen and one molecule of oxygen form two molecules of water.  Another example: 2Al 2 O 3  Al + 3O 2 2formula unitsAl 2 O 3 form4 atoms Al and3moleculesO2O2 Now read this: 2Na + 2H 2 O  2NaOH + H 2

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

20. Mole Ratios u These mole ratios can be used to calculate the moles of one chemical from the given amount of a different chemical u 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 ? mol Cl 2 X 2 mol Na 1 mol Cl 2 = 2.5 moles Cl 2

21. #3. In terms of Mass u The Law of Conservation of Mass applies u We can check mass by using moles. 2H 2 + O 2   2H 2 O 2 moles H 2 2 g H 2 1 mole H 2 = 4 g H 2 1 mole O 2 32 g O 2 1 mole O 2 = 32 g O 2 36 g H 2 + O 2 + reactants

22. In terms of Mass (for products) 2H 2 + O 2   2H 2 O 2 moles H 2 O 18 g H 2 O 1 mole H 2 O = 36 g H 2 O 36 g H 2 + O 2 = 36 g H 2 O The mass of the reactants must equal the mass of the products. 36 grams reactant = 36 grams product

23. Practice: u Show that the following equation follows the Law of Conservation of Mass (show the atoms balance, and the mass on both sides is equal) 2Al 2 O 3  Al + 3O 2

24. Chemical Calculations u OBJECTIVES: Calculate stoichiometric quantities from balanced chemical equations using units of moles, mass, and representative particles

25. Chemical Calculations Basic equation for all of these: Given amount Unknown amount ------------------- X ----------------------- Recipe amount Recipe amount Cross multiply and you got this!

26. Mole to Mole conversions u How many moles of O 2 are produced when 3.34 moles of Al 2 O 3 decompose? 2Al 2 O 3  Al + 3O 2 3.34 mol Al 2 O 3 ? O 2 ----------------------------- X ----------------- 2 mol Al 2 O 3 3 mol O 2 = 5.01 mol O 2 If you know the amount of ANY chemical in the reaction, you can find the amount of ALL the other chemicals!

27. How do you get good at this?

28. Practice: 2C 2 H 2 + 5 O 2  4CO 2 + 2 H 2 O If 3.84 moles of C 2 H 2 are burned, how many moles of O 2 are needed? How many moles of C 2 H 2 are needed to produce 8.95 mole of H 2 O? If 2.47 moles of C 2 H 2 are burned, how many moles of CO 2 are formed?

29. Practice: 2C 2 H 2 + 5 O 2  4CO 2 + 2 H 2 O If 3.84 moles of C 2 H 2 are burned, how many moles of O 2 are needed?

30. Practice: 2C 2 H 2 + 5 O 2  4CO 2 + 2 H 2 O If 3.84 moles of C 2 H 2 are burned, how many moles of O 2 are needed? How many moles of C 2 H 2 are needed to produce 8.95 mole of H 2 O? If 2.47 moles of C 2 H 2 are burned, how many moles of CO 2 are formed?

31. Practice: 2C 2 H 2 + 5 O 2  4CO 2 + 2 H 2 O If 3.84 moles of C 2 H 2 are burned, how many moles of O 2 are needed? (9.6 mol) How many moles of C 2 H 2 are needed to produce 8.95 mole of H 2 O? (8.95 mol) If 2.47 moles of C 2 H 2 are burned, how many moles of CO 2 are formed? (4.94 mol)

32. How do you get good at this?

33. Steps to Calculate Stoichiometric Problems 1. Correctly balance the equation. 2. Place the given amount over the recipe amount. 3. Place the unknown amount over the recipe amount. 4. Cross multiply to solve

34. Another example: u If 10.1 g of Fe are added to a solution of copper (II) sulfate, how many grams of solid copper would form? 2Fe + 3CuSO 4  Fe 2 (SO 4 ) 3 + 3Cu

35. Another example: u If 10.1 g of Fe are added to a solution of copper (II) sulfate, how many grams of solid copper would form?

36. Another example: u If 10.1 g of Fe are added to a solution of copper (II) sulfate, how many grams of solid copper would form? 2Fe + 3CuSO 4  Fe 2 (SO 4 ) 3 + 3Cu Answer = 17.2 g Cu

37. The Concept of: A little different type of yield than you learned in Driver’s Ed…

38. What is Yield? u Yield is the amount of product made in a chemical reaction. u There are three types: 1. Actual yield- what you actually get in the lab when the chemicals are mixed 2. Theoretical yield- what the balanced equation tells should be made 3. Percent yield 3. Percent yield = Actual Theoretical x 100

39. Example: u 6.78 g of copper is produced when 3.92 g of Al are reacted with excess copper (II) sulfate. 2Al + 3 CuSO 4  Al 2 (SO 4 ) 3 + 3Cu u What is the actual yield? u What is the theoretical yield? u What is the percent yield? = 6.78 g Cu = 13.8 g Cu = 49.1 %

40. Details on Yield u Percent yield tells us how “efficient” a reaction is. u Percent yield can not be bigger than 100 %. u Theoretical yield will always be larger than actual yield! Why? Due to impure reactants; competing side reactions; loss of product in filtering or transferring between containers; measuring

41. Limiting Reactant & Percent Yield u OBJECTIVES: Identify the limiting reactant in a reaction.

42. Limiting Reactant & Percent Yield u OBJECTIVES: Calculate theoretical yield, percent yield, and the amount of excess reactant that remains unreacted given appropriate information.

43. Limiting Reactant: 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?

44. How do you find out which is limited? u The chemical that makes the least amount of product is the “limiting reactant”. u You can recognize limiting reactant problems because they will give you 2 amounts of chemical u Do two stoichiometry problems, one for each reactant you are given.

45. “Limiting” Reactant u If you are given one dozen loaves of bread and three pieces of salami, how many salami sandwiches can you make? u The limiting reactant is the reactant you run out of first. u The excess reactant is the one you have left over. u The limiting reactant determines how much product you can make

46. u If 10.6 g of copper reacts with 3.83 g sulfur, how many grams of the product (copper (I) sulfide) will be formed? = 13.3 g Cu 2 S = 19.0 g Cu 2 S = 13.3 g Cu 2 S Cu is the Limiting Reactant, since it produced less product.

47. Another example: u If 10.3 g of aluminum are reacted with 51.7 g of CuSO 4 how much copper (grams) will be produced? Al + CuSO 4 → Cu + Al 2 (SO 4 ) 3

48. Another example: u If 10.3 g of aluminum are reacted with 51.7 g of CuSO 4 how much copper (grams) will be produced? 2Al + 3CuSO 4 → 3Cu + Al 2 (SO 4 ) 3

49. Another example: In the preceding reaction, how much of the excess reactant will remain? Remember: the CuSO 4 is limited, so the Cu produced was 20.6 g. Flip the equation around to see how much Al was used to make this much Cu. In other words…

50. Another example: 20.6g of Cu was produced in the following reaction. How much Al was required? 2Al + 3CuSO 4 → 3Cu + Al 2 (SO 4 ) 3

51. Another example: Next, subtract the needed amount from the given amount from the first problem. If 10.3 g of aluminum are reacted with 51.7g of CuSO 4 how much copper (grams) will be produced?

52. Another example: u If 10.3 g of aluminum are reacted with 51.7 g of CuSO 4 how much copper (grams) will be produced? 2Al + 3CuSO 4 → 3Cu + Al 2 (SO 4 ) 3 the CuSO 4 is limited, so Cu = 20.6 g u How much excess reagent will remain? Excess = 4.47 grams