1. USING THE CHEMICAL EQUATION LIKE A RECIPE

2. Stoichiometry  What is It?  Quantitative relationships  In chemical reactions  Based on law of Conservation of Mass  Why/How is it Useful?

3. Straight from Wikipedia  Stoichiometry (pron.: / ˌ st ɔɪ ki ˈɒ m ɨ tri/) is a branch of chemistry that deals with the relative quantities of reactants and products in chemical reactions. In a balanced chemical reaction, the relations among quantities of reactants and products typically form a ratio of positive integers. For example, in a reaction that forms ammonia (NH 3 ), exactly one molecule of nitrogen (N 2 ) reacts with three molecules of hydrogen (H 2 ) to produce two molecules of NH 3 :/ ˌ st ɔɪ ki ˈɒ m ɨ tri/chemistryreactantsproductschemical reactions  N 2 + 3H 2 → 2NH 3

4. Stoichiometry is a lot like making tacos. Taco Recipe What if I need to make tacos for 50 people?  1o shells  1 lb ground beef  1 pkg of taco seasoning  ¾ c water  ½ cup salsa  1 cup cheddar cheese  1 cup of lettuce Makes 10 tacos.  You know how much product you need (tacos)  You have to calculate how much of each ingredient you need.  That’s stoichiometry!!!!

5.  Equations are a chemist’s recipe.  Equations tell chemists  what amounts of reactants to mix and what amounts of products to expect.  If you know the quantity of one substance in a reaction  you can calculate the quantity of any other substance consumed or created in the reaction  Quantity meaning the amount of a substance in grams, liters, molecules, or moles.

6. Balanced Chemical Equations  The coefficients in a balanced chemical equation shows how many moles/particles of one reactant are needed to react with other reactants and how many moles/particles of product will be formed.  Based on Law of Conservation of Mass 4Fe + 3O 2  2Fe 2 O 3

7.  Write the relationships we can derive from this equation.

8. C 3 H 8 + 5O 2  3CO 2 + 4H 2 O  The combustion of propane provides energy for heating homes, cooking food, and soldering metal parts. Interpret the equation for the combustion of propane in terms of particles, moles, and mass.  Show that the law of conservation of mass is observed

9. C 3 H 8 + 5O 2  3CO 2 + 4H 2 O

10.  The calculation of quantities in chemical reactions is called stoichiometry.  Imagine you are in charge of manu- facturing for Rugged Rider Bicycle Company.  The business plan for Rugged Rider requires the production of 128 custom- made bikes each day.  You are responsible for insuring that there are enough parts at the start of each day.

11. An Analogy for Stoichiometry  Assume that the major components of the bike are the frame (F), the seat (S), the wheels (W), the handlebars (H), and the pedals (P).  The finished bike has a “formula” of FSW 2 HP 2.  The balanced equation for the production of 1 bike is. F +S+2W+H+2P  FSW 2 HP 2

12.  Now in a 5 day workweek, Rugged Riders is scheduled to make 640 bikes. How many wheels should be in the plant on Monday morning to make these bikes?  What do we know?  Number of bikes = 640 bikes  1 FSW 2 HP 2 =2W (balanced eqn)  What is unknown?  # of wheels = ? wheels

13.  The connection between wheels and bikes is 2 wheels per bike. We can use this information as a conversion factor to do the calculation. 640 FSW 2 HP 2 1 FSW 2 HP 2 2 W = 1280 Wheels We can make the same kinds of connections from a chemical rxn eqn. N 2 (g) + 3H 2 (g)  2NH 3 (g) The key is the “coefficient ratio”.

14.  The coefficients of the balanced chemical equation indicate the numbers of moles of reactants and products in a chemical reaction.  1 mole of N 2 reacts with 3 moles of H 2 to produce 2 moles of NH 3.  N 2 and H 2 will always react to form ammonia in this 1:3:2 ratio of moles.  So if you started with 10 moles of N 2 it would take 30 moles of H 2 and would produce 20 moles of NH 3

15.  Using the coefficients, from the balanced equation as ratios to make connections between reactants and products, is the most important information that an equation provides.  Using this information, you can calculate the amounts of the reactants involved and the amount of product you might expect.

16. Mole Ratios Shows the mole-to-mole ratio between two of the substances in a balanced equation Written from the coefficients of any two substances in the equation This is a very important conversion and the key to solving stoichiometry problems.

17. 2Al + 3Br 2  2AlBr 3

18. 4 Fe + 3 O 2  2 Fe 2 O 3

19. 3 H 2 (g) + N 2 (g)  2 NH 3 (g) A mol ratio for H 2 and N 2 A mol ratio for NH 3 and H 2

20. Stoichiometric Calculations  How much product can be obtained from a given reaction?  How much reactant is needed to produce this much product?

21. Review of Mole Conversions  Three mole equalities 1 mol = 6.02 x 10 23 particles 1 mol = (molar mass) (from periodic table) 1 mol = 22.4 L for a gas at STP

22. Review practice  How many moles are in 5.0 x 10 25 molecules of CO2?  What is the mass in grams of 0.250 moles of CO2?  How many moles of CO2 gas are in 100.0 L at STP?

23. Types of Stoichiometry Calculations  Mole - Mole Calculations  Moles of Known  Moles of Unknown  Mole-Mass Calculations  Moles of Known  Mass of Unknown  Mass-Mole Calculations  Mass of Known  Moles of Unknown  Mass-Mass Calculations  Mass of Known  Mass of Unknown

24. MOLE – MOLE EXAMPLE  The following reaction shows the synthesis of aluminum oxide. 3O 2 (g) + 4Al(s)  2Al 2 O 3 (s) If you only had 1.8 mols of Al how much product could you make? 3O 2 (g) + 4Al(s)  2Al 2 O 3 (s)

25.  Solve for the unknown: 1.8 mol Al 4 mol Al 2 mol Al 2 O 3 = 0.90mol Al 2 O 3 3O 2 (g) + 4Al(s)  2Al 2 O 3 (s) Mole Ratio

26. 3O 2 (g) + 4Al(s)  2Al 2 O 3 (s)  If you wanted to produce 24 moles of product how many moles of each reactant would you need?

27.  Solve for the unknowns: 24 mol Al 2 O 3 2 mol Al 2 O 3 4 mol Al = 48 mol Al 3O 2 (g) + 4Al(s)  2Al 2 O 3 (s) 24 mol Al 2 O 3 2 mol Al 2 O 3 3 mol O 2 = 36 mol O 2

28. Practice Problem How many moles of hydrogen will be produced if 0.44 mol of CaH 2 reacts according to the following equation? CaH 2 + 2H 2 O  Ca(OH) 2 + 2H 2

29. Iron will react with oxygen to produce Iron III oxide. How many moles of Iron (III) oxide will be produced if 0.18 mol of Iron reacts? 4 4 Fe + 3 O 2  2 Fe 2 O 3

30. 4 Fe + 3 O 2  2 Fe 2 O 3 How many moles of Fe 2 O 3 are produced when 6.0 moles O 2 react?

31. 31 How many moles of Fe are needed to react with 12.0 moles of O 2 ?

32. MASS – MASS CALCULATIONS  No lab balance measures moles directly, generally mass is the unit of choice.  From the mass of 1 reactant or product, the mass of any other reactant or product in a given chemical equation can be calculated, provided you have a balanced equation.  As in mole-mole calculations, the unknown can be either a reactant or a product.

33. Acetylene gas (C 2 H 2 ) is produced by adding water to calcium carbide (CaC 2 ). How many grams of C 2 H 2 are produced by adding water to 5.00 g CaC 2 ? CaC 2 + 2H 2 O  C 2 H 2 + Ca(OH) 2 Example # 1 Mass-Mass CaC 2 + 2H 2 O  C 2 H 2 + Ca(OH) 2

34. Step 1: “Get to Moles!” in this case that can be done by using the Molar Mass of your given compound. 5.0 g CaC 2 64.0 g CaC 2 1 mol CaC 2 =.07813 mol CaC 2 Step 2: Now we are ready for the KEY step…converting from mols of our given to mols of unknown using the mole ratio..07813 mol CaC 2 1 mol CaC 2 1 mol C 2 H 2 =.07813 mol C 2 H 2

35. Step 3: Since we are asked for mass of our unknown in this problem, we need to use our molar mass of our unknown and convert our newly calculated mols into grams..07813 mol C 2 H 2 1 mol C 2 H 2 26.0 g C 2 H 2 = 2.03 g C 2 H 2

36. Summary of 3 Steps of Stoichiometry Problems 1.Get to Moles of Given 2.Mole Ratio to calculate moles of unknown 3.Get to wanted final unit

37. The double replacement reaction between Lead II nitrate and Potassium Iodide produces a bright yellow precipitate that can be used as a color additive in paint. How many grams of potassium iodide would we need to completely react 25.3 g of lead (II) nitrate? mass A  mols A  mols B  mass B Pb(NO 3 ) 2 + 2 KI  PbI 2 + 2 KNO 3

38. 25.3 g Pb(NO 3 ) 2 331.2g Pb(NO 3 ) 2 1mol Pb(NO 3 ) 2 = 25.4 g KI 1mol Pb(NO 3 ) 2 2mol KI 1mol KI 166 g KI

39. Practice Problems What mass of Barium chloride is needed to react completely with 46.8 g of Sodium phosphate according to the following equation? BaCl 2 + Na 3 PO 4  Ba 3 (PO 4 ) 2 + NaCl

40. Another problem Use the equation to determine what mass of FeS must react to form 326g of FeCl 2. FeS + 2HCl  H 2 S + FeCl 2

41. Moles of Known  Mass of Unknown  Find moles of unknown using the mole ratio.  Convert moles of unknown to mass of unknown.  Calculate the mass of carbon dioxide produced by the decomposition of 2.5 moles of copper(II) carbonate. CuCO 3  CuO + CO 2

42. Mass to Moles Calculations  Mass of known  moles of unknown  Convert mass of known to moles of known.  Use mole ratio to find moles of unknown.  Calculate the moles of water produced by the reaction of 20 grams of hydrogen gas with excess oxygen gas.

43.  A balanced reaction equation indicates the relative numbers of moles of reactants and products.  We can expand our stoichiometric calculations to include any unit of measure that is related to the mole.  The given quantity can be expressed in numbers of particles, units of mass, or volumes of gases at STP.  The problems can include mass-volume, volume-volume, and particle-mass calculations.

44.  In any of these problems 1. the given quantity is first converted to moles. 2. Then the mole ratio from the balanced equation is used to convert from the moles of given to the number of moles of the unknown 3. Then the moles of the unknown are converted to the units that the problem requests.  The next slide summarizes these steps for all typical stoichiometric problems

45. Moles Given Particles G Mass G Volume G Stoichiometry Roadmap Moles Unknown Coefficient mol U Coefficient mol G Balanced Equation Mass UVolume U Particles U Coefficient part U Coefficient part G Molar mass 1 mole Molar mass 1 mole 22.4 L

46. How many molecules of O 2 are produced when a sample of 29.2 g of H 2 O is decomposed by electrolysis according to this balanced equation: 2H 2 O  2H 2 + O 2

47. The last step in the production of nitric acid is the reaction of NO 2 with H 2 O. How many liters of NO 2 must react with water to produce 5.00x10 22 molecules of NO? 3NO 2 +H 2 O  2HNO 3 +NO

48. molecules A  mols A  mols B  volume B 5.0x10 22 mol- ecules NO 6.02x10 23 mol- ecules NO 1 mol NO 3 mol NO 2 = 5.58 L NO 2 1 mol NO 2 22.4 L NO 2

49. Aspirin can be made from a chemical reaction between the reactants salicylic acid and acetic anhydride. The products of the reaction are acetylsalicylic acid (aspirin) and acetic acid (vinegar). Our factory makes 125,000 100- count bottles of Bayer Aspirin/day. Each bottle contains 100 tablets, and each tablet contains 325mg of aspirin. How much in kgs + 10% for production problems, of each reactant must we have in order to meet production? C 7 H 6 O 3 + C 4 H 6 O 3  C 9 H 8 O 4 + HC 2 H 3 O 2 Salicylic acid Acetic anhydride aspirinvinegar

50. 22,549.4 mols aspirin 1 mol asp 1 mol C 7 H 6 O 3 136.10g C 7 H 6 O 3 1000 g 1 kg = 3068.97 kg salicylic acid + (306.897 g) Salicylic Acid: = 3380 kg of salicylic acid

51. 22,549.4 mols aspirin 1 mol asp 1 mol C 4 H 6 O 3 102.06g C 4 H 6 O 3 1000 g 1 kg = 2301.39 kg Acetic anhydride + 230.139 kg Acetic Anhydride: = 2530 kg Acetic anhydride

52. Example 4 Fe + 3 O 2 2 Fe 2 O 3 How many grams of O 2 are needed to produce 0.400 mol of Fe 2 O 3 ? 52

53. Calculating Mass of A Substance Balance equation Convert starting amount to moles Use coefficients to write a mol-mol ratio Convert from moles of known to moles of unknown Convert moles of unknown to grams 53

54. Calculation The reaction between H 2 and O 2 produces 13.1 g of water. How many grams of O 2 reacted? Write the equation H 2 (g) + O 2 (g)H 2 O (g) Balance the equation 2 H 2 (g) + O 2 (g)2 H 2 O (g) 54

55. 2 H 2 (g) + O 2 (g)2 H 2 O (g) ? g 13.1 g 55

56. Points to Remember 1. Read an equation in moles 2. Convert given amount to moles 3. Use mole ratio to find unknown moles 4. Convert moles to grams moles (given)moles (unknown) grams (given)grams (unknown) 56

57. How many O 2 molecules will react with 505 grams of Na to form Na 2 O? 4 Na + O 2 2 Na 2 O 57

58. Acetylene gas C 2 H 2 burns in the oxyactylene torch for welding. How many grams of C 2 H 2 are burned if the reaction produces 75.0 g of CO 2? 2 C 2 H 2 + 5 O 2 4 CO 2 + 2 H 2 O 58

59. Mass of KnownMass of Unknown Moles of Unknown Convert grams to moles Convert moles known to moles of unknown Use Mole Ratio from Coefficients Convert moles to grams Moles of Known

60. Limiting Reactants  The reactant that is not completely used up in the reaction  The limiting reactant is not present in sufficient quantity to react with all other reactants.  The reaction stops when the limiting reactant is completely consumed.  Any remaining reactants are considered "excess reactants".  The amount of product formed is determined by the "limiting reactant"

61. Steps in solving a limiting reactant problem  1. Write a balanced equation for the reaction.  2. Convert both reactant quantities to moles.  3. Determine the moles of product that could be formed by each reactant.  4. The least amount in step #3 identifies the limiting reactant.  5. Use that number of moles of product to determine the mass produced.

62. An example of a limiting reactant problem  What mass of water can be produced by 4 grams of H 2 reacting with 16 grams of O 2 ?

63. Remember the following things about limiting reactants:  The reaction will stop when the reactants are used up.  If one reactant is used up before the other, the reaction stops then.  The first reactant used up is the limiting reactant, use it for the calculation.  The other reactant is the excess reactant.

64.  Step 1. Write a balanced equation for the reaction.  2 H 2 + O 2 2 H 2 O  Step 2. Convert both reactant quantities to moles.

65.  Step 3. Using the mole ratio from the equation, determine the moles of water that could be formed by each reactant.

66.  Step 4. Oxygen produces the least amount of water.  16 grams of oxygen cannot produce as much water as 4 grams of hydrogen. In other words, 16 grams of oxygen will be used up in the reaction before 4 grams of hydrogen.  Oxygen is the "limiting" reactant.  Use oxygen for the calculation of product amount.

67.  Step 5.Complete the problem by converting moles of H 2 O to mass of H 2 O.

68. Analogy Suppose you are preparing cheese sandwiches. Each sandwich requires 2 pieces of bread and 1 slice of cheese. If you have 4 slices of cheese and 10 pieces of bread, how many cheese sandwiches can you make? 68

69. Cheese Sandwich Products Sandwich 1 + + = Sandwich 2 + + = 69

70. How many sandwiches can you make? ____ slices of bread + ____ slices of cheese = ____ sandwiches What is left over? ________________ What is the limiting reactant? 70

71. How many sandwiches can you make? __10__ slices of bread + __4__ slices of cheese = __4__ sandwiches What is left over? _2 slices of bread What is the limiting reactant? cheese 71

72. Hints for LR Problems 1. For each reactant amount given, calculate the moles (or grams) of a product it could produce. 2.The reactant that produces the smaller amount of product is the limiting reactant. 3. The number of moles of product produced by the limiting reactant is ALL the product possible. There is no more limiting reactant left. 72

73. Example 25 grams of hydrogen and 25 grams of oxygen react to produce water. How much water is produced?

74. Percent Yield  Theoretical yield: Maximum amount of product that can be obtained from a given amount of reactant.  A reaction rarely produces maximum amount. Things do not go perfectly.  From the calculation (equation)  Actual Yield: The amount actually produced.

75. Percent Yield  In all of the calculations we have done thus far the amount of product we have calculated has been determined in perfect lab conditions. This means everything goes exactly right. Everything is ideal.  This yield of product is called Theoretical Yield  Theoretical yield is obtained from stoichiometric calculations.

76.  However, in real lab situations, conditions don’t always work out perfectly.  When an experiment is carried out it is nice to be able to determine how efficiently a reaction has proceeded.  This can be done by determining the actual yield produced in the experiment and then using an equation called percent yield.  Percent yield is the ratio of the actual amount of product to the theoretical amount of product.

77. Equation  Percent Yield = Actual Yield/Theoretical Yield x 100

78. Example  You produced 2.5 grams of magnesium oxide in a reaction of magnesium with excess oxygen. Calculate the percent yield for this reaction.

79. Percent Yield You prepared cookie dough to make 5 dozen cookies. The phone rings while a sheet of 12 cookies is baking. You talk too long and the cookies burn. You throw them out (or give them to your dog.) The rest of the cookies are okay. How many cookies could you have made (theoretical yield)? How many cookies did you actually make to eat? (Actual yield) 79

80. Vocabulary Actual yield is the amount of product actually recovered from an experiment Theoretical (possible) yield is the maximum amount of product that could be produced from the reactant. Calculated Yield. Percent Yield is the actual yield compared to the maximum (theoretical yield) possible. 80

81. Percent Yield Calculation What is the percent yield of cookies? Percent Yield = Actual Yield (g) recovered X 100 Possible Yield (g) % cookie yield = 48 cookies x 100 = 80% yield 60 cookies 81