2. Stoichiometry The study of quantities of materials consumed and produced in chemical reactions.

3. Stoichiometry Stoichiometry is the branch of chemistry that deals with the quantities of substances that enter into (reactants), and are produced by (products), chemical reactions. Stoichiometry is the branch of chemistry that deals with the quantities of substances that enter into (reactants), and are produced by (products), chemical reactions. For example, when methane unites with oxygen in complete combustion: For example, when methane unites with oxygen in complete combustion: –16g of methane require 64g of oxygen (reactants) –16g of methane require 64g of oxygen (reactants) –At the same time 44g of carbon dioxide and 36g of water are formed as reaction products. (products)

4. Why is Stoichiometry Important? Just one example….. Just one example….. In industry, profits depend on the amount of products made from a given amount of materials. Therefore, manufacturers need to be able to calculate the amounts of reactants that will result in product. In industry, profits depend on the amount of products made from a given amount of materials. Therefore, manufacturers need to be able to calculate the amounts of reactants that will result in product.

5. Hamburger Analogy My recipe for a bacon double cheeseburger is: My recipe for a bacon double cheeseburger is: 1 hamburger bun 2 hamburger patties 2 slices of cheese 4 strips of bacon Based on this recipe: Based on this recipe: If I have five bacon double cheeseburgers: If I have five bacon double cheeseburgers: –How many hamburger buns do I have? –How many hamburger patties do I have? –How many slices of cheese do I have? How many bacon double cheeseburgers can you make if you start with: How many bacon double cheeseburgers can you make if you start with: –2 bun, 2 patties, 2 slices of cheese, 6 strips of bacon –2 bun, 4 patties, 4 slices of cheese, 8 strips of bacon

6. When making hamburgers, we simply used the relationship between the numbers of each item necessary. The numbers represented our quantities. When making hamburgers, we simply used the relationship between the numbers of each item necessary. The numbers represented our quantities. When dealing with chemical reactions, we use the same basic principle but must use balanced equations and conversion bridges to enable us to convert between quantities of chemical substances. When dealing with chemical reactions, we use the same basic principle but must use balanced equations and conversion bridges to enable us to convert between quantities of chemical substances.

7. Chemical Reaction Review You will often have to create formula equations given only partial information. You will often have to create formula equations given only partial information. In order to obtain a balanced equation, which is necessary for all stoic problems, let us review Rx types: In order to obtain a balanced equation, which is necessary for all stoic problems, let us review Rx types: 1. Synthesis Rx: A + B  AB 2. Decomposition Rx: AB  A + B 3. Single Replacement: A + XY  AY + X 4. Double Replacement: AB + XY  AY + XB 5. Combustion Rx: C x H x + O 2  CO 2 + H 2 O

8. Conversion Bridges 2C 2 H 2 (g) + 5O 2 (g)  4CO 2 (g) + 2 H 2 O(l) Molar mass: Remember, molar mass is the mass of 1 mole of any substance. Molar mass: Remember, molar mass is the mass of 1 mole of any substance. C 2 H 2 = 26.04 g/mol, O 2 = 32.00 g/mol Molar ratios: Use the coefficients of the balanced equation to establish mole ratios for any of the substances. Molar ratios: Use the coefficients of the balanced equation to establish mole ratios for any of the substances. C 2 H 2 to CO 2 = 2:4 = 1:2 H 2 O to O 2 = 2:5

9. Ratios are found in the chemical equations 2 HCl + 1 Ba(OH) 2  2 H 2 O + 1 BaCl 2 Coefficients give molar ratios Coefficients give molar ratios 2 moles of HCl react with 1 mole of Ba(OH) 2 to form 2 moles of H 2 O and 1 mole of BaCl 2 2 moles of HCl react with 1 mole of Ba(OH) 2 to form 2 moles of H 2 O and 1 mole of BaCl 2

10. Interpreting balanced chemical equations The coefficients in a chemical equation correlate to the number of moles of each substance in the reaction. The coefficients in a chemical equation correlate to the number of moles of each substance in the reaction. N 2 H 4 + 2H 2 O 2 ----> N 2 + 4H 2 O N 2 H 4 + 2H 2 O 2 ----> N 2 + 4H 2 O Same as Same as 1 mole N 2 H 4 + 2 moles H 2 O 2 ----> 1 mole N 2 + 4 moles H 2 O

11. 8Al + 3Fe 3 O 4  4Al 2 O 3 + 9 Fe Using the balanced equation: Write the molar ratios for: Al to Fe Fe 3 O 4 to Fe Al 2 O 3 to Al Fe 3 O 4 to Al 2 O 3

12. Types of Stoic Problems There are several types of stoichiometry problems based on what is given and unknown in a problem. There are several types of stoichiometry problems based on what is given and unknown in a problem. 1. Mol  Mol  Mass  Mass 2. Limiting and Excess Reactants 3. Theoretical Yield of Products

13. Conversion Factors 1 mol 6.02 x 10 23 particles 1 mol Molar mass in grams # mol A # mol B

14. Mol  Mol These are the easiest type of conversion problem. Just be careful to not do extra steps! These are the easiest type of conversion problem. Just be careful to not do extra steps! Na 2 O + H 2 O  NaOH How many moles of NaOH are produced if 6.5 moles of sodium oxide is combined with excess water? How many moles of NaOH are produced if 6.5 moles of sodium oxide is combined with excess water?

15. Step 1: Balanced equation. Step 2: Identify given and unknown. Step 3: Place given over 1. Step 4: Convert from moles of given to moles of unknown using molar ratio. Step 5: Perform calculation and express answer in correct sig figs, unit, and substance.

16. Step 1: Na 2 O + H 2 O  2NaOH Step 2: Given is 6.5 mol Na 2 O, Unknown is mol NaOH Step 3-5: 6.5 mol Na 2 O ( 2 mol NaOH ) 6.5 mol Na 2 O ( 2 mol NaOH ) 1 ( 1 mol Na 2 O ) = 13 mol NaOH 1 ( 1 mol Na 2 O ) = 13 mol NaOH

17. Given the equation: Given the equation: CH 4 (g) + O 2 (g)  CO 2 (g) + H 2 O(g) 1. Calculate moles of water produced by the combustion of 500.0 moles of methane. 2. How many moles of carbon dioxide is produced if 5 moles of water is produced? 3. How many moles of oxygen are needed to produce 65.5 moles of carbon dioxide?

18. Mole  Mass Given the equation: H 2 (g) + O 2 (g)  H 2 O(g) Given the equation: H 2 (g) + O 2 (g)  H 2 O(g) Calculate moles of H 2 O produced if 25 grams of oxygen is reacted with excess hydrogen. Step 1: Start with a balanced equation. Step 2: Identify given and unknown. Step 3: Place given over 1. Step 4: Convert from grams of given to moles of given using molar mass.

19. Step 5: Convert from moles of given to moles of unknown. *You may now stop because you are in moles of unknown. No need to use the molar mass of unknown. *You may now stop because you are in moles of unknown. No need to use the molar mass of unknown. Step 6: Perform calculation and express answer in correct sig figs, unit, and substance.

20. Step 1: 2H 2 (g) + O 2 (g)  2H 2 O(g) Step 2: Given is 25 g of O 2, Unknown is mol of H 2 O Step 3-5 25 g O 2 ( 1 mol ) ( 2 mol H 2 O ) 1 ( 32.00 g ) ( 1 mol O 2 ) = 1 ( 32.00 g ) ( 1 mol O 2 ) = 1.6 mol H 2 O produced

21. KClO 3 (aq)  KCl(aq) + O 2 (g) Using the above reaction: Using the above reaction: 1. How much potassium chloride is produced from the decomposition of 5.0 mol of potassium chlorate? 2. How many moles of oxygen are produced if 45.09 g of potassium chloride are produced?

22. Mass  Mass (Also sometimes called Gram   Gram) Given the equation: H 2 (g) + O 2 (g)  H 2 O(g) Given the equation: H 2 (g) + O 2 (g)  H 2 O(g) Calculate grams of H 2 O produced if 25 grams of oxygen is reacted with excess hydrogen. Step 1: Start with a balanced equation. Step 2: Identify given and unknown. Step 3: Complete table.

23. 2H 2 (g) + O 2 (g)  2H 2 O(g) Given/UnknownAmountMoles (based on coefficients) MolarMass O2O2O2O2 25 g 1 32.00 g/mol H2OH2OH2OH2OX2 18.02 g/mol *There are only two substances that will be calculated in a stoic problem, the given and unknown. *When identifying given and unknown, make sure the chemical formulas are correct. EX: H 2 not H

24. Step 4: Place given over 1. Step 5: Convert from grams of given to moles of given using molar mass. Step 6: Convert from moles of given to moles of unknown using molar ratio from balanced equation. Step 7: Convert from moles of unknown to grams of unknown using molar mass. Step 8: Perform calculation and express answer in correct sig figs, unit, substance, and produced or needed. *Units must be placed so that they cancel throughout the solving process.

25. 2H 2 (g) + O 2 (g)  2H 2 O(g) 25 g O 2 1 mol O 2 2 mol H 2 O 18.02 g H 2 O 1 32.00g O 2 1 mol O 2 1 mol H 2 O = 1 32.00g O 2 1 mol O 2 1 mol H 2 O = 28 g H 2 O produced

26. Aluminum is an active metal that when placed in hydrochloric acid produces hydrogen gas and aluminum chloride. How many grams of aluminum chloride can be produced when 3.45 grams of aluminum are reacted with an excess of hydrochloric acid? First write a balanced equation. Al(s) + HCl(aq)  AlCl 3 (aq) + H 2 (g) 2 6 2 3 Gram to Gram Conversions Example #1

27. Aluminum is an active metal that when placed in hydrochloric acid produces hydrogen gas and aluminum chloride. How many grams of aluminum chloride can be produced when 3.45 grams of aluminum are reacted with an excess of hydrochloric acid? Now let’s get organized. Write the information below the substances. Al(s) + HCl(aq)  AlCl 3 (aq) + H 2 (g) 2 6 2 3 3.45 g ? grams Gram to Gram Conversions Example #1

28. Aluminum is an active metal that when placed in hydrochloric acid produces hydrogen gas and aluminum chloride. How many grams of aluminum chloride can be produced when 3.45 grams of aluminum are reacted with an excess of hydrochloric acid? Let’s work the problem. = g AlCl 3 produced 3.45 g Al We must always convert to moles.Now use the molar ratio.Now use the molar mass to convert to grams. 17.0 Units match 1 Al(s) + HCl(aq)  AlCl 3 (aq) + H 2 (g) 2 6 2 3 3.45 g ? grams

29. Gram to Gram Conversions Example #2 EX: How much hydrogen is produced from the reaction of 25.5 g of magnesium metal with excess hydrochloric acid? Step 1: Balanced equation. We must write a reaction based on reactants, predict products, then we can balance. Mg + HCl  MgCl 2 + H 2 Mg + HCl  MgCl 2 + H 2 Balance: Mg + 2HCl  H 2 + MgCl 2

30. Mg + 2HCl  H 2 + MgCl 2 Step 2: Given is 25.5 g Mg, Unknown is g H 2 At this point, you may skip step 3 (table) and solve the problems using the remaining steps. At this point, you may skip step 3 (table) and solve the problems using the remaining steps. Steps 4-8: 25.5 g Mg 1 mol Mg 1 mol H 2 2.02 g H 2 ) = 1 24.31g Mg 1 mol Mg 1 mol H 2 ) 1 24.31g Mg 1 mol Mg 1 mol H 2 ) 2.12 g H 2 produced 2.12 g H 2 produced

31. Example #3 6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed? 4 Al + 3 O 2  2Al 2 O 3 = 6.50 g Al ? g Al 2 O 3 1 mol Al 26.98 g Al 4 mol Al 2 mol Al 2 O 3 1 mol Al 2 O 3 101.96 g Al 2 O 3 12.30 g Al 2 O 3 produced 1

32. Practice: Given the following equation: Fe + S 8  FeS a. What mass of iron is needed to react with 16.0 grams of sulfur?

33. Practice: Given the following equation: Fe + S 8  FeS b. How many grams of FeS are produced?

34. Given the following equation: NaClO 3  NaCl + O 2 a. 12.0 g of NaClO 3 will produce how many grams of O 2 ?

35. Given the following equation: NaClO 3  NaCl + O 2 b. How many grams of NaCl are produced when 80.0 grams of O 2 are produced?

36. LIMITING AND EXCESS REACTANTS 10,000 Na + 1Cl  NaCl + 9,999 Na NaCl

37. Limiting and Excess Reactants In the hamburger analogy, it would take 1 bun, 2 hamburger patties, 2 slices cheese, and 4 strips bacon to make 1 bacon double cheeseburger. In the hamburger analogy, it would take 1 bun, 2 hamburger patties, 2 slices cheese, and 4 strips bacon to make 1 bacon double cheeseburger. If given certain amount of each item, you could calculate the number of cheeseburgers you could produce. If given certain amount of each item, you could calculate the number of cheeseburgers you could produce. This is the basis for limiting reactant problems. This is the basis for limiting reactant problems. One reactant will always limit the amount of product that can be produced in a chemical rx. One reactant will always limit the amount of product that can be produced in a chemical rx.

38. Refer to the diagram demonstrating the reaction between nitrogen and hydrogen. 1. Write the balanced equation for the reaction. 2. Which of the reactants is limiting the amount of product that is made? 3. How many molecules of product are made?

39. How many grams of water can be produced if 50.0 grams of hydrogen combines with 50.0 grams of oxygen? How many grams of water can be produced if 50.0 grams of hydrogen combines with 50.0 grams of oxygen? Step 1: Write a balanced equation. Step 2: Perform 2 mass to mass conversions, one for each reactant. Step 3: The conversion with the least amount of product produced represents that of the limiting reactant. *You now know the limiting reactant and the amount of product that can be produced.

40. Step 1: 2H 2 (g) + O 2 (g)  2H 2 O Step 2: 50.0 g H 2 1 mol H 2 2 mol H 2 O 18.02 g H 2 O 1 2.02 g H 2 2 mol H 2 1 mol H 2 O = 1 2.02 g H 2 2 mol H 2 1 mol H 2 O = 446 g H 2 O produced 446 g H 2 O produced 50.0 g O 2 1 mol O 2 2 mol H 2 O 18.02 g H 2 O 1 32.00 g O 2 1 mol O 2 1 mol H 2 O = 1 32.00 g O 2 1 mol O 2 1 mol H 2 O = 56.3 g H 2 O produced 56.3 g H 2 O produced Step 3: O 2 is the limiting reactant because only 56.3 grams of water is produced.

41. Calculate grams of copper produced if 25 grams of zinc is reacted with 50. grams of copper(I) carbonate. Calculate grams of copper produced if 25 grams of zinc is reacted with 50. grams of copper(I) carbonate. Step 1: Zn + Cu 2 CO 3  ZnCO 3 + 2Cu Step 2: 25 g Zn 1 mol Zn 2 mol Cu 63.55 g Cu 1 65.37 g Zn 1 mol Zn 1 mol Cu = 1 65.37 g Zn 1 mol Zn 1 mol Cu = 49 g Cu produced 50.g Cu 2 CO 3 1mol Cu 2 CO 3 2 mol Cu 63.55 g Cu 1 187.11g Cu 2 CO 3 1 mol Cu 2 CO 3 1 mol Cu = 1 187.11g Cu 2 CO 3 1 mol Cu 2 CO 3 1 mol Cu = 34 g Cu produced

42. EX: Calculate grams of product if 80. g of aluminum reacts with 80. g of chlorine to form aluminum chloride. Which reactant is limiting the amount of product being made?

43. EXCESS REACTANTS 10,000 Na + 1Cl  NaCl + 9,999 Na NaCl

44. Excess Reactant In order to calculate how much excess reactant remains, follow these steps: In order to calculate how much excess reactant remains, follow these steps: Step 1: Find the limiting reactant. Step 2: Convert from grams of limiting reactant to grams of the other reactant. This is the amount of that reactant consumed. Step 3: Subtract the amount of other reactant consumed from the original quantity in the problem to find the amount in excess. * Original grams – consumed grams = grams excess

45. How much of the excess reactant remains if 100. g of bromine reacts with 75.0 grams of magnesium ribbon? How much of the excess reactant remains if 100. g of bromine reacts with 75.0 grams of magnesium ribbon? Step 1: Br 2 (g) + Mg(s)  MgBr 2 100. g Br 2 1 mol Br 2 1 mol MgBr 2 184.11g MgBr 2 1 159.80g Br 2 1 mol Br 2 1 mol MgBr 2 = 1 159.80g Br 2 1 mol Br 2 1 mol MgBr 2 = 115.21 g MgBr 2 115.21 g MgBr 2 75.0 g Mg 1 mol Mg 1 mol MgBr 2 184.11g MgBr 2 1 24.31 g Mg 1 mol Mg 1 mol MgBr 2 = 1 24.31 g Mg 1 mol Mg 1 mol MgBr 2 = 568.01 g MgBr 2 568.01 g MgBr 2 *Br 2 is the limiting reactant.

46. Step 2: 100. g Br 2 1 mol Br 2 1 mol Mg 24.31 g Mg 1 159.80g Br 2 1 mol Br 2 1 mol Mg = 1 159.80g Br 2 1 mol Br 2 1 mol Mg = 15.2 g Mg consumed Step 3: 75.0 g Mg - 15.2 g Mg = 59.8 g Mg excess 75.0 g Mg - 15.2 g Mg = 59.8 g Mg excess Original Quantity From Problem Quantity Consumed From Calculation

47. Calculate grams of excess reactant if 80. g of aluminum reacts with 35g of copper(II) chloride. Calculate grams of excess reactant if 80. g of aluminum reacts with 35g of copper(II) chloride.

48. Theoretical Yield & Percent Yield

49. Theoretical Yield and Percent Yield Thus far in all our calculations we assumed that the reaction conditions were ideal and led to reactions that went to 100% completion. Thus far in all our calculations we assumed that the reaction conditions were ideal and led to reactions that went to 100% completion. Calculation of product mass with these ideal conditions in mind are known as the "theoretical yield". Calculation of product mass with these ideal conditions in mind are known as the "theoretical yield". However, in reality reaction are influenced by many "external" factors (i.e., factors apart from the reactants). Some of these factors are: However, in reality reaction are influenced by many "external" factors (i.e., factors apart from the reactants). Some of these factors are: *Temp, Pressure, Accuracy of Equipment

50. Percent Yield In most cases the reaction does not go to completion. In most cases the reaction does not go to completion. In this case the mass of products formed (the actual yield) is less than the theoretical yield. In this case the mass of products formed (the actual yield) is less than the theoretical yield. A quantity that describes this less-than-ideal yield is known as the "percent yield": A quantity that describes this less-than-ideal yield is known as the "percent yield": The actual amount is either given in the problem or obtained through experimentation. The actual amount is either given in the problem or obtained through experimentation.

51. EX: In a reaction between hydrogen and oxygen, an experiment yielded 25.6 grams of water. Calculate the percent yield of the product if the theoretical yield for this reaction (based on a stoich gram-gram problem) is 28.8 grams of water could be formed. EX: In a reaction between hydrogen and oxygen, an experiment yielded 25.6 grams of water. Calculate the percent yield of the product if the theoretical yield for this reaction (based on a stoich gram-gram problem) is 28.8 grams of water could be formed.

52. EX: Given the reaction of 65.5g of sodium chloride reacting with 45.3g of zinc hydroxide forming zinc chloride and sodium hydroxide: a. Determine the limiting reactant. b. Determine the mass of both products. c. Determine the amount of excess reactant. d. Calculate % yield of an experiment if a student obtained 34.6g of NaOH.