1. Stoichiometry

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

3. 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

4. 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.

5. 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

6. Identify each rx type and predict products for: Identify each rx type and predict products for: 1. Magnesium ribbon and oxygen gas react: 2. Silver nitrate combines with iron(III) sulfate: 3. The complete combustion of C 2 H 6 : 4. The decomposition of sodium chlorate upon heating: 5. Sodium metal and chlorine gas react: 6. Aluminum reacts with hydrochloric acid: 7. Silver acetate reacts with magnesium ribbon: 8. The decomposition of calcium hydroxide:

7. 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

8. 8Al + 3Fe 3 O 4  4Al 2 O 3 + 9 Fe Using the balanced equation: Using the balanced equation: 1. Calculate the molar masses of each of the substances. 2. 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

9. 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. Mass  Mass 2. Mole  Mass 3. Mole  Mole 4. Limiting and Excess Reactants 5. Theoretical Yield of Products

10. Mass  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 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.

11. 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

12. 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, and substance. *Units must be placed so that they cancel throughout the solving process.

13. 25 g O 2 ( 1 mol ) ( 2 mol H 2 O ) ( 18.02 g ) 1 ( 32.00g) ( 1 mol O 2 ) ( 1 mol ) = 1 ( 32.00g) ( 1 mol O 2 ) ( 1 mol ) = 28 g H 2 O produced Giv: 25 g O 2 Unk: g H 2 O

14. C 4 H 10 + O 2  CO 2 + H 2 O How much (g) of carbon dioxide gas is produced from the complete combustion of 50.0 g butane (C 4 O 10 ) and excess oxygen? How much (g) of carbon dioxide gas is produced from the complete combustion of 50.0 g butane (C 4 O 10 ) and excess oxygen? Step 1: Balance equation: 2C 4 H 10 + 13O 2  8CO 2 + 10H 2 O 2C 4 H 10 + 13O 2  8CO 2 + 10H 2 O Step 2: Identify given and unknown. Given is 50.0 g C 4 H 10, Unknown is g of CO 2 (Excess oxygen is background information at this point. We will deal with it during limiting and excess reactant calculations) (Excess oxygen is background information at this point. We will deal with it during limiting and excess reactant calculations)

15. Step 3: Given/UnknownAmountMoles Molar Mass C 4 H 10 50.0 g 2 58.14 g/mol CO 2 X8 44.01 g/mol Steps 4-8 50.0g C 4 H 10 ( 1 mol )( 8 mol CO 2 )(44.01 g) 1 (58.14g)( 2 mol C 4 H 10 )( 1 mol ) = 151 g CO 2 produced

16. 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  MgCl 2 + H 2

17. 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 ) (1 mol H 2 ) ( 2.02 g ) 1 (24.31g) (1 mol Mg) ( 1 mol ) = 1 (24.31g) (1 mol Mg) ( 1 mol ) = 2.12 g H 2 produced Giv: 25.5 g Mg Unk: g H 2

18. Practice: Given the following equation: Fe + S 8  FeS a. What mass of iron is needed to react with 16.0 grams of sulfur? b. How many grams of FeS are produced? Given the following equation: NaClO 3  NaCl + O 2 a. 12.0 g of NaClO 3 will produce how many grams of O 2 ? b. How many grams of NaCl are produced when 80.0 grams of O 2 are produced? Given the following equation: Cu + AgNO 3  Cu(NO 3 ) 2 + Ag a. How much copper is needed to react with 350 g of AgNO 3 ? b. If 115.5 grams of copper(II) nitrate were produced, how many grams of silver nitrate reacted?

19. 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.

20. 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.

21. 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

22. 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?

23. 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?

24. 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. Step 5: Perform calculation and express answer in correct sig figs, unit, and substance.

25. 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

26. 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?

27. Mixed Stoic Problems 1. How much magnesium oxide is produced from 7.5 moles of magnesium ribbon? 2. How many moles of sulfuric acid is needed to react with 100. moles of copper(II) nitrate? 3. If you combust 1500 g of propane (C 3 H 8 ), how much carbon dioxide can be produced? 4. How many moles of aluminum is produced by the complete decomposition of 45 g of aluminum oxide? 5. How much copper is produced if zinc reacts with 35.5 moles of copper(I) sulfate?

28. 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. You could also calculate how much of each substance would be left over if one of your items was limited. You could also calculate how much of each substance would be left over if one of your items was limited. 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.

30. Refer to the diagram demonstrating the reaction between nitrogen and hydrogen. 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 this limiting reactant is in excess? 4. How many molecules of product are made?

31. How many grams of product can be produced if 50.0 grams of hydrogen combines with 50.0 grams of oxygen? How many grams of product 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. The amount of excess will be calculated later.

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

33. 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 )( 2 mol Cu )( 63.55 g ) 1 ( 65.37 g)( 1 mol Zn )( 1 mol ) = 1 ( 65.37 g)( 1 mol Zn )( 1 mol ) = 49 g Cu produced 50.g Cu 2 CO 3 ( 1mol )( 2 mol Cu )( 63.55 g ) 1 (187.11g)(1mol Cu 2 CO 3 )( 1 mol ) = 1 (187.11g)(1mol Cu 2 CO 3 )( 1 mol ) = 34 g Cu produced

34. 50. grams of cuprous carbonate is the limiting reactant. 50. grams of cuprous carbonate is the limiting reactant. You can also calculate moles of product produced from given quantities of reactants. You can also calculate moles of product produced from given quantities of reactants. -Just convert grams of each reactant into moles. The one with the least amount of moles is the limiting reactant.

35. 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

36. 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 ) ( 1 mol MgBr 2 ) 1 ( 159.80 g) ( 1 mol Br 2 ) = 1 ( 159.80 g) ( 1 mol Br 2 ) = 0.626 mol MgBr 2 0.626 mol MgBr 2 75.0 g Mg ( 1 mol ) ( 1 mol MgBr 2 ) 1 ( 24.31 g ) ( 1 mol Mg ) = 1 ( 24.31 g ) ( 1 mol Mg ) = 3.09 mol MgBr 2 3.09 mol MgBr 2 *Br 2 is the limiting reactant.

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

38. Calculate grams of product and grams of excess reactant if 80. g of aluminum reacts with 80. g of chlorine. Calculate grams of product and grams of excess reactant if 80. g of aluminum reacts with 80. g of chlorine.

39. Your Challenge: If given 25.0 grams of each reactant in the equation: If given 25.0 grams of each reactant in the equation: 2CH 2 CHCH 3 + 2NH 3 + 3O 2  2CH 2 CHCN + 6H 2 O 1. Find the limiting reactant. 2. Calculate grams of each product produced. 3. Calculate grams of each excess reactant. *Hint: Follow steps for finding the limiting reactant. Just do 3 mass to mass problems instead of 2.

40. 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

41. You have been calculating the theoretical yield for every stoic problem so far. You have been calculating the theoretical yield for every stoic problem so far. Calculate theoretical yield of product if 25 g of hydrogen combines with 25 g of oxygen. Step 1: Find the limiting. Step 2: Identify that the least amount of product produced not only indicates your limiting reactant, it indicates the theoretical yield of product that could be produced if the experimental conditions were 100% accurate.

42. 2H 2 (g) + O 2 (g)  2H 2 O 25 g H 2 ( 1 mol ) ( 2 mol H 2 O ) (18.02 g ) 1 (2.02 g ) ( 2 mol H 2 ) ( 1 mol ) = 1 (2.02 g ) ( 2 mol H 2 ) ( 1 mol ) = 220 g H 2 O produced 25 g O 2 ( 1 mol ) ( 2 mol H 2 O ) ( 18.02 g ) 1 (32.00g) ( 1 mol O 2 ) ( 1 mol ) = 1 (32.00g) ( 1 mol O 2 ) ( 1 mol ) = 28 g H 2 O produced *Oxygen is the limiting reactant and 28 g of H 2 O is the theoretical yield of product that can be made if everything went 100%.

43. 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":

44. 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. In the previous reaction between hydrogen and oxygen, an experiment yielded 25.6 grams of water. Calculate the percent yield of the product. In the previous reaction between hydrogen and oxygen, an experiment yielded 25.6 grams of water. Calculate the percent yield of the product. % yield = 25.6 g x 100 = 28 g 28 g 91% yield

45. 50. g of silver nitrate is mixed with 50. g of hydrochloric acid in a water based solution. A white precipitate forms (silver chloride). The solution is filtered and the white precipitate collected and dried. The dried precipitate is measured to have a mass of 40.3 g. What is the theoretical and percent yield? 50. g of silver nitrate is mixed with 50. g of hydrochloric acid in a water based solution. A white precipitate forms (silver chloride). The solution is filtered and the white precipitate collected and dried. The dried precipitate is measured to have a mass of 40.3 g. What is the theoretical and percent yield?