1. TOPIC IV: Chemical Stoichiometry LECTURE SLIDES Balancing Equations Simple Stoichiometry Limiting reagent % Yield Combustion Analysis Kotz &Treichel, Chapter 4

2. When a substance reacts with another substance or interacts with energy to form one or more different substances, a chemical reaction occurs. The changes that have taken place are represented by the chemical equation, which keeps track of all the atoms involved in the reactants and the products. The chemical equation must be carefully balanced to insure that it adequately describes the reaction without loosing or gaining any atoms in the process!

3. One cannot create or destroy matter in a chemical reaction, so all reactant atoms must show up somewhere in the product(s)... To completely describe the reaction that has taken place, the equation should include not only the formula or symbol for each reactant and product, but also their physical state, and the energy process involved. We will indicate physical state by subscripts (s) solid, (l)liquid, (g) gas, (aq) aqueous, in water solution.

4. We will not at this point add the energy component to our chemical equations but we must keep in mind that all reactions involve energy in some form: Either: The reaction requires some form of heat and is termed “endothermic” or The reaction gives off heat and is described as “exothermic”

5. Typical chemical reactions include predictable types: Combination: A + B --> C Decomposition: AB ---> A + B Double Replacement in Aqueous solution: AB + CD --> AD + CB Combustion: C x H y + O 2 --> CO 2 + H 2 O (We’ll consider another type, oxidation/reduction later in the unit!)

6. As we review balancing equations, let’s describe them by type: Combination or “Synthesis”: A + B --> C H 2(g) + O 2(g) --> H 2 O (g) 2 atoms O 1 atom O UNBALANCED!

7. 2 H 2(g) + O 2(g) --> 2 H 2 O (g) We have now accounted for all atoms 4 atoms H, left and right 2 atoms O, left and right Equation is balanced... H 2(g) + O 2(g) --> 2 H 2 O (g) Rebalance: To balance O’s, place coefficient before formula: Now 4 atoms H

8. NaCl (l) --> Na (l) + Cl 2(g) 2NaCl (l) --> Na (l) + Cl 2(g) 2NaCl (l) --> 2Na (l) + Cl 2(g) Balancing process, Decomposition Reactions: energy AB ----------> A + B Al 2 (CO 3 ) 3(s) --> Al 2 O 3(s) + CO 2(g) Al 2 (CO 3 ) 3(s) --> Al 2 O 3(s) + 3CO 2(g) Balance Cl Balance Na Balance C, then O ok

9. Balancing Combustion Reactions: (C,H) + O 2 --> CO 2 + H 2 O + heat (C,H,O) + O 2 --> CO 2 + H 2 O + heat C 5 H 12 + O 2 --> CO 2 + H 2 O Methodology: a) Balance C b) Balance H c) Balance O Organic molecules: hydrocarbons, carbohydrates, alcohols: “fuels”

10. c) Balance O: C 5 H 12 + 8 O 2 --> 5 CO 2 + 6 H 2 O a) Balance C: C 5 H 12 + O 2 --> 5 CO 2 + H 2 O b) Balance H: C 5 H 12 + O 2 --> 5 CO 2 + 6 H 2 O 10 O 6 O

11. Now, a more interesting one: C 8 H 18 + O 2 --> CO 2 + H 2 O a) C 8 H 18 + O 2 --> 8 CO 2 + H 2 O b) C 8 H 18 + O 2 --> 8 CO 2 + 9 H 2 O c) C 8 H 18 + 12.5 O 2 --> 8 CO 2 + 9 H 2 O 16 O + 9 O = 25 O 25 O 16 O 9 O

12. At this point we have: C 8 H 18 + 25/2 or 12.5 O 2 --> 8 CO 2 + 9 H 2 O Should have whole number coefficients, so MULTIPLY ALL SPECIES BY 2: 2 C 8 H 18 + 25 O 2 --> 16 CO 2 + 18 H 2 O Checking for balance: 16 C 36 H 50 O -->; 16 C 36 H; 32 O +18 O

13. GROUP WORK 4.1, BALANCE: C 6 H 12 O 6 + O 2 --> CO 2 + H 2 O (sugar) C 5 H 11 OH + O 2 --> CO 2 + H 2 O (alcohol) a) Balance C b) Balance H c) Balance O d) Check

14. Balancing Double Replacement Reactions : AB + CD --> AD + CB General directions: save H for next to last, O for last: a) balance P: 2 H 3 PO 4 + Ca(OH) 2 --> Ca 3 (PO 4 ) 2 + H 2 O b) balance Ca: 2H 3 PO 4 + 3 Ca(OH) 2 --> Ca 3 (PO 4 ) 2 + H 2 O H 3 PO 4 + Ca(OH) 2 --> Ca 3 (PO 4 ) 2 + H 2 O

15. c) balance H 2 H 3 PO 4 + 3 Ca(OH) 2 --> Ca 3 (PO 4 ) 2 + H 2 O 6H 6H 2H 2 H 3 PO 4 + 3 Ca(OH) 2 --> Ca 3 (PO 4 ) 2 + 6 H 2 O 6H 6H 12H d) checkout O’s: 8O + 6O ---> 8O + 6O BALANCED!

16. Stoichiometry: Introduction Let us balance one more equation, using techniques just introduced: Fe 2 O 3(s) + HCl (aq) ---> H 2 O + FeCl 3(aq) a) Fe 2 O 3(s) + HCl (aq) ---> H 2 O + 2FeCl 3(aq) b) Fe 2 O 3(s) + 6HCl (aq) ---> H 2 O + 2FeCl 3(aq) c) Fe 2 O 3(s) + 6HCl (aq) ---> 3H 2 O + 2FeCl 3(aq)

17. Now let us go one step further and determine the quantitative relationships a BALANCED equation implies, utilizing the below: Fe 2 O 3(s) + 6HCl (aq) ---> 3H 2 O + 2FeCl 3(aq) On the most basic level, the equation counts number of “molecules” or ionic “formula units”of each species are needed in order for this reaction to occur: 1 “formula unit” Fe 2 O 3 + 6 molecules HCl---> 3 molecules H 2 O + 2 “formula units” FeCl 3

19. But, we know we can’t reach into a box and pull out molecules: we have to weigh out a specific mass of the molecules and then figure out how many we have because they are too tiny to count. To count by weighing, we use the mole, knowing that 1 MOLE of molecules = 6.022 X 10 23 molecules and we can work with our equation in terms of moles...

20. The relationship: 1 “formula unit” Fe 2 O 3 + 6 molecules HCl---> 3 molecules H 2 O + 2 “formula units” FeCl 3 BECOMES: 1 mole Fe 2 O 3 + 6 moles HCl---> 3 moles H 2 O + 2 moles FeCl 3 and...

21. When we talk moles, we can talk mass in grams by calculating the molar masses of the compounds involved: Fe 2 O 3 : 2 Fe= 2X 55.85 = 111.70 3O = 3X 16.00 = 48.00 159.70 g/mol Fe 2 O 3 HCl: 1H = 1.01 1Cl = 35.45 36.46 g/mol HCl

22. FeCl 3 : 1 Fe = 1 X 55.847 = 55.85 3 Cl = 3 X 35.45 = 106.35 162.20 g/mol FeCl 3 H 2 O: 2H = 2X1.01 = 2.02 1O = 1X 16.00 = 16.00 18.02 g/mol H 2 O

23. 1 mole Fe 2 O 3 = 1 mol X 159.70 g/mol Fe 2 O 3 = 159.70 g 6 moles HCl = 6 mol X 36.46 g/mol HCl = 218.76 g 3 moles H 2 O = 3 mol X 18.02 g/mol H 2 O = 54.06 g 2 moles FeCl 3 = 2 mol X 162.20 g/mol FeCl 3 = 324.40 g Equivalent: 1 mol Fe 2 O 3 + 6 mol HCl---> 3 mol H 2 O + 2 mol FeCl 3 159.70 g Fe 2 O 3 + 218.76 g HCl ---> 54.06 g H 2 O + 324.40 g FeCl 3

24. 159.70 g Fe 2 O 3 + 218.76 g HCl ---> 54.06 g H 2 O + 324.40 g FeCl 3 Note how the Law Of Conservation of Matter has been upheld: 159.70 g + 218.76 g ---> 54.06 g + 324.40 g 378.46 g total mass reactants -----> 378.46 g total mass products

25. Let us now apply our new knowledge to problem solving situations: In all cases, we will use the equation we have developed: Fe 2 O 3(s) + 6HCl (aq) ---> 3H 2 O + 2FeCl 3(aq) 159.70 g/mol 36.46 g/mol 18.02 g/mol 162.20 g/mol If we start with 25.00 g of Fe 2 O 3, how many moles and how many g of HCl are required for the reaction, and how many g of FeCl 3 and moles of water will we theoretically obtain?

26. To solve, copy balanced equation, and insert under each reactant and product the molar mass and any question or information given in the problem: Fe 2 O 3(s) + 6HCl (aq) ---> 3H 2 O + 2FeCl 3(aq) 159.70 g/mol 36.46 g/mol 18.02 g/mol 162.20 g/mol 25.00g ?g, mol ?moles ?g We can now proceed to solve each question:

27. Fe 2 O 3(s) + 6HCl (aq) ---> 3H 2 O + 2FeCl 3(aq) 159.70 g/mol 36.46 g/mol 18.02 g/mol 162.20 g/mol 25.00g ?g, mol Question #1: 25.00 g Fe 2 O 3 = ? mol HCl Relationships: 159.70 g Fe 2 O 3 = 1 mol Fe 2 O 3 1 mol Fe 2 O 3 = 6 mol HCl

28. PATHWAY: g Fe 2 O 3 ---> mol Fe 2 O 3 ---> mol HCl Conversion factor: M Conversion factor: balanced equation

29. Conversion factor: M Conversion factor: M Conversion factor: balanced equation Question #2: 25.00 g Fe 2 O 3 = ? g HCl

30. Fe 2 O 3(s) + 6HCl (aq) ---> 3H 2 O + 2FeCl 3(aq) 159.70 g/mol 36.46 g/mol 18.02 g/mol 162.20 g/mol 25.00g ?g, mol ?moles ?g answer:.9393 mol, 34.25 g To finish: Question #3: 25.00 g Fe 2 O 3 = ? mol H 2 O Question #4: 25.00 g Fe 2 O 3 = ? g FeCl 3

31. Fe 2 O 3(s) + 6HCl (aq) ---> 3H 2 O + 2FeCl 3(aq Question #3: 25.00 g Fe 2 O 3 = ? mol H 2 O Question #4: 25.00 g Fe 2 O 3 = ? g FeCl 3

32. To prove we did everything in a correct fashion, let’s see if it all adds up: Fe 2 O 3(s) + 6HCl (aq) ---> 3H 2 O + 2FeCl 3(aq) 159.70 g/mol 36.46 g/mol 18.02 g/mol 162.20 g/mol 25.00g ?g, mol ?moles ?g answer:.9393 mol,.4696 mol 34.25 g ---> 8.462 g 50.78 g 25.00 + 34.25 = 59.25 g reactants 8.462 + 50.78 = 59.245 g products

33. GROUP WORK 4.2: How many g of Fe 2 O 3 are required to make 75.00 g of FeCl 3 ? Fe 2 O 3(s) + 6HCl (aq) ---> 3H 2 O + 2FeCl 3(aq) 159.70 g/mol 36.46 g/mol 18.02 g/mol 162.20 g/mol ? g 75.00 g PATHWAY: g FeCl 3 --->mol FeCl 3 ---> mol Fe 2 O 3 ---> g Fe 2 O 3

34. Next, let’s consider problems we’ll call “LIMITING REAGENT PROBLEMS” In this type of problem, amounts OF BOTH reactants will be given to you, and you’ll be asked to calculate how much of one or more products can be formed. In most cases, one given value will be in excess, to insure that all of the other reagent is used up.

35. We can call the the reactant present in excessive amount “the excess reagent” and we can call the reactant we expect to get used up “the limiting reagent” because it limits the amount of product which we can form..... Never try to guess from amounts given which reactant is which; you should always calculate to determine the nature of both given reactants. There are several ways this can be accomplished; I will present the method I am most comfortable with:

36. Let’s use the same equation we have been considering today: Fe 2 O 3(s) + 6HCl (aq) ---> 3H 2 O + 2FeCl 3(aq) 159.70 g/mol 36.46 g/mol 18.02 g/mol 162.20 g/mol 35.00 g 100.0 g ? g Now, suppose you were given 100.0 g HCl and 35.00 g Fe 2 O 3. How many g of FeCl 3 will you theoretically obtain? Amounts of both: Limiting Reagent problem

37. My method of choice: Calculate g of product from each reactant. The reactant which gives you the larger amount of product is the excess reagent. The amount of product calculated from this reagent answer is incorrect, too much... The reactant that gives you the smaller amount of product is the limiting reagent. (The calculation from the limiting reagent gives is the correct answer!).....

38. Fe 2 O 3(s) + 6HCl (aq) ---> 3H 2 O + 2FeCl 3(aq) 159.70 g/mol 36.46 g/mol 18.02 g/mol 162.20 g/mol 35.00 g 100.0 g ? g The question becomes: 35.00 g Fe 2 O 3 = ? g FeCl 3 100.0 g HCl = ? g FeCl 3 which answer is correct? GROUP WORK 4.3: CALCULATE BOTH, DECIDE!

39. Theoretical, Actual and Percent Yield So far we have considered how much product we could theoretically obtain from a given reaction by calculating from a balanced equation. We have been working with the following equation, so let’s finish up the topic with it: Fe 2 O 3(s) + 6HCl (aq) ---> 3H 2 O + 2FeCl 3(aq) 159.70 g/mol 36.46 g/mol 18.02 g/mol 162.20 g/mol

40. “theoretical yield” is the amount of any product you calculate from your balanced equation, using the limiting reagent when amounts of all reactants are given. “actual yield” is the amount of product you actually obtain from a given experiment using the reaction described in the balanced equation; this amount is rarely as large as the theoretical yield. Percent yield = actual yield X 100% theoretical yield

41. Let’s recall the last problem we did: If 35.00 g Fe 2 O 3 were reacted with 100.0 g HCl, how many g of FeCl 3 would theoretically be obtained? We calculated the amount of product from both reactants, and decided that 71.10 g of FeCl 3 would be the “expected” amount of product, based on the limiting reagent and balanced equation. (See next slide)

42. Fe 2 O 3(s) + 6HCl (aq) ---> 3H 2 O + 2FeCl 3(aq) 159.70 g/mol 36.46 g/mol 18.02 g/mol 162.20 g/mol ? g, theoretical 35.00 g 100.0 g ans: 71.10 g Too much

43. Now suppose you ran the reaction in the laboratory and discovered that you could only isolate 65.88 g of FeCl 3. The next question asked would be, “what is your percent yield for this experiment?” Fe 2 O 3(s) + 6HCl (aq) ---> 3H 2 O + 2FeCl 3(aq) 159.70 g/mol 36.46 g/mol 18.02 g/mol 162.20 g/mol 71.10 g theoretical 35.00 g 100.0 g 65.88 g actual % Yield = ?

44. Percent yield = actual yield X 100 theoretical yield % Yield = 65.88 g obtained or “actual” X 100 71.10 g calculated or “theoretical” = 92.66 %

45. Combustion Analysis To round off Chapter 4, let’s consider the analysis of organic hydrocarbons (C x H y ) and compounds C x H y O z : This type of analysis is done by the complete combustion of these compounds with O 2 and an analysis of the CO 2 and H 2 O formed. All the carbon in the original compound will be contained in the CO 2, all the hydrogen in the water, and O can be determined by difference. Let’s see how that works: C x H y or C x H y O z + O 2 ---> CO 2 + H 2 O

46. C x H y : Since all the C in the original compound is in the CO 2, we can get moles of C from grams of CO 2 produced: g, CO 2  mol, CO 2  mol, C Since all the H in the original compound is in the H 2 O produced, we can get moles of H from moles of H 2 O, g, H 2 O  mol H 2 O  mol H To get formula, obtain simplest mole ratio, C, H

47. A 0.523 g sample of unknown compound C x H y was burned in air to give 1.612 g CO 2 and 0.7425 g H 2 O. Its molar mass was determined to be 114 g/mol. What is its empirical and molecular formula? 1.612 g CO 2 = ? Moles C CO 2, 44.010 g/mol.7425 g H 2 O = ? Moles H H 2 O, 18.015 g/mol

49. g CO 2, H 2 O ---> mol CO 2, H 2 O ---> mol C,H ---> g C,H g original compound - g C,H = g O in original compound g C,H,O ---> mol C,H,O ---> simplest mole ratio If there is oxygen in the original compound, the problem is more complicated: the amount of oxygen must be detected by difference: CHECKOUT EXAMPLE 4.7 IN TEXT!