1. Chapter 9 “Stoichiometry” Modified & adapted from: Pre-AP Chemistry Charles Page High School Stephen L. Cotton Mr. Mole

2. The Arithmetic of Equations  OBJECTIVES: Explain how balanced equations apply to both chemistry and everyday life.

3. The Arithmetic of Equations  OBJECTIVES: Interpret balanced chemical equations in terms of: a) moles b) representative particles c) mass d) gas volume (Liters) at STP

4. The Arithmetic of Equations  OBJECTIVES: Identify the quantities that are always conserved in chemical reactions.

5. Let’s make some Cookies!  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  Thus, a recipe is much like a balanced equation.

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

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

8. 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 2 formula units Al 2 O 3 form 4 atoms Al and 3 molecules O 2 Now read this: 2Na + 2H 2 O  2NaOH + H 2

9. #2. In terms of Moles  The coefficients tell us how many moles of each substance 2Al 2 O 3  Al + 3O 2  2 moles Al 2 O 3 form 4 moles Al and 3 moles O 2 2Na + 2H 2 O  2NaOH + H 2  Remember: A balanced equation is a Molar Ratio

10. #3. In terms of Mass  The Law of Conservation of Mass applies  We can check mass by using moles. 2H 2 + O 2   2H 2 O 2 moles H 2 2.02 g H 2 1 mole H 2 = 4.04 g H 2 1 mole O 2 32.00 g O 2 1 mole O 2 = 32.00 g O 2 36.04 g H 2 + O 2 + reactants

11. In terms of Mass (for products) 2H 2 + O 2   2H 2 O 2 moles H 2 O 18.02 g H 2 O 1 mole H 2 O = 36.04 g H 2 O 36.04 g H 2 + O 2 = 36.04 g H 2 O The mass of the reactants must equal the mass of the products. 36.04 grams reactant = 36.04 grams product

12. #4. In terms of Volume  At STP, 1 mol of any gas = 22.4 L 2H 2 + O 2   2H 2 O (2 x 22.4 L H 2 ) + (1 x 22.4 L O 2 )  (2 x 22.4 L H 2 O) NOTE: mass and atoms are ALWAYS conserved - however, molecules, formula units, moles, and volumes will not necessarily be conserved! 67.2 Liters of reactant ≠ 44.8 Liters of product!

13. Practice:  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

14. Chemical Calculations  OBJECTIVES: Construct “mole ratios” from balanced chemical equations, and apply these ratios in mole-mole stoichiometric calculations.

15. Chemical Calculations  OBJECTIVES: Calculate stoichiometric quantities from balanced chemical equations using units of moles, mass, representative particles, and volumes of gases at STP.

16. Steps to Calculate Stoichiometric Problems 1. Correctly balance the equation. 2. Convert the given amount into moles. 3. Set up mole ratios. 4. Use mole ratios to calculate moles of desired chemical. 5. Convert moles back into final unit.

17. Mole to Mole conversions 2Al 2 O 3  Al + 3O 2 each time we use 2 moles of Al 2 O 3 we will also make 3 moles of O 2 2 moles Al 2 O 3 3 mole O 2 or 2 moles Al 2 O 3 3 mole O 2 These are the two possible conversion factors to use in the solution of the problem.

18. Mole to Mole conversions  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 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! Conversion factor from balanced equation

19. 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)

20. How do you get good at this?

21. Steps to Calculate Stoichiometric Problems 1. Correctly balance the equation. 2. Determine the mole ratio. 3. Determine where you are starting and where you are finishing. 4. Use mole ratios to calculate moles of desired chemical. 5. Check that units cancel

22. Mass-Mass Problem: 6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed? 4Al + 3O 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 (6.50 x 1 x 2 x 101.96) ÷ (26.98 x 4 x 1) = 12.3 g Al 2 O 3 are formed

23. Another example:  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

24. Volume-Volume Calculations:  How many liters of CH 4 at STP are required to completely react with 17.5 L of O 2 ? CH 4 + 2O 2  CO 2 + 2H 2 O 17.5 L O 2 22.4 L O2O2 1 mol O2O2 2 O2O2 1 CH 4 1 mol CH 4 22.4 L CH 4 = 8.75 L CH 4 22.4 L O 2 1 mol O 2 1 mol CH 4 22.4 L CH 4 Notice anything relating these two steps?

25. Avogadro told us:  Equal volumes of gas, at the same temperature and pressure contain the same number of particles.  Moles are numbers of particles  You can treat reactions as if they happen liters at a time, as long as you keep the temperature and pressure the same. 1 mole = 22.4 L @ STP

26. Shortcut for Volume-Volume?  How many liters of CH 4 at STP are required to completely react with 17.5 L of O 2 ? CH 4 + 2O 2  CO 2 + 2H 2 O 17.5 L O2O2 2 L O 2 1 L CH 4 = 8.75 L CH 4 Note: This only works for Volume-Volume problems.

28. Limiting Reagent & Percent Yield  OBJECTIVES: Identify the limiting reagent in a reaction.

29. Limiting Reagent & Percent Yield  OBJECTIVES: Calculate theoretical yield, percent yield, and the amount of excess reagent that remains unreacted given appropriate information.

30. “Limiting” Reagent  If you are given one dozen loaves of bread, a gallon of mustard, and three pieces of salami, how many salami sandwiches can you make?  The limiting reagent is the reactant you run out of first.  The excess reagent is the one you have left over.  The limiting reagent determines how much product you can make

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

32.  If 10.6 g of copper reacts with 3.83 g sulfur, how many grams of the product (copper (I) sulfide) will be formed? 2Cu + S  Cu 2 S 10.6 g Cu 63.55g Cu 1 mol Cu 2 mol Cu 1 mol Cu 2 S 1 mol Cu 2 S 159.16 g Cu 2 S = 13.3 g Cu 2 S 3.83 g S 32.06g S 1 mol S 1 S 1 Cu 2 S 1 mol Cu 2 S 159.16 g Cu 2 S = 19.0 g Cu 2 S = 13.3 g Cu 2 S Cu is the Limiting Reagent, since it produced less product.

33. Another example:  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  How much excess reagent will remain? Excess = 4.47 grams

34. The Concept of: A little different type of yield than you had in Driver’s Education class.

35. What is Yield?  Yield is the amount of product made in a chemical reaction.  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

36. Example:  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  What is the actual yield?  What is the theoretical yield?  What is the percent yield? = 6.78 g Cu = 13.8 g Cu = 49.1 %

37. Details on Yield  Percent yield tells us how “efficient” a reaction is.  Percent yield can not be bigger than 100 %.  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

38. Because most reactions occur in solutions, scientists must express the relationship between the solute and the solvent. This expression is called: Because most reactions occur in solutions, scientists must express the relationship between the solute and the solvent. This expression is called:MOLARITY

39. 39 Molarity Molarity is a term used to express concentration. The units of molarity are moles per liter (It is abbreviated as a capital M) When working problems, it is a good idea to change M into its units.

40. Molarity Solving Process: Convert volume cm 3 to dm 3 Convert volume cm 3 to dm 3 Convert grams to moles Convert grams to moles Divide moles by dm 3 Divide moles by dm 3

41. ex. Calculate the molarity of 2.50 x 10 2 cm 3 of solution containing 9.46 g of CsBr. = 0.178 M CsBr

42. What if you wanted to make a solution? use the molarity equation and solve for grams. use the molarity equation and solve for grams. ex. How would you make 5.00x 10 2 cm 3 of a 0.133M solution of MnSeO 4 ? = 13.2 g of MnSeO4

43. Boiling Point Elevation & Freezing Point Depression The temperature at which a substance changes state (melting, freezing, boiling) at standard conditions is a definite fixed quantity, characteristic of that substance. However, if a second substance is mixed with the original substance, the change of state temperature of the original gets altered. This is particularly important when dealing with aqueous solutions.

44. Boiling Point Elevation & Freezing Point Depression For example, dissolving a solute in water causes the freezing point to be lowered, as the molecules of water have more difficulty aligning with each other to form a pure crystal. On the other hand, the boiling point of an aqueous solution is higher than the boiling point of pure water.

45. Boiling Point Elevation & Freezing Point Depression When a pure substance like water freezes, the temperature stays constant throughout the phase change. However with a solution, the temperature continues to drop as more water freezes and is removed from the solution, making the remaining solution more concentrated. (In a similar manner, the boiling point of a solution continues to rise as more water evaporates.)

46. Boiling Point Elevation & Freezing Point Depression In addition, when a solution freezes, supercooling often occurs. This is a situation in which the temperature briefly drops below the freezing point, then rises slightly when the crystals actually form. The temperature you need to record is not the lowest temperature reached, but the temperature at which the ice crystals first start forming.

47. Boiling Point Elevation & Freezing Point Depression The concentration of a solution used in freezing point depression experiments must be expressed in molality, not molarity. The molality of a solution is the number of moles of solute added to 1 kg of solvent. The symbol used for molality is a small case m, italicized: m.

48. Boiling Point Elevation & Freezing Point Depression The freezing point depression of a solute (ΔT) is given by the equation: ΔT = Kf × m × i Where: Kf = freezing pt. constant (°C/m), (Kb = boiling pt. constant (°C/m) m = the molality of the solution, i = the # of particles (molecules or ions) produced by one molecule of the solute.

49. Boiling Point Elevation & Freezing Point Depression Molality 1. Convert grams of solvent to kilograms 2. Convert grams solute to moles 3. Calculate: (moles / Kg) = m OR m = grams | 1 mole | 1 | 1000 g |molar mass | grams of | 1 kg solvent (water)

50. Boiling Point Elevation & Freezing Point Depression Boiling Point/Freezing Point 1. Calculate the molality of the solution 2. Determine which constant to use: *Freezing Point Depression: Kf = 1.853 ⁰ C/m *Boiling Point Elevation : Kb = 0.515 ⁰ C/m 3. Calculate: ΔT = Kf × m × i 4. Then subtract from the normal FPt of water OR add to the normal BPt of water