1. Chapter 8 “Stoichiometry” Mr. Mole

2. Section 8.2 The Arithmetic of Equations u OBJECTIVES: Interpret balanced chemical equations in terms of: a) moles, b) representative particles, c) mass, and d) gas volume (Liters) at STP.

3. Section 8.2 The Arithmetic of Equations u OBJECTIVES: Identify the quantities that are always conserved in chemical reactions.

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

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

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

7. 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 2formula unitsAl 2 O 3 form4 atoms Al and3moleculesO2O2 Now read this: 2Na + 2H 2 O  2NaOH + H 2

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

9. #3. In terms of Mass u The Law of Conservation of Mass applies u 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

10. 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

11. #4. In terms of Volume u 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, volumes will not necessarily be conserved! 67.2 Liters of reactant ≠ 44.8 Liters of product!

12. Practice: u 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 Reactant: 2[2(27g)+3(16g)] = 204g Products: 4(27g) + 3(32g) = 204 g

13. Homework u u Do practice problems 8A u Due Monday 12/15

14. Section 8.3 Chemical Calculations u OBJECTIVES: Construct “mole ratios” from balanced chemical equations, and apply these ratios in mole-mole stoichiometric calculations.

15. Section 8.4 & 8.5 Chemical Calculations u OBJECTIVES: Calculate stoichiometric quantities from balanced chemical equations using units of moles, mass, representative particles, and volumes of gases at STP.

16. 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.

17. Mole to Mole conversions u 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

18. 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? (5 mol O 2 /2 mol C 2 H 4 )3.84 mol C 2 H 4 = (9.6 mol) How many moles of C 2 H 2 are needed to produce 8.95 mole of H 2 O? (2 mol C 2 H 4 /2 mol H 2 O )8.95 mol 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 mol CO 2 /2 mol C 2 H 4 )2.47 mol C 2 H 4 = (4.94 mol)

19. How do you get good at this?

20. 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.

21. 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

22. Another example: u If 10.1 g of Fe are added to a solution of Copper (II) Sulfate, how many grams of solid copper would form? Fe + CuSO 4  FeSO 4 + Cu 10.1g Fe(1 mol Fe/56 g Fe) = 0.18 mol Fe 0.18 mol Fe ( 1 mol Cu/1 mol Fe)(63.5g Cu/mol Cu) Answer = 11.43 g Cu

23. Volume-Volume Calculations: u 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?

24. Avogadro told us: u Equal volumes of gas, at the same temperature and pressure contain the same number of particles. u Moles are numbers of particles u 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

25. Shortcut for Volume-Volume? u 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.

26. Use coefficients from balanced equation MOLE (mol) Mass (g) Particle (at. or m’c) 1 mol = molar mass (in g) Volume (L or dm 3 ) 1 mol = 22.4 L 1 mol = 22.4 dm 3 1 mol = 6.02 x 10 23 particles SUBSTANCE “A” Stoichiometry Island Diagram MOLE (mol) Mass (g) 1 mol = molar mass (in g) Volume (L or dm 3 ) 1 mol = 22.4 L 1 mol = 22.4 dm 3 1 mol = 6.02 x 10 23 particles SUBSTANCE “B” Particle (at. or m’c)

27. 2 __TiO 2 + __Cl 2 + __C __TiCl 4 + __CO 2 + __CO How many mol chlorine will react with 4.55 mol carbon? 3 mol C 4 mol Cl 2 4.55 mol C = 6.07 mol Cl 2 What mass titanium (IV) oxide will react with 4.55 mol carbon? () = 242 g TiO 2 43221 () CCl 2 CTiO 2 3 mol C 2 mol TiO 2 4.55 mol C () 1 mol TiO 2 79.9 g TiO 2

28. 2 __TiO 2 + __Cl 2 + __C __TiCl 4 + __CO 2 + __CO How many mol chlorine will react with 4.55 mol carbon? 3 mol C 4 mol Cl 2 4.55 mol C = 6.07 mol Cl 2 What mass titanium (IV) oxide will react with 4.55 mol carbon? () = 242 g TiO 2 43221 () CCl 2 CTiO 2 3 mol C 2 mol TiO 2 4.55 mol C () 1 mol TiO 2 79.9 g TiO 2

29. 3. The units on the islands at each end of the bridge being crossed appear in the conversion factor for that bridge. How many molecules titanium (IV) chloride can be made from 115 g titanium (IV) oxide? TiCl 4 TiO 2 () = 8.7 x 10 23 m’c TiCl 4 115 g TiO 2 1 mol TiO 2 79.9 g TiO 2 () 2 mol TiO 2 2 mol TiCl 4 () 1 mol TiCl 4 6.02 x 10 23 m’c TiCl 4 Island Diagram helpful reminders: 2. The middle bridge conversion factor is the only one that has two different substances in it. The conversion factors for the other six bridges have the same substance in both the numerator and denominator. 1. Use coefficients from the equation only when crossing the middle bridge. The other six bridges always have “1 mol” before a substance’s formula. 1 mol coeff. 1 mol

30. 2 mol Ir How many moles of nickel are produced if 8.7 x 10 25 atoms of iridium are consumed? IrNi = 217 mol Ni () 3 mol Ni () 1 mol Ir 6.02 x 10 23 at. Ir 8.7 x 10 25 at. Ir 2 Ir + Ni 3 P 2 3 Ni + 2 IrP iridium (Ir)nickel (Ni)

31. Section 8.6 Limiting Reagent & Percent Yield u OBJECTIVES: Identify the limiting reagent in a reaction.

32. Section 8.7 Limiting Reagent & Percent Yield u OBJECTIVES: Calculate theoretical yield, percent yield, and the amount of excess reagent that remains unreacted given appropriate information.

33. “Limiting” Reagent u 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? u The limiting reagent is the reactant you run out of first. u The excess reagent is the one you have left over. u The limiting reagent determines how much product you can make

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

35. u 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.

36. Another example: u 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 u How much excess reagent will remain? Excess = 4.47 grams

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

38. What is Yield? u Yield is the amount of product made in a chemical reaction. u 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

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

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