2. Stoichiometric Calculations Stoichiometry – Ch. 9

3. Stoichiometry Greek for “measuring elements” The calculations of quantities in chemical reactions based on a balanced equation. We can interpret balanced chemical equations several ways.

4. Stoichiometry Definition Composition stoichiometry deals with the mass relationships of elements in compounds. Reaction stoichiometry involves the mass relationships between reactants and products in a chemical reaction.

5. Click below to watch the Visual Concept. Visual ConceptStoichiometry

6. A. Proportional Relationships I have 5 eggs. How many cookies can I make? 3/4 c. brown sugar 1 tsp vanilla extract 2 eggs 2 c. chocolate chips Makes 5 dozen cookies. 2 1/4 c. flour 1 tsp. baking soda 1 tsp. salt 1 c. butter 3/4 c. sugar 5 eggs5 doz. 2 eggs = 12.5 dozen cookies Ratio of eggs to cookies

7. A. Proportional Relationships Mole Ratio Mole Ratio is a conversion factor that relates the amounts in moles of any two substances involved in a chemical reaction Example: 2Al 2 O 3 (l) → 4Al(s) + 3O 2 (g) Mole Ratios: 2 mol Al 2 O 3 2 mol Al 2 O 3 4 mol Al 4 mol Al 3 mol O 2 3 mol O 2

8. B. Stoichiometry Steps 1. Write a balanced equation. 2. Identify known & unknown. 3. Line up conversion factors. ◦ Mole ratio - moles  moles ◦ Molar mass -moles  grams ◦ Molarity - moles  liters soln ◦ Molar volume -moles  liters gas 4. Check answer.

9. C. Molar Conversions Molar Mass (g/mol) 6.02  10 23 particles/mol MOLES NUMBER OF PARTICLES Mass In Grams

10. Converting Between Amounts in Moles

11. Conversions of Quantities in Moles

12. D. Stoichiometry Problems How many moles of KClO 3 must decompose in order to produce 9 moles of oxygen gas? 9 mol O 2 2 mol KClO 3 3 mol O 2 = 6 mol KClO 3 2KClO 3  2KCl + 3O 2 ? mol9 mol

13. We can’t measure moles!! What can we do? We can convert grams to moles. ◦ Periodic Table Then use moles to change chemicals ◦ Balanced equation Then turn the moles back to grams. ◦ Periodic table

14. Periodic Table MolesAMolesBMass g B Periodic Table Balanced Equation Mass g A Decide where to start based on the units you are given Stop based on what unit you are asked for

15. D. Stoichiometry Problems How many grams of silver will be formed from 12.0 g copper? 12.0 g Cu 1 mol Cu 63.55 g Cu = 40.7 g Ag Cu + 2AgNO 3  2Ag + Cu(NO 3 ) 2 2 mol Ag 1 mol Cu 107.87 g Ag 1 mol Ag 12.0 g? g

16. For example... If 10.1 g of Fe are added to a solution of Copper (II) Sulfate, how much solid copper would form? 2Fe + 3CuSO 4  Fe 2 (SO 4 ) 3 + 3Cu 10.1 g Fe 55.85 g Fe 1 mol Fe 2 mol Fe 3 mol Cu 1 mol Cu 63.55 g Cu =17.2 g Cu

17. How do you get good at this?

18. A. Limiting Reactants Available Ingredients Available Ingredients ◦ 4 slices of bread ◦ 1 jar of peanut butter ◦ 1/2 jar of jelly b Limiting Reactant bread b Excess Reactants peanut butter and jelly

19. Limiting Reactants The limiting reactant is the reactant that limits the amount of the other reactant that can combine and the amount of product that can form in a chemical reaction. The excess reactant is the substance that is not used up completely in a reaction.

20. How do you find out? Do two stoichiometry problems. The one that makes the least product is the limiting reagent.

21. Click below to watch the Visual Concept. Visual Concept Limiting Reactants and Excess Reactants

22. If 10.6 g of copper reacts with 3.83 g S. How many grams of product 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 Limiting Reagent

24. Percentage Yield The theoretical yield is the maximum amount of product that can be produced from a given amount of reactant. The actual yield of a product is the measured amount of that product obtained from a reaction.. Section 3 Limiting Reactants and Percentage Yield Percent yield Percent yield = Actual x 100 % Theoretical

25. Click below to watch the Visual Concept. Visual Concept Comparing Actual and Theoretical Yield

26. Percent Yield When 45.8 g of K 2 CO 3 react with excess HCl, 46.3 g of KCl are formed. Calculate the theoretical and % yields of KCl. K 2 CO 3 + 2HCl  2KCl + H 2 O + CO 2 45.8 g? g actual: 46.3 g

27. Percent Yield 45.8 g K 2 CO 3 1 mol K 2 CO 3 138.21 g K 2 CO 3 = 49.4 g KCl 2 mol KCl 1 mol K 2 CO 3 74.55 g KCl 1 mol KCl K 2 CO 3 + 2HCl  2KCl + H 2 O + CO 2 45.8 g? g actual: 46.3 g Theoretical Yield:

28. Percent Yield Theoretical Yield = 49.4 g KCl % Yield = 46.3 g 49.4 g  100 = 93.7% K 2 CO 3 + 2HCl  2KCl + H 2 O + CO 2 45.8 g49.4 g actual: 46.3 g