1. Chapter 9 – STOICHIOMETRY The MATH of a CHEMICAL REACTION

2. 9.1 – Reaction Stoichiometry -Ratio of compounds in a reaction -YOU NEED A BALANCED CHEMICAL REACTION!! Examples: Al + O 2  Al 2 O 3 What this means: 4 MOLES Al react with 3 MOLES O 2 to form 2 MOLES Al 2 O 3

3. Example: 9.1 – Reaction Stoichiometry C 2 H 6 + O 2  CO 2 + H 2 O 2 moles C 2 H 6 reacts with 7 moles O 2 to form 4 mole CO 2 and 6 moles H 2 O

4. 9.2 – Stoichiometry Calculations -You need to be sure your units are canceling - algebra!! CAN ONLY SWITCH FROM ONE SUBSTANCE TO ANOTHER BY MOLES!!

5. 9.2 – Stoichiometry Calculations Types of Calculations: 1. mole of one substance to mole of another Moles AMoles B Ratio from Balanced Equation

6. Example: 4 NH 3 + 3 O 2  2 N 2 + 6 H 2 O 9.4 mole NH 3 = ? mole N 2 1.Start with given 2.Multiply by a conversion factor 3.Conversion factor – mole ratio between substances 4.Plug into calculator 5.Remember units!!! Work:

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8. 9.2 – Stoichiometry Calculations Types of Calculations: 2. mole of one substance to gram of another Moles AMoles B Ratio from Balanced Equation Molar Mass from Periodic Table Grams B

9. Example: 4 NH 3 + 3 O 2  2 N 2 + 6 H 2 O 4.35 mole NH 3 = ? g H 2 O Work:

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11. 9.2 – Stoichiometry Calculations Types of Calculations: 3. gram of one substance to mole of another Grams AMoles A Ratio from Balanced Equation Molar Mass from Periodic Table Moles B

12. Example: 4 NH 3 + 3 O 2  2 N 2 + 6 H 2 O 19.3 g NH 3 = ? mol O 2 Work:

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14. 9.2 – Stoichiometry Calculations Types of Calculations: 4. gram of one substance to gram of another Grams A Moles A Ratio from Balanced Equation Molar Mass from Periodic Table Moles B Grams B Molar Mass from Periodic Table

15. Example: 4 NH 3 + 3 O 2  2 N 2 + 6 H 2 O 18.6 g O 2 = ? g N 2 Work:

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18. Limiting Reactants -Need a balanced equation! -You will be given information about more than one starting material -Limiting Reactant = The reactant which will run out first -Excess Reactant = The reactant which you have “more than enough of”

19. Examples: -Making a Bicycle: Limiting Reactants 12 1 1 ++ frametireshandlebarsbike 1 + 2 + 1 1 Stockroom Inventory: 101813

20. Limiting Reactants Examples: -Making a Bicycle: How many bikes can we make??? 12 1 1 ++ frametireshandlebarsbike 1 + 2 + 1 1 101813 How many bikes COULD we make from each component? 10 13 9

21. How many maximum could be made?? Limiting Reactants 9 *We picked the smallest value of the “could make” answers!! Which was the limiting reactant??tires What are the excess reactants?? frames and handlebars

22. Now a Chemistry Example: Limiting Reactants 3Fe + 4H 2 O  Fe 3 O 4 + 4H 2 Starting materials: 24 g Fe and 20 g H 2 O 1.) What is the LR? 2.) How many grams Fe 3 O 4 will be produced? Work:

23. Percent Yield Compares the amount actually received in lab to the amount that was expected from math problem. Actual Yield = The amount actually obtained (will be given!!!) Theoretical Yield = The amount expected. The number you calculate with pen and paper (stoich problem.) % Yield =

24. Percent Yield Example: Easy 3Fe + 4H 2 O  Fe 3 O 4 + 4H 2 Starting materials: 14.2 g Fe and 27.8 g H 2 O After the experiment is complete: 17.3 g Fe 3 O 4 are actually obtained What is % yield? Harder:

25. Percent Yield Work: 3Fe + 4H 2 O  Fe 3 O 4 + 4H 2 14.2 g Fe 27.8 g H2O % Yield