Chapter 4: Chemical Quantities & Aqueous Reactions

1. Chapter 4: Chemical Quantities & Aqueous Reactions CHE 123: General Chemistry I Dr. Jerome Williams, Ph.D. Saint Leo University

2. Overview • Stoichiometry • Percent Yield

3. Stoichiometry • Stoichiometrydeals with the quantity of materials consumed and or produced in chemical reactions. • Stoichiometry deals specifically with mass-mole relationships between reactants and products found in a chemical equation.

4. Stoichiometry • Two Fundamental Principles • Must always work from a balanced reaction. • Axiom #3 is always foremost in out minds. When in doubt, convert to moles for they guide us everywhere we wish to go in chemistry.

6. Stoichiometry • How to Solve Stoichiometry Problems? • Use Handy-Dandy Five Step Method • 1. Write out balanced chemical reaction. Place the given and unknown quantities (with units) above item in reaction. • 2. Calculate molar masses for items needed to solve problem. Place these below items in reaction.

7. Stoichiometry • Handy Dandy Five Step Method • 3. Convert given quantity into moles (Axiom #3). • 4. Using the balanced chemical equation (step 1), convert moles obtained in step 3 into moles of quantity you are after. • 5. Convert moles of final quantity into the desired unit asked for in problem.

8. Stoichiometry: Concept Map

9. Stoichiometry • How many grams of water are produced from the combustion of 32.75 g of butane (C4H10)? The balanced chemical reaction for this process is shown below. 2 C4H10 + 13 O2→ 8 CO2 + 10 H2O

10. Stoichiometry • Apply Handy Dandy Five Step Method • Step 1: Equation is already balanced. Known and unknown values are placed above item in equation. 32.75 g ? g 2 C4H10 + 13 O2→ 8 CO2 + 10 H2O

11. Stoichiometry • Apply Handy Dandy Five Step Method • Step 2: Calculate Molar Masses for known and unknown. Values are placed below item in equation. C4H10 = (4 x 12.01 g/mol) + (10 x 1.01 g/mol) = 58.14 g/mol H2O = (2 x 1.01 g/mol) + (1 x 16.00 g/mol) = 18.02 g/mol

12. Stoichiometry • Apply Handy Dandy Five Step Method • Step 2: Calculate Molar Masses for known and unknown. Values are placed below item in equation. 32.75 g ? g 2 C4H10 + 13 O2→ 8 CO2 + 10 H2O 58.14 g/mol 18.02 g/mol

13. Stoichiometry • Apply Handy Dandy Five Step Method • Step 3: Convert given quantity into moles (Axiom #3) 32.75 g C4H10 X ( 1 mol C4H10 / 58.14 g C4H10 ) = 0.56329 mol C4H10

14. Stoichiometry • Apply Handy Dandy Five Step Method • Step 4: Using the balanced chemical equation (see Step 1), convert moles obtained in step 3 into moles of quantity you are after. • Balanced reaction shows relationship between water and butane (2 C4H10 + 13 O2→ 8 CO2 + 10 H2O) 0.56329 mol C4H10(10 moles H2O / 2 moles C4H10) = 2.8164 mol H2O

15. Stoichiometry • Apply Handy Dandy Five Step Method • Step 5: Convert moles of final quantity into the desired unit asked for in problem. 2.81647 mol H2O x (18.02 g H2O / 1 mol H2O) = 50.75 g H2O • Check: Answer is reasonable with correct units.

16. Percent Yield • Percent yield is a measure of how completely reactants are turned into products. • Most reactions do not go to completion (i.e., reach 100% conversion) due to side reactions, transfer losses, etc.

17. Percent Yield • Percent yield is calculated using the following equation. % yield = (actual yield / theoretical yield) x 100% • actual yield is amount obtained in the analysis. • theoretical yield is amount obtained assuming 100 % complete reaction.

18. A Second Example for Practice • According to the U.S. Dept. Energy if 3.4 x 1015 g of octane was burned in 2004, how many grams of carbon dioxide was released into the atmosphere that year? • Use the Handy-Dandy Five Step Method to Solve this problem.

21. Practice Problem • Answer to Problem 1.0 x 1016 g CO2 • This value is the maximum amount of carbon dioxide that would be produced assuming complete reaction (hence, this is the theoretical yield).