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Chapter 9: Stoichiometry. 9.1 Mole to Mole Objective: To perform mole to mole conversion problems.

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Presentation on theme: "Chapter 9: Stoichiometry. 9.1 Mole to Mole Objective: To perform mole to mole conversion problems."— Presentation transcript:

1 Chapter 9: Stoichiometry

2 9.1 Mole to Mole Objective: To perform mole to mole conversion problems.

3 Stoichiometry Stoichiometry is the branch of chemistry that deals with quantities of substances in chemical reactions. Stoichiometry is the branch of chemistry that deals with quantities of substances in chemical reactions.

4 Mole to Mole Problems: The Steps 1. Write chemical equation 2. Balance chemical equation using coefficients 3. Use the following setup to perform calculation: A is the known quantity B is the unknown quantity 4. Don’t forget units on your final answer!! Mole Ratio

5 9.2 Mole to Mass and Mass to Mole Objective: To perform mole to mass and mass to mole conversion problems. Objective: To perform mole to mass and mass to mole conversion problems.

6 Mole to Mass Moles Given Given Moles (Eqn.) Unknown Moles (Eqn.) Molar Mass Unknown 1 mole Unknown Grams Unknown Mole Ratio

7 Mass to Mole Grams Given Molar Mass Given Given Moles (Eqn.) 1 mol Given Unknown Moles (Eqn.) Moles Unknown Mole Ratio

8 9.3 Mass to Mass Objective: To perform mass to mass conversion problems. Objective: To perform mass to mass conversion problems.

9 Stoichiometry Roadmap Grams Given Molar Mass Given Given Moles (Eqn.) 1 mol Given Unknown Moles (Eqn.) Molar Mass Unknown 1 mole Unknown Grams UnknownMole Ratio

10 9.4 Limiting Reactant Objectives: Objectives: (1) To calculate the theoretical yield of a chemical reactions. (2) To determine the limiting reactant and excess reactant in a chemical reaction.

11 Limiting Reactant Any reactant that is used up first in a chemical reaction. Any reactant that is used up first in a chemical reaction. It determines the amount of product that can be formed in the reaction. It determines the amount of product that can be formed in the reaction.

12 Excess Reactant The reactant that is not completely used up in a reaction. The reactant that is not completely used up in a reaction.

13 Limiting Reactant Problems Use the mass to mass conversions Use the mass to mass conversions Grams Given1 mol Given Unknown Moles Molar Mass of Unknown Molar Mass GivenGiven Moles1 mol of Unknown

14 Example Copper reacts with sulfur to form copper(I) sulfide according to the following balanced equation: Copper reacts with sulfur to form copper(I) sulfide according to the following balanced equation: 2Cu + S  Cu 2 S What is the limiting reactant when 80.0 grams of Cu reacts with 25.0 grams of S?

15 Example 2Cu + S  Cu 2 S The general equation is: Grams Given1 mol Given Unknown Moles Molar Mass of Unknown Molar Mass GivenGiven Moles1 mol of Unknown 80.0 g Cu1 mol Cu 1 mol Cu 2 S g Cu 2 S g Cu2 mol Cu1 mol of Cu 2 S 25.0 g S1 mol S 1 mol Cu 2 S g Cu 2 S g S1 mol S1 mol of Cu 2 S Start with Copper: Now use Sulfur: = g Cu 2 S = g Cu 2 S

16 Example The limiting reactant is copper. The limiting reactant is copper. The excess reactant is sulfur. The excess reactant is sulfur. The amount of Cu 2 S that is produced is g Cu 2 S. The amount of Cu 2 S that is produced is g Cu 2 S.

17 9.5 Percent Yield Objective: To calculate percent yield. Objective: To calculate percent yield.

18 Introduction to Percent Yield… If you get 15 out of 20 questions correct on a test, what percentage did you receive on the test? How did you figure this out? If you get 15 out of 20 questions correct on a test, what percentage did you receive on the test? How did you figure this out? 15/20 x 100 = 75% If Sammy Sosa gets 25 hits in the month of May and has 113 bats, what is his batting average for the month of May? Explain how you arrived at your answer. If Sammy Sosa gets 25 hits in the month of May and has 113 bats, what is his batting average for the month of May? Explain how you arrived at your answer. 25/113 = As a percentage, this is written, 25/113 x 100 = 22.1%

19 Percent Yield · Percent yield is the ratio of the actual yield to the theoretical yield for a chemical reaction expressed as a percentage. · It is a measure of efficiency of a reaction. Percent Yield = actual yieldx 100% theoretical yield theoretical yield

20 Actual Yield The amount of product that forms when a reaction is carried out in the laboratory. The amount of product that forms when a reaction is carried out in the laboratory.

21 Theoretical Yield The amount of product that could form during a reaction calculated from a balanced chemical equation. The amount of product that could form during a reaction calculated from a balanced chemical equation. It represents the maximum amount of product that could be formed from a given amount of reactant. It represents the maximum amount of product that could be formed from a given amount of reactant.

22 Example #1 The equation for the complete combustion of ethene (C2H2) is 2C2H2 + 5O2  4CO2 + 2H2O 1. I f 0.10 g of C2H2 is reacted with g of O2, identify the limiting reactant. 2. W hat is the theoretical yield of H2O? 3. I f the actual yield of H2O is 0.05 g, calculate the percent yield.

23 Example #1 2C 2 H 2 + 5O 2  4CO 2 + 2H 2 O 0.10 g C 2 H 2 1 mol C 2 H 2 2 mol H 2 O18.00 g H 2 O g C 2 H 2 2 mol C 2 H 2 1 mol H 2 O g O 2 1 mol O 2 2 mol H 2 O g H 2 O g O 2 5 mol O 2 1 mol H 2 O Start with C 2 H 2 : Now start with O 2 : = 0.07 g H 2 O = g H 2 O Theoretical Yield Limiting Reactant = C 2 H 2

24 Example #1 2C 2 H 2 + 5O 2  4CO 2 + 2H 2 O Actual Yield = 0.05 g H 2 O Theoretical Yield = 0.07 g H 2 O Percent Yield = actual yieldx 100% theoretical yield theoretical yield Percent Yield = 0.05 g H 2 O x 100% = % 0.07 g H 2 O

25 Example #2 Determine the percent yield for the reaction between 3.74 g of Na and excess O 2 if 5.34 g of Na 2 O 2 is recovered? Determine the percent yield for the reaction between 3.74 g of Na and excess O 2 if 5.34 g of Na 2 O 2 is recovered? First, write the chemical equation: First, write the chemical equation: Na + O 2  Na 2 O 2 Second, balance the chemical equation: Second, balance the chemical equation: 2Na + O 2  Na 2 O 2

26 Example #2 2Na + O 2  Na 2 O 2 Solve the mass-mass problem, starting with Na: Solve the mass-mass problem, starting with Na: Actual Yield = 5.34 g Na 2 O 2 Theoretical Yield = 6.34 g Na 2 O 2 Percent Yield = actual yield x 100% theoretical yield Percent Yield = 5.34 g x 100% = 84.23% 6.34 g 3.74 g Na1 mol Na 1 mol Na 2 O g Na 2 O g Na2 mol Na1 mol Na 2 O 2 = 6.34 g Na 2 O 2


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