Chapter 9 Stoichiometry 9 1 Mole to Mole
- Slides: 26
Chapter 9: Stoichiometry
9. 1 Mole to Mole Objective: To perform mole to mole conversion problems.
Stoichiometry n Stoichiometry is the branch of chemistry that deals with quantities of substances in chemical reactions.
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 Mole Ratio B is the unknown quantity 4. Don’t forget units on your final answer!!
9. 2 Mole to Mass and Mass to Mole n Objective: To perform mole to mass and mass to mole conversion problems.
Mole to Mass Moles Given Unknown Moles (Eqn. ) Molar Mass Unknown Given Moles 1 mole (Eqn. ) Unknown Mole Ratio Grams Unknown
Mass to Mole Grams Given Unknown 1 mol Given Moles (Eqn. ) Molar Mass Given Moles Unknown (Eqn. ) Mole Ratio
9. 3 Mass to Mass n Objective: To perform mass to mass conversion problems.
Stoichiometry Roadmap Grams Given Unknown 1 mol Given Moles (Eqn. ) Molar Mass Unknown Molar Mass Given 1 mole Unknown Given Moles (Eqn. ) Grams Unknown Mole Ratio
9. 4 Limiting Reactant Objectives: (1) To calculate theoretical yield of a chemical reactions. (2) To determine the limiting reactant and excess reactant in a chemical reaction. n
Limiting Reactant n n Any reactant that is used up first in a chemical reaction. It determines the amount of product that can be formed in the reaction.
Excess Reactant n The reactant that is not completely used up in a reaction.
Limiting Reactant Problems Use the mass to mass conversions n Grams Given 1 mol Given Molar Mass Given Unknown Moles Molar Mass of Unknown Given Moles 1 mol of Unknown
Example n Copper reacts with sulfur to form copper(I) sulfide according to the following balanced equation: 2 Cu + S Cu 2 S What is the limiting reactant when 80. 0 grams of Cu reacts with 25. 0 grams of S?
Example 2 Cu + S Cu 2 S The general equation is: Grams Given 1 mol Given Molar Mass Given Unknown Moles Molar Mass of Unknown Given Moles 1 mol of Unknown Start with Copper: 80. 0 g Cu 1 mol Cu 2 S 159. 17 g Cu 2 S 63. 55 g Cu 2 mol Cu 1 mol of Cu 2 S 159. 17 g Cu 2 S 1 mol of Cu 2 S = 100. 19 g Cu 2 S Now use Sulfur: 25. 0 g S 1 mol Cu 2 S 32. 07 g S 1 mol S = 124. 08 g Cu 2 S
Example n The limiting reactant is copper. n The excess reactant is sulfur. n The amount of Cu 2 S that is produced is 100. 19 g Cu 2 S.
9. 5 Percent Yield n Objective: To calculate percent yield.
Introduction to Percent Yield… n 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. 25/113 = 0. 221 As a percentage, this is written, 25/113 x 100 = 22. 1% n
Percent Yield ·Percent yield is the ratio of the actual yield to theoretical yield for a chemical reaction expressed as a percentage. · It is a measure of efficiency of a reaction. Percent Yield = actual yield theoretical yield x 100%
Actual Yield n The amount of product that forms when a reaction is carried out in the laboratory.
Theoretical Yield n n 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.
Example #1 The equation for the complete combustion of ethene (C 2 H 2) is 2 C 2 H 2 + 5 O 2 4 CO 2 + 2 H 2 O 1. If 0. 10 g of C 2 H 2 is reacted with 201. 60 g of O 2, identify the limiting reactant. 2. What is theoretical yield of H 2 O? 3. If the actual yield of H 2 O is 0. 05 g, calculate the percent yield.
Example #1 2 C 2 H 2 + 5 O 2 4 CO 2 + 2 H 2 O Start with C 2 H 2: 0. 10 g C 2 H 2 1 mol C 2 H 2 2 mol H 2 O 26. 02 g C 2 H 2 2 mol C 2 H 2 Limiting Reactant = C 2 H 2 18. 00 g H 2 O 1 mol H 2 O = 0. 07 g H 2 O Theoretical Yield Now start with O 2: 201. 60 g O 2 1 mol O 2 32. 00 g O 2 2 mol H 2 O 5 mol O 2 18. 00 g H 2 O 1 mol H 2 O = 45. 36 g H 2 O
Example #1 2 C 2 H 2 + 5 O 2 4 CO 2 + 2 H 2 O Actual Yield = 0. 05 g H 2 O Theoretical Yield = 0. 07 g H 2 O Percent Yield = actual yield x 100% theoretical yield Percent Yield = 0. 05 g H 2 O x 100% = 71. 43 % 0. 07 g H 2 O
Example #2 n n n 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: Na + O 2 Na 2 O 2 Second, balance the chemical equation: 2 Na + O 2 Na 2 O 2
Example #2 2 Na + O 2 Na 2 O 2 n Solve the mass-mass problem, starting with Na: 3. 74 g Na 1 mol Na 22. 99 g Na 1 mol Na 2 O 2 2 mol Na Actual Yield = 5. 34 g Na 2 O 2 Theoretical Yield = 6. 34 g Na 2 O 2 Percent Yield = actual yield theoretical yield x 100% Percent Yield = 5. 34 g x 100% = 84. 23% 6. 34 g 77. 98 g Na 2 O 2 1 mol Na 2 O 2 = 6. 34 g Na 2 O 2
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