2 Objective: To perform mole to mole conversion problems.
3 StoichiometryStoichiometry 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.
6 Mole to Mass Molar Mass Unknown Moles Given Unknown Moles (Eqn.) Grams UnknownGiven Moles (Eqn.)1 mole UnknownMole Ratio
7 Mass to Mole Unknown Moles (Eqn.) Grams Given 1 mol Given Moles UnknownMolar Mass GivenGivenMoles (Eqn.)Mole Ratio
8 9.3 Mass to MassObjective: To perform mass to mass conversion problems.
9 Stoichiometry Roadmap Unknown Moles (Eqn.)Molar Mass UnknownGrams Given1 mol GivenMolar Mass GivenGivenMoles (Eqn.)1 mole UnknownGrams UnknownMole Ratio
10 9.4 Limiting Reactant Objectives: To calculate the theoretical yield of a chemical reactions.To determine the limiting reactant and excess reactant in a chemical reaction.
11 Limiting ReactantAny reactant that is used up first in a chemical reaction.It determines the amount of product that can be formed in the reaction.
12 Excess ReactantThe reactant that is not completely used up in a reaction.
13 Limiting Reactant Problems Use the mass to mass conversionsGrams Given1 mol GivenUnknown MolesMolar Mass of UnknownMolar Mass GivenGiven Moles1 mol of Unknown
14 ExampleCopper reacts with sulfur to form copper(I) sulfide according to the following balanced equation:2Cu + S Cu2SWhat is the limiting reactant when 80.0 grams of Cu reacts with 25.0 grams of S?
15 Example 2Cu + S Cu2S The general equation is: Start with Copper: Grams Given1 mol GivenUnknown MolesMolar Mass of UnknownMolar Mass GivenGiven Moles1 mol of UnknownStart with Copper:80.0 g Cu1 mol Cu1 mol Cu2Sg Cu2S63.55 g Cu2 mol Cu1 mol of Cu2S= g Cu2SNow use Sulfur:25.0 g S1 mol S1 mol Cu2Sg Cu2S32.07 g S1 mol of Cu2S= g Cu2S
16 Example The limiting reactant is copper. The excess reactant is sulfur.The amount of Cu2S that is produced is g Cu2S.
17 9.5 Percent YieldObjective: 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?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 = 0.221As a percentage, this is written, 25/113 x 100 = 22.1%
19 · It is a measure of efficiency of a reaction. 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 yield x 100%theoretical yield
20 Actual YieldThe amount of product that forms when a reaction is carried out in the laboratory.
21 Theoretical YieldThe 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.
22 Example #1The equation for the complete combustion of ethene (C2H2) is2C2H O2 4CO H2OIf 0.10 g of C2H2 is reacted with g of O2, identify the limiting reactant.What is the theoretical yield of H2O?If the actual yield of H2O is 0.05 g, calculate the percent yield.
23 Example #1 2C2H2 + 5O2 4CO2 + 2H2O Start with C2H2: 0.10 g C2H21 mol C2H22 mol H2O18.00 g H2O26.02 g C2H22 mol C2H21 mol H2O= 0.07 g H2OTheoretical YieldLimiting Reactant = C2H2Now start with O2:g O21 mol O22 mol H2O18.00 g H2O32.00 g O25 mol O21 mol H2O= g H2O
24 Example #1 2C2H2 + 5O2 4CO2 + 2H2O Actual Yield = 0.05 g H2O Theoretical Yield = 0.07 g H2OPercent Yield = actual yield x 100%theoretical yieldPercent Yield = g H2O x 100% = %0.07 g H2O
25 Example #2Determine the percent yield for the reaction between 3.74 g of Na and excess O2 if 5.34 g of Na2O2 is recovered?First, write the chemical equation:Na + O2 Na2O2Second, balance the chemical equation:2Na + O2 Na2O2
26 Example #2 2Na + O2 Na2O2 3.74 g Na 1 mol Na 1 mol Na2O2 Solve the mass-mass problem, starting with Na:Actual Yield = 5.34 g Na2O2Theoretical Yield = 6.34 g Na2O2Percent Yield = actual yield x 100%theoretical yieldPercent Yield = g x 100% = 84.23%6.34 g3.74 g Na1 mol Na1 mol Na2O277.98 g Na2O222.99 g Na2 mol Na= 6.34 g Na2O2