Chapter 9: Stoichiometry

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

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

Mole to Mole Problems: The Steps1. Write chemical equation2. Balance chemical equation using coefficients3. Use the following setup to perform calculation:

A is the known quantityB is the unknown quantity4. Dont forget units on your final answer!!Mole Ratio

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

Mole to MassMoles GivenGiven Moles (Eqn.)Unknown Moles (Eqn.)Molar Mass Unknown1 mole UnknownGrams UnknownMole Ratio

Mass to MoleGrams GivenMolar Mass GivenGiven Moles (Eqn.)1 mol GivenUnknown Moles (Eqn.)Moles UnknownMole Ratio

9.3 Mass to MassObjective: To perform mass to mass conversion problems.

Stoichiometry RoadmapGrams GivenMolar Mass GivenGiven Moles (Eqn.)1 mol GivenUnknown Moles (Eqn.)Molar Mass Unknown1 mole UnknownGrams UnknownMole Ratio

9.4 Limiting ReactantObjectives: To calculate the theoretical yield of a chemical reactions.To determine the limiting reactant and excess reactant in a chemical reaction.

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.

Excess ReactantThe reactant that is not completely used up in a reaction.

Limiting Reactant ProblemsUse the mass to mass conversions

Grams Given1 mol GivenUnknown MolesMolar Mass of UnknownMolar Mass GivenGiven Moles1 mol of Unknown

ExampleCopper reacts with sulfur to form copper(I) sulfide according to the following balanced equation:2Cu + S Cu2S

What is the limiting reactant when 80.0 grams of Cu reacts with 25.0 grams of S?

Example2Cu + S Cu2SThe general equation is:

Start with Copper:Now use Sulfur:= 100.19 g Cu2S= 124.08 g Cu2S

Grams Given1 mol GivenUnknown MolesMolar Mass of UnknownMolar Mass GivenGiven Moles1 mol of Unknown

80.0 g Cu 1 mol Cu1 mol Cu2S159.17 g Cu2S

63.55 g Cu2 mol Cu1 mol of Cu2S

25.0 g S1 mol S1 mol Cu2S159.17 g Cu2S32.07 g S1 mol S1 mol of Cu2S

ExampleThe limiting reactant is copper.

The excess reactant is sulfur.

The amount of Cu2S that is produced is 100.19 g Cu2S.

9.5 Percent YieldObjective: To calculate percent yield.

Introduction to Percent YieldIf 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%

Percent YieldPercent 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

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

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.

Example #1The equation for the complete combustion of ethene (C2H2) is 2C2H2 + 5O2 4CO2 + 2H2O

If 0.10 g of C2H2 is reacted with 201.60 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.

Example #12C2H2 + 5O2 4CO2 + 2H2O

Start with C2H2:Now start with O2:= 0.07 g H2O= 45.36 g H2OTheoretical YieldLimiting Reactant = C2H2

0.10 g C2H21 mol C2H22 mol H2O18.00 g H2O26.02 g C2H22 mol C2H21 mol H2O

201.60 g O21 mol O22 mol H2O18.00 g H2O

32.00 g O25 mol O21 mol H2O

Example #12C2H2 + 5O2 4CO2 + 2H2O

Actual Yield = 0.05 g H2OTheoretical Yield = 0.07 g H2O

Percent Yield = actual yieldx 100% theoretical yield

Percent Yield = 0.05 g H2O x 100% = 71.43 %0.07 g H2O

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

Example #22Na + O2 Na2O2

Solve the mass-mass problem, starting with Na:

Actual Yield = 5.34 g Na2O2Theoretical Yield = 6.34 g Na2O2

Percent Yield = actual yield x 100%theoretical yield

Percent Yield = 5.34 g x 100% = 84.23%6.34 g

= 6.34 g Na2O2

3.74 g Na1 mol Na1 mol Na2O277.98 g Na2O2 22.99 g Na2 mol Na1 mol Na2O2

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