Chapter 9
Stoichiometry
Unit Essential Question:
What numerical information can we find
from a balanced chemical reaction?
Lesson Essential Question:
How can we tell the amounts of substances
involved in a reaction?
Section 1: Calculating Quantities in
Reactions
 Coefficients of balanced equations show proportions
(ratios) of substances in reactions.
 Similar to amounts in a recipe.
 Use this information to adjust amount of
product(s) made or reactants needed.
 Relative amounts (coefficients) in equation = moles.
 Example: 2 C8H18 + 25 O2  16 CO2 + 18 H2O
 2 moles of C8H18 react with 25 moles of O2 to form 16
moles of CO2 and 18 moles of H2O
Stoichiometry
 Deals with quantities of substances in chemical
reactions.
 Compare amounts (moles) and also masses.
 Limiting reactants and percent yield
can be calculated.
 Can be useful for everyday activities such as
baking.
Mole Ratio
 Needed in order to perform calculations!
 Allows you to convert between substances.
 Set up using fractions and conversion factors.
 Use dimensional analysis.
 Get ratios from balanced equation.
 Ex: 2H2 + O2  2H2O
 Mole ratio examples: 2mol H2/1mol O2,
1mol O2/2mol H2O, or 2molH2/2molH2O
Steps to Converting Between
Amounts of Substances
 Step 1: Balance the chemical equation.
 Step 2: Convert given amount to moles.
 Remember- always start with your given!
 Step 3: Use a mole ratio from the balanced
equation to convert from the amount of one
substance to the amount of another substance.
 Use dimensional analysis for all conversions!
 Step 4: Convert out of moles if asked for
grams or the number of particles.
Sample Problem #1
 Consider the reaction for the commercial preparation
of ammonia:
 N2 + 3 H2  2 NH3
 How many moles of hydrogen are needed to prepare
312 moles of ammonia?
 Notice that mole ratios convert between substances
and units stay the same (all moles)!
 Moles of nitrogen needed?
Sample Problem #2
 What mass of NH3 can be made from 1,221 g of H2
and excess N2?
 N2 + 3 H2  2 NH3
Sample Problem #3
 How many grams of C5H8 form from 1.89 x 1024
molecules C5H12?
 C5H12 (l)  C5H8 (l) + 2 H2 (g)
Lesson Essential Question:
How can we calculate the amount of
products made in a reaction?
Section 2: Limiting Reactants and Percent Yield
 Limiting reactant – substances that limit the
amount of product that can be formed.
 Limiting reactants run out first!
 Excess reactant – substances that are not used
up in the reaction. There are some left over.
 Do not limit the amount of product that can be
formed.
 Think back to the sandwich activity- what were...
 LR [limiting reagent(s)]?
 ER [excess reagent(s)]?
Theoretical &
Actual Yield
This is what we have been
calculating in practice
problems so far!
 Theoretical: the amount that should be able to be
produced based on the limiting reactant.
 If everything in the reaction went according to plan,
and all of the reactant(s) reacted, this is how much
product should be made.
 This is NOT the same as the actual yield- amount
that is produced based on an experiment
 Error occurs, so actual yield is less than the
theoretical yield.
 Can never be greater than the theoretical yield.
Determining the Limiting Reactant
 You will be given amounts of two or more reactants.
 Convert all amounts to moles.
 Use mole ratios in the balanced chemical equation
to determine the number of moles of a product that
will be produced.
 Convert all reactant amounts into the same product!
 Convert the moles of product to grams of product.
 Do this for each reactant amount given and see how
much product each produces.
 Whichever produces LESS product is your limiting
reactant- it will be completely consumed first.
Sample Problem #5
 Identify the limiting reactant and the theoretical
yield of phosphorus acid, H3PO3, if 225 g of PCl3 is
mixed with 123 g of H2O
 PCl3 + 3 H2O  H3PO3 + 3 HCl
Sample Problem #6
 Determine the limiting reactant if 14.0g N2 are
mixed with 9.0g H2. What is the theoretical
yield in grams of the product?
 N2 + 3H2  2NH3
Application
 Cheapest ingredient = excess reactant.
 Most expensive = limiting reactant.
 Cost effective!
 Examples:
 Cider vinegar
 ER = oxygen; LR = ethanol
 Banana Flavoring
 ER = acetic acid; LR = isopentyl alcohol
Percent Yield
 Actual yield is always less than the theoretical yield.
 LR is not always 100% used up.
 Can be due to human errors, not reacting to
completion, reverse reactions and/or unwanted side
reactions occurring, and loss during purification.
 Efficiency is another name for percent yield.
 Percent Yield:
(actual yield/theoretical yield) x 100 = % yield
 If your percent yield is greater than 100%, there could
be unreacted or unwanted materials in your product.
Sample Problem #7
 Using your answers from Sample Problem #6,
calculate the percent yield if 16.1 g NH3 are
formed in an experiment.
 N2 + 3 H2  2 NH3
Sample Problem #8
 How many grams of CH3COOC5H11 should form
if 4808 g are theoretically possible and the percent
yield for the reaction is 80.5%?