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Stoichiometry
Mr. Mole
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• First
• A bit of review

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The Arithmetic of Equations

• OBJECTIVES
• Interpret balanced chemical equations in terms
  of
• a) moles
• b) representative particles
• c) mass

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Remember Avocado, I mean, AVOGADRO?
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Remember Avocado, I mean, AVOGADRO?
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Flowchart
Atoms or Molecules
Divide by 6.02 X 1023
Multiply by 6.02 X 1023
Moles
Multiply by atomic/molar mass from periodic table
Divide by atomic/molar mass from periodic table
Mass (grams)
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Remember calculating Molar Mass?
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Molar Mass

• Molar mass is determined by adding the atomic
  masses for the atoms, molecules or compounds you
  are working with
• Ex. Molar mass of CaCl2
• Avg. Atomic mass of Calcium 40g
• Avg. Atomic mass of Chlorine 35.5g
• Molar Mass of calcium chloride 40 g/mol Ca
  (2 X 35.5) g/mol Cl ? 111 g/mol CaCl2

20 Ca  40
17Cl 35.5
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Practice

• Calculate the Molar Mass of calcium phosphate
• Formula
• Masses elements
• Ca 3 Cas X 40
• P 2 Ps X 31
• O 8 Os X 16
• Molar Mass

Ca3(PO4)2
120 g
62 g
128 g
120g Ca 62g P 128g O
310 g/mol
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Lets make some Cookies!

• When baking cookies, a recipe is usually used,
  telling the exact amount of each ingredient.
• If you need more, you can double or triple the
  amount
• Thus, a recipe is much like a balanced equation.

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Chocolate Chip Cookies!!

• 1 cup butter
• 1/2 cup white sugar
• 1 cup packed brown sugar
• 1 teaspoon vanilla extract
• 2 eggs
• 2 1/2 cups all-purpose flour
• 1 teaspoon baking soda
• 1 teaspoon salt
• 2 cups semisweet chocolate chips
• Makes 3 dozen

How many eggs are needed to make 3 dozen
cookies? How much butter is needed for the
amount of chocolate chips used? How many eggs
would we need to make 9 dozen cookies? How much
brown sugar would I need if I had 1 ½ cups white
sugar?
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Cookies and ChemistryHuh!?!?

• Just like chocolate chip cookies have recipes,
  chemists have recipes as well
• Instead of calling them recipes, we call them
  reaction equations
• Furthermore, instead of using cups and teaspoons,
  we use moles
• Lastly, instead of eggs, butter, sugar, etc. we
  use chemical compounds as ingredients

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Practice

• Write the balanced reaction for hydrogen gas
  reacting with oxygen gas.
• 2 H2 O2 ? 2 H2O
• How many moles of reactants are needed?
• What if we wanted 4 moles of water?
• What if we had 3 moles of oxygen, how much
  hydrogen would we need to react and how much
  water would we get?
• What if we had 50 moles of hydrogen, how much
  oxygen would we need and how much water produced?

2 mol H2 1 mol O2
4 mol H22 mol O2
6 mol H2, 6 mol H2O
25 mol O2, 50 mol H2O
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Stoichiometry is

• Greek for measuring elements
• Pronounced stoy kee ahm uh tree
• Defined as calculations of the quantities in
  chemical reactions, based on a balanced equation.
• There are 4 ways to interpret a balanced chemical
  equation

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1. In terms of Particles

• An Element is made of atoms
• A Molecular compound (made of only nonmetals) is
  made up of molecules (Dont forget the diatomic
  elements)
• Ionic Compounds (made of a metal and nonmetal
  parts) are made of formula units

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Example 2H2 O2 ? 2H2O

• Two molecules of hydrogen and one molecule of
  oxygen form two molecules of water.
• Another example 2Al2O3 4Al 3O2

2
formula units
Al2O3
form
4
atoms
Al
and
3
molecules
O2
Now read this 2Na 2H2O 2NaOH H2
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2. In terms of Moles

• The coefficients tell us how many moles of each
  substance
• 2Al2O3 4Al 3O2
• 2Na 2H2O 2NaOH H2
• Remember A balanced equation is a Molar Ratio

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Mole Ratios

• These mole ratios can be used to calculate the
  moles of one chemical from the given amount of a
  different chemical
• Example How many moles of chlorine are needed to
  react with 5 moles of sodium (without any sodium
  left over)?
• 2 Na Cl2 ? 2 NaCl

5 moles Na ? mol Cl2 X 2 mol Na 1 mol
Cl2
2.5 moles Cl2
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3. In terms of Mass

• The Law of Conservation of Mass applies
• We can check mass by using moles.
• 2H2 O2 2H2O

2 g H2
2 moles H2
4 g H2

1 mole H2

32 g O2
1 mole O2
32 g O2

1 mole O2
36 g H2 O2
36 g H2 O2
reactants
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In terms of Mass (for products)

• 2H2 O2 2H2O

18 g H2O
36 g H2O

2 moles H2O
1 mole H2O
36 g H2 O2

36 g H2O
36 grams reactant 36 grams product
The mass of the reactants must equal the mass of
the products.
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Practice

• Show that the following equation follows the Law
  of Conservation of Mass (show the atoms balance,
  and the mass on both sides is equal)
• 2Al2O3 4Al 3O2

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Chemical Calculations

• OBJECTIVES
• Calculate stoichiometric quantities from balanced
  chemical equations using units of moles, mass,
  and representative particles

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Chemical Calculations

• Basic equation for all of these
• Given amount Unknown amount
• ------------------- X
  -----------------------
• Recipe amount Recipe amount
• Cross multiply and you got this!

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Mole to Mole conversions

• How many moles of O2 are produced when 3.34 moles
  of Al2O3 decompose?
• 2Al2O3 4Al 3O2

3.34 mol Al2O3 ? O2 ------------------
----------- X ----------------- 2
mol Al2O3 3 mol O2
5.01 mol O2
If you know the amount of ANY chemical in the
reaction, you can find the amount of ALL the
other chemicals!
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How do you get good at this?
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Practice

• 2C2H2 5 O2 4CO2 2 H2O
• If 3.84 moles of C2H2 are burned, how many moles
  of O2 are needed?

• How many moles of C2H2 are needed to produce
  8.95 mole of H2O?

• If 2.47 moles of C2H2 are burned, how many moles
  of CO2 are formed?

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Practice

• 2C2H2 5 O2 4CO2 2 H2O
• If 3.84 moles of C2H2 are burned, how many moles
  of O2 are needed?

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Practice

• 2C2H2 5 O2 4CO2 2 H2O
• If 3.84 moles of C2H2 are burned, how many moles
  of O2 are needed?

• How many moles of C2H2 are needed to produce
  8.95 mole of H2O?

• If 2.47 moles of C2H2 are burned, how many moles
  of CO2 are formed?

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Practice

• 2C2H2 5 O2 4CO2 2 H2O
• If 3.84 moles of C2H2 are burned, how many moles
  of O2 are needed?

(9.6 mol)

• How many moles of C2H2 are needed to produce
  8.95 mole of H2O?

(8.95 mol)

• If 2.47 moles of C2H2 are burned, how many moles
  of CO2 are formed?

(4.94 mol)
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How do you get good at this?
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Steps to Calculate Stoichiometric Problems

1. Correctly balance the equation.
2. Place the given amount over the recipe amount.
3. Place the unknown amount over the recipe amount.
4. Cross multiply to solve

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Another example

• If 10.1 g of Fe are added to a solution of copper
  (II) sulfate, how many grams of solid copper
  would form?
• 2Fe 3CuSO4 Fe2(SO4)3 3Cu

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Another example

• If 10.1 g of Fe are added to a solution of copper
  (II) sulfate, how many grams of solid copper
  would form?

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Another example

• If 10.1 g of Fe are added to a solution of copper
  (II) sulfate, how many grams of solid copper
  would form?
• 2Fe 3CuSO4 Fe2(SO4)3 3Cu

Answer 17.2 g Cu
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The Concept of
A little different type of yield than you learned
in Drivers Ed
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What is Yield?

• Yield is the amount of product made in a chemical
  reaction.
• There are three types
• 1. Actual yield- what you actually get in the lab
  when the chemicals are mixed
• 2. Theoretical yield- what the balanced equation
  tells should be made
• 3. Percent yield Actual
  Theoretical

x 100
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Example

• 6.78 g of copper is produced when 3.92 g of Al
  are reacted with excess copper (II) sulfate.
• 2Al 3 CuSO4 Al2(SO4)3 3Cu
• What is the actual yield?
• What is the theoretical yield?
• What is the percent yield?

6.78 g Cu
13.8 g Cu
49.1
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Details on Yield

• Percent yield tells us how efficient a reaction
  is.
• Percent yield can not be bigger than 100 .
• Theoretical yield will always be larger than
  actual yield!
• Why? Due to impure reactants competing side
  reactions loss of product in filtering or
  transferring between containers measuring

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Limiting Reactant Percent Yield

• OBJECTIVES
• Identify the limiting reactant in a reaction.

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Limiting Reactant Percent Yield

• OBJECTIVES
• Calculate theoretical yield, percent yield, and
  the amount of excess reactant that remains
  unreacted given appropriate information.

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Limiting Reactant Cookies

• 1 cup butter
• 1/2 cup white sugar
• 1 cup packed brown sugar
• 1 teaspoon vanilla extract
• 2 eggs
• 2 1/2 cups all-purpose flour
• 1 teaspoon baking soda
• 1 teaspoon salt
• 2 cups semisweet chocolate chips
• Makes 3 dozen

If we had the specified amount of all ingredients
listed, could we make 4 dozen cookies? What if we
had 6 eggs and twice as much of everything else,
could we make 9 dozen cookies? What if we only
had one egg, could we make 3 dozen cookies?
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How do you find out which is limited?

• The chemical that makes the least amount of
  product is the limiting reactant.
• You can recognize limiting reactant problems
  because they will give you 2 amounts of chemical
• Do two stoichiometry problems, one for each
  reactant you are given.

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Limiting Reactant

• If you are given one dozen loaves of bread and
  three pieces of salami, how many salami
  sandwiches can you make?
• The limiting reactant is the reactant you run out
  of first.
• The excess reactant is the one you have left
  over.
• The limiting reactant determines how much product
  you can make

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• If 10.6 g of copper reacts with
  3.83 g sulfur, how many grams of the product
  (copper (I) sulfide) will be formed?

Cu is the Limiting Reactant, since it produced
less product.
13.3 g Cu2S
13.3 g Cu2S
19.0 g Cu2S
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Another example

• If 10.3 g of aluminum are reacted with 51.7 g of
  CuSO4 how much copper (grams) will be produced?
• Al CuSO4 ? Cu Al2(SO4)3

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Another example

• If 10.3 g of aluminum are reacted with 51.7 g of
  CuSO4 how much copper (grams) will be produced?
• 2Al 3CuSO4 ? 3Cu Al2(SO4)3

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Another example

• In the preceding reaction, how much of the excess
  reactant will remain?
• Remember the CuSO4 is limited, so the Cu
  produced was 20.6 g.
• Flip the equation around to see how much Al was
  used to make this much Cu. In other words

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Another example

• 20.6g of Cu was produced in the following
  reaction. How much Al was required?
• 2Al 3CuSO4 ? 3Cu Al2(SO4)3

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Another example

• Next, subtract the needed amount from the given
  amount from the first problem.
• If 10.3 g of aluminum are reacted with 51.7g of
  CuSO4 how much copper (grams) will be produced?

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Another example

• If 10.3 g of aluminum are reacted with 51.7 g of
  CuSO4 how much copper (grams) will be produced?
• 2Al 3CuSO4 ? 3Cu Al2(SO4)3
• the CuSO4 is limited, so Cu 20.6 g
• How much excess reagent will remain?

Excess 4.47 grams
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Since you now know all you need to do, GO DO!!!