Stoichiometry and the Mole

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Stoichiometry and the Mole

• SOL Review

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Stoichiometry and the Mole
The Mole
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The Mole

• Not everything that counts can be counted, and
  not everything that can be counted counts.
• Albert Einstein

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What is a mole?

• The mole is a counting unit.
• Like . . .
• 1 dozen 12 each
• 1 yard 3 feet
• 1 cup 8 ounce
• So then ...
• 1 mol 6.022 x 1023 particles
• Thats Avogadros Number!

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Where did it come from?

• Mole (n) is the SI unit for the number of
  particles
• Amedo Avogadro determined the number of particles
  in a mole
• The mole is the measure of the amount of a
  substance whose number of particles is the same
  as 12 grams of Carbon - 12

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Calculations

• Using dimensional analysis you can determine the
  number of particles in a mole
• 1 mole 6.022 x 1023 particles, molecules, etc.
• 6.022 x 1023 particles 1 mole

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So, lets count . . .

• 1 mol of Ag
• 6.022 x 1023 atoms Ag
• 1 mol of CO2
• 6.022 x 1023 molecules of CO2
• 1 mol of pizza
• 6.022 x 1023 pizzas!

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The Mole Road Map
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Moles to Mass Conversions

• 1 Moles Molar Mass (g)
• Molar mass/Mole or (g/mol)
• Now lets apply that knowledge!

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The Mole Molar Mass Calculations
Reminder Finding Molar Mass The molar mass
the sum of all the atomic masses.
You try one What is the gram formula mass (molar
mass) of Mg3(PO4)2? Mg 3(24.305) P
2(30.97376) O 8(15.9994) 262.86
grams
Example Ca(NO3)2 Ca 40.08 N 2(14.01) O
6(16.00) 164.10 grams
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Molar or Formula Mass

• Chemical compounds are written as an empirical
  formula.
• Ex. H2SO4 is Sulfuric Acid
• Calculating atomic mass, add each atom.
• H 1.008 x 2 2.016
• S 32.07 x 1 32.07
• O 15.999 x 4 63.996
• Total Atomic Mass
• 2.01632.0763.996 98.08 amu

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The Mole and Mole Calculations
One mole 6.02 x 1023 representative
particles One mole 22.4 Liters of gas at 0C
and one atmosphere of pressure One mole the
atomic mass listed on the periodic table. For
example one mole of Helium contains 6.02 x 1023
atoms of Helium and it has a mass of 4.00260
grams. At 0C and one atmosphere of pressure, it
would occupy 22.4 Liters. Sample problem How
many liters would 2.0 moles of Neon occupy? 2.0
moles Ne x 22.4 Liters Ne 44.8 Liters Ne
1.0 moles Ne
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Calculating MolesOne Step

• How many moles are in 3.011 x 1023 atoms of
  Oxygen?

3.011 x 1023 atoms O2 1 mol Cu
6.02 X 1023 atoms

0.5 moles of Oxygen
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More One-Step Conversions

• Ex 1) Convert 4.3 grams of NaCl to moles.
• Mass ? mol
• 4.3 g NaCl x 1 mol NaCl 7.4 x 10-2 mol NaCl
• 58.45 g NaCl
• Ex 2) Convert 0.00563 mol NH3 to grams.
• Mol -gt mass
• 0.00563 mol NH3 x 17 g NH3 9.57 x 10 2 g NH3
  1 mol NH3

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The Mole and Mole Calculations
Sample problem How many moles are in 15.2
grams of Lithium? Answer 15.2 g Li x 1 mole
Li 2.19 mole Li 6.941 g Li

• REMINDER
• One mole 6.02 x 1023 representative particles
• One mole 22.4 Liters of gas at 0C and one
  atmosphere of pressure
• One mole the atomic mass listed on the periodic
  table.

Sample problem How many liters would 14 grams
of Helium occupy? Answer 14 g He x 1 mole
He x 22.4 L He 78 Liters He
4.0026 g He 1 mole He
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Mole Calculations

• How many atoms of Cu are present in 35.4 g of Cu?
• Molar mass of Cu is 63.55 g/mol

35.4 g Cu 1 mol Cu 6.02 X 1023 atoms
Cu 63.5 g Cu 1 mol Cu
3.4 X 1023 atoms Cu
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And another . . .

• What mass would 4.52 x 1024 molecules of water
  have?
• 135 g H2O

4.52 x 1024 mlcs of H20 1 mol
18.02 g
6.02 X 1023 mlcs 1 mol
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Stoichiometry and the Mole
Stoichiometry
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Stoichiometry

• Stoichiometry means that if you know one piece of
  information about ONE compound in an equation,
  you can determine EVERYTHING else!
• If you have 3L of Nitrogen, how many liters of
  ammonia will you produce?
• N2 3H2 ?? 2NH3

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Stoichiometry

• Lets look at that last reaction again.
• N2(g) 3H2(g) ?? 2NH3(g)
• If you start out with 1 mole of Nitrogen gas and
  3 moles of Hydrogen gas, you will make 2 moles of
  Ammonia gas.
• It is important in industry to know the exact
  proportions of your ingredients so that you will
  not have excess waste in your product.

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How many grams of silver chloride can be produced
from the reaction of 17.0 g silver nitrate with
excess sodium chloride solution?
Stoichiometry
1. Write the balanced equation
17.0g
?g
AgNO3 NaCl ? AgCl NaNO3
2. Given and asked for
3. Moles of given
17.0g AgNO3 x 1 mol
170 g
0.100 mol AgNO3
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StoichiometryMass-Mass Problem
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StoichiometryMass-Mass Problem
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StoichiometryMass-Mass Problem
AgNO3 NaCl ? AgCl NaNO3
4. Moles asked for
0.100 mols AgNO3 x 1 mol AgCl
1 mol AgNO3
0.100 mol AgCl
5. Convert your answer
0.100 mol AgCl x 144 g AgCl
1 mol AgCl
14.4 g AgCl
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C. Molar Volume at STP
1 mol of a gas22.4 L at STP
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Stoichiometry Volume

• How many grams of KClO3 are reqd to produce 9.00
  L of O2 at STP?

2KClO3 ? 2KCl 3O2
9.00 L
? g
9.00 L O2
1 mol O2 22.4 L O2
2 mol KClO3 3 mol O2
122.55 g KClO3 1 mol KClO3
32.8 g KClO3
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Stoichiometry Concentrations

• How many grams of Cu are required to react with
  1.5 L of 0.10M AgNO3?

Cu 2AgNO3 ? 2Ag Cu(NO3)2
1.5L 0.10M
? g
63.55 g Cu 1 mol Cu
1.5 L
.10 mol AgNO3 1 L
1 mol Cu 2 mol AgNO3
4.8 g Cu
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Stoichiometry and the Mole
Limiting Reactant Problems
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Limiting Reactant Problems
Given the following reaction 2Cu S ? Cu2S

• What is the limiting reactant when 82.0 g of Cu
  reacts with 25.0 g S?
• What is the maximum amount of Cu2S that can be
  formed?
• How much of the other reactant is wasted?

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Limiting Reactant Problems

• Our 1st goal is to calculate how much S would
  react if all of the Cu was reacted.
• From that we can determine the limiting reactant
  (LR).
• Then we can use the Limiting Reactant to
  calculate the amount of product formed and the
  amount of excess reactant left over.

82g Cu?
mol Cu?
mol S?
g S
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2Cu S ? Cu2S
1molCu
1mol S
32.1g S
82.0gCu
63.5gCu
2molCu
1mol S
20.7 g S

• So if all of our 82.0g of Copper were reacted
  completely it would require only 20.7 grams of
  Sulfur.
• Since we initially had 25g of S, we are going to
  run out of the Cu, the limiting reactant) end
  up with 4.3 grams of S

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Limiting Reactant Problems

• Copper being our Limiting Reactant is then used
  to determine how much product is produced.
• The amount of Copper we initially start with
  limits the amount of product we can make.

1molCu
159gCu2S
2molCu2S
82.0gCu
1molCu2S
63.5gCu
1molCu2S
103 g Cu2S
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Limiting Reactant Problems

• So the reaction between 82.0g of Cu and 25.0g of
  S can only produce 103g of Cu2S.
• The Cu runs out before the S and we will end up
  wasting 4.7 g of the S.

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Stoichiometry and the Mole
Percent Yield
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Calculating Percent Yield

• In theory, when a teacher gives an exam to the
  class, every student should get a grade of 100.
  Sadly, this is not always true. The calculation
  for percent yield is similar.
• We already know that we do not get a 100 yield
  of products in an reaction.

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Calculating Percent Yield

• Consider the Following Reactions
• Mg 2HCl ? MgCl2 H2
• 5.0 g Mg is reacted with an excess of HCL. How
  much MgCl2 will be produced.
• 21.6 g of MgCl2

5.0g Mg
1molMg
1mol MgCl2
105.2 g MgCl2
24.3 g Mg
1mol MgCl2
1mol Mg
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Calculating Percent Yield

• You might assume that using stoichiometry to
  calculate that our reaction will produce 21.6 g
  of MgCl2, but we will actually only recover 15.2
  g of MgCl2 in the lab.
• 21.6 g of MgCl2 is the value representing the
  theoretical yield(theoretical yield is the
  maximum amount of product that could be formed).
• The 15.2 g of MgCl2 is called the actual yield
  (the actual yield is less than the theoretical
  yield).

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Calculating Percent Yield

• The percent yield is the ratio of the actual
  yield to the theoretical yield as a percent
• It measures the measures the efficiency of the
  reaction.

measured in lab
actual yield
Percent yield
x 100
theoretical yield
calculated on paper
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Calculating Percent Yield

• Why do reactions not go to completion.
• Impure reactants and competing side rxns may
  cause unwanted products to form.
• Actual yield can also be lower than the
  theoretical yield due to a loss of product during
  filtration or transferring between containers.
• If a wet precipitate is recovered it might weigh
  heavy due to incomplete drying, etc.

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Calculating Percent Yield
Calcium carbonate is synthesized by heating, as
shown in the following equation CaO CO2 ?
CaCO3

• What is the theoretical yield of CaCO3 if 24.8 g
  of CaO is heated with 43.0 g of CO2?
• What is the percent yield if 33.1 g of CaCO3 is
  produced?
• Determine which reactant is the limiting and then
  decide what the theoretical yield is.

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24.8gCaO?
molCaO?
mol CO2?
gCO2
24.8 g CaO
1molCaO
1mol CO2
44 g CO2
56g CaO
1mol CaO
1molCO2
LR
19.5gCO2
1mol CaO
100g CaCO3
24.8 g CaO
1molCaCO3
56g CaO
1mol CaO
1molCaCO3
44.3 g CaCO3
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Calculating Percent Yield

• CaO is our LR, so the reaction should
  theoretically produce 44.3 g of CaCO3 (How
  efficient were we?)
• Our percent yield is

33.1 g CaCO3
Percent yield
x 100
44.3 g CaCO3
Percent yield 74.7
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GOOD LUCK!!
Mrs. Armani Mr. Buchanan Ms. Nichols Mr. Smith