# The Mole

```					It’s time to learn about . . .
What's The Difference Between Roast
Beef And Pea Soup?

Anyone Can Roast Beef.
Stoichiometry
Stoichiometry : Mole Ratios to
Determining Grams of Product
At the conclusion of our time together, you
should be able to:

1. Review the conversion of particles or grams to
moles
2. Determine mole ratios from a balanced chemical
equation
3. Determine the amount of product produced when
given the amount of reactants
Review the Molar Mass of Compounds

   The molar mass (MM) of a compound is determined
by adding up all the atomic masses for the molecule
(or compound)
20
Ca
◦   Ex. Molar mass of CaCl2                 40.08
◦   Avg. Atomic mass of Calcium = 40.08g
◦   Avg. Atomic mass of Chlorine = 35.45g
◦   Molar Mass of calcium chloride =
40.08 g/mol Ca + (2 X 35.45) g/mol Cl    17
 110.98 g/mol CaCl2                    Cl
35.45
Practice

   Calculate the Molar Mass of
calcium phosphate
◦ Formula = Ca3(PO4)2
◦ Masses elements:

◦ Molar Mass =
310.18 g
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Chapter 5 - When to Give the Ranger the
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Chapter 14 – The Best Time to Let a
Calculations

Grams        Moles                Particles

Everything must go through Moles!!!
Atoms or             Flowchart
Molecules

Divide by 6.02 X 1023
Multiply by
6.02 X 1023                           Multiply by
Moles              atomic/molar
mass from
periodic
Divide by            table
atomic/molar
mass from
periodic table       Mass
(grams)
1 cup butter                 2 eggs
1/2 cup white sugar          1 teaspoon salt
1 cup packed brown sugar
1 teaspoon vanilla extract
2 1/2 cups all-purpose flour
1 teaspoon baking soda
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 much brown sugar would I need if I had 1½ cups
white sugar?
 Just like chocolate chip cookies
have recipes, chemists have
recipes as well
 Instead of calling them recipes,
we call them chemical equations
cups and teaspoons, we use
moles
 Lastly, instead of eggs, butter,
sugar, etc. we use chemical
compounds as ingredients
Chemistry Recipes

 Looking at a reaction tells us how much of
something you need to react with something else to
get a product (like the cookie recipe)
 Be sure you have a balanced reaction before you
start! Balance #1 on your worksheet!!
 Example: 2 Ag+ Cl2  2 AgCl
 This reaction tells us that by mixing 2 moles of
silver with 1 mole of chlorine we will get 2 moles
of silver chloride
 What if we wanted 4 moles of AgCl?
 10 moles?
 50 moles?
Let’s sit down and figure this out!!
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 would
be produced?
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 silver
(without any silver left over)?
2 Ag +      Cl2  2 AgCl
5 Ag + 2.5 Cl2  5 AgCl
2 Ag+ Cl2  2 AgCl

# moles of Cl2 =

5 mol Ag   x 1 mol Cl2
2 mol Ag

= 2.5 mol Cl2
Mole-Mole Conversions
   How many moles of silver chloride will be
produced if you react 2.6 moles of chlorine gas
with an excess (more than you need) of silver
metal?
2 Ag + Cl2  2 AgCl
# moles of AgCl =

2.6 mol Cl2 x 2 mol AgCl
1 mol Cl2

= 5.2 mol AgCl
Mass-Mole
 We can also start with mass and convert to moles of
product or another reactant
 We use molar mass and the mole ratio to get to moles
of the compound of interest
◦ Calculate the number of moles of the combustion of
ethane (C2H6) needed to produce 10.0 g of water

◦   2 C2H6 + 7 O2  4 CO2 + 6 H20
Mass-Mole
    2 C2H6 + 7 O2  4 CO2 + 6 H20

10.0 g H2O   x     1 mol H2O   x    2 mol C2H6
18.02 g H2O        6 mol H2O

= 0.185 mol C2H6
A Comparison of the Arousal of the Sexes
Stoichiometry Continued:
Stoichiometry : Mole Ratios to
Determining Grams of Product
At the conclusion of our time together, you
should be able to:

1. Review the conversion of particles or grams to
moles
2. Determine mole ratios from a balanced chemical
equation
3. Determine the amount of product produced when
given the amount of reactants
Mass-Mass Conversions

 Most often we are given a starting mass and want
to find out the mass of a product we will get (called
theoretical yield) or how much of another reactant
we need to completely react with it (no leftover
ingredients!)
 Now we must go from grams to moles, to mole
ratios, and back to grams of the compound we are
interested in.
Mass-Mass Conversion

 Ex. Calculate how many grams of ammonia are
produced when you react 2.00 g of nitrogen
with excess hydrogen.
 N2 + 3 H2  2 NH3

2.00 g N2    1 mol N2    2 mol NH3   17.06 g NH3
28.02 g N2   1 mol N2    1 mol NH3

= 2.4 g NH3
Practice Mass-Mole-Mass

   How many grams of calcium nitride are produced
when 2.00 g of calcium reacts with an excess of
nitrogen?

   Ca + N2  Ca3N2

   3 Ca + N2  Ca3N2
Mass-Mole-Mass
      3 Ca + N2  Ca3N2

2.00 g Ca       x     1 mol Ca    x   1 mol Ca3N2
40.08 g Ca        3 mol Ca

x 148.26 g Ca3N2
= 2.47 g Ca3N2
1 mol Ca3N2
Some Things Just Take a Lot of Work!!!
Stoichiometry : Mole Ratios to
Determining Grams of Product
Let’s see if you can:

1. Review the conversion of particles or grams to
moles
2. Determine mole ratios from a balanced chemical
equation
3. Determine the amount of product produced when
given the amount of reactants
Mass-Mole-Mass

    NaN3  Na + N2
    Synthesis
     2 NaN3  2 Na + 3 N2
      65.02 g 22.99 g 28.02 g

94.5 g NaN3 x     1 mol NaN3   x   3 mol N2
40.08 g NaN3       2 mol NaN3
x 28.02 g N2
= 99.1 g N2
1 mol N2
Don’t you just love these types of
problems???
Review Mass-Mole-Molecules: Determine
the number of molecules in 73 g of water

# H2O molecules =

73 g H2O   x 1 mol H2O    x   6.02x1023 molecules
18.02 g H2O          1 mol H2O

= 2.4 x 1024 molecules H2O
Try this one:

Calculate the mass in grams of iodine required to react
completely with 0.50 moles of aluminum.
Al + I2  AlI3
2 Al + 3 I2  2 AlI3

0.50 mol Al       x 3 mol I2
                                2 mol Al
x    253.80 g I2
1 mol I2
= 190 g I2
Try this one:

Calculate the mass in grams of iodine required to react
completely with 0.50 g of aluminum.
Al + I2  AlI3
2 Al + 3 I2  2 AlI3

0.50 g Al    x 1 mol Al        x 3 mol I2
26.98 g Al        2 mol Al

x   253.80 g I2
= 7.1 g I2
1 mol I2

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