SOL Review Moles and Stoichiometry

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Stoichiometry
and the Mole
SOL Review
Stoichiometry
and the Mole
The Mole
The Mole
Not everything that counts can
be counted, and not everything
that can be counted counts.
Albert Einstein
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
That’s Avogadro’s Number!
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

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

So, let’s count . . .

1 mol of Ag =


1 mol of CO2 =


6.022 x 1023 atoms Ag
6.022 x 1023 molecules of CO2
1 mol of pizza =

6.022 x 1023 pizzas!
The Mole
Road Map
Moles to Mass
Conversions

1 Moles = Molar Mass (g) =
Molar mass/Mole or (g/mol)
Now let’s apply that knowledge!
The Mole:
Molar Mass Calculations
Reminder – Finding Molar Mass:
The molar mass = the sum of all the atomic masses.
Example: Ca(NO3)2
Ca = 40.08
N = 2(14.01)
O = 6(16.00)
164.10 grams
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
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 =
2.016+32.07+63.996 = 98.08 amu
The Mole and
Mole Calculations
One mole = 6.02 x 1023 representative particles
One mole = 22.4 Liters of gas at 0°C 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 0°C 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
Calculating Moles
One 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
=
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 -> mass
0.00563 mol NH3 x 17 g NH3 =9.57 x 10 –2 g NH3
1 mol NH3
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 0°C 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
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
63.5 g Cu
6.02 X 1023 atoms Cu
1 mol Cu
= 3.4 X 1023 atoms Cu
And another . . .

What mass would 4.52 x 1024 molecules of
water have?
4.52 x 1024 mlcs of H20
1 mol
6.02 X 1023 mlcs
= 135 g H2O
18.02 g
1 mol
Stoichiometry
and the Mole
Stoichiometry
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

Stoichiometry

Let’s 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.

Stoichiometry
How many grams of silver chloride can be
produced from the reaction of 17.0 g silver nitrate
with excess sodium chloride solution?
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
Stoichiometry
Mass-Mass Problem
Stoichiometry
Mass-Mass Problem
Stoichiometry
Mass-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
C. Molar Volume at STP
1 mol of a gas=22.4 L
at STP
Standard Temperature & Pressure
0°C and 1 atm
Stoichiometry
and the Mole
Limiting Reactant
Problems
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?
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
2Cu + S  Cu2S
82.0gCu
 So
1molCu
1mol S
63.5gCu
2molCu
32.1g S
1mol S
=20.7 g S
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
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 63.5gCu 1molCu S 1molCu2S
2
= 103 g Cu2S
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.
Stoichiometry
and the Mole
Percent Yield
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.
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.
5.0g Mg 1molMg 1mol MgCl2 105.2 g MgCl2
24.3 g Mg 1mol Mg
= 21.6 g of MgCl2
1mol MgCl2
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).

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
Percent yield=
calculated on paper
actual yield
theoretical yield
x 100
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.
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.
24.8gCaOmolCaO mol CO2 gCO2
24.8 g 1molCaO 1mol CO2
CaO 56g CaO 1mol CaO
LR
44 g CO2
1molCO2
= 19.5gCO2
24.8gCaOmolCaO mol CaCO3gCaCO3
24.8 g 1mol CaO 1molCaCO3 100g CaCO3
CaO 56g CaO 1mol CaO 1molCaCO3
= 44.3 g CaCO3
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%
GOOD LUCK!!
Mrs. Armani
Mr. Buchanan
Ms. Nichols
Mr. Smith
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