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Let’s 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.
Stoichiometry
CH 2 (Vol. 2), page 38
I. Stoichiometry
 Greek
Mr. Mole
for “measuring elements”
 Defined:
the study of quantitative relationship
between the amts of reactants used &
amts of products formed by a chem. rxn.
• based on the Law of Conservation of Mass
 (mostly)
REVIEW: There are 4 ways to
interpret a balanced chemical equation
#1. In terms of Particles
An Element is made of atoms
A Molecular compound (made
of only nonmetals) is made up
of molecules (Don’t forget the diatomic elements!!!)
Ionic Compounds (made of a
metal and nonmetal parts) are
made of formula units
a. Example: 2H2 + O2 → 2H2O
Two molecules of hydrogen and one
molecule of oxygen form two molecules of
water. Balanced!
b. Another example: 2Al2O3 Al + 3O2
2 formula units Al2O3 form 4 atoms Al
and 3 molecules O2
Now read this: 2Na + 2H2O  2NaOH + H2
#2. In terms of Moles
a. The coefficients tell us how
many moles of each substance
2Al2O3 Al + 3O2
2Na + 2H2O  2NaOH + H2
b. A balanced equation is a
Molar Ratio (more on this later)
#3. In terms of Mass
The Law of Conservation of Mass applies
a. We can check mass by using moles (reactants).

2H2 + O2  2H2O
2 moles H2
1 mole O2
2.02 g H2
1 mole H2
32.00 g O2
1 mole O2
= 4.04 g H2
+
= 32.00 g O2
36.04 gg H
H22 ++ O2
36.04
reactants
b. In terms of Mass (for products)
2H2 + O2  2H2O
2 moles H2O
18.02 g H2O
= 36.04 g H2O
1 mole H2O
36.04 g H2 + O2 = 36.04 g H2O
36.04 grams reactant = 36.04 grams product
The mass of the reactants must
equal the mass of the products.
#4. In terms of Volume
At
STP, 1 mol of any gas = 22.4 L
2H2
+ O2
 2H2O
(2 x 22.4 L H2) + (1 x 22.4 L O2)  (2 x 22.4 L H2O)
A. Mole Ratio – a ratio between the numbers
of moles of any two of the substances in a
balanced chemical equation
1. Example: Write all possible mole ratios for:
2HgO (s)  2Hg(l) + O2(g)
2 mol HgO
2 mol HgO
2 mol Hg
1 mol O2
2 mol Hg
2 mol Hg
2 mol HgO
1 mol O2
1 mol O2
1 mol O2
2 mol HgO
2 mol Hg
Practice Problem
Write all the possible mole ratios for the
following equation:
4 Al + 3 O2  2 Al2O3
4 Al + 3 O2  2 Al2O3
Answer:
4 mol Al
3 mol O2
3 mol O2
4 mol Al
2 mol Al2O3
4 mol Al
4 mol Al
2 mol Al2O3
3 mol O2
2 mol Al2O3
2 mol Al2O3
3 mol O2
Using Compound Masses
II. Stoichiometric Calculations – p. 40


All stoichiometric calculations begin with a balanced
chemical equation.
Mole ratios are needed, as well as mass-to-mole
conversions (using molar mass)
A. stoichiometric mole-to-mole conversion
2K(s) + 2H20 (l)  2KOH (aq) + H2(g)
• From the balanced equation we know that two moles of
potassium yields one mole of hydrogen.
• But how many moles of hydrogen is produced if we have
only 0.0400 mol of K?
Mole to Mole conversions
2Al2O3 Al + 3O2
• each time we use 2 moles of Al2O3
we will also make 3 moles of O2
2 moles Al2O3
3 mole O2
or
Mole
Ratio
3 mole O2
2 moles Al2O3
These are the two possible conversion
factors to use in the solution of the problem.
Mole to Mole conversions
How
many moles of O2 are
produced when 3.34 moles of Al2O3
decompose?
2Al2O3 Al + 3O2
3.34 mol Al2O3
3 mol O2
2 mol Al2O3
= 5.01 mol O2
Conversion factor from balanced equation
If you know the amount of ANY chemical in the reaction,
you can find the amount of ALL the other chemicals!
Steps to Calculate
Stoichiometric Problems
1. Balance the equation!
2. Convert the given amount into moles (if
not given in moles)
3. Set up mole ratios (to use to calculate
moles of desired chemical)
4. If needed, convert to grams (using Molar
Mass)
Practice:
2C2H2 + 5 O2  4CO2 + 2 H2O
• If 3.84 moles of C2H2 are burned, how
many moles of O2 are needed? (9.60 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)
Practice Problem
Methane gas and sulfur (S8) react to produce
carbon disulfide (a liquid often used in the
production of cellophane) and a colorless,
poisonous gas of hydrogen sulfide (rotten egg
smell).
a. Write chemical equation (don’t forget to balance!)
b. Calculate moles of CS2 produced when
1.50 mol S8 is used
c. How many moles of H2S is produced?
Practice Problem - ANSWER
a. 2CH4(g) + S8(s)  2CS2(l) + 4H2S(g)
b. find: moles of CS2
1.50 mol S8 2 mol CS2
= 3.00 mol CS2
1 mol S8
c. Find moles of H2S
1.50 mol S8 4 mol H2S = 6.00 mol H S
2
1 mol S8
Steps to Calculate
Stoichiometric Problems
1. Balance the equation!
2. Convert to moles (if needed)
3. Set up mole ratios
4. If needed, convert to grams (using MM)
B. Stoichiometric mole-to-mass calculations

Suppose you know the number of moles of a reactant or product
in a reaction & you want to calculate the mass of another
product or reactant.
Example:
Determine the mass of NaCl produced when
1.25 mol of chlorine gas reacts with excess Na.
2 Na(s) + Cl2 (g)  2 NaCl (s)
1.25 mol Cl2
2 mol NaCl
1 mol Cl2
Mole
Ratio!!!
58.44 g NaCl
1 mol NaCl
Molar
Mass!!!
= 146 g NaCl
Practice Problem
Sodium chloride is decomposed into the elements
sodium and chlorine gas by means of electrical
energy. How much chlorine gas, in grams, is
obtained when you have 2.50 mol NaCl?
Answer: 88.6 g Cl2
Steps to Calculate
Stoichiometric Problems
1. Balance the equation!
2. Convert to moles (if needed)
3. Set up mole ratios
4. If needed, convert to grams (using MM)
C. Stoichiometric mass-to-mass
• If you were preparing to carry out a chemical
reaction in the laboratory, you would
need to know how much of each
reactant to use in order to produce
the mass of product required.
Mass-Mass Problem:
6.50 grams of aluminum reacts with an excess of
oxygen. How many grams of aluminum oxide are
formed?
4Al + 3O2  2Al2O3
6.50 g Al
1 mol Al
2 mol Al2O3 101.96 g Al2O3
26.98 g Al
4 mol Al
1 mol Al2O3
(6.50 x 1 x 2 x 101.96) ÷ (26.98 x 4 x 1) =
= ? g Al2O3
12.3 g Al2O3
are formed
How do you get good at this?
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
Practice Problem
One of the reactions used to inflate automobile air
bags involves solid sodium azide (NaN3), which
produces sodium and nitrogen gas.
a.
b.
Write the chemical equation.
Determine the mass of N2 produced from the
decomposition of 100.0 g NaN3.
Answer: 64.64 g N2
D. Volume Stoichiometry
At STP, 1 mol of any gas = 22.4 L
2H2
+ O2
 2H2O
(2 x 22.4 L H2) + (1 x 22.4 L O2)  (2 x 22.4 L H2O)
67.2 Liters of reactant ≠ 44.8 Liters of product!
NOTE: mass and atoms are
ALWAYS conserved - however,
molecules, formula units, moles, and
volumes will not necessarily be conserved!
Avogadro told us:
 Equal
volumes of gas, at the same
temperature and pressure contain the same
number of particles.
 Moles are numbers of particles
 You can treat reactions as if they happen
liters at a time, as long as you keep the
temperature and pressure the same.
1 mole = 22.4 L @ STP
Page 135 - 139 - Chapter 3: The KMT
(Section entitled: “Stoichiometry Involving Gases”)
Volume-Volume Calculations:
How many liters of CH4 at STP are required to
completely react with 17.5 L of O2 ?
CH4 + 2O2  CO2 + 2H2O
1 mol O2 1 mol CH4 22.4 L CH4
17.5 L O2
22.4 L O2 2 mol O2 1 mol CH4
= 8.75 L CH4
Notice anything relating these two steps?
Shortcut for Volume-Volume?
 How
many liters of CH4 at STP are required
to completely react with 17.5 L of O2?
CH4 + 2O2  CO2 + 2H2O
17.5 L O2
1 L CH4
2 L O2
= 8.75 L CH4
Note: This only works for VolumeVolume problems.
Calculations
Grams
molar mass
Moles
Avogadro’s number
particles
Everything must go through
Moles!!!
III. “Limiting” Reactants – p. 59
A. Why do reactions stop?

If you are given one dozen loaves of
bread, a gallon of mustard, and three
pieces of salami, how many salami
sandwiches can you make?
1. The limiting reactant is the reactant you
run out of first. Determines how much
product you can make!
2. The excess reactant is the one you have
left over.
• Often referred to as limiting reagent or excess reagent
Limiting Reagents - Combustion
How do you find out which is limited?
3. 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
4. Do two stoichiometry problems, one for
each reactant given
 If
10.6 g of copper reacts with
Cugrams
is the of the
3.83 g sulfur, how many
Limiting
product (copper (I) sulfide)
will be formed?
Reagent,
2Cu + S  Cu2S
since it
1 mol Cu2S 159.16 g Cu2S
1
mol
Cu
produced less
10.6 g Cu
Cu
63.55g Cu 2 mol
1 mol Cu2S
product.
= 13.3 g Cu2S
1
mol
S
3.83 g S
32.06g S
1 mol Cu2S 159.16 g Cu2S
1 mol S
1 mol Cu2S
= 19.0 g Cu2S
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
10.3 g Al 1 mol Al
3 mol Cu 63.546 g Cu
= 36.4 g Cu
26.981 g Al 2 mol Al 1 mol Cu
51.7 g CuSO4 1 mol CuSO4
3 mol Cu
63.546 g Cu =
159.60 g CuSO4 3 mol CuSO4 1 mol Cu
= 20.6 g Cu
the CuSO4 is limited, so Cu = 20.6 g
Another example:
2Al + 3CuSO4 → 3Cu + Al2(SO4)3
10.3 g Al 1 mol Al
3 mol Cu 63.546 g Cu
= 36.4 g Cu
26.981 g Al 2 mol Al 1 mol Cu
51.7 g CuSO4 1 mol CuSO4
3 mol Cu
63.546 g Cu
=
159.60 g CuSO4 3 mol CuSO4 1 mol Cu
= 20.6 g Cu
 How
much excess reactant will remain?
• Means how much excess Al will remain!!!
36.4 g Cu – 20.6 g Cu = 15.8 g Cu
 How
much excess reactant will remain?
36.4 g Cu – 20.6 g Cu = 15.8 g Cu
1
mol
Cu
15.8 g Cu
63.55g Cu
2 mol Al
3 mol Cu
26.981 g Al
1 mol Al
= 4.47 g Al
ANOTHER Example
Photosynthesis reactions in green plants use
carbon dioxide and water to produce glucose and
oxygen. A plant has 88.0 g of carbon dioxide and
64.0 g of water available for photosynthesis.
1. Determine the limiting reactant.
2. Determine the excess reactant.
3. Determine the amount of the reactant in
excess.
Write a balanced chemical reaction:
6CO2(g) + 6H2O(l)  C6H12O6(aq) + 6O2(g)
A plant has 88.0 g of carbon dioxide and
64.0 g of water available for photosynthesis.
 1. Find limiting reactant

6CO2(g) + 6H2O(l)  C6H12O6(aq) + 6O2(g)
88.0 g CO2
1 mol CO2
6 mol O2
31.988 g O2
43.99 g CO2
6 mol CO2
1 mol O2
== 64.0
64.0 gg O
O22
64.O g H2O
1 mol H2O
6 mol O2
18.01 g H2O
6 mol H20
Limiting Reactant is CO2!
31.988 g O2
1 mol O2
= 114 g O2
 2.
Find excess reactant
6CO2(g) + 6H2O(l)  C6H12O6(aq) + 6O2(g)
88.0 g CO2
1 mol CO2
6 mol O2
31.988 g O2
43.99 g CO2
6 mol CO2
1 mol O2
= 64.0 g O2
64.O g H2O
1 mol H2O
6 mol O2
18.01 g H2O
6 mol H20
EXCESS Reactant is H2O!
31.988 g O2
1 mol O2
= 113.71 g O2
6CO2(g) + 6H2O(l)  C6H12O6(aq) + 6O2(g)
3. Determine the mass (of the reactant) in
excess.
114 g O2 – 64.0 g O2 = 50.0 g O2
50.O g O2
1 mol O2
31.998 g O2
6 mol H20
18.02 g H2O
6 mol O2
1 mol H2O
= 28.2 g H2O
Practice Problem
The reaction between solid sodium and iron (III)
oxide yields solid sodium oxide and iron.
(Another reaction, in a series, that inflates an
automobile airbag). If 100.0 g of Na and 100.0 g
of iron (III) oxide are used in this reaction,
determine the following:
a. Limiting reactant
b. Excess reactant
c. Mass of solid iron produced
d. Mass of excess reactant that remains after the
reaction is complete
Practice Problem - ANSWERS
6Na(s) + Fe2O3(s)  3Na2O(s) + 2Fe(s)
a. Fe2O3
b. Na
c. 69.92 g Fe
d. 13.6 g Na
The Concept of:
A little different type of yield than you had in Driver’s
Education class.
What is YIELD?
IV. Percent Yield – p. 67
A. 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
• Or the amount of product produced when the chemical
reaction is carried out in an experiment
2. Theoretical yield- what the balanced
equation tells should be made
• The maximum amount of product that can be produced
from a given amount of reactant
3. Percent yield:
Actual
x 100
Theoretical
• Percent yield of the product is the ratio of the actual
yield to the theoretical yield, expressed as a percent
– Percent yield tells us how “efficient” a reaction is.
• A way of measuring the efficiency of the reaction in
producing a product
– 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
Example:
6.78 g of copper is produced when 3.92 g of Al
are reacted with excess copper (II) sulfate.
 Write balanced equation:
2Al + 3 CuSO4  Al2(SO4)3 + 3Cu
 What is the actual yield?
 What is the theoretical yield?
 What is the percent yield?

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? = 6.78 g Cu
 What is the theoretical yield?


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? = 6.78 g Cu
What is the theoretical yield?
3.92 g Al
1 mol Al
3 mol Cu
63.546 g Cu
26.982 g Al
2 mol Al
1 mol Cu
= 13.8481 g Cu
= 13.8 g Cu
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? = 6.78 g Cu
 What is the theoretical yield? = 13.8 g Cu
 What is the percent yield?

actual yield
X 100
Percent Yield =
theoretical yield
6.78 g Cu
13.8 g Cu
X 100
= 49.1 %
Your Turn:
Aluminum hydroxide is often present in antacids to
neutralize stomach acid (HCl). The reaction is as
follows:
Al(OH)3(s) + 3HCl(aq)  AlCl3(aq) + 3H2O(l)
a.
b.
If 14.0 g of Al(OH)3 is present in an antacid
tablet, determine the theoretical yield of AlCl3
produced when the tablet reacts with HCl.
What is the percent yield if 20.5 g is recovered?
Practice Problem - ANSWER
Al(OH)3(s) + 3HCl(aq)  AlCl3 (aq) + 3H2O(l)
If 14.0 g of Al(OH)3 is present, determine the theoretical yield
of AlCl3 produced.
14.0 g Al(OH)3
1 mol Al(OH)3
1 mol AlCl3 133.34 g AlCl3
78.004 g Al(OH)3 1 mol
Al(OH)3
1 mol AlCl3
= 23.9 g AlCl3
(20.5 g / 23.9 g ) x 100% = 85.7%