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Stoichiometry: the study of the quantitative relationships that exist between
the amounts of reactants and products in a chemical reaction.
Stoichiometry is based on the Law of Conservation of Mass which says
that mass can not be created or destroyed in a chemical reaction.
In equation form:
Mass reactants = Mass products
If 433.2 g of HgO react, we should obtain 401.2 g of Hg and 32.0 g of O2.
This leads to the concept of a balanced chemical equation!
2 HgO (s)
→
2 Hg (l)
+
O2 (g)
Notice that the balanced equation has no 2 to 1 ratio between the mass of
HgO and the mass of O2. The coefficients in chemical equations are based
on relative numbers of atoms not on relative masses.
1
CH4 (g)
+
H2O (g)
→
CO (g)
+
3 H2 (g)
There is a 1 to 1 ratio of molecules between CH4 and H2O.
CH4 is methane and H2O is water.
1 methane molecule and 1 water molecule
1 to 1 ratio between CH4 and H2O.
2 methane molecules and 2 water molecules
2 to 2 is also a 1 to 1 ratio between CH4 and H2O.
7 methane molecules and 7 water
molecules
7 to 7 is also a
1 to 1 ratio between CH4 and H2O.
2
CH4 (g)
+
H2O (g)
→
CO (g)
+
3 H2 (g)
6.022X1023 methane molecules and 6.022X1023 water molecules
6.022X1023 to 6.022X1023 is also a 1 to 1 ratio between CH4 and H2O
Realizing that 6.022X1023 molecules is also 1 mole leads to the idea that
1 mole methane to 1 mole water is also a 1 to 1 ratio.
0.5 mole to 0.5 mole is also a 1 to 1 ratio.
We will use this language from now on with chemical equations.
The balanced chemical equation above now reads:
1 mole of methane reacts with 1 mole of water
to produce 1 mole of carbon monoxide and 3 moles of hydrogen.
3
CH4 (g)
+
H2O (g)
→
CO (g)
+
3 H2 (g)
Therefore the Stoichiometry of the above reaction is:
1 to 1
to 1
to 3
These coefficients can be used to make conversion factors if we write
them in the form of ratios (called mole ratios).
(
(
1 molCH4
1 mol H2O
1 mol H2O
1 molCH4
) (
) (
3 mol H2
1 molCH4
1 molCH4
3 mol H2
) (
) (
1 mol H2O
1 mol CO
1 mol CO
1 mol H2O
)
)
Any Others?
4
Write all the possible mole ratios for each of the reactions below.
2 HgO (s) →
2 Hg (l)
4 NH3 (g)
6 NO (g)
+
+
O2 (g)
→
5 N2 (g)
+
6 H2O (g)
5
Mole ratios from balanced chemical equations allow us to convert from the
moles of one substance in a reaction to the needed number of moles of another
substance in a reaction.
For example using the reaction of iron and oxygen below:
4 Fe (s)
+
3 O2 (g)
→
2 Fe2O3 (s)
If we use 1.00 mole of oxygen, how many moles of iron can be used?
(
)(
1.00 mol O2
4 mol Fe
3 mol O2
)
= 1.33 mol Fe
This mole ratio converts from oxygen into iron!
6
How many moles of Fe2O3 could be produced if 0.623 mole of O2 are used?
4 Fe (s)
(
+
)(
0.623 mol O2
3 O2 (g)
→
2 Fe2O3 (s)
)
= 0.415 mol Fe2O3
2 mol Fe2O3
3 mol O2
Using the same chemical equation, how many moles of Fe would be required
to react with 2.67 mole of O2?
7
The molar mass problems from chapter 7 can be combined with the
stoichiometry concepts from chapter 9 to make extended problems.
If 1.50 mole of iron react with excess oxygen, how many grams of Fe2O3
can be produced?
4 Fe (s)
(
1.50 mol Fe
)(
+
3 O2 (g)
)(
2 mol Fe2O3
4 mol Fe
Mole Ratio
→
)
159.7 g Fe2O3
1 mol Fe2O3
= 120. g Fe2O3
Molar Mass
Mole
Mol A
2 Fe2O3 (s)
Ratio
Molar
Mol B
Mass
grams B
8
When magnesium burns in air, it combines with oxygen to form magnesium oxide.
What mass in grams of magnesium oxide can be produced when 1.46 moles of
magnesium burns in air?
First write the chemical equation based on the word equation that was given in
the problem.
Click to see Solution
2 Mg (s)
+
O2 (g)
→
2 MgO (s)
Use the chemical equation to create the needed mole ratio (from Mg to MgO).
2 mol MgO
2 mol Mg
Use the periodic table to find the molar mass of MgO
1 mol MgO = 40.31 g MgO
Start with what was given and convert the units “mol Mg” to “g MgO”
(
)(
1.46 mol Mg
2 mol MgO
2 mol Mg
)(
)
40.31g MgO
1 mol MgO
= 58.85 g MgO
9
These problems can be extend one more step if we start with mass of Fe and
ask about mass of Fe2O3.
4 Fe (s)
+
3 O2 (g)
→
2 Fe2O3 (s)
If 22.3 g of iron react with excess oxygen, how many grams of Fe2O3 can
be produced?
The thought Process is:
Mass
Fe
Molar
Mass
( )(
22.3 g Fe
Mol
Fe
1 mol Fe
Mole
Ratio
)(
55.85 g Fe
Molar
Mol
Fe2O3
Mass
)(
2 mol Fe2O3
4 mol Fe
grams
Fe2O3
)
159.7 g Fe2O3
1 mol Fe2O3
= 31.9 g Fe2O3
10
What mass of aluminum can be produced from the decomposition of 647.2 g of
aluminum oxide?
First write the chemical equation based on the word equation that was given in
the problem.
Click to see Solution
2 Al2O3 (s) →
4 Al (s)
+
3 O2 (g)
Use the chemical equation to create the needed mole ratio (from Al2O3 to Al).
4 mol Al
2 molAl2O3
Use the periodic table to find the molar mass of Al2O3 and Al
1 mol Al2O3 = 102.0 g Al2O3
and
1 mol Al = 26.98 g Al
Start with what was given and convert the units “g Al2O3” to “g Al”
(
)(
647.2 g Al2O3
1 mol Al2O3
)(
102.0 g Al2O3
4 mol Al
2 mol Al2O3
)(
)
26.98 g Al
1 mol Al
= 342.4 g Al
11
Problem Type Recognition Diagram
Avogadro’s
Particles
Molar
Moles
Mass
Number
Mass
Type A: particles A to moles A (or moles A to particles A)
Type B: moles A to mass A (or mass A to moles A)
Type C: particles A to mass A (or mass A to particles A)
Molar
Mass A
Mole
Moles A
Mass A
Moles B
Ratio
Type 1:
Type 2:
Type 3:
Type 4:
Molar
Mass B
Mass B
moles A to moles B
moles A to mass B
mass A to moles B
mass A to mass B
If you do not like these names, make up your own!
12
How many grams of CO2 can be made when 38.5 g of C2H6 react with excess O2?
C2H6
+
O2
CO2
+
H2O
First: balance the equation so that you can find any needed mole ratios.
2 C2H6
+
7 O2
4 CO2
+
6 H2 O
Second: excess O2 means that you can ignore how much O2 is used.
Third: complete the problem like previous examples.
(
38.5 g C2H6
)(
1 mol C2H6
30.07 g C2H6
)(
4 mol CO2
2 mol C2H6
)(
)
44.01 g CO2
1 mol CO2
= 113 g CO2
13
BLT Sandwich Recipe:
2 pieces of bread
5 slices of bacon
3 slices of tomato
1 piece of lettuce
If your kitchen has the following list of
items, how many BLT sandwiches could
you make that follow the recipe?
Kitchen List:
Bread: 52 pieces
Bacon: 70 slices
Tomato: 30 slices
Lettuce: 25 pieces
(52 bread)(1 sandwich/2 bread) = 26 sandwiches
(70 bacon)(1 sandwich/5 bacon) = 14 sandwiches
(30 tomato)(1 sandwich/3 tomato) = 10 sandwiches
(25 lettuce)(1 sandwich/1 lettuce) = 25 sandwiches
Since tomato ran out after 10 sandwiches, 10 sandwiches is the most that
can be made using the recipe.
The item that “runs out” first is called the limiting reactant because it limits
the amount of product that can be made.
14
Determine the amount of P4O10 produced when 25.0 g of P4 reacts with 50.0 g of O2.
First write the chemical equation (balanced!).
Click to see Solution
P4 (s)
+
5 O2 (g)
→
P4O10 (s)
Second identify important equivalence statements (molar masses and mole ratios).
1 mole P4 = 123.9 g P4
1 mole O2 = 32.00 g O2
5 mole O2 = 1 mole P4O10
1 mole P4 = 1 mole P4O10
Third, set up the calculations (like type 4) but start two separate
problems (one starting with each reactant given in the question).
Fourth, evaluate your answers. The smallest amount of product
produced by the two reactants is the theoretical yield (the maximum
amount of product that can be produced in the reaction).
15
Determine the amount of P4O10 produced when 25.0 g of P4 reacts with 25.0 g of O2.
Continued:
P4 (s)
+
5 O2 (g)
→
P4O10 (s)
Click to see Solution
(
(
)(
25.0 g P4
123.9 g P4
)(
25.0 g O2
1 mol P4
1 mol O2
32.00 g O2
)(
)(
1 mol P4O10
1 mol P4
1 mol P4O10
5 mol O2
)(
)
= 57.3 g P4O10
)
= 44.4 g P4O10
283.9 g P4O10
1 mol P4O10
)(
283.9 g P4O10
1 mol P4O10
So, how much P4O10 could we actually make?
16
Percent Yield
Percent Yield =
(
Actual Yield
)
Theoretical Yield
X 100
When we perform a reaction, we often fail to collect all of the product that
could have been produced because of difficulties with separation of the
product from excess reactants or other products.
The amount of product we should produce based on the starting amounts of
reactants is called the theoretical yield and is the maximum amount of
product that we could produce from that reaction.
The actual yield is the amount of purified product that is actually collectedavailable for use for another purpose.
The percent yield is the actual yield divided by the theoretical yield times 100.
17
When a student reacted 25.0 g of P4 with 25.0 g of O2, a total of 36.8 g of P4O10 was
collected. What is the percent yield for the reaction?
Note: this is the same set of starting conditions that was used in the previous example.
After completing the limiting reactant problem (previous example)
the
Clicktotodetermine
see Solution
theoretical yield for this reaction, we can used the formula for percent yield.
Actual Yield = 36.8 g P4O10
Percent Yield =
Percent Yield =
Percent Yield =
Theoretical Yield = 44.4 g P4O10 (see previous work)
Actual Yield
Theoretical Yield
36.8 g P4O10
44.4 g P4O10
X 100
X 100
82.9 %
18