Chapter #8
Chemical Quantities
Chapter Outline
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8.2 Mole relationships defined in balanced equations
8.3 Mole conversions- Stoichiometry
8.4 Mass Conversions Stoichiometry
8.5 Excess and Limiting Reactants
8.5 Theoretical Yield
8.6 Theoretical Yield From Initial Reactant Masses
8.7 Thermochemical Equations
8.3-8.4 STOICHIOMETRY
Stoichiometry is the use of balanced chemical
equations in the conversion process.
8.4 STOICHIOMETRY
Stoichiometry is the use of balanced chemical
equations in the conversion process.
Examples
Calculate the mass of water formed from 6.33 g of
hydrogen. A balanced equation is required.
2 H2 + O 2
2H2O
8.4 STOICHIOMETRY
Stoichiometry is the use of balanced chemical
equations in the conversion process.
Examples
Calculate the mass of water formed from 6.33 g of
hydrogen. A balanced equation is required.
2 H2 + O 2
2H2O
6.33 g H2
8.4 STOICHIOMETRY
Stoichiometry is the use of balanced chemical
equations in the conversion process.
Examples
Calculate the mass of water formed from 6.33 g of
hydrogen. A balanced equation is required.
2 H2 + O 2
2H2O
6.33 g H2
mole H2
2.016 g H2
8.4 STOICHIOMETRY
Stoichiometry is the use of balanced chemical
equations in the conversion process.
Examples
Calculate the mass of water formed from 6.33 g of
hydrogen. A balanced equation is required.
2 H2 + O 2
2H2O
6.33 g H2
Mole H2
2.016 g H2
8.4 STOICHIOMETRY
Stoichiometry is the use of balanced chemical
equations in the conversion process.
Examples
Calculate the mass of water formed from 6.33 g of
hydrogen. A balanced equation is required.
2 H2 + O 2
2H2O
6.33 g H2
Mole H2
2.016 g H2O
2 mole H2O
2 Mole H2
8.4 STOICHIOMETRY
Stoichiometry is the use of balanced chemical
equations in the conversion process.
Examples
Calculate the mass of water formed from 6.33 g of
hydrogen. A balanced equation is required.
2 H2 + O 2
2 H2O
6.33 g H2
Mole H2
2.016 g H2
2 mole H2O 18.02 g H2O
2 Mole H2 Mole H2O
8.4 STOICHIOMETRY
Stoichiometry is the use of balanced chemical
equations in the conversion process.
Examples
Calculate the mass of water formed from 6.33 g of
hydrogen. A balanced equation is required.
2 H2 + O 2
2 H2O
6.33 g H2
Mole H2
2.016 g H2
2 mole H2O 18.02 g H2O
= 56.6 g H2O
2 Mole H2 Mole H2O
8.5 Excess and Limiting Reactants
Reactants are substances that can be changed into
something else. For example, nails and boards are
reactants for carpenters, while thread and fabric are
reactants for the seamstress. And for a chemist
hydrogen and oxygen are reactants for making water.
8.5 Building Houses
Ok, we want to build some houses, so we order
2 truck loads of boards and 2 truck loads of
nails. If two truck loads of boards make one
house and two truck loads of nails make 10
houses, then how many houses can we make?
8.5 Building Houses
Ok, we want to build some houses, so we order
2 truck loads of boards and 2 truck loads of
nails. If two truck loads of boards make one
house and two truck loads of nails make 10
houses, then how many houses can we make?
Yes, only one house!
8.5 Building Houses
Ok, we want to build some houses, so we order
2 truck loads of boards and 2 truck loads of
nails. If two truck loads of boards make one
house and two truck loads of nails make 10
houses, then how many houses can we make?
What reactant is in excess? And how many more
houses could we use if we had enough boards?
8.5 Building Houses
Ok, we want to build some houses, so we order
2 truck loads of boards and 2 truck loads of
nails. If two truck loads of boards make one
house and two truck loads of nails make 10
houses, then how many houses can we make?
What reactant is in excess? And how many more
houses could we use if we have enough boards?
8.5 Building Houses
Ok, we want to build some houses, so we order
2 truck loads of boards and 2 truck loads of
nails. If two truck loads of boards make one
house and two truck loads of nails make 10
houses, then how many houses can we make?
What reactant is in excess? And how many more
houses could we use if we have enough boards?
Yes, nails are in excess!
8.5 Building Houses
Ok, we want to build some houses, so we order
2 truck loads of boards and 2 truck loads of
nails. If two truck loads of boards make one
house and two truck loads of nails make 10
houses, then how many houses can we make?
What reactant is in excess? And how many more
houses could we use if we have enough boards?
Yes, nails are in excess! Nine more houses if we
have an adequate amount of boards.
8.5 Excess and Limiting Example
If we react 10.0g of hydrogen with 10.0g of oxygen,
which, if any, reactant will be in excess?
8.5 Excess and Limiting Example
If we react 10.0g of hydrogen with 10.0g of oxygen,
which, if any, reactant will be in excess?
Our conversion process can easily determine the excess
reactant. We can convert 10.0 g of oxygen to grams of
hydrogen to determine if there is enough hydrogen to
consume the oxygen.
8.5 Excess and Limiting Example
If we react 10.0g of hydrogen with 10.0g of oxygen,
which, if any, reactant will be in excess?
Our conversion process can easily determine the excess
reactant. We can convert 10.0 g of oxygen to grams of
hydrogen to determine if there is enough hydrogen to
consume the oxygen.
2 H2 + O2
10.0 g O2
2 H2O
8.5 Excess and Limiting Example
If we react 10.0g of hydrogen with 10.0g of oxygen,
which, if any, reactant will be in excess?
Our conversion process can easily determine the excess
reactant. We can convert 10.0 g of oxygen to grams of
hydrogen to determine if there is enough hydrogen to
consume the oxygen.
2 H2 + O2
10.0 g O2 mole O2
32.0 g O2
2 H2O
8.5 Excess and Limiting Example
If we react 10.0g of hydrogen with 10.0g of oxygen,
which, if any, reactant will be in excess?
Our conversion process can easily determine the excess
reactant. We can convert 10.0 g of oxygen to grams of
hydrogen to determine if there is enough hydrogen to
consume the oxygen.
2 H2 + O2
10.0 g O2 mole O2 2 mole H2
32.0 g O2 mole O2
2 H2O
8.5 Excess and Limiting Example
If we react 10.0g of hydrogen with 10.0g of oxygen,
which, if any, reactant will be in excess?
Our conversion process can easily determine the excess
reactant. We can convert 10.0 g of oxygen to grams of
hydrogen to determine if there is enough hydrogen to
consume the oxygen.
2 H2 + O2
2 H2O
10.0 g O2 mole O2 2 mole H2 2.02 g H2
32.0 g O2 mole O2 mole H2
8.5 Excess and Limiting Example
If we react 10.0g of hydrogen with 10.0g of oxygen,
which, if any, reactant will be in excess?
Our conversion process can easily determine the excess
reactant. We can convert 10.0 g of oxygen to grams of
hydrogen to determine if there is enough hydrogen to
consume the oxygen.
2 H2 + O2
2 H2O
10.0 g O2 mole O2 2 mole H2 2.02 g H2
= 1.26 g H2
32.0 g O2 mole O2 mole H2
8.5 Excess and Limiting Example
Only 1.26 g of hydrogen are required to react with 10.0 g
of oxygen. Since there are 10.0 g of hydrogen available,
then hydrogen must be the excess reactant and oxygen is
the limiting reactant. The remainder of hydrogen 10.0 1.26 = 8.7 g is called the amount in excess. The amount of
water produced is determined by using the limiting
reactant and converting it into water.
8.5 Excess and Limiting Example
Only 1.26 g of hydrogen are required to react with 10.0 g
of oxygen. Since there are 10.0 g of hydrogen available,
then hydrogen must be the excess reactant and oxygen is
the limiting reactant. The remainder of hydrogen 10.0 1.26 = 8.7 g is called the amount in excess. The amount of
water produced is determined by using the limiting
reactant and converting it into water.
10.0 g O2 mole O2
32.0 g O2
8.5 Excess and Limiting Example
Only 1.26 g of hydrogen are required to react with 10.0 g
of oxygen. Since there are 10.0 g of hydrogen available,
then hydrogen must be the excess reactant and oxygen is
the limiting reactant. The remainder of hydrogen 10.0 1.26 = 8.7 g is called the amount in excess. The amount of
water produced is determined by using the limiting
reactant and converting it into water.
10.0 g O2 mole O2 2 mole H2O
32.0 g O2 mole O2
8.5 Excess and Limiting Example
Only 1.26 g of hydrogen are required to react with 10.0 g
of oxygen. Since there are 10.0 g of hydrogen available,
then hydrogen must be the excess reactant and oxygen is
the limiting reactant. The remainder of hydrogen 10.0 1.26 = 8.7 g is called the amount in excess. The amount of
water produced is determined by using the limiting
reactant and converting it into water.
10.0 g O2 mole O2 2 mole H2O 18.0 g H2O
mole H2O
32.0 g O2 mole O2
8.5 Excess and Limiting Example
Only 1.26 g of hydrogen are required to react with 10.0 g
of oxygen. Since there are 10.0 g of hydrogen available,
then hydrogen must be the excess reactant and oxygen is
the limiting reactant. The remainder of hydrogen 10.0 1.26 = 8.7 g is called the amount in excess. The amount of
water produced is determined by using the limiting
reactant and converting it into water.
10.0 g O2 mole O2 2 mole H2O 18.0 g H2O = 11.3 g H O
2
mole H2O
32.0 g O2 mole O2
8.5 Percentage Yield
The percent yield is a comparison of the laboratory
answer to the correct answer which is determined by the
conversion process. Suppose a student combined 10.0 g
of oxygen and 10.0 g of hydrogen in the lab and
recovered 8.66 g of water. What would be the percent
yield?
8.5 Percentage Yield
The percent yield is a comparison of the laboratory
answer to the correct answer which is determined by the
conversion process. Suppose a student combined 10.0 g
of oxygen and 10.0 g of hydrogen in the lab and
recovered 8.66 g of water. What would be the percent
yield?
Yield (the lab amount)
X 100
percent yield =
Theoretical Yield (by conversions)
percent yield = 8.66 X 100 = 76.6%
11.3
8.7 Thermochemistry
Eact
- ΔH
Exothermic Reaction
+ ΔH
Endothermic Reaction
8.7 Thermochemistry
When a chemical or physical change takes place
energy is either lost of gained. A Thermochemical
equation describes this change. Equations gaining
energy are called endothermic and equations losing
energy are called exothermic.
8.7 Thermochemical Equations
When a chemical or physical change takes place
energy is either lost of gained. A Thermochemical
equation describes this change. Equations gaining
energy are called endothermic and equations losing
energy are called exothermic.
Examples:
C3H6O (l ) + 4O2 (g)
3CO2(g) + 3 H2O (g) ΔH = -1790 kj
Exothermic
H2O (l)
H2O (g) ΔH = 44.01 kj
Endothermic
8.7 Thermochemical Equations
How many kj of heat are released when 709 g of C3H6O
are burned?
8.7 Thermochemical Equations
How many kj of heat are released when 709 g of C3H6O
are burned?
C3H6O (l ) + 4O2 (g)
709 g C3H6O
3CO2(g) + 3 H2O (g) ΔH = -1790 kj
8.7 Thermochemical Equations
How many kj of heat are released when 709 g of C3H6O
are burned?
C3H6O (l ) + 4O2 (g)
3CO2(g) + 3 H2O (g) ΔH = -1790 kj
709 g C3H6O mole C3H6O
58.1 g C3H6O
8.7 Thermochemical Equations
How many kj of heat are released when 709 g of C3H6O
are burned?
C3H6O (l ) + 4O2 (g)
709 g C3H6O mole C3H6O
58.1 g C3H6O
3CO2(g) + 3 H2O (g) ΔH = -1790 kj
8.7 Thermochemical Equations
How many kj of heat are released when 709 g of C3H6O
are burned?
C3H6O (l ) + 4O2 (g)
709 g C3H6O mole C3H6O
3CO2(g) + 3 H2O (g) ΔH = -1790 kj
1790 kj
58.1 g C3H6O mole C3H6O
= 21800 kj
The End