### Chapter #9: Chemical Quantities

```Chapter #9
Chemical Quantities
BALANCED EQUATION RELATIONS
Consider the balanced equation below. What
relations between reactants and products exist?
2 H2 + O2
2H2O
BALANCED EQUATION RELATIONS
Consider the balanced equation below. What
relations between reactants and products exist?
2 H2 + O2
2H2O
This equation states that two molecules will combine
with one oxygen molecule to produce two molecules of
water. Since, one mole is equal to 6.022X1023
molecules, then two moles of hydrogen combine with
one mole of oxygen to produce two moles of water.
BALANCED EQUATION RELATIONS
Conversion problems using molar relations from a balanced
chemical equation are referred to as stoichiometry problems.
Stoichiometry is of industrial importance and used routinely
by the overpaid chemical engineers. Examples of this process
are found on the following slides.
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.
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.
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
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
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
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
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
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
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.
Building Houses
Ok, we want to build some houses, so we order
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?
Building Houses
Ok, we want to build some houses, so we order
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!
Building Houses
Ok, we want to build some houses, so we order
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?
Building Houses
Ok, we want to build some houses, so we order
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?
Building Houses
Ok, we want to build some houses, so we order
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!
Building Houses
Ok, we want to build some houses, so we order
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.
Making Water
If we react 10.0g of hydrogen with 10.0g of oxygen,
which, if any, reactant will be in excess?
Making Water
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.
Making Water
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
Making Water
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
Making Water
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
Making Water
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
Making Water
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
Making Water
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.
Making Water
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
Making Water
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
Making Water
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
Making Water
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
Percentage Yield
The percent yield is a comparison of the laboratory
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?
Percentage Yield
The percent yield is a comparison of the laboratory
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
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.
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 )
H2O (l)
4O2 (g)
3CO2(g) + 3 H2O (g)
Exothermic
H2O (g)
ΔH = 44.01 kj
Endothermic
ΔH = -1790 kj
Thermochemical Conversions
How many kj of heat are released when 709 g of C3H6O
are burned?
Thermochemical Conversions
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
Thermochemical Conversions
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
Thermochemical Conversions
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
```