Chemical Reactions

A physical change alters the physical state of
a substance without changing its composition
◦ Examples

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Boiling
melting
Freezing
Vaporization
Condensation
Sublimation
Breaking a bottle
Crushing a can

A chemical change (a chemical reaction)
converts one substance into another
◦ Breaking bonds in the reactants (starting materials)
◦ Forming new bonds in the products
aA (physical state) + bB (state)  cC (state) + dD (state)
A, B = reactants
C, D = products
a, b, c, d = coefficients to indicate
molar ratios of reactants and products
CH4 (g) + 2O2 (g)  CO2 (g) + 2H2O (g)
CH4 and O2
CO2 and H2O
Balancing Chemical Equations:
Unbalanced equation:
C4H10 + O2  CO2 + H2O
Balanced equation:
2C4H10 + 13O2  8CO2 + 10H2O
2 molecules of
C4H10
13 molecules of O2
8 molecules of
CO2
10 molecules of
C4H10
H2 + O2 → H20

Reactants are on the left
◦ Things that are used
◦ H2 + O2

Product(s) are on the right
◦ Things that are made
◦ H20

This equation is not yet balanced

Do the number of atoms of each element on
either side of the arrow balance?
◦ Compounds consist of more than one element
◦ Examples: NaCl, H2SO4, H20
◦ Look at numbers of each atoms within the
compounds

Correct any imbalances with a
coefficient
◦ Coefficient = the large number to
the left of a substance in the
equation
◦ Don’t change subscripts
 This will change what the molecule is

Example: to balance an equation you
need two atoms of oxygen from water
(H2O)
◦ If you change the subscript you change the
compound
◦ H2O2 does provide two atoms of oxygen but
 H2O (water) ≠ H2O2 (hydrogen peroxide)
◦ 2 H2O provides 2 atoms of oxygen and keeps
the compound as water
◦ It also gives you 4 atoms of hydrogen that
you should then make sure is balanced

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

Relax and calmly go through what is on
either side of the equation
There are many different ways to start
balancing the equation
Once you start the remaining coefficients
should fall into place
Where to start?
◦ Compounds (NaCl, H2SO4, H20)
 See what elements they have in common
◦ Molecular elements (O2, H2)
 More than one atom present
◦ Monoatomic elements (Ca, Cl)
 Only one atom present
H2 + O2 → H20


Left
Right
2 hydrogen
2 hydrogen
2 oxygen
1 oxygen
List out what is on each side
Oxygen does not balance
H2 + O2 → 2H20


Left
Right
2 hydrogen
4 hydrogen
2 oxygen
2 oxygen
Multiply right side by 2 to balance oxygen
Now hydrogen does not balance
2H2 + O2 → 2H20


Left
Right
4 hydrogen
4 hydrogen
2 oxygen
2 oxygen
Multiply hydrogen by 2 to balance hydrogen
Equation is now balanced
C6H12O6 + O2 → CO2 + H2O


Left
Right
6 carbon
1 carbon
12 hydrogen
2 hydrogen
8 oxygen
3 oxygen
List out what is on each side
Carbon does not balance
C6H12O6 + O2 → 6CO2 + H2O



Left
Right
6 carbon
6 carbon
12 hydrogen
2 hydrogen
8 oxygen
13 oxygen
Look at compounds first
Multiply CO2 on right by 6 to balance C
Hydrogen does not balance
C6H12O6 + O2 → 6CO2 + 6H2O


Left
Right
6 carbon
6 carbon
12 hydrogen
12 hydrogen
8 oxygen
18 oxygen
Multiply water on right side by 6 to balance
hydrogen
Oxygen does not balance
C6H12O6 + 6O2 → 6CO2 + 6H2O


Left
Right
3 carbon
6 carbon
8 hydrogen
12 hydrogen
2 oxygen
18 oxygen
Multiply oxygen on left side by 6 to balance
oxygen
Equation is now balanced
Propane + oxygen → carbon dioxide + water
C3H8 + O2 → CO2 + H2O


Left
Right
3 carbon
1 carbon
8 hydrogen
2 hydrogen
2 oxygen
3 oxygen
List out what is on each side
Carbon does not balance
Propane + oxygen → carbon dioxide + water
C3H8 + O2 → 3CO2 + H2O



Left
Right
3 carbon
3 carbon
8 hydrogen
2 hydrogen
2 oxygen
7 oxygen
Look at compounds first
Multiply CO2 on right side by 3 to balance
carbon
Hydrogen does not balance
Propane + oxygen → carbon dioxide + water
C3H8 + O2 → 3CO2 + 4H2O


Left
Right
3 carbon
3 carbon
8 hydrogen
8 hydrogen
2 oxygen
10 oxygen
Multiply water on right side by 4 to balance
hydrogen
Oxygen does not balance
Propane + oxygen → carbon dioxide + water
C3H8 + 5O2 → 3CO2 + 4H2O


Left
Right
3 carbon
3 carbon
8 hydrogen
8 hydrogen
10 oxygen
10 oxygen
Multiply oxygen on left side by 5 to balance
oxygen
Equation is now balanced
Sodium azide → Sodium + nitrogen
NaN3 → Na + N2
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

Left
Right
1 sodium
1 sodium
3 nitrogen
2 nitrogen
Identify what you have
Nitrogen does not balance
How do we balance nitrogen when left side has 3 and the
right has 2?
o
o
o
find the lowest common multiple for both
In this case 6
Multiply each side to make 6 atoms
Sodium azide → Sodium + nitrogen
2NaN3 → Na + N2


Left
Right
2 sodium
1 sodium
6 nitrogen
2 nitrogen
Multiply sodium azide on left side by 2 to get 6
nitrogen atoms
Still need to make 6 atoms of nitrogen on right
side
Sodium azide → Sodium + nitrogen
2NaN3 → Na + 3N2


Left
Right
2 sodium
1 sodium
6 nitrogen
6 nitrogen
Multiply nitrogen by 3 on left side to get 6 nitrogen
atoms
Sodium is not balanced
Sodium azide → Sodium + nitrogen
2NaN3 → 2Na + 3N2


Left
Right
2 sodium
2 sodium
6 nitrogen
6 nitrogen
Multiply sodium on right side by 2 to balance
sodium
Equation is now balanced

A mole is a quantity that contains 6.02 X 1023
items (usu. atoms, molecules or ions)
◦ An amount of a substance whose weight, in grams is
numerically equal to what its molecular weight was in
amu
◦ Just like a dozen is a quantity that contains 12 items
◦ 1 mole of C atoms = 6.02 x 1023 C atoms
◦ 1 mole of CO2 molecules = 6.02 x 1023 CO2 molecules
◦ 1 mole of H2O molecules = 6.02 x 1023 H2O
molecules

The number 6.02 X 1023 is Avogadro’s number
1 mol
6.02 x 1023 molecules
1 mol
6.02 x 1023 atoms

How many molecules are in 2.5 moles of
penicillin
2.5 moles penicillin X 6.02 x 1023 molecules =
1 mole
1.5 X 1024 molecules


The formula weight is the sum of the atomic
weights of all the atoms in a compound,
reported in atomic mass units
The molar mass is the mass of one mole of
any substance, reported in grams
◦ The value of the molar mass of a compound in
grams equals the value of its formula weight in
amu.

The sum of the atomic weights of all the atoms
in a compound, reported in atomic mass units
◦ This may be called molecular weight for covalent
compounds

Example
◦ H2O

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Contains 2 Hydrogen atoms and 1 Oxygen atom
Hydrogen weighs 1.01 amu (1.01 g H/mole)
Oxygen weighs 16.0 amu (16.0 g O/mole)
Formula/molecular weight = 2(1.01 amu) +16.0 amu =
18.0 amu
 Looking only at one molecule
 Molar mass = 2(1.01 g/mol) +16.0 g/mol = 18.0 g/mol
 Looking at one mole of the substance


Stoichiometry is the study of the quantitative
relationships that exist between substances
involved in a chemical reaction
Mole ratios within molecules:
AxBy
Mole ratio of A:B = x:y
Example: H2O2 ⇒ H:O = 2:2 = 1:1
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Mole ratios between molecules
A balanced equation tells us the number of
moles of each reactant that combine and
number of moles of each product formed
aA + bB  cC + dD
Mole ratio of A:B:C:D = a:b:c:d
Example: 2H2 + O2 → 2H20
H2:O2:H20 = 2:1:2


The coefficients in the balanced chemical
equation can represent the ratio of molecules
of the substances that are consumed or
produced
The coefficients in the balanced chemical
equation can represent the ratio of moles of
the substances that are consumed or
produced
2H2 + O2 → 2H20

There are 4 basic types of stoichiometry
problems:
◦
◦
◦
◦
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Moles to moles
Moles to grams
Grams to moles
Grams to grams
However, all stoichiometry problems are
really very similar, and the same general
approach can be used to solve any of them
◦ So really only one type of problem

Grams to moles to moles to grams
mol
g
mol
g
N2 +3H2→2NH3

mol
mol
g
How many moles of H2 are required to
produce 3.89 mol of NH3?
◦ Equation says H2:NH3 is 3:2
◦ 3.89 mol NH3 * 2 mol H2 = 5.84 moles H2
3 mol NH3
g
N2 +3H2→2NH3

mol
mol
g
g
How many grams of NH3 are produced from
3.44 mol of N2?
◦ Use mole ratio between NH3 and N2
◦ Then use molar mass to convert moles to g
◦ 3.44 mol N2 * 2 mol NH3 * 17 g NH3 =117 g NH3
1 mol N2 1 mol NH3
mol
N2 +3H2→2NH3

g
mol
g
How many moles H2 react with 6.77 g of N2?


Convert grams to moles using molar mass
Use molar ratio between N2 and H2
◦ 6.77 g N2 * 1 mol N2 * 3 mol H2 = 0.725 mol H2
28.0 g N2 1 mol N2
N2 +3H2→2NH3

mol
g
mol
g
How many grams of NH3 are produced from 8.23
g of H2?
 Use molar mass to convert g to moles
 Use molar ratio to convert between moles
 Use molar mass to convert moles to g
8.23 g H2 * 1 mol H2 * 2 mol NH3 * 17.0 g NH3= 46.2 g NH3
2.02 g H2 3 mol H2
mol NH3
mol
unit given


mol
unit requested
Once you have converted things into moles
you can use molar ratios in balanced
equations to convert between different
elements, compounds, and molecules
Problems may ask for g, molecules, atoms, or
many other units

If I gave you a value in g how would you convert it to
mol?
◦ Use formula weight (mass)
 Calculate using the periodic table
◦ Example: 53.21 g C6H12O6
Formula weight/molecular mass:
6*(12.01 g/mol) + 12*(1.01 g/mol) +6*(16.00 g/mol) = 180.18 g/mol
◦ Use value to calculate moles (may have to invert value)
53.21 g C6H12O6 * 1 mol C6H12O6
180.18 g C6H12O6
= 0.2953 mol C6H12O6

If I give you a value in mol how would you
compare it to mol of another substance?
◦ Use a balanced equation

Example 5 mol CO2 to mol O2
◦ 6CO2 + 6H2O  C6H12O6 + 6O2
◦ Use the stoichiometric coefficients to convert
between any of the substances in this equation
6CO2
6CO2
6CO2
6H2O
C6H12O6
6O2
◦ In our case only interested in the relationship with O2
5 mol CO2 * 6CO2 = 5 mol O2
6O2

All questions will build on the examples on
the previous two slides
◦ May ask you to go g  mol  mol  g
◦ May ask you to convert mol to number of atoms or
number of molecules
 Use Avagadro’s number 6.02 X 1023
◦ May ask you to use some other value in the future,
but it will be something that you can relate to g or
to mol

INGREDIENTS:
◦
◦
◦
◦
◦
◦
◦
◦

3 cups all-purpose flour
1 teaspoon salt
1 cup shortening
1/2 cup cold water
2 cups pumpkin
2 eggs, beaten
3/4 cup packed brown sugar
2 teaspoon spices
If you want to make multiple pies - the amount
of pie that you can bake depends on which of the
ingredients you have the “least” of – what will you
run out of first?
Ingredient
Recipe
In pantry
# of recipes
it can make
flour
3 cups
14 cups
4.67 x
salt
1 tsp
20 tsp
20 x
shortening
1 cup
7 cup
7x
water
½ cup
∞∞
∞
pumpkin
2 cups
19 cups
9.5 x
eggs
2
18
9x
sugar
¾ cup
3 cups
4X
spice
2 tsp
21 tsp
10.5 x
Limiting
ingredient
N2 +3H2→2NH3


You have 4 moles N2 and 9 moles H2
How many moles of NH3 could be produced?
Actual
Required
Rx
N2
4
1
4
H2
9
3
3
H2 is the limiting reactant and limits how
much NH3 can be made
9 moles H2 * 2 mol NH3 = 6 mol NH3
3 mol H2



Compare the actual amount of each reactant
to the amount required in the balanced
equation to determine how many times the
“reaction can be run”
Use the amount of the limiting reactant to
calculate how much product can be produced

How many g of S can be produced if we attempt to react 9.00g
of Bi2S3 with 16.00g of HNO3?

9.00 g Bi2S3 * 1 mol Bi2S3 = 0.0175 mol Bi2S3
514.3 g Bi2S3
0.0175 mol Bi2S3 *

16.00 g HNO3 * 1 mol HNO3 = 0.254 mol HNO3
63.0 g HNO3
0.254 mol HNO3 *

1 Rx
= 0.0175  allows Rx to run 0.0175 times
1 mol Bi2S3
1 Rx
= 0.3175  allows Rx to run 0.0318 times
8 mol HNO3
Therefore Bi2S3 is the limiting reactant

Use Bi2S3 - the limiting reactant to calculate S
◦ How many g of S can be produced if we attempt to react
9.00 g of Bi2S3 with 16.00 g of HNO3?



To see how much S can be produced this is gram
to mole to mole to gram problem
9.00 g Bi2S3 * 1 mol Bi2S3 = 0.0175 mol Bi2S3
514.3 g Bi2S3
0.0175 mol Bi2S3 * 3 mol S * 32.1 g S = 1.69 g S
mol Bi2S3
mol S


If 3.00 mol H2 reacts with 2.00 mol O2 how many
moles of H2O will form?
Solution 1:
◦ 3.00 mol H2 *
1 Rx = 1.50 Rx
2 mol H2
◦ 2.00 mol O2 *
1 Rx = 2.00 Rx
1 mol O2
◦ H2 is limiting reactant so use it to calculate mol of H2O
◦ 3.00 mol H2 * 2 mol H2O = 3.00 mol H2O
2 mol H2

H2 is the limiting reactant so 3.00 mol H2O will
theoretically form


If 3.00 mol H2 reacts with 2.00 mol O2 how
many moles of H2O will form?
Solution 2:
◦ 3.00 mol H2 * 2 mol H2O = 3.00 mol H2O
2 mol H2
◦ 2.00 mol O2 * 2 mol H2O = 4.00 mol H2O
1 mol O2

H2 is the limiting reactant so 3.00 mol H2O
will theoretically form
 actual yield
percentage yield  
 theoretica l yield



 100%

Actual yield is determined experimentally, it
is the mass of the product that is measured
Theoretical yield is the calculated mass of the
products based on the initial mass or number
of moles of the reactants



If 3.00 mol H2 reacts with 2.00 mol O2 how
many moles of H2O will form?
We just saw that H2 is the limiting reactant so
3.00 mol H2O will theoretically form
If we run this in the lab and end up producing
1.75 mol H2O what is the percent yield?
1.75 mol H2O * 100% = 58.3 %
3.00 mol H2O



Both processes occur together in a single
reaction called an oxidation−reduction or
redox reaction.
Thus, a redox reaction always has two
components, one that is oxidized and one
that is reduced
A redox reaction involves the transfer of
electrons from one element to another.

Oxidation is the loss of electrons from an
atom.
◦ Reducing agents are oxidized

Reduction is the gain of electrons by an
atom
◦ Oxidizing agents are reduced.
LEO says GER
Cu2+ gains 2 e−
Zn + Cu2+
Zn2+ + Cu
Zn loses 2 e–
•Zn loses 2 e− to form Zn2+, so Zn is oxidized.
•Cu2+ gains 2 e− to form Cu, so Cu2+ is reduced.
Cu2+ gains 2 e−
Zn2+ + Cu
Zn + Cu2+
Zn loses 2 e–
Each of these processes can be written as an
individual half reaction:
Zn2+ + 2 e−
loss of e−
Oxidation half reaction:
Zn
Reduction half reaction:
Cu2+ + 2e−
gain of e−
Cu
Zn
oxidized


+ Cu2+
Zn2+ + Cu
reduced
A compound that is oxidized while causing
another compound to be reduced is called a
reducing agent
Zn acts as a reducing agent because it causes
Cu2+ to gain electrons and become reduced
58
Zn
oxidized


+ Cu2+
Zn2+ + Cu
reduced
A compound that is reduced while causing
another compound to be oxidized is called an
oxidizing agent
Cu2+ acts as an oxidizing agent because it
causes Zn to lose electrons and become
oxidized
59
60
Iron Rusting
O gains e– and is reduced.
4 Fe(s) + 3 O2(g)
neutral Fe
neutral O
2 Fe2O3(s)
Fe3+
O2–
Fe loses e– and is oxidized.
61
Inside an Alkaline Battery
Mn4+ gains e− and is reduced.
Zn + 2 MnO2
neutral Zn
Mn4+
ZnO + Mn2O3
Zn2+
Mn3+
Zn loses e− and is oxidized.
62
Zn + 2 MnO2
ZnO + Mn2O3
63
Oxidation results in the:
Reduction results in the:
•Gain of oxygen atoms
•Loss of oxygen atoms
•Loss of hydrogen atoms
•Gain of hydrogen atoms
64