Chemical Composition

1
Chapter 3

• Chemical Composition

2
Chemical Reactions

• A chemical reactions is an abbreviated way to
  show a physical orchemical change
• A chemical change alters the physical and
  chemical properties of a substance
• Factors that indicate a chemical change
• Change in color
• Temperature change
• Change in odor
• Change in taste (we do not taste chemicals)
• Reactions always contain an arrow that separates
  the reactants from the products
• Reactants Products

3
Types of Chemical Reactions

• Combination reaction (synthesis)
• Elements for reactants
• Examples
• H2 O2 H2O
• N2 H2 NH3
• Al O2 Al2O3
• The Law of Conservation of matter, states
  matter cannot be created nor destroyed, that
  means equations must be balanced.

4
Types of Chemical Reactions
Combination reaction Continued

• Balance the first equation
• H2 O2 H2O
• Note two oxygen atoms on the reactant side and
  only one on the product side, therefore place a
  two in front of water

5
Types of Chemical Reactions

• Balance the first equation
• H2 O2 2H2O
• Note two oxygen atoms on the reactant side and
  only one on the product side, therefore place a
  two in front of water
• The two now doubles everything in water, thus 4
  hydrogen and 2 oxygen. Now place a 2 in front of
  hydrogen.

6
Types of Chemical Reactions

• Balance the first equation
• 2H2 O2 2H2O
• Note two oxygen atoms on the reactant side and
  only one on the product side, therefore place a
  two in front of water
• The two now doubles everything in water, thus 4
  hydrogen and 2 oxygen. Now place a 2 in front of
  hydrogen.

7
Types of Chemical Reactions

• Now balance the second equation
• N2 H2 NH3
• Note two nigrogen atoms on the reactant side and
  only one on the product side.
• Place a 2 in front of ammonia

8
Types of Chemical Reactions

• Now balance the second equation
• N2 H2 2NH3
• Note two nitrogen atoms on the reactant side and
  only one on the product side.
• Place a 2 in front of ammonia. This makes 2
  nitrogen atoms and 6 hydrogen atoms. Now place a
  3 in front of hydrogen to balance hydrogen atoms.

9
Types of Chemical Reactions

• Now balance the second equation
• N2 3 H2 2NH3
• Note two nitrogen atoms on the reactant side and
  only one on the product side.
• Place a 2 in front of ammonia. This makes 2
  nitrogen atoms and 6 hydrogen atoms. Now place a
  3 in front of hydrogen to balance hydrogen atoms.

10
Types of Chemical Reactions

• Decomposition Reaction
• Compounds form simpler compounds or elements.
• Examples
• H2O H2 O2

11
Types of Chemical Reactions

• Decomposition Reaction
• Compounds form simpler compounds or elements.
• Examples
• 2H2O H2 O2

12
Types of Chemical Reactions

• Decomposition Reaction
• Compounds form simpler compounds or elements.
• Examples
• 2H2O 2H2 O2
• Notice decomposition reactions are the opposite
  of combination reactions

13
Types of Chemical Reactions

• Single Replacement reactions have an element and
  a compound for reactants.
• Example
• Zn HCl
• How do we predict the products? Trade places
  with the metal or nonmetal with the metal or
  nonmetal in the compound

14
Types of Chemical Reactions

• Single Replacement reactions have an element and
  a compound for reactants.
• Example
• Zn HCl
• How do we predict the products? Trade places
  with the metal or nonmetal with the metal or
  nonmetal in the compound

15
Types of Chemical Reactions

• Single Replacement reactions have an element and
  a compound for reactants.
• Example
• Zn HCl ZnCl H
• Now make the products stable. Slide with Clyde

16
Types of Chemical Reactions

• Single Replacement reactions have an element and
  a compound for reactants.
• Example
• Zn HCl ZnCl2 H2
• Now make the products stable. Slide with Clyde

17
Types of Chemical Reactions

• Single Replacement reactions have an element and
  a compound for reactants.
• Example
• Zn HCl ZnCl2 H2
• Now make the products stable.
• Now Balance

18
Types of Chemical Reactions

• Single Replacement reactions have an element and
  a compound for reactants.
• Example
• Zn 2HCl ZnCl2 H2
• Now make the products stable.
• Now Balance

19
Types of Chemical Reactions

• Single Replacement reactions have an element and
  a compound for reactants.
• Another Example
• Cl2 MgBr2
• How do we predict the products? Trade places
  with the metal or nonmetal with the metal or
  nonmetal in the compound. In this case we are
  trading nonmetals

20
Types of Chemical Reactions

• Single Replacement reactions have an element and
  a compound for reactants.
• Another Example
• Cl2 MgBr2 Br MgCl
• How do we predict the products? Trade places
  with the metal or nonmetal with the metal or
  nonmetal in the compound. In this case we are
  trading nonmetals

21
Types of Chemical Reactions

• Single Replacement reactions have an element and
  a compound for reactants.
• Another Example
• Cl2 MgBr2 Br2 MgCl2
• How do we predict the products? Trade places
  with the metal or nonmetal with the metal or
  nonmetal in the compound. In this case we are
  trading nonmetals

22
Types of Chemical Reactions

• Double Replacement reactions contain compounds as
  reactants.
• HCl Ca(OH)2 CaCl HOH
• Check formulas, and slide with Clyde when
  necessary

23
Types of Chemical Reactions

• Double Replacement reactions contain compounds as
  reactants.
• HCl Ca(OH)2 CaCl 2 HOH
• Check formulas, and slide with Clyde when
  necessary

24
Types of Chemical Reactions

• Double Replacement reactions contain compounds as
  reactants.
• 2HCl Ca(OH)2 CaCl 2 2HOH
• Check formulas, and slide with Clyde when
  necessary
• Now Balance!

25
Types of Chemical Reactions

• Combustion Reactions occur when an element or
  compound combine with oxygen to produce oxides of
  each element.
• H2 O2
• CH4 O2
• What is the oxide of hydrogen?

26
Types of Chemical Reactions

• Combustion Reactions occur when an element or
  compound combine with oxygen to produce oxides of
  each element.
• H2 O2
• CH4 O2
• What is the oxide of hydrogen? Water

27
Types of Chemical Reactions

• Combustion Reactions occur when an element or
  compound combine with oxygen to produce oxides of
  each element.
• H2 O2 H2O
• CH4 O2
• What is the oxide of hydrogen? Water
• And the oxide of carbon?

28
Types of Chemical Reactions

• Combustion Reactions occur when an element or
  compound combine with oxygen to produce oxides of
  each element.
• H2 O2 H2O
• CH4 O2 CO2 H2O
• What is the oxide of hydrogen? Water
• And the oxide of carbon? Carbon dioxide

29
Types of Chemical Reactions

• Combustion Reactions occur when an element or
  compound combine with oxygen to produce oxides of
  each element.
• 2H2 O2 2H2O
• CH4 O2 CO2 H2O
• Now balance

30
Types of Chemical Reactions

• Combustion Reactions occur when an element or
  compound combine with oxygen to produce oxides of
  each element.
• 2H2 O2 2H2O
• CH4 O2 CO2 2H2O
• Now balance

31
Types of Chemical Reactions

• Combustion Reactions occur when an element or
  compound combine with oxygen to produce oxides of
  each element.
• 2H2 O2 2H2O
• CH4 2O2 CO2 2H2O
• Now balance

32
Ionic Solution Formation
KCN (S)
K (aq) CN- (aq)
Ionic equation
Note Not all ionic solutes are soluble in
water. How can we tell if an ionic solute is
soluble in water?
33
Ionic Solution Formation
KCN (S)
K (aq) CN- (aq)
Ionic equation
Note Not all ionic solutes are soluble in
water. How can we tell if an ionic solute is
soluble in water? The solubility rules gives
ionic solubility.
34
Solubility Rules
There are some more specific rules that allows us
to better estimate the solubility of ionic
compounds.
You will be given these if you need them.
35
Ionic Equations
Using the solubility rules write the formula
equation, the ionic equation and the net ionic
equation when aqueous silver nitrate is combined
with aqueous sodium chloride.
AgNO3 (aq) NaCl (aq) AgCl (s)
NaNO3 (aq)
Formula Equation
36
Ionic Equations
Using the solubility rules write the formula
equation, the ionic equation and the net ionic
equation when aqueous silver nitrate is combined
with aqueous sodium chloride.
AgNO3 (aq) NaCl (aq) AgCl
(s) NaNO3 (aq)
Formula Equation
Ag (aq) NO3- (aq) Na (aq) Cl- (aq)
AgCl (s) Na (aq) NO3- (aq)
Ionic Equation
37
Ionic Equations
Using the solubility rules write the formula
equation, the ionic equation and the net ionic
equation when aqueous silver nitrate is combined
with aqueous sodium chloride.
AgNO3 (aq) NaCl (aq) AgCl
(s) NaNO3 (aq)
Formula Equation
Ag (aq) NO3- (aq) Na (aq) Cl- (aq)
AgCl (s) Na (aq) NO3- (aq)
Ionic Equation
Spectator ions are ions that are identical on the
reactants and products side of the equation.
38
Ionic Equations
Using the solubility rules write the formula
equation, the ionic equation and the net ionic
equation when aqueous silver nitrate is combined
with aqueous sodium chloride.
AgNO3 (aq) NaCl (aq) AgCl
(s) NaNO3 (aq)
Formula Equation
Ag (aq) NO3- (aq) Na (aq) Cl- (aq)
AgCl (s) Na (aq) NO3- (aq)
Ionic Equation
Spectator ions are ions that are identical on the
reactants and products side of the equation.
Place a around the spectator ions.
39
Ionic Equations
Using the solubility rules write the formula
equation, the ionic equation and the net ionic
equation when aqueous silver nitrate is combined
with aqueous sodium chloride.
AgNO3 (aq) NaCl (aq) AgCl
(s) NaNO3 (aq)
Formula Equation
Ag (aq) NO3- (aq) Na (aq) Cl- (aq)
AgCl (s) Na (aq) NO3- (aq)
Ionic Equation
Spectator ions are ions that are identical on the
reactants and products side of the equation.
Place a around the spectator ions.
40
Ionic Equations
Using the solubility rules write the formula
equation, the ionic equation and the net ionic
equation when aqueous silver nitrate is combined
with aqueous sodium chloride.
AgNO3 (aq) NaCl (aq) AgCl
(s) NaNO3 (aq)
Formula Equation
Ag (aq) NO3- (aq) Na (aq) Cl- (aq)
AgCl (s) Na (aq) NO3- (aq)
Ionic Equation
Spectator ions are ions that are identical on the
reactants and products side of the equation.
Eliminating the spectator ions produces the
netionic equation.
Ag (aq) Cl- (aq) AgCl (s)
Net ionic equation
41
Yet Another Ionic Equation

• Write the formula, ionic and net ionic equations
• when aqueous sodium chloride combines with
• aqueous calcium bromide.

42
Yet Another Ionic Equation

• Write the formula, ionic and net ionic equations
• when aqueous sodium chloride combines with
• aqueous calcium bromide.

NaCl(aq) CaBr2 CaCl2
NaBr
Now balance
43
Yet Another Ionic Equation

• Write the formula, ionic and net ionic equations
• when aqueous sodium chloride combines with
• aqueous calcium bromide.

NaCl(aq) CaBr2 CaCl2
NaBr(aq)
2
2
Now balance
44
Yet Another Ionic Equation

• Write the formula, ionic and net ionic equations
• when aqueous sodium chloride combines with
• aqueous calcium bromide.

NaCl(aq) CaBr2 CaCl2(aq)
NaBr(aq)
2
2
Now balance
45
Yet Another Ionic Equation

• Write the formula, ionic and net ionic equations
• when aqueous sodium chloride combines with
• aqueous calcium bromide.

NaCl(aq) CaBr2 CaCl2(aq)
NaBr(aq)
2
2
Now balance
2Na(aq) 2Cl-(aq) Ca2(aq) 2Br-(aq)
Ca2(aq) 2Cl-(aq) 2Na(aq) 2 Br-(aq)
Ionic equation
46
Yet Another Ionic Equation

• Write the formula, ionic and net ionic equations
• when aqueous sodium chloride combines with
• aqueous calcium bromide.

NaCl(aq) CaBr2 CaCl2(aq)
NaBr(aq)
2
2
Now balance
2Na(aq) 2Cl-(aq) Ca2(aq) 2Br-(aq)
Ca2(aq) 2Cl-(aq) 2Na(aq) 2 Br-(aq)
Ionic equation
No net ionic equation No Reaction (NR)
47
Types of Chemical Reactions

• REDOX reactions where the oxidation number
  changes from reactants to products.
• Oxidation is when the oxidation number increases,
  by losing of electrons.
• Reduction is when the oxidation number decreases
  by gaining electrons.
• Consider the following equation
• H2 O2 H2O
• What are the oxidation numbers of hydrogen and
  oxygen?

0
0
48
Types of Chemical Reactions

• REDOX reactions where the oxidation number
  changes from reactants to products.
• Oxidation is when the oxidation number increases,
  by losing of electrons.
• Reduction is when the oxidation number decreases
  by gaining electrons.
• Consider the following equation
• H2 O2 H2O
• What are the oxidation numbers of hydrogen and
  oxygen?

49
REDOX REACTIONS
0
0
2(1)
2-
0

• H2 O2 H2O
• How about hydrogen and oxygen in water?

50
REDOX REACTIONS
0
0
2(1)
2-
0

• H2 O2 H2O
• How about hydrogen and oxygen in water?
• Oxidation is caused by the oxygen molecule, so it
  is referred to as the oxidizing agent (OA)
• Reduction is caused by the hydrogen molecule, so
  it is referred to as the reducing agent (RA)

reduced
oxidized
51
REDOX REACTIONS

• Note
• All of the previously discussed reactions are
  REDOX except the double replacement reactions.
• The number of electrons lost is equal to the
  number of electrons gained in a reaction. Why?
• Most elements have variable oxidation numbers,
  except for hydrogen, oxygen, and the memorized
  polyatomic ions.

52
REDOX REACTIONS

• Oxidation numbers for a compound must add up to
  equal zero, while the oxidation numbers for a
  polyatomic ion must up to equal the charge of
  that ion.
• Consider the following chlorine compounds
• HClO4, HClO3, HClO2, HClO, Cl2, HCl
• What is the oxidation number of chlorine in each
  of these compounds, assuming
• H 1 and oxygen is 2-

1
4(2-)0
53
REDOX REACTIONS

• Oxidation numbers for a compound must add up to
  equal zero, while the oxidation numbers for a
  polyatomic ion must up to equal the charge of
  that ion.
• Consider the following chlorine compounds
• HClO4, HClO3, HClO2, HClO, Cl2, HCl
• What is the oxidation number of chlorine in each
  of these compounds, assuming H is 1 and oxygen
  is 2-

1
4(2-)0
7
54
REDOX REACTIONS

• Oxidation numbers for a compound must add up to
  equal zero, while the oxidation numbers for a
  polyatomic ion must up to equal the charge of
  that ion.
• Consider the following chlorine compounds
• HClO4, HClO3, HClO2, HClO, Cl2, HCl
• What is the oxidation number of chlorine in each
  of these compounds, assuming H is 1 and oxygen
  is 2-

1
4(2-)0
7
5
55
REDOX REACTIONS

• Oxidation numbers for a compound must add up to
  equal zero, while the oxidation numbers for a
  polyatomic ion must up to equal the charge of
  that ion.
• Consider the following chlorine compounds
• HClO4, HClO3, HClO2, HClO, Cl2, HCl
• What is the oxidation number of chlorine in each
  of these compounds, assuming H is 1 and oxygen
  is 2-

1
4(2-)0
7
5
3
56
REDOX REACTIONS

• Oxidation numbers for a compound must add up to
  equal zero, while the oxidation numbers for a
  polyatomic ion must up to equal the charge of
  that ion.
• Consider the following chlorine compounds
• HClO4, HClO3, HClO2, HClO, Cl2, HCl
• What is the oxidation number of chlorine in each
  of these compounds, assuming H is 1 and oxygen
  is 2-

1
4(2-)0
7
5
3
1
57
REDOX REACTIONS

• Oxidation numbers for a compound must add up to
  equal zero, while the oxidation numbers for a
  polyatomic ion must up to equal the charge of
  that ion.
• Consider the following chlorine compounds
• HClO4, HClO3, HClO2, HClO, Cl2, HCl
• What is the oxidation number of chlorine in each
  of these compounds, assuming H is 1 and oxygen
  is 2-

1
4(2-)0
7
5
3
1
0
58
REDOX REACTIONS

• Oxidation numbers for a compound must add up to
  equal zero, while the oxidation numbers for a
  polyatomic ion must up to equal the charge of
  that ion.
• Consider the following chlorine compounds
• HClO4, HClO3, HClO2, HClO, Cl2, HCl
• What is the oxidation number of chlorine in each
  of these compounds, assuming H is 1 and oxygen
  is 2-

1
4(2-)0
7
5
3
1
0
1-
59
REDOX REACTIONS
3(2-)2-

• How about sulfur in SO3 2-

60
REDOX REACTIONS
3(2-)2-
4

• How about sulfur in SO3 2-
• How about carbon in C6H12O6

12(1)
6(2-)0
61
REDOX REACTIONS
3(2-)2-
4

• How about sulfur in SO3 2-
• How about carbon in C6H12O6

12(1)
6(2-)0
0
62
Types of Chemical Reactions

• REDOX reactions where the oxidation number
  changes from reactants to products.
• Oxidation is when the oxidation number increases,
  by losing of electrons.
• Reduction is when the oxidation number decreases
  by gaining electrons.
• Consider the following equation
• H2 O2 H2O
• What are the oxidation numbers of hydrogen and
  oxygen?

63
Types of Chemical Reactions

• REDOX reactions where the oxidation number
  changes from reactants to products.
• Oxidation is when the oxidation number increases,
  by losing of electrons.
• Reduction is when the oxidation number decreases
  by gaining electrons.
• Consider the following equation
• H2 O2 H2O
• What are the oxidation numbers of hydrogen and
  oxygen?

0
0
64
REDOX REACTIONS
0
0
2(1-)
2-
0

• H2 O2 H2O
• How about hydrogen and oxygen in water?

65
REDOX REACTIONS
0
0
2(1)
2-
0

• H2 O2 H2O
• How about hydrogen and oxygen in water?
• Oxidation is caused by the oxygen molecule, so it
  is referred to as the oxidizing agent (OA)
• Reduction is caused by the hydrogen molecule, so
  it is referred to as the reducing agent (RA)

reduced
oxidized
66
REDOX REACTIONS

• Note
• All of the previously discussed reactions are
  REDOX except the double replacement reactions.
• The number of electrons lost is equal to the
  number of electrons gained in a reaction. Why?
• Most elements have variable oxidation numbers,
  except for hydrogen, oxygen, and the memorized
  polyatomic ions.

67
Balancing Redox Reactions

• I. Oxidation Number Method
• a. Assign oxidation numbers to each element
• b. Determine the elements oxidized and reduced
• c. Balance the atoms that are oxidized and
  reduced
• d. Balance the electrons lost or gained, to
  conform to the Law of Conservation of Matter, by
  placing coefficients in front of the formulas
  containing the atoms oxidized and reduced to both
  sides of the equation.
• e. The remaining atoms are balanced by
  inspection
• f. Balance oxygen, or hydrogen by adding H2O
• Balance remaining hydrogen atoms by adding H
• Simplify
• i. For basic reactions add the same number of
  OH- ions to both sides of the equation as there
  are H ions.
• j. Combine H and OH- ions to make water
• k. Simplify again if necessary.

HNO3 Cu2O ? Cu(NO3)2 NO
H2O
68
Balancing REDOX Reactions

• I. Oxidation Number Method
• a. Assign oxidation numbers to each element
• b. Determine the elements oxidized and reduced
• c. Balance the atoms that are oxidized and
  reduced
• d. Balance the electrons lost or gained, to
  conform to the Law of Conservation of Matter, by
  placing coefficients in front of the formulas
  containing the atoms oxidized and reduced to both
  sides of the equation.
• e. The remaining atoms are balanced by
  inspection
• f. Balance oxygen, or hydrogen by adding H2O
• Balance remaining hydrogen atoms by adding H
• Simplify
• i. For basic reactions add the same number of
  OH- ions to both sides of the equation as there
  are H ions.
• j. Combine H and OH- ions to make water
• k. Simplify again if necessary.

1
3(2-)0
?
HNO3 Cu2O ? Cu(NO3)2 NO
H2O
69
Balancing REDOX Reactions

• I. Oxidation Number Method
• a. Assign oxidation numbers to each element
• b. Determine the elements oxidized and reduced
• c. Balance the atoms that are oxidized and
  reduced
• d. Balance the electrons lost or gained, to
  conform to the Law of Conservation of Matter, by
  placing coefficients in front of the formulas
  containing the atoms oxidized and reduced to both
  sides of the equation.
• e. The remaining atoms are balanced by
  inspection
• f. Balance oxygen, or hydrogen by adding H2O
• Balance remaining hydrogen atoms by adding H
• Simplify
• i. For basic reactions add the same number of
  OH- ions to both sides of the equation as there
  are H ions.
• j. Combine H and OH- ions to make water
• k. Simplify again if necessary.

1
3(2-)0
5
2-0
2(?)
HNO3 Cu2O ? Cu(NO3)2 NO
H2O
70
Balancing REDOX Reactions

• I. Oxidation Number Method
• a. Assign oxidation numbers to each element
• b. Determine the elements oxidized and reduced
• c. Balance the atoms that are oxidized and
  reduced
• d. Balance the electrons lost or gained, to
  conform to the Law of Conservation of Matter, by
  placing coefficients in front of the formulas
  containing the atoms oxidized and reduced to both
  sides of the equation.
• e. The remaining atoms are balanced by
  inspection
• f. Balance oxygen, or hydrogen by adding H2O
• g. Balance remaining hydrogen atoms by adding H
  
• h. Simplify
• i. For basic reactions add the same number of
  OH- ions to both sides of the equation as there
  are H ions.
• j. Combine H and OH- ions to make water
• k. Simplify again if necessary.

1
3(2-)0
5
2-0
2(1)
HNO3 Cu2O ? Cu(NO3)2 NO
H2O
71
Balancing Redox Reactions

• I. Oxidation Number Method
• a. Assign oxidation numbers to each element
• b. Determine the elements oxidized and reduced
• c. Balance the atoms that are oxidized and
  reduced
• d. Balance the electrons lost or gained, to
  conform to the Law of Conservation of Matter, by
  placing coefficients in front of the formulas
  containing the atoms oxidized and reduced to both
  sides of the equation.
• e. The remaining atoms are balanced by
  inspection
• f. Balance oxygen, or hydrogen by adding H2O
• g. Balance remaining hydrogen atoms by adding H
  
• h. Simplify
• i. For basic reactions add the same number of
  OH- ions to both sides of the equation as there
  are H ions.
• j. Combine H and OH- ions to make water
• k. Simplify again if necessary.

2(1-)0
?
1
3(2-)0
5
2-0
2(1)
HNO3 Cu2O ? Cu(NO3)2 NO
H2O
72
Balancing REDOX Reactions

• I. Oxidation Number Method
• a. Assign oxidation numbers to each element
• b. Determine the elements oxidized and reduced
• c. Balance the atoms that are oxidized and
  reduced
• d. Balance the electrons lost or gained, to
  conform to the Law of Conservation of Matter, by
  placing coefficients in front of the formulas
  containing the atoms oxidized and reduced to both
  sides of the equation.
• e. The remaining atoms are balanced by
  inspection
• f. Balance oxygen, or hydrogen by adding H2O
• g. Balance remaining hydrogen atoms by adding H
  
• h. Simplify
• i. For basic reactions add the same number of
  OH- ions to both sides of the equation as there
  are H ions.
• j. Combine H and OH- ions to make water
• k. Simplify again if necessary.

2(1-)0
2
1
3(2-)0
5
2-0
2(1)
HNO3 Cu2O ? Cu(NO3)2 NO
H2O
73
Balancing REDOX Reactions

• I. Oxidation Number Method
• a. Assign oxidation numbers to each element
• b. Determine the elements oxidized and reduced
• c. Balance the atoms that are oxidized and
  reduced
• d. Balance the electrons lost or gained, to
  conform to the Law of Conservation of Matter, by
  placing coefficients in front of the formulas
  containing the atoms oxidized and reduced to both
  sides of the equation.
• e. The remaining atoms are balanced by
  inspection
• f. Balance oxygen, or hydrogen by adding H2O
• g. Balance remaining hydrogen atoms by adding H
  
• h. Simplify
• i. For basic reactions add the same number of
  OH- ions to both sides of the equation as there
  are H ions.
• j. Combine H and OH- ions to make water
• k. Simplify again if necessary.

2(1-)0
? 2- 0
1
3(2-)0
5
2-0
2(1)
HNO3 Cu2O ? Cu(NO3)2 NO
H2O
74
Balancing REDOX Reactions

• I. Oxidation Number Method
• a. Assign oxidation numbers to each element
• b. Determine the elements oxidized and reduced
• c. Balance the atoms that are oxidized and
  reduced
• d. Balance the electrons lost or gained, to
  conform to the Law of Conservation of Matter, by
  placing coefficients in front of the formulas
  containing the atoms oxidized and reduced to both
  sides of the equation.
• e. The remaining atoms are balanced by
  inspection
• f. Balance oxygen, or hydrogen by adding H2O
• g. Balance remaining hydrogen atoms by adding H
  
• h. Simplify
• i. For basic reactions add the same number of
  OH- ions to both sides of the equation as there
  are H ions.
• j. Combine H and OH- ions to make water
• k. Simplify again if necessary.

2(1-)0
2 2- 0
1
3(2-)0
5
2-0
2(1)
HNO3 Cu2O ? Cu(NO3)2 NO
H2O
75
Balancing REDOX Reactions

• I. Oxidation Number Method
• a. Assign oxidation numbers to each element
• b. Determine the elements oxidized and reduced
• c. Balance the atoms that are oxidized and
  reduced
• d. Balance the electrons lost or gained, to
  conform to the Law of Conservation of Matter, by
  placing coefficients in front of the formulas
  containing the atoms oxidized and reduced to both
  sides of the equation.
• e. The remaining atoms are balanced by
  inspection
• f. Balance oxygen, or hydrogen by adding H2O
• g. Balance remaining hydrogen atoms by adding H
  
• h. Simplify
• i. For basic reactions add the same number of
  OH- ions to both sides of the equation as there
  are H ions.
• j. Combine H and OH- ions to make water
• k. Simplify again if necessary.

2(1-)0
2 2- 0
2
1
3(2-)0
5
2-0
2(1)
HNO3 Cu2O ? Cu(NO3)2 NO
H2O
oxidized
reduced
76
Balancing REDOX Reactions

• I. Oxidation Number Method
• a. Assign oxidation numbers to each element
• b. Determine the elements oxidized and reduced
• c. Balance the atoms that are oxidized and
  reduced
• d. Balance the electrons lost or gained, to
  conform to the Law of Conservation of Matter, by
  placing coefficients in front of the formulas
  containing the atoms oxidized and reduced to both
  sides of the equation.
• e. The remaining atoms are balanced by
  inspection
• f. Balance oxygen, or hydrogen by adding H2O
• g. Balance remaining hydrogen atoms by adding H
  
• h. Simplify
• i. For basic reactions add the same number of
  OH- ions to both sides of the equation as there
  are H ions.
• j. Combine H and OH- ions to make water
• k. Simplify again if necessary.

2(1-)0
2 2- 0
2
1
3(2-)0
5
2-0
2(1)
HNO3 Cu2O ? 2 Cu(NO3)2 NO
H2O
oxidized
reduced
77
Balancing REDOX Reactions

• I. Oxidation Number Method
• a. Assign oxidation numbers to each element
• b. Determine the elements oxidized and reduced
• c. Balance the atoms that are oxidized and
  reduced
• d. Balance the electrons lost or gained, to
  conform to the Law of Conservation of Matter, by
  placing coefficients in front of the formulas
  containing the atoms oxidized and reduced to both
  sides of the equation.
• e. The remaining atoms are balanced by
  inspection
• f. Balance oxygen, or hydrogen by adding H2O
• g. Balance remaining hydrogen atoms by adding H
  
• h. Simplify
• i. For basic reactions add the same number of
  OH- ions to both sides of the equation as there
  are H ions.
• j. Combine H and OH- ions to make water
• k. Simplify again if necessary.

2(1-)0
2 2- 0
2
1
3(2-)0
5
2-0
2(1)
HNO3 3 Cu2O ? 3 (2) Cu(NO3)2
NO H2O
Oxidized3( -2e)
Reduced 2(3)e
78
Balancing REDOX Reactions

• I. Oxidation Number Method
• a. Assign oxidation numbers to each element
• b. Determine the elements oxidized and reduced
• c. Balance the atoms that are oxidized and
  reduced
• d. Balance the electrons lost or gained, to
  conform to the Law of Conservation of Matter, by
  placing coefficients in front of the formulas
  containing the atoms oxidized and reduced to both
  sides of the equation.
• e. The remaining atoms are balanced by
  inspection
• f. Balance oxygen, or hydrogen by adding H2O
• g. Balance remaining hydrogen atoms by adding H
  
• h. Simplify
• i. For basic reactions add the same number of
  OH- ions to both sides of the equation as there
  are H ions.
• j. Combine H and OH- ions to make water
• k. Simplify again if necessary.

2(1-)0
2 2- 0
2
1
3(2-)0
5
2-0
2(1)
2HNO3 3 Cu2O ? 3 (2) Cu(NO3)2
2NO H2O
Oxidized3( -2e)
Reduced 2(3)e
79
Balancing REDOX Reactions

• I. Oxidation Number Method
• a. Assign oxidation numbers to each element
• b. Determine the elements oxidized and reduced
• c. Balance the atoms that are oxidized and
  reduced
• d. Balance the electrons lost or gained, to
  conform to the Law of Conservation of Matter, by
  placing coefficients in front of the formulas
  containing the atoms oxidized and reduced to both
  sides of the equation.
• e. The remaining atoms are balanced by
  inspection
• f. Balance oxygen, or hydrogen by adding H2O
• g. Balance remaining hydrogen atoms by adding H
  
• h. Simplify
• i. For basic reactions add the same number of
  OH- ions to both sides of the equation as there
  are H ions.
• j. Combine H and OH- ions to make water
• k. Simplify again if necessary.

1
3(2-)0
5
2-0
2(1)
2(1-)0
2 2- 0
2
2HNO3 3 Cu2O ? 3 (2) Cu(NO3)2
2NO H2O
Oxidized3( -2e)
Reduced 2(3)e
80
Balancing REDOX Reactions

• I. Oxidation Number Method
• a. Assign oxidation numbers to each element
• b. Determine the elements oxidized and reduced
• c. Balance the atoms that are oxidized and
  reduced
• d. Balance the electrons lost or gained, to
  conform to the Law of Conservation of Matter, by
  placing coefficients in front of the formulas
  containing the atoms oxidized and reduced to both
  sides of the equation.
• e. The remaining atoms are balanced by
  inspection
• f. Balance oxygen, or hydrogen by adding H2O
• g. Balance remaining hydrogen atoms by adding H
  
• h. Simplify
• i. For basic reactions add the same number of
  OH- ions to both sides of the equation as there
  are H ions.
• j. Combine H and OH- ions to make water
• k. Simplify again if necessary.

1
3(2-)0
5
2-0
2(1)
2(1-)0
2 2- 0
2
14HNO3 3 Cu2O ? 3 (2) Cu(NO3)2
2NO 7 H2O
Oxidized3( -2e)
Reduced 2(3)e
81
Balancing REDOX Reactions

• I. Oxidation Number Method
• a. Assign oxidation numbers to each element
• b. Determine the elements oxidized and reduced
• c. Balance the atoms that are oxidized and
  reduced
• d. Balance the electrons lost or gained, to
  conform to the Law of Conservation of Matter, by
  placing coefficients in front of the formulas
  containing the atoms oxidized and reduced to both
  sides of the equation.
• e. The remaining atoms are balanced by
  inspection
• f. Balance oxygen, or hydrogen by adding H2O
• g. Balance remaining hydrogen atoms by adding H
  
• h. Simplify
• i. For basic reactions add the same number of
  OH- ions to both sides of the equation as there
  are H ions.
• j. Combine H and OH- ions to make water
• k. Simplify again if necessary.

2 2- 0
1
3(2-)0
5
2-0
2(1)
2(1-)0
2
14 HNO3 3 Cu2O ? 6 Cu(NO3)2
2 NO 7 H2O
Oxidized3( -2e)
Reduced 2(3)e
82
OX BALANCING EXAMPLE

• MnO4 - Cl- ? Mn2 Cl2

83
OX BALANCING EXAMPLE
7

• MnO4 - Cl- ? Mn2 Cl2

84
OX BALANCING EXAMPLE
7
1-

• MnO4 - Cl- ? Mn2 Cl2

85
OX BALANCING EXAMPLE
7
1-
0

• MnO4 - Cl- ? Mn2 Cl2

86
OX BALANCING EXAMPLE
1-
0
7

• MnO4 - Cl- ? Mn2
  Cl2

87
OX BALANCING EXAMPLE
1-
0
7

• MnO4 - Cl- ? Mn2 Cl2

oxidized
reduced
88
OX BALANCING EXAMPLE
1-
0
7

• MnO4 - Cl- ? Mn2 Cl2

oxidized
reduced
Step C, balance atoms oxidized or reduced
89
OX BALANCING EXAMPLE
1-
0
7

• MnO4 - Cl- ? Mn2
  Cl2

2
oxidized
reduced
Step C, balance atoms oxidized or reduced
90
OX BALANCING EXAMPLE
1-
0
7

• MnO4 - Cl- ? Mn2 Cl2

2
oxidized
- 2 e-
reduced
5 e-
Step d, balance electrons lost or gained.
common denominator between 5 and 2 is
10. Therefore multiply Mn on both
sides of the equation by two and Cl on
both sides by 5.
91
OX BALANCING EXAMPLE
1-
0
7

• MnO4 - Cl- ? Mn2 Cl2

5(2)
5
2
2
oxidized
- 2 e-
reduced
5 e-
Step d, balance electrons lost or gained. The
common denominator between 5 and 2
is 10. Therefore multiply Mn on both
sides of the equation by 2 and Cl on
both sides by 5.
92
OX BALANCING EXAMPLE
1-
0
7

• MnO4 - Cl- ? Mn2 Cl2

5(2)
5
2
2
oxidized
- 2 e-
reduced
5 e-
Step e, balance remaining elements by inspection.
There are 8 oxygen atoms on the left. Oxygen is
balanced by adding water to the appropriate side.
In this case since there are 8 oxygen atoms on
the reactant side which requires adding 8 water
molecules to the product side of the equation.
93
OX BALANCING EXAMPLE
1-
7
0

• MnO4- Cl- ? Mn2 Cl2

2
10
2
5
8H2O
oxidized
- 2 e-
reduced
5 e-
Step e, balance remaining elements by inspection.
There are 8 oxygen atoms on the left. Oxygen is
balanced by adding water to the appropriate side.
In this case since there are 8 oxygen atoms on
the reactant side which requires adding 8 water
molecules to the product side of the equation.
Now the hydrogen atoms need to be balanced by
adding 16 H to the reactant side.
94
OX BALANCING EXAMPLE
7
0
1-

• MnO4- Cl- ? Mn2 Cl2

8H2O
10
2
2
5
16 H
oxidized
- 2 e-
reduced
5 e-
Step e, balance remaining elements by inspection.
There are 8 oxygen atoms on the left. Oxygen is
balanced by adding water to the appropriate side.
In this case since there are 8 oxygen atoms on
the reactant side which requires adding 8 water
molecules to the product side of the equation.
Now the hydrogen atoms need to be balanced by
adding 16 H to the reactant side.
95
Balancing REDOX EquationsbyThe Half Reaction
Method
96
Half Reaction Steps
1. Write separate equations (Half-reactions) for
oxidized and reduced substances. 2. For each
half-reaction balance all elements, except
hydrogen and oxygen a. Balance oxygen using
H2O b. Balance hydrogen using H c. Balance
charge in each half-reaction by adding electrons
(reduction), or removing electrons
(oxidation) to the appropriate half reaction. 3.
Multiply each half-reaction by an integer so that
the number of electrons lost equal the number of
electrons gained a. Add half-reactions, and
simplify b. For basic reactions add the same
number of OH- ions to both sides of the equation
as there are H ions. c. Combine H and OH- ions
to make water d. Simplify again if necessary.
97
Half Reaction Example
MnO4- Fe2 ? Mn2
Fe3
98
Half Reaction Example
MnO4- Fe2 ? Mn2
Fe3
Step 1, Write half reactions
99
Half Reaction Example
MnO4- Fe2 ? Mn2
Fe3
Step 1, Write half reactions
MnO4-
?
Mn2
?
Fe3
Fe2
100
Half Reaction Example
MnO4- Fe2 ? Mn2
Fe3
Step 2a, Balance Oxygen by adding water.
MnO4-
?
Mn2
4 H2O
?
Fe3
Fe2
101
Half Reaction Example
MnO4- Fe2 ? Mn2
Fe3
Step 2b, Balance hydrogen by adding H.
MnO4-
?
Mn2
4 H2O
8 H
?
Fe3
Fe2
102
Half Reaction Example
MnO4- Fe2 ? Mn2
Fe3
Step 2c, Balance charge by adding/removing es
MnO4-
?
Mn2
4 H2O
8 H
?
Fe3
Fe2
In the top half equation the reactants have 7
and the products 2, adding 5 es to the reactant
side gives 2 on both sides. In the bottom half
equation the reactants have 2 and the products
have 2, removing 1 e from the reactant side
gives 2 on both sides.
103
Half Reaction Example
MnO4- Fe2 ? Mn2
Fe3
Step 2c, Balance charge by adding/removing es
MnO4-
?
Mn2
4 H2O
8 H
5e-
?
Fe3
Fe2
In the top half equation the reactants have 7
and the products 2, adding 5 es to the reactant
side gives 2 on both sides. In the bottom half
equation the reactants have 2 and the products
have 2, removing 1 e from the reactant side
gives 2 on both sides.
104
Half Reaction Example
MnO4- Fe2 ? Mn2
Fe3
Step 2c, Balance charge by adding/removing es
MnO4-
?
Mn2
4 H2O
8 H
5e-
?
Fe3
- e-
Fe2
In the top half equation the reactants have 7
and the products 2, adding 5 es to the reactant
side gives 2 on both sides. In the bottom half
equation the reactants have 2 and the products
have 2, removing 1 e from the reactant side
gives 2 on both sides.
105
Half Reaction Example
MnO4- Fe2 ? Mn2
Fe3
Step 2c, Balance charge by adding/removing es
MnO4-
?
Mn2
4 H2O
8 H
5e-
?
Fe3
- e-
Fe2
In the top half equation the reactants have 7
and the products 2, adding 5 es to the reactant
side gives 2 on both sides. In the bottom half
equation the reactants have 2 and the products
have 2, removing 1 e from the reactant side
gives 2 on both sides.
106
Half Reaction Example
MnO4- Fe2 ? Mn2
Fe3
Step 3, The common denominator between 5 and 1 is
5. Multiply the bottom half equation by 5
MnO4-
?
Mn2
4 H2O
8 H
5e-
?
Fe3
- e-
Fe2
107
Half Reaction Example
MnO4- Fe2 ? Mn2
Fe3
Step 4, Add the two half equations together
MnO4-
?
Mn2
4 H2O
8 H
5e-
)
?
Fe3
- e-
5(
Fe2
108
Half Reaction Example
MnO4- Fe2 ? Mn2
Fe3
Step 4, Add the two half equations together
MnO4-
?
Mn2
4 H2O
8 H
5e-
)
?
Fe3
- e-
5(
Fe2
?
8 H
MnO4-
5 Fe3
5 Fe2
Mn2
4 H2O
109
Other REDOX Examples
HNO2 Cr2O72- ? Cr2
NO3- (acidic) CN- MnO4- ? CNO-
MnO2 (basic) Al(s) OH- (aq) ?
Al(OH)4- (aq) H2 (g) (acidic or
basic)
110
Real Life Examples of REDOX

• REDOX reactions can be used to generate
  electricity.
• REDOX reactions can be used to protect metals
  from oxidation.
• REDOX reactions can be used to plate metals on to
  other metals or surfaces.

111
The Chemical Package
About Packages

• The baker uses a package called the dozen. All
  dozen packages contain 12 objects.
• The stationary store uses a package called a
  ream, which contains 500 sheets of paper.
• So what is the chemistry package?

112
The Chemical Package
About Packages

• The baker uses a package called the dozen. All
  dozen packages contain 12 objects.
• The stationary store uses a package called a
  ream, which contains 500 sheets of paper.
• So what is the chemistry package? Well, it is
  called the mole (Latin for heap).

Each of the above packages contain a number of
objects that are convenient to work with, for
that particular discipline.
113
The Mole
A mole contains 6.022X1023 particles, which is
the number of carbon-12 atoms that will give a
mass of 12.00 grams, which is a convenient number
of atoms to work with in the chemistry laboratory.
The atomic weights listed on the periodic chart
are the weights of a mole of atoms. For example
a mole of hydrogen atoms weighs 1.00797 g and a
mole of carbon atoms weighs 12.01 g which are
weighted averages of the natural abundance of
isotopes for that element.
114
Moles of Objects
Suppose we order a mole of marshmallows for a
chemistry party. How much space here at Central
would be required to store the marshmallows?
115
Moles of Objects
Suppose we order a mole of marshmallows for a
chemistry party. How much space here at Central
would be required to store the marshmallows? Would
cover the entire 50 states 60 miles deep
116
Moles of Objects
Suppose we order a mole of marshmallows for a
chemistry party. How much space here at Central
would be required to store the marshmallows? Would
cover the entire 50 states 60 miles deep
How about a mole of computer paper instead of a
ream of computer paper, how far would that
stretch?
117
Moles of Objects
Suppose we order a mole of marshmallows for a
chemistry party. How much space here at Central
would be required to store the marshmallows? Would
cover the entire 50 states 60 miles deep
How about a mole of computer paper instead of a
ream of computer paper, how far would that
stretch? Way past the planet Pluto!
118
Formula Weight Calculation
To calculate the molar mass of a compound we sum
together the atomic weights of the atoms that
make up the formula of the compound. This is
called the formula weight (MW, M).
Formula weights are the sum of atomic weights of
atoms making up the formula.
The following outlines how to find the formula
weight of water
symbol
weight
number

2.02
1.01
2
H
X
16.0
1

X
O
16.0
g/mole
18.0
119
Percent Composition

• Find the formula weight and the percent
  composition of
• glucose (C6H12O6)

weight
number
symbol
72.0
6

x
12.0
C
12

12.12
x
H
1.01
96.0
O
x

16.0
6
180.1 g/mole
72.0
40.0 C
X
C
180.1
12.12
6.73 H
X
H
180.1
96.0
53.3 O
X
O
180.1
120
Mole Concepts
A mole of glucose (C6H12O6) contains 6.022 X 1023
molecules of glucose. And 6 X 6.022 X 1023 atoms
of C. Since a mole is 6.022 X 1023 particles
then a mole of glucose must contain 6 moles of C
atoms. How many moles of hydrogen atoms are
contained in a mole of glucose? In 5 moles of
H2SO4 how many moles of oxygen atoms is there?
121
Mole Concepts
A mole of glucose (C6H12O6) contains 6.022 X 1023
molecules of glucose. And 6 X 6.022 X 1023 atoms
of C. Since a mole is 6.022 X 1023 particles
then a mole of glucose must contain 6 moles of C
atoms. How many moles of hydrogen atoms are
contained in a mole of glucose? 12 Moles of
hydrogen. How many moles of oxygen and hydrogen
are in one mole of H2O contains
122
Mole Concepts
A mole of glucose (C6H12O6) contains 6.022 X 1023
molecules of glucose. And 6 X 6.022 X 1023 atoms
of C. Since a mole is 6.022 X 1023 particles
then a mole of glucose must contain 6 moles of C
atoms. How many moles of hydrogen atoms are
contained in a mole of glucose? 12 Moles of
hydrogen. How many moles of oxygen and hydrogen
are in one mole of H2O contains One mole of
oxygen atoms Two moles of hydrogen atoms
123
Mole Concepts
A mole of glucose (C6H12O6) contains 6.022 X 1023
molecules of glucose. And 6 X 6.022 X 1023 atoms
of C. Since a mole is 6.022 X 1023 particles
then a mole of glucose must contain 6 moles of C
atoms. How many moles of hydrogen atoms are
contained in a mole of glucose? 12 Moles of
hydrogen. How many moles of oxygen and hydrogen
are in one mole of H2O contains One mole of
oxygen atoms Two moles of hydrogen atoms In 5
moles of H2SO4 how many moles of oxygen atoms is
there?
124
Mole Concepts
A mole of glucose (C6H12O6) contains 6.022 X 1023
molecules of glucose. And 6 X 6.022 X 1023 atoms
of C. Since a mole is 6.022 X 1023 particles
then a mole of glucose must contain 6 moles of C
atoms. How many moles of hydrogen atoms are
contained in a mole of glucose? 12 Moles of
hydrogen. How many moles of oxygen and hydrogen
are in one mole of H2O contains One mole of
oxygen atoms Two moles of hydrogen atoms In 5
moles of H2SO4 how many moles of oxygen atoms is
there?
20 moles of O atoms.
125
Mole Conversions
In 50.0g of H2SO4 how many moles of sulfuric acid
are there?
50.0g of H2SO4
126
Mole Conversions
In 50.0g of H2SO4 how many moles of oxygen atoms
are there?
mole H2SO4
50.0g of H2SO4

98.0g of H2SO4
127
Mole Conversions
In 50.0g of H2SO4 how many moles of oxygen atoms
are there?
mole H2SO4
50.0g of H2SO4
0.510 mole H2SO4
98.0g of H2SO4
128
Mole Conversions
In 50.0g of H2SO4 how many moles of oxygen atoms
are there?
129
Mole Conversions
In 50.0g of H2SO4 how many moles of oxygen atoms
are there?
50.0g of H2SO4

130
Mole Conversions
In 50.0g of H2SO4 how many moles of oxygen atoms
are there?
mole H2SO4
50.0g of H2SO4

98.0g of H2SO4
131
Mole Conversions
In 50.0g of H2SO4 how many moles of oxygen atoms
are there?
mole H2SO4
50.0g of H2SO4
4mole O

mole H2SO4
98.0g of H2SO4
132
Mole Conversions
In 50.0g of H2SO4 how many moles of oxygen atoms
are there?
mole H2SO4
50.0g of H2SO4
4mole O

2.04 mole O
mole H2SO4
98.0g of H2SO4
133
Mole Conversions
In 5 moles of H2SO4 how many atoms of oxygen are
present?
134
Mole Conversions
In 5 moles of H2SO4 how many atoms of oxygen are
present?
5 moles H2SO4

135
Mole Conversions
In 5 moles of H2SO4 how many atoms of oxygen are
present?
5 moles H2SO4
4 mole O
mole H2SO4
136
Mole Conversions
In 5 moles of H2SO4 how many atoms of oxygen are
present?
5 moles H2SO4
6.02 x 1023 atoms O
4 mole O

mole O
mole H2SO4
1.20 x 1025 atoms
137
Empirical Formulas

• Empirical formula is the smallest whole number
  ratio between atoms and can be calculated from
  the percent composition.
• Molecular formulas happen to be the exact number
  of atoms making up a molecule, and may or may no
  be the simplest whole number ratio. Molecular
  formulas are whole number multiples of the
  empirical formula.

138
Empirical Formula Steps

• Assume 100 g of compound.
• Convert percent to a mass number.
• Convert the mass to moles.
• Divide each mole number by the smallest mole
  number.
• Rounding
• If the decimal is 0.1, then drop the decimals
• If the decimal is 0.9, then round up.
• All other decimal need to be multiplied by a
  whole number until roundable.

139
Empirical Formula Example

• A compound is composed of 75.0 C and 15.0 H.
  Find its empirical formula.

Step 1 Assume 100 g of compound
75.0 g C
25.0 g H
140
Empirical Formula Example

• A compound is composed of 75.0 C and 15.0 H.
  Find its empirical formula.

Step 2 Convert grams to moles.
Mole C
75.0 g C
6.225 mole C
12.01 g
25.0 g H
Mole H
24.802 mole
1.008 g H
141
Empirical Formula Example

• A compound is composed of 75.0 C and 15.0 H.
  Find its empirical formula.

Step 3 Divide each mole number by the smallest.
Mole C
75.0 g C
6.225 mole C
12.01 g
25.0 g H
Mole H
24.802 mole
1.008 g H
24.802
6.225
1.00
3.98
6.225
6.225
142
Empirical Formula Example

• A compound is composed of 75.0 C and 15.0 H.
  Find its empirical formula.

Step 4 Rounding Decimal 0.1, drop decimals
Mole C
75.0 g C
6.225 mole C
12.01 g
25.0 g H
Mole H
24.802 mole
1.008 g H
24.802
6.225
1.00
3.98
6.225
6.225
143
Empirical Formula Example

• A compound is composed of 75.0 C and 15.0 H.
  Find its empirical formula.

Step 4 Rounding Decimal 0.1, drop decimals
Mole C
75.0 g C
6.225 mole C
12.01 g
25.0 g H
Mole H
24.802 mole
1.008 g H
24.802
6.225
1 C
3.98
6.225
6.225
144
Empirical Formula Example

• A compound is composed of 75.0 C and 15.0 H.
  Find its empirical formula.

Step 4 Rounding Decimal 0.9, round up
Mole C
75.0 g C
6.225 mole C
12.01 g
25.0 g H
Mole H
24.802 mole
1.008 g H
24.802
6.225
1 C
3.98
6.225
6.225
145
Empirical Formula Example

• A compound is composed of 75.0 C and 15.0 H.
  Find its empirical formula.

Step 4 Rounding Decimal 0.9, round up
Mole C
75.0 g C
6.225 mole C
12.01 g
25.0 g H
Mole H
24.802 mole
1.008 g H
24.802
6.225
1 C
3.98
6.225
6.225
146
Empirical Formula Example

• A compound is composed of 75.0 C and 15.0 H.
  Find its empirical formula.

Step 4 Rounding Decimal