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Chapter #3
Chemical Composition
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
Types of Chemical Reactions
•
Combination reaction (synthesis)
– Elements for reactants
– Examples:
H2 + O2
N2 + H2
Al + O2
H2O
NH3
Al2O3
The Law of Conservation of matter,
states matter cannot be created nor
destroyed, that means equations must
be balanced.
Types of Chemical Reactions
Combination reaction Continued
Balance the first equation
H 2 + O2
H 2O
Note two oxygen atoms on the reactant side
and only one on the product side,
therefore place a two in front of water
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.
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.
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
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.
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.
Types of Chemical Reactions
• Decomposition Reaction
– Compounds form simpler compounds or
elements.
– Examples
H2O
H2 + O2
Types of Chemical Reactions
• Decomposition Reaction
– Compounds form simpler compounds or
elements.
– Examples
2H2O
H2 + O2
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
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
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
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
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
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
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
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
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
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
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
Types of Chemical Reactions
Double Replacement reactions contain
compounds as reactants.
HCl + Ca(OH)2
CaCl2 + HOH
Check formulas, and slide with Clyde when
necessary
Types of Chemical Reactions
Double Replacement reactions contain
compounds as reactants.
2HCl + Ca(OH)2 CaCl2 + 2HOH
Check formulas, and slide with Clyde when
necessary
Now Balance!
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?
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
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?
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
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
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
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
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?
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.
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.
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
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
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.
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.
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.
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
Yet Another Ionic Equation
Write the formula, ionic and net ionic equations
when aqueous sodium chloride combines with
aqueous calcium bromide.
Yet Another Ionic Equation
Write the formula, ionic and net ionic equations
when aqueous sodium chloride combines with
aqueous calcium bromide.
NaCl(aq) + CaBr2
Now balance
CaCl2 +
NaBr
Yet Another Ionic Equation
Write the formula, ionic and net ionic equations
when aqueous sodium chloride combines with
aqueous calcium bromide.
2 NaCl(aq) + CaBr2
Now balance
CaCl2
+
2 NaBr(aq)
Yet Another Ionic Equation
Write the formula, ionic and net ionic equations
when aqueous sodium chloride combines with
aqueous calcium bromide.
2 NaCl(aq) + CaBr2
Now balance
CaCl2(aq) + 2 NaBr(aq)
Yet Another Ionic Equation
Write the formula, ionic and net ionic equations
when aqueous sodium chloride combines with
aqueous calcium bromide.
2 NaCl(aq) + CaBr2
CaCl2(aq) + 2 NaBr(aq)
Now balance
2Na+(aq) + 2Cl-(aq) + Ca2+(aq) + 2Br-(aq)
Ca2+(aq) + 2Cl-(aq) + 2Na+(aq) +2 Br-(aq)
Ionic equation
Yet Another Ionic Equation
Write the formula, ionic and net ionic equations
when aqueous sodium chloride combines with
aqueous calcium bromide.
2 NaCl(aq) + CaBr2
CaCl2(aq) + 2 NaBr(aq)
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)
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:
0
0
H2 + O2
H2 O
What are the oxidation numbers of hydrogen and
oxygen?
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
H2 O
What are the oxidation numbers of hydrogen and
oxygen?
REDOX REACTIONS
0
0
H2 + O2
2(1+) 2- = 0
H2O
How about hydrogen and oxygen in water?
REDOX REACTIONS
0
0
H2 + O2
oxidized
reduced
2(1+) 2- = 0
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)
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.
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
1+
4(2-)=0
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-
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
1+ 7+ 4(2-)=0
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-
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
1+ 7+ 4(2-)=0
5+
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-
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
1+ 7+ 4(2-)=0
5+
3+
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-
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
1+ 7+ 4(2-)=0
5+
3+
1+
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-
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
1+ 7+ 4(2-)=0
5+
3+
1+
0
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-
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
1+ 7+ 4(2-)=0
5+
3+
1+
0
1HClO4, 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-
REDOX REACTIONS
3(2-)=2-
How about sulfur in SO3 2-
REDOX REACTIONS
4+ 3(2-)=2-
How about sulfur in SO3 212(1+) +6(2-)=0
How about carbon in C6H12O6
REDOX REACTIONS
4+ 3(2-)=2-
How about sulfur in SO3 20 + 12(1+) +6(2-)=0
How about carbon in C6H12O6
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
H2 O
What are the oxidation numbers of hydrogen and
oxygen?
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:
0
0
H2 + O2
H2 O
What are the oxidation numbers of hydrogen and
oxygen?
REDOX REACTIONS
0
0
H2 + O2
2(1-) 2- = 0
H2O
How about hydrogen and oxygen in water?
REDOX REACTIONS
0
0
H2 + O2
oxidized
2(1+) 2- = 0
H2O
reduced
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)
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.
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.
HNO3
+
Cu2O
→
Cu(NO3)2
+
NO
+
H 2O
Balancing REDOX Reactions
I. Oxidation Number Method
a.
b.
c.
d.
Assign oxidation numbers to each element
Determine the elements oxidized and reduced
Balance the atoms that are oxidized and reduced
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
HNO3 + Cu2O
→
Cu(NO3)2
+
NO
+
H2O
Balancing REDOX Reactions
I. Oxidation Number Method
a.
b.
c.
d.
Assign oxidation numbers to each element
Determine the elements oxidized and reduced
Balance the atoms that are oxidized and reduced
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+ 5+ 3(2-)=0 2(?)+2-=0
HNO3 + Cu2O →
Cu(NO3)2
+
NO
+
H2O
Balancing REDOX Reactions
I. Oxidation Number Method
a.
b.
c.
d.
Assign oxidation numbers to each element
Determine the elements oxidized and reduced
Balance the atoms that are oxidized and reduced
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+ 5+ 3(2-)=0 2(1+)+2-=0
HNO3 + Cu2O →
Cu(NO3)2
+
NO
+
H2O
Balancing Redox Reactions
I. Oxidation Number Method
a.
b.
c.
d.
Assign oxidation numbers to each element
Determine the elements oxidized and reduced
Balance the atoms that are oxidized and reduced
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+ 5+ 3(2-)=0 2(1+)+2-=0
HNO3 + Cu2O →
? + 2(1-)=0
Cu(NO3)2 +
NO
+
H2O
Balancing REDOX Reactions
I. Oxidation Number Method
a.
b.
c.
d.
Assign oxidation numbers to each element
Determine the elements oxidized and reduced
Balance the atoms that are oxidized and reduced
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+ 5+ 3(2-)=0 2(1+)+2-=0
HNO3 + Cu2O →
2+ + 2(1-)=0
Cu(NO3)2 +
NO
+
H2O
Balancing REDOX Reactions
I. Oxidation Number Method
a.
b.
c.
d.
Assign oxidation numbers to each element
Determine the elements oxidized and reduced
Balance the atoms that are oxidized and reduced
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+ 5+ 3(2-)=0 2(1+)+2-=0
HNO3 + Cu2O →
2(1-)=0
? + 2- =0
Cu(NO3)2 + NO + H2O
Balancing REDOX Reactions
I. Oxidation Number Method
a.
b.
c.
d.
Assign oxidation numbers to each element
Determine the elements oxidized and reduced
Balance the atoms that are oxidized and reduced
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+ 5+ 3(2-)=0 2(1+)+2-=0
HNO3 + Cu2O →
2(1-)=0
2 + 2- =0
Cu(NO3)2 + NO + H2O
Balancing REDOX Reactions
I. Oxidation Number Method
a.
b.
c.
d.
Assign oxidation numbers to each element
Determine the elements oxidized and reduced
Balance the atoms that are oxidized and reduced
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+ 5+ 3(2-)=0 2(1+)+2-=0
HNO3 + Cu2O →
oxidized
reduced
2+ 2(1-)=0
2 + 2- =0
Cu(NO3)2 + NO + H2O
Balancing REDOX Reactions
I. Oxidation Number Method
a.
b.
c.
d.
Assign oxidation numbers to each element
Determine the elements oxidized and reduced
Balance the atoms that are oxidized and reduced
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+ 5+ 3(2-)=0 2(1+)+2-=0 2 2(1 )=0
HNO3 + Cu2O → 2 Cu(NO3)2 + NO +
oxidized
reduced
H2O
Balancing REDOX Reactions
I. Oxidation Number Method
a.
b.
c.
d.
Assign oxidation numbers to each element
Determine the elements oxidized and reduced
Balance the atoms that are oxidized and reduced
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+ 5+ 3(2-)=0 2(1+)+2-=0 2 2(1 )=0
HNO3 + 3 Cu2O → 3 (2) Cu(NO3)2
Oxidized3( -2e)
Reduced 2(+3)e
2 + 2- =0
+ NO + H2O
Balancing REDOX Reactions
I. Oxidation Number Method
a.
b.
c.
d.
Assign oxidation numbers to each element
Determine the elements oxidized and reduced
Balance the atoms that are oxidized and reduced
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+ 5+ 3(2-)=0 2(1+)+2-=0 2 2(1 )=0
2HNO3 + 3 Cu2O → 3 (2) Cu(NO3)2
Oxidized3( -2e)
Reduced 2(+3)e
2 + 2- =0
+ 2NO +
H2 O
Balancing REDOX Reactions
I. Oxidation Number Method
a.
b.
c.
d.
Assign oxidation numbers to each element
Determine the elements oxidized and reduced
Balance the atoms that are oxidized and reduced
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+ 5+ 3(2-)=0 2(1+)+2-=0 2+ 2(1-)=0
2 + 2- =0
2HNO3
+
3 Cu2O
→ 3 (2) Cu(NO3)2
Oxidized3( -2e)
Reduced 2(+3)e
+
2NO + H2O
Balancing REDOX Reactions
I. Oxidation Number Method
a.
b.
c.
d.
Assign oxidation numbers to each element
Determine the elements oxidized and reduced
Balance the atoms that are oxidized and reduced
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+ 5+ 3(2-)=0 2(1+)+2-=0 2+ 2(1-)=0
2 + 2- =0
14HNO3
+
3 Cu2O
→ 3 (2) Cu(NO3)2
Oxidized3( -2e)
Reduced 2(+3)e
+
2NO + 7 H2O
Balancing REDOX Reactions
I. Oxidation Number Method
a.
b.
c.
d.
Assign oxidation numbers to each element
Determine the elements oxidized and reduced
Balance the atoms that are oxidized and reduced
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+ 5+ 3(2-)=0 2(1+)+2-=0 2+ 2(1-)=0
14 HNO3 +
3 Cu2O
→
6 Cu(NO3)2
Oxidized3( -2e)
Reduced 2(+3)e
+
2 NO + 7 H2O
OX # BALANCING EXAMPLE
MnO4 -
+
Cl-
→
Mn2+
+
Cl2
OX # BALANCING EXAMPLE
7+
MnO4 -
+
Cl-
→
Mn2+
+
Cl2
OX # BALANCING EXAMPLE
7+
MnO4 -
1-
+
Cl-
→
Mn2+
+
Cl2
OX # BALANCING EXAMPLE
7+
MnO4 -
1-
+
Cl-
→
Mn2+
0
+
Cl2
OX # BALANCING EXAMPLE
1-
7+
MnO4 -
+
Cl-
→ Mn2+
0
+
Cl2
OX # BALANCING EXAMPLE
1-
7+
MnO4 -
+
reduced
Cl-
→ Mn2+ +
oxidized
0
Cl2
OX # BALANCING EXAMPLE
1-
7+
MnO4 -
+
Cl-
→
Mn2+
0
+
oxidized
Cl2
reduced
Step C, balance atoms oxidized or reduced
OX # BALANCING EXAMPLE
1-
7+
MnO4 -
+
2 Cl-
→
Mn2+
0
+
oxidized
Cl2
reduced
Step C, balance atoms oxidized or reduced
OX # BALANCING EXAMPLE
1-
7+
MnO4 -
+
2 Cl-
→
- 2 e-
Mn2+ +
oxidized
0
Cl2
+ 5 e- reduced
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.
OX # BALANCING EXAMPLE
1-
7+
2 MnO4 -
+
5(2) Cl-
→
2 Mn2+ +
- 2 eoxidized
0
5 Cl2
+ 5 e- reduced
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.
OX # BALANCING EXAMPLE
1-
7+
2 MnO4 -
+
5(2)Cl-
→
2
- 2 e-
Mn2+
oxidized
0
+ 5 Cl2
+ 5 e- reduced
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.
OX # BALANCING EXAMPLE
7+
2
1-
MnO4- +10 Cl- →
2 Mn2+
- 2 e- oxidized
+ 5 e- reduced
+
0
5
Cl2+ 8H2O
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.
OX # BALANCING EXAMPLE
7+
1-
0
16 H++ 2 MnO4-+ 10 Cl- → 2 Mn2++ 5 Cl2 + 8H2O
- 2 e- oxidized
+ 5 e- reduced
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.
Balancing REDOX Equations
by
The Half Reaction Method
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.
Half Reaction Example
MnO4-
+
Fe2+
→
Mn2+
+
Fe3+
Half Reaction Example
MnO4-
+
Fe2+
→
Mn2+
Step 1, Write half reactions
+
Fe3+
Half Reaction Example
MnO4-
+
Fe2+
→
Mn2+
Step 1, Write half reactions
MnO4-
→
Mn2+
Fe2+
→
Fe3+
+
Fe3+
Half Reaction Example
MnO4-
+
Fe2+
→
Mn2+
+
Fe3+
Step 2a, Balance Oxygen by adding water.
MnO4-
→
Mn2+ + 4 H2O
Fe2+
→
Fe3+
Half Reaction Example
MnO4-
+
Fe2+
→
Mn2+
+
Fe3+
Step 2b, Balance hydrogen by adding H+.
8 H+ + MnO4-
→
Mn2+ + 4 H2O
Fe2+
→
Fe3+
Half Reaction Example
MnO4-
+
Fe2+
→
Mn2+
+
Fe3+
Step 2c, Balance charge by adding/removing e’s
8 H+ + MnO4-
→
Mn2+ + 4 H2O
Fe2+
→
Fe3+
In the top half equation the reactants have 7+ and the products
2+, adding 5 e’s 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.
Half Reaction Example
MnO4-
+
Fe2+
→
Mn2+
+
Fe3+
Step 2c, Balance charge by adding/removing e’s
8 H+ + MnO4-+ 5e-→
Fe2+
→
Mn2+ + 4 H2O
Fe3+
In the top half equation the reactants have 7+ and the products
2+, adding 5 e’s 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.
Half Reaction Example
MnO4-
+
Fe2+
→
Mn2+
+
Fe3+
Step 2c, Balance charge by adding/removing e’s
8 H+ + MnO4-+ 5e-→
Fe2+ - e →
Mn2+ + 4 H2O
Fe3+
In the top half equation the reactants have 7+ and the products
2+, adding 5 e’s 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.
Half Reaction Example
MnO4-
+
Fe2+
→
Mn2+
+
Fe3+
Step 2c, Balance charge by adding/removing e’s
8 H+ + MnO4-+ 5e-→
Fe2+ - e →
Mn2+ + 4 H2O
Fe3+
In the top half equation the reactants have 7+ and the products
2+, adding 5 e’s 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.
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
8 H+ + MnO4-+ 5e-→
Fe2+ - e →
Mn2+ + 4 H2O
Fe3+
Half Reaction Example
MnO4- + Fe2+ → Mn2+ + Fe3+
Step 4, Add the two half equations together
8 H+ + MnO4-+ 5e-→
2+ - e- →
Fe
5(
Mn2+ + 4 H2O
Fe3+)
Half Reaction Example
MnO4- + Fe2+ → Mn2+ + Fe3+
Step 4, Add the two half equations together
8 H+ + MnO4-+ 5e-→
2+ - e- →
Fe
5(
Mn2+ + 4 H2O
Fe3+)
8 H+ + MnO4- + 5 Fe2+ → 5 Fe3+ + Mn2+ + 4 H2O
Other REDOX Examples
HNO2
CN-
+
+
→
Cr2O72MnO4-
→
Cr2+
CNO-
+
+
NO3- (acidic)
MnO2 (basic)
Al(s) + OH- (aq) → Al(OH)4- (aq) + H2 (g) (acidic
or basic)
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.
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?
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.
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.
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?
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
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?
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!
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
H
O
number
weight
= 2.02
1.01 X
2
1
= 16.0
X
16.0
18.0 g/mole
Percent Composition
Find the formula weight and the percent composition of
glucose (C6H12O6)
symbol
C
H
O
%C =
weight
number
12.0 x 6 = 72.0
1.01 x 12 = 12.12
16.0 x 6 = 96.0
180.1 g/mole
72.0
X = 40.0 %C
180.1
%H = 12.12
180.1
96.0
%O =
180.1
X = 6.73 %H
X =
53.3 %O
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?
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:
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
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?
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.
Mole Conversions
In 50.0g of H2SO4 how many moles of sulfuric acid are there?
50.0g of H2SO4
Mole Conversions
In 50.0g of H2SO4 how many moles of sulfuric acid are there?
50.0g of H2SO4 mole H2SO4
98.0g of H2SO4
=
Mole Conversions
In 50.0g of H2SO4 how many moles of sulfuric acid are there?
50.0g of H2SO4 mole H2SO4
98.0g of H2SO4
= 0.510 mole H2SO4
Mole Conversions
In 50.0g of H2SO4 how many moles of oxygen atoms are there?
Mole Conversions
In 50.0g of H2SO4 how many moles of oxygen atoms are there?
50.0g of H2SO4
=
Mole Conversions
In 50.0g of H2SO4 how many moles of oxygen atoms are there?
50.0g of H2SO4 mole H2SO4
98.0g of H2SO4
=
Mole Conversions
In 50.0g of H2SO4 how many moles of oxygen atoms are there?
50.0g of H2SO4 mole H2SO4
4mole O
98.0g of H2SO4 mole H2SO4
=
Mole Conversions
In 50.0g of H2SO4 how many moles of oxygen atoms are there?
50.0g of H2SO4 mole H2SO4
4mole O
98.0g of H2SO4 mole H2SO4
= 2.04 mole O
Mole Conversions
In 5 moles of H2SO4 how many atoms of oxygen are present?
Mole Conversions
In 5 moles of H2SO4 how many atoms of oxygen are present?
5 moles H2SO4
=
Mole Conversions
In 5 moles of H2SO4 how many atoms of oxygen are present?
5 moles H2SO4 4 mole O
mole H2SO4
Mole Conversions
In 5 moles of H2SO4 how many atoms of oxygen are present?
5 moles H2SO4 4 mole O
6.02 x 1023 atoms O
=
mole
O
mole H2SO4
1.20 x 1025 atoms
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.
Empirical Formula Steps
1.
2.
3.
4.
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.
5. Rounding:
a. If the decimal is ≤ 0.1, then drop the decimals
b. If the decimal is ≥0.9, then round up.
c. All other decimal need to be multiplied by a whole
number until roundable.
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
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.
75.0 g C Mole C
= 6.225 mole C
12.01 g
25.0 g H Mole H
= 24.802 mole
1.008 g H
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.
75.0 g C Mole C
= 6.225 mole C
12.01 g
25.0 g H Mole H
= 24.802 mole
1.008 g H
6.225
= 1.00
6.225
24.802
6.225
= 3.98
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
75.0 g C Mole C
= 6.225 mole C
12.01 g
25.0 g H Mole H
= 24.802 mole
1.008 g H
6.225
= 1.00
6.225
24.802
6.225
= 3.98
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
75.0 g C Mole C
= 6.225 mole C
12.01 g
25.0 g H Mole H
= 24.802 mole
1.008 g H
6.225
=1C
6.225
24.802
6.225
= 3.98
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
75.0 g C Mole C
= 6.225 mole C
12.01 g
25.0 g H Mole H
= 24.802 mole
1.008 g H
6.225
=1C
6.225
24.802
6.225
= 3.98
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
75.0 g C Mole C
= 6.225 mole C
12.01 g
25.0 g H Mole H
= 24.802 mole
1.008 g H
6.225
=1C
6.225
24.802
6.225
= 3.98
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
75.0 g C Mole C
= 6.225 mole C
12.01 g
25.0 g H Mole H
= 24.802 mole
1.008 g H
6.225
=1C
6.225
24.802
6.225
=4H
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
75.0 g C Mole C
= 6.225 mole C
12.01 g
25.0 g H Mole H
= 24.802 mole
1.008 g H
6.225
=1C
6.225
24.802
6.225
Empirical Formula = CH4
=4H
Molecular Formulas
Empirical formula, is the smallest ratio
between atoms in a molecular or formula
unit.
Molecular formula, is the exact number of
atoms in a molecule; a whole number
multiple of an empirical formula
Possible Molecular Formulas
Assume an empirical formula of C3H5O
Empirical formula
Integer
C3H5O
1
C3H5O
2
C3H5O
3
C3H5O
4
C3H5O
5
Molecular Formula
C3H5O
Possible Molecular Formulas
Assume an empirical formula of C3H5O
Empirical formula
Integer
Molecular Formula
C3H5O
1
C3H5O
C3H5O
2
C6H10O
C3H5O
3
C3H5O
4
C3H5O
5
Possible Molecular Formulas
Assume an empirical formula of C3H5O
Empirical formula
Integer
Molecular Formula
C3H5O
1
C3H5O
C3H5O
2
C6H10O2
C3H5O
3
C9H15O3
C3H5O
4
C3H5O
5
Possible Molecular Formulas
Assume an empirical formula of C3H5O
Empirical formula
Integer
Molecular Formula
C3H5O
1
C3H5O
C3H5O
2
C6H10O2
C3H5O
3
C9H15O3
C3H5O
4
C12H20O4
C3H5O
5
C15H25O5
Sample Problem
Calculate the molecular formula of a molecule
composed of 83.7%C and 16.3% H, with a molar
mass of 86.0 g/mole
Sample Problem
Calculate the molecular formula of a molecule
composed of 83.7%C and 16.3% H, with a molar
mass of 86.0 g/mole
Step #1 Assume 100g of compound
83.6 g C
16.3 g H
Sample Problem
Calculate the molecular formula of a molecule
composed of 83.7%C and 16.3% H, with a molar
mass of 86.0 g/mole
Step #2 Convert grams to moles
83.6 g C mole
12.01 g C
16.3 g H mole
1.008 g H
Sample Problem
Calculate the molecular formula of a molecule
composed of 83.7%C and 16.3% H, with a molar
mass of 86.0 g/mole
Step #2 Convert grams to moles
83.6 g C mole
12.01 g C
16.3 g H mole
1.008 g H
= 6.961 mole
= 16.17 mole
Sample Problem
Calculate the molecular formula of a molecule
composed of 83.7%C and 16.3% H, with a molar
mass of 86.0 g/mole
Step #3 Divide each mole number by the smallest.
83.6 g C mole
12.01 g C
16.3 g H mole
1.008 g H
= 6.961 mole
= 16.17 mole
Sample Problem
Calculate the molecular formula of a molecule
composed of 83.7%C and 16.3% H, with a molar
mass of 86.0 g/mole
Step #3 Divide each mole number by the smallest.
83.6 g C mole
12.01 g C
16.3 g H mole
1.008 g H
6.961 = 1.00
= 6.961 mole
6.961
= 16.17 mole
= 2.32
Sample Problem
Calculate the molecular formula of a molecule
composed of 83.7%C and 16.3% H, with a molar
mass of 86.0 g/mole
Step #4 Round if---Not Roundable
6.961 = 1.00
83.6 g C mole
= 6.961 mole
6.961
12.01 g C
16.3 g H mole
= 16.17 mole
= 2.32
1.008 g H
Step #4, Multiply by an integer until roundable
1.00 X 3 = 3
Empirical formula C3H7
2.32 X 3 = 7
Molecular Formula Integer
Divide empirical weight into molecular weight
3x12 + 7x1 =43
43
2
86
Now multiply the empirical formula by 2
Molecular Formula Integer
Divide empirical weight into molecular weight
3x12 + 7x1 =43
43
2
86
Now multiply the empirical formula by 2
Molecular Formula is C6 H14
Stoichiometry
Sotichiometery is the process of converting
quantities of reactants or products to other
participants of a chemical or physical change using
the coefficients of a balanced equation.
STOICHIOMETRY
Stoichiometry is the use of balanced chemical
equations in the conversion process.
Example
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.
Example
Calculate the mass of water formed from 6.33 g
of hydrogen. A balanced equation is required.
2 H2 + O2
2 H2O
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 + O2
2H2O
6.33 g H2
STOICHIOMETRY
Stoichiometry is the use of balanced chemical
equations in the conversion process.
Example
Calculate the mass of water formed from 6.33 g
of hydrogen. A balanced equation is required.
2 H2 + O2
2 H2O
6.33 g H2 Mole H2
2.016 g H2O
STOICHIOMETRY
Stoichiometry is the use of balanced chemical
equations in the conversion process.
Example
Calculate the mass of water formed from 6.33 g
of hydrogen. A balanced equation is required.
2 H2 + O2
2 H2O
6.33 g H2 Mole H2
2 Mole H2O
2.016 g H2O 2 Mole H2
STOICHIOMETRY
Stoichiometry is the use of balanced chemical
equations in the conversion process.
Example
Calculate the mass of water formed from 6.33 g
of hydrogen. A balanced equation is required.
2 H2 + O2
2 H2O
6.33 g H2 Mole H2
2 Mole H2O 18.02 g H2O
2.016 g H2O 2 Mole H2 Mole H2O
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 + O2
2 H2O
6.33 g H2 Mole H2
2 Mole H2O 18.02 g H2O
= 56.6 g H2O
2.016 g H2O 2 Mole H2 Mole H2O
Excess and Limiting
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
Suppose, 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?
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!
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?
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?
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!
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.
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
2 H2O
10.0 g O2
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
32.0 g O2
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
32.0 g O2 mole O2
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
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?
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
Combustion Analysis
Empirical formulas of hydrocarbons can be
determined by combustion analysis. The complete
combustion of a hydrocarbon produces carbon
dioxide and water. Measuring the mass of the
carbon dioxide and water produced can give the
mass of hydrogen and carbon present in the
compound. Subtracting the mass of carbon and
hydrogen from the weight of the starting
hydrocarbon gives the mass of the oxygen.
Ascarite™ a commercial name for sodium or
potassium hydroxide absorbs between 0-1 ppm of
the carbon dioxide.
Combustion Analysis
Ascarite the weight increase of Ascarite gives the
mass of the carbon dioxide according to the following
equation.
CO2 + 2 KOH
K2CO3 +
2 H 2O
Vitamin C is essential for the prevention of scurvy.
Combustion of a 0.2000 g sample of this
hydrocarbon, which may or may not contain oxygen,
gave 0.2998 g of carbon dioxide and 0.0819 g of
water. What is the empirical formula of vitamin C?
Combustion Analysis
Vitamin C is essential for the prevention of scurvy.
Combustion of a 0.2000 g sample of this
hydrocarbon, which may or may not contain
oxygen, gave 0.2998 g of carbon dioxide and
0.0819 g of water. What is the empirical formula of
vitamin C?
First covert the mass of carbon dioxide and water
into grams of carbon and of carbon and hydrogen.
Subtract these masses from the sample weight, if
the difference is zero then vitamin C does not
contain any oxgen.
Combustion Analysis
0.2298 g CO2
Combustion Analysis
0.2298 g CO2 Mole CO2
44.010 g CO2
Combustion Analysis
0.2298 g CO2 Mole CO2
Mole C
44.010 g CO2 Mole CO2
Combustion Analysis
0.2298 g CO2 Mole CO2
Mole C
12.011 g C
44.010 g CO2 Mole CO2 Mole C
= 0.08182 g C
Combustion Analysis
0.2298 g CO2 Mole CO2
Mole C
12.011 g C
44.010 g CO2 Mole CO2 Mole C
0.0819 g H2O mole H2O
18.02 g H2O
2 mole H 1.008 g H
mole H2O mole H
= 0.08182 g C
= 0.00916 g H
Combustion Analysis
0.2298 g CO2 Mole CO2
Mole C
12.011 g C
44.010 g CO2 Mole CO2 Mole C
0.0819 g H2O mole H2O
18.02 g H2O
2 mole H 1.008 g H
mole H2O mole H
0.2000 - 0.08182 - 0.00916 = 0.1090 g O
= 0.08182 g C
= 0.00916 g H
Combustion Analysis
0.2298 g CO2 Mole CO2
Mole C
12.011 g C
44.010 g CO2 Mole CO2 Mole C
0.0819 g H2O mole H2O
2 mole H 1.008 g H
18.02 g H2O
mole H2O mole H
0.2000 - 0.08182 - 0.00916 = 0.1090 g O
0.08182 g C Mole C
12.011 g C
0.00916 g H mole H
1.008 g H
0.1090 g O Mole O
16.00 g O
= 0.006812 mole
= 0.00909 mole
= 0.006813 mole
= 0.08182 g C
= 0.00916 g H
Combustion Analysis
0.2298 g CO2 Mole CO2
Mole C
12.011 g C
44.010 g CO2 Mole CO2 Mole C
0.0819 g H2O mole H2O
2 mole H 1.008 g H
18.02 g H2O
mole H2O mole H
= 0.08182 g C
= 0.00916 g H
0.2000 - 0.08182 - 0.00916 = 0.1090 g O
0.006812
= 0.006812 mole 0.006812 = 1.000 X 3 = 3
12.011 g C
0.00916 g H mole H
0.00909
= 0.00909 mole
= 1.333 X 3 = 4
1.008 g H
0.006812
0.08182 g C Mole C
0.1090 g O Mole O
16.00 g O
= 0.006813 mole
0.006813
= 1.000 X 3 = 3
0.006812
Combustion Analysis
0.2298 g CO2 Mole CO2
Mole C
12.011 g C
44.010 g CO2 Mole CO2 Mole C
0.0819 g H2O mole H2O
2 mole H 1.008 g H
18.02 g H2O
mole H2O mole H
= 0.08182 g C
= 0.00916 g H
0.2000 - 0.08182 - 0.00916 = 0.1090 g O
0.006812
= 0.006812 mole 0.006812 = 1.000 X 3 = 3
12.011 g C
0.00916 g H mole H
0.00909
= 0.00909 mole
= 1.333 X 3 = 4
1.008 g H
0.006812
0.08182 g C Mole C
0.1090 g O Mole O
16.00 g O
= 0.006813 mole
0.006813
= 1.000 X 3 = 3
0.006812
C 3H 4O 3
The End