John E. McMurry • Robert C. Fay
C H E M I S T R Y
Fifth Edition
Chapter 3
Formulas, Equations, and Moles
Lecture Notes
Alan D. Earhart
Southeast Community College • Lincoln, NE
Copyright © 2008 Pearson Prentice Hall, Inc.
Balancing Chemical Equations
A balanced chemical equation shows that the law of
conservation of mass is adhered to.
In a balanced chemical equation, the numbers and
kinds of atoms on both sides of the reaction arrow are
identical.
2Na(s) + Cl2(g)
2NaCl(s)
left side:
right side:
2 Na
2 Cl
2 Na
2 Cl
Copyright © 2008 Pearson Prentice Hall, Inc.
Chapter 3/2
Balancing Chemical Equations
Hg(NO3)2(aq) + 2KI(aq)
HgI2(s) + 2KNO3(aq)
left side:
right side:
1 Hg
2N
6O
2K
2I
1 Hg
2I
2K
2N
6O
Copyright © 2008 Pearson Prentice Hall, Inc.
Chapter 3/3
Balancing Chemical Equations
1. Write the unbalanced equation using the correct
chemical formula for each reactant and product.
H2(g) + O2(g)
H2O(l)
2. Find suitable coefficients—the numbers placed
before formulas to indicate how many formula
units of each substance are required to balance
the equation.
2H2(g) + O2(g)
2H2O(l)
Copyright © 2008 Pearson Prentice Hall, Inc.
Chapter 3/4
Balancing Chemical Equations
3. Reduce the coefficients to their smallest wholenumber values, if necessary, by dividing them
all by a common denominator.
4H2(g) + 2O2(g)
4H2O(l)
divide all by 2
2H2(g) + O2(g)
2H2O(l)
Copyright © 2008 Pearson Prentice Hall, Inc.
Chapter 3/5
Balancing Chemical Equations
4. Check your answer by making sure that the
numbers and kinds of atoms are the same on
both sides of the equation.
2H2(g) + O2(g)
2H2O(l)
left side:
right side:
4H
2O
4H
2O
Copyright © 2008 Pearson Prentice Hall, Inc.
Chapter 3/6
Balancing Chemical Equations
Do not change subscripts when you balance a
chemical equation. You are only allowed to change
the coefficients.
H2(g) + O2(g)
2H2(g) + O2(g)
2H2O(l)
Balanced properly
H2O(l)
unbalanced
H2(g) + O2(g)
H2O2(l)
Chemical equation changed!
Copyright © 2008 Pearson Prentice Hall, Inc.
Chapter 3/7
Chemical Symbols on Different
Levels
2H2(g) + O2(g)
2H2O(l)
microscopic:
2 molecules of hydrogen gas react with
1 molecule of oxygen gas to yield 2
molecules of liquid water.
macroscopic:
0.56 kg of hydrogen gas react with
4.44 kg of oxygen gas to yield 5.00 kg
of liquid water.
How can we relate the two to each other?
Copyright © 2008 Pearson Prentice Hall, Inc.
Chapter 3/8
Avogadro’s Number and the
Mole
Molecular Mass: Sum of atomic masses of all atoms
in a molecule.
Formula Mass: Sum of atomic masses of all atoms in
a formula unit of any compound, molecular or ionic.
HCl:
C2H4:
1.0 amu + 35.5 amu = 36.5 amu
2(12.0 amu) + 4(1.0 amu) = 28.0 amu
Copyright © 2008 Pearson Prentice Hall, Inc.
Chapter 3/9
Avogadro’s Number and the
Mole
Avogadro’s Number (NA): One mole of any
substance contains 6.022 x 1023 formula units.
One mole of any substance is equivalent to its
molecular or formula mass.
HCl:
1 mole = 36.5 g
6.022 x 1023 molecules = 36.5 g
C2H4:
1 mole = 28.0 g
6.022 x 1023 molecules = 28.0 g
Copyright © 2008 Pearson Prentice Hall, Inc.
Chapter 3/10
Avogadro’s Number and the
Mole
Avogadro’s Number (NA): One mole of any
substance contains 6.022 x 1023 formula units.
One mole of any substance is equivalent to its
molecular or formula mass.
HCl:
1 mole = 36.5 g
C2H4:
1 mole = 28.0 g
Copyright © 2008 Pearson Prentice Hall, Inc.
Chapter 3/11
Stoichiometry: Chemical
Arithmetic
Stoichiometry: The relative proportions in which
elements form compounds or in which substances
react.
aA + bB
Grams of
A
Moles of
A
Molar Mass
of A
cC + dD
Moles of
B
Mole Ratio
Between A
and B
(Coefficients)
Grams of
B
Molar Mass
of B
Copyright © 2008 Pearson Prentice Hall, Inc.
Chapter 3/12
Stoichiometry: Chemical
Arithmetic
Aqueous solutions of sodium hypochlorite (NaOCl), best
known as household bleach, are prepared by reaction of
sodium hydroxide with chlorine gas:
2NaOH(aq) + Cl2(g)
NaOCl(aq) + NaCl(aq) + H2O(l)
How many grams of NaOH are needed to react with 25.0 g
Cl2?
Grams of
Cl2
Moles of
Cl2
Molar Mass
Mole Ratio
Moles of
NaOH
Grams of
NaOH
Molar Mass
Copyright © 2008 Pearson Prentice Hall, Inc.
Chapter 3/13
Stoichiometry: Chemical
Arithmetic
Aqueous solutions of sodium hypochlorite (NaOCl), best
known as household bleach, are prepared by reaction of
sodium hydroxide with chlorine gas:
2NaOH(aq) + Cl2(g)
NaOCl(aq) + NaCl(aq) + H2O(l)
How many grams of NaOH are needed to react with 25.0 g
Cl2?
25.0 g Cl2
x
1 mol Cl2
70.9 g Cl2
2 mol NaOH
x
1 mol Cl2
40.0 g NaOH
x
1 mol NaOH
= 28.2 g NaOH
Copyright © 2008 Pearson Prentice Hall, Inc.
Chapter 3/14
Yields of Chemical Reactions
Actual Yield: The amount actually formed in a reaction.
Theoretical Yield: The amount predicted by calculations.
Percent Yield =
actual yield
theoretical yield
x 100%
Copyright © 2008 Pearson Prentice Hall, Inc.
Chapter 3/15
Reactions with Limiting
Amounts of Reactants
Limiting Reactant: The reactant that is present in
limiting amount. The extent to which a chemical
reaction takes place depends on the limiting reactant.
Excess Reactant: Any of the other reactants still
present after determination of the limiting reactant.
Copyright © 2008 Pearson Prentice Hall, Inc.
Chapter 3/16
Reactions with Limiting
Amounts of Reactants
At a high temperature, ethylene oxide reacts with water to
form ethylene glycol which is an automobile antifreeze and
a starting material in the preparation of polyester polymers:
C2H4O(aq)
+
H2O(l)
C2H6O2(l)
Because water is so cheap and abundant, it is used in
excess when compared to ethylene oxide. This ensures
that all of the relatively expensive ethylene oxide is entirely
consumed.
Copyright © 2008 Pearson Prentice Hall, Inc.
Chapter 3/17
Reactions with Limiting
Amounts of Reactants
At a high temperature, ethylene oxide reacts with water to
form ethylene glycol which is an automobile antifreeze and
a starting material in the preparation of polyester polymers:
C2H4O(aq)
+
H2O(l)
C2H6O2(l)
If 3 moles of ethylene oxide react with 5 moles of water,
which reactant is limiting and which reactant is present in
excess?
Copyright © 2008 Pearson Prentice Hall, Inc.
Chapter 3/18
Reactions with Limiting
Amounts of Reactants
At a high temperature, ethylene oxide reacts with water to
form ethylene glycol which is an automobile antifreeze and
a starting material in the preparation of polyester polymers:
C2H4O(aq)
+
H2O(l)
C2H6O2(l)
Copyright © 2008 Pearson Prentice Hall, Inc.
Chapter 3/19
Reactions with Limiting
Amounts of Reactants
Lithium oxide is used aboard the space shuttle to remove
water from the air supply according to the equation:
Li2O(s) + H2O(g)
2LiOH(s)
If 80.0 g of water are to be removed and 65.0 g of Li2O are
available, which reactant is limiting? How many grams of
excess reactant remain? How many grams of LiOH are
produced?
Copyright © 2008 Pearson Prentice Hall, Inc.
Chapter 3/20
Reactions with Limiting
Amounts of Reactants
Li2O(s) + H2O(g)
2LiOH(s)
Which reactant is limiting?
Amount of H2O that will react with 65.0 g Li2O:
65.0 g Li2O
x
1 mol Li2O
29.9 g Li2O
x
1 mol H2O
1 mol Li2O
= 2.17 moles H2O
Amount of H2O given:
80.0 g H2O
x
1 mol H2O
18.0 g H2O
= 4.44 moles H2O
Li2O is limiting
Copyright © 2008 Pearson Prentice Hall, Inc.
Chapter 3/21
Reactions with Limiting
Amounts of Reactants
Li2O(s) + H2O(g)
2LiOH(s)
How many grams of excess H2O remain?
2.17 mol H2O
x
18.0 g H2O
1 mol H2O
= 39.1 g H2O (consumed)
80.0 g H2O - 39.1 g H2O = 40.9 g H2O
initial
consumed
remaining
Copyright © 2008 Pearson Prentice Hall, Inc.
Chapter 3/22
Reactions with Limiting
Amounts of Reactants
Li2O(s) + H2O(g)
2LiOH(s)
How many grams of LiOH are produced?
2.17 mol H2O
2 mol LiOH
x
1 mol H2O
23.9 g LiOH
x
1 mol LiOH
Copyright © 2008 Pearson Prentice Hall, Inc.
= 104 g LiOH
Chapter 3/23
Concentrations of Reactants in
Solution: Molarity
Molarity: The number of moles of a substance
dissolved in each liter of solution. In practice, a
solution of known molarity is prepared by weighing an
appropriate amount of solute, placing it in a container
called a volumetric flask, and adding enough solvent
until an accurately calibrated final volume is reached.
Solution: A homogeneous mixture.
Solute: The dissolved substance in a solution.
Solvent: The major component in a solution.
Copyright © 2008 Pearson Prentice Hall, Inc.
Chapter 3/24
Concentrations of Reactants in
Solution: Molarity
Molarity converts between mole of solute and liters of
solution:
moles of solute
molarity =
liters of solution
1.00 mol of sodium chloride placed in enough water to
make 1.00 L of solution would have a concentration
equal to:
1.00 mol
1.00 L
= 1.00
mol
L
or 1.00 M
Copyright © 2008 Pearson Prentice Hall, Inc.
Chapter 3/25
Concentrations of Reactants in
Solution: Molarity
How many grams of solute would you use to prepare
1.50 L of 0.250 M glucose, C6H12O6?
Molar mass C6H12O6 = 180.0 g/mol
1.50 L 0.250 mol
x
= 0.275 mol
1L
0.275 mol
x
180.0 g
1 mol
= 49.5 g
Copyright © 2008 Pearson Prentice Hall, Inc.
Chapter 3/26
Diluting Concentrated
Solutions
concentrated solution + solvent
dilute solution
initial final
Mi Vi = Mf Vf
Since the number of moles of solute remains constant,
all that changes is the volume of solution by adding
more solvent.
Copyright © 2008 Pearson Prentice Hall, Inc.
Chapter 3/27
Diluting Concentrated
Solutions
Sulfuric acid is normally purchased at a concentration of
18.0 M. How would you prepare 250.0 mL of 0.500 M
aqueous H2SO4?
Mi = 18.0 M
Mf = 0.500 M
Vi = ? mL
Vf = 250.0 mL
Vi =
Mf Vf
Mi
0.500 M
=
18.0 M
250.0 mL
x
= 6.94 mL
Add 6.94 mL 18.0 M sulfuric acid to enough water to
make 250.0 mL of 0.500 M solution.
Copyright © 2008 Pearson Prentice Hall, Inc.
Chapter 3/28
Solution Stoichiometry
aA + bB
Volume of
Solution of A
Moles of
A
Molarity of
A
cC + dD
Moles of
B
Mole Ratio
Between A
and B
(Coefficients)
Volume of
Solution of B
Molar Mass
of B
Copyright © 2008 Pearson Prentice Hall, Inc.
Chapter 3/29
Solution Stoichiometry
What volume of 0.250 M H2SO4 is needed to react with
50.0 mL of 0.100 M NaOH?
H2SO4(aq) + 2NaOH(aq)
Volume of
Solution of H2SO4
Moles of
H2SO4
Molarity of
H2SO4
Na2SO4(aq) + 2H2O(l)
Moles of
NaOH
Mole Ratio
Between H2SO4
and NaOH
Volume of
Solution of NaOH
Molarity of
NaOH
Copyright © 2008 Pearson Prentice Hall, Inc.
Chapter 3/30
Solution Stoichiometry
H2SO4(aq) + 2NaOH(aq)
Na2SO4(aq) + 2H2O(l)
Moles of NaOH available:
50.0 mL NaOH
0.100 mol
x
1L
1L
x
1000 mL
= 0.00500 mol NaOH
Volume of H2SO4 needed:
0.00500 mol NaOH 1 mol H2SO4
1 L solution
1000 mL
x
x
x
2 mol NaOH 0.250 mol H2SO4
1L
10.0 mL solution (0.250 M H2SO4)
Copyright © 2008 Pearson Prentice Hall, Inc.
Chapter 3/31
Titration
Titration: A procedure for determining the concentration
of a solution by allowing a carefully measured volume to
react with a solution of another substance (the standard
solution) whose concentration is known.
HCl(aq) + NaOH(aq)
NaCl(aq) + 2H2O(l)
Once the reaction is complete you can calculate the
concentration of the unknown solution.
How can you tell when the reaction is complete?
Copyright © 2008 Pearson Prentice Hall, Inc.
Chapter 3/32
Titration
buret
Erlenmeyer
flask
standard solution
(known concentration)
unknown concentration solution
An indicator is added which changes
color once the reaction is complete
Copyright © 2008 Pearson Prentice Hall, Inc.
Chapter 3/33
Titration
48.6 mL of a 0.100 M NaOH solution is needed to react
with 20.0 mL of an unknown HCl concentration. What is
the concentration of the HCl solution?
HCl(aq) + NaOH(aq)
Volume of
Solution of NaOH
Moles of
NaOH
Molarity of
NaOH
NaCl(aq) + 2H2O(l)
Moles of
HCl
Mole Ratio
Between NaOH
and HCl
Volume of
Solution of HCl
Molarity of
HCl
Copyright © 2008 Pearson Prentice Hall, Inc.
Chapter 3/34
Titration
HCl(aq) + NaOH(aq)
NaCl(aq) + 2H2O(l)
Moles of NaOH available:
48.6 mL NaOH 0.100 mol
1L
x
x
= 0.00486 mol NaOH
1L
1000 mL
Moles of HCl reacted:
0.00486 mol NaOH 1 mol HCl
= 0.00486 mol HCl
x
1 mol NaOH
Concentration of HCl solution:
0.00486 mol HCl
20.0 mL solution
1000 mL
x
1L
= 0.243 M HCl
Copyright © 2008 Pearson Prentice Hall, Inc.
Chapter 3/35
Percent Composition and
Empirical Formulas
Percent Composition: Expressed by identifying the
elements present and giving the mass percent of each.
Empirical Formula: It tells only the ratios of the atoms in
a compound.
Molecular Formula: It tells the actual numbers of atoms
in a compound. It can be either the empirical formula or a
multiple of it.
multiple =
molecular mass
empirical formula mass
Copyright © 2008 Pearson Prentice Hall, Inc.
Chapter 3/36
Percent Composition and
Empirical Formulas
A colorless liquid has a composition of 84.1 % carbon
and 15.9 % hydrogen by mass. Determine the empirical
formula. Also, assuming the molar mass of this
compound is 114.2 g/mol, determine the molecular
formula of this compound.
Mass percents
Moles
Molar
masses
Mole ratios
Subscripts
Relative
mole ratios
Copyright © 2008 Pearson Prentice Hall, Inc.
Chapter 3/37
Percent Composition and
Empirical Formulas
Assume 100.0 g of the substance:
Mole of carbon:
84.1 g C
1 mol C
x
12.0 g C
= 7.01 mol C
Mole of hydrogen:
15.9 g H
1 mol H
x
1.0 g H
= 15.9 mol H
Copyright © 2008 Pearson Prentice Hall, Inc.
Chapter 3/38
Percent Composition and
Empirical Formulas
Empirical formula:
C7.01H15.9
C7.01H15.9 = C1H2.27
smallest value for the ratio
C1H2.27
7.01
7.01
C1x4H2.27x4 = C4H9
need whole numbers
Molecular formula:
multiple =
114.2
57.0
=2
C4x2H9x2 = C8H18
Copyright © 2008 Pearson Prentice Hall, Inc.
Chapter 3/39
Determining Empirical
Formulas: Elemental Analysis
Combustion Analysis: A compound of unknown
composition (containing a combination of carbon,
hydrogen, and possibly oxygen) is burned with oxygen to
produce the volatile combustion products CO2 and H2O,
which are separated and weighed by an automated
instrument called a gas chromatograph.
hydrocarbon + O2(g)
xCO2(g) + yH2O(g)
carbon
hydrogen
Copyright © 2008 Pearson Prentice Hall, Inc.
Chapter 3/40
Determining Molecular
Masses: Mass Spectrometry
Copyright © 2008 Pearson Prentice Hall, Inc.
Chapter 3/41