John E. McMurry • Robert C. Fay
General Chemistry: Atoms First
Chapter 6
Mass Relationships in Chemical
Reactions
Lecture Notes
Alan D. Earhart
Southeast Community College • Lincoln, NE
Copyright © 2010 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
Chapter 6/2
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
Chapter 6/3
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)
Chapter 6/4
Balancing Chemical Equations
3. Reduce the coefficients to their smallest wholenumber values, if necessary, by dividing them
all by a common divisor.
4H2(g) + 2O2(g)
4H2O(l)
divide all by 2
2H2(g) + O2(g)
2H2O(l)
Chapter 6/5
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
Chapter 6/6
Chemical Symbols on a
Different Level
2H2(g) + O2(g)
microscopic:
2H2O(l)
2 molecules of hydrogen gas react with
1 molecule of oxygen gas to yield 2
molecules of liquid water.
Chapter 6/7
Chemical Symbols on a
Different Level
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.
Chapter 6/8
Chemical Arithmetic:
Stoichiometry
Chapter 6/9
Chemical Arithmetic:
Stoichiometry
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.
C2H4:
HCl:
2(12.0 amu) + 4(1.0 amu) = 28.0 amu
1.0 amu + 35.5 amu = 36.5 amu
Chapter 6/10
Chemical Arithmetic:
Stoichiometry
One mole of any substance is equivalent to its
molecular or formula mass.
C2H4:
1 mole = 28.0 g
6.022 x 1023 molecules = 28.0 g
HCl:
1 mole = 36.5 g
6.022 x 1023 molecules = 36.5 g
Chapter 6/11
Chemical Arithmetic:
Stoichiometry
How many moles of chlorine gas, Cl2, are in 25.0 g?
25.0 g Cl2
x
1 mol Cl2
70.9 g Cl2
= 0.353 mol Cl2
Chapter 6/12
Chemical Arithmetic:
Stoichiometry
How many grams of sodium hypochlorite, NaOCl, are in
0.705 mol?
0.705 mol NaOCl
40.0 g NaOCl
x
1 mol NaOCl
= 28.2 g NaOCl
Chapter 6/13
Chemical Arithmetic:
Stoichiometry
Stoichiometry: The relative proportions in which
elements form compounds or in which substances
react.
aA + bB
Grams of
A
Molar Mass
of A
Moles of
A
cC + dD
Moles of
B
Mole Ratio
Between A
and B
(Coefficients)
Grams of
B
Molar Mass
of B
Chapter 6/14
Chemical Arithmetic:
Stoichiometry
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
Molar Mass
Moles of
Cl2
Mole Ratio
Moles of
NaOH
Grams of
NaOH
Molar Mass
Chapter 6/15
Chemical Arithmetic:
Stoichiometry
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
Chapter 6/16
Yields of Chemical Reactions
Actual Yield: The amount actually formed in a reaction.
Theoretical Yield: The amount predicted by calculations.
Percent Yield =
Actual yield of product
Theoretical yield of product
x 100%
Chapter 6/17
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.
Chapter 6/18
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)
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.
Chapter 6/19
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?
Chapter 6/20
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)
Chapter 6/21
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?
Chapter 6/22
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
Chapter 6/23
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
Chapter 6/24
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
= 104 g LiOH
Chapter 6/25
Concentrations of Reactants in
Solution: Molarity
Molarity (M): 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.
Chapter 6/26
Chapter 6/27
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
Chapter 6/28
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
Chapter 6/29
Chapter 6/30
Diluting Concentrated
Solutions
concentrated solution + solvent
dilute solution
initial
final
Mi x Vi = Mf x Vf
Since the number of moles of solute remains constant,
all that changes is the volume of solution by adding
more solvent.
Chapter 6/31
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?
Vi =
Mi = 18.0 M
Mf = 0.500 M
Vi = ? mL
Vf = 250.0 mL
Mf
Mi
x
Vf
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.
Chapter 6/32
Solution Stoichiometry
aA + bB
Volume of
Solution of A
Molarity of
A
Moles of
A
cC + dD
Moles of
B
Mole Ratio
Between A
and B
(Coefficients)
Volume of
Solution of B
Molar Mass
of B
Chapter 6/33
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
Molarity of
H2SO4
Moles of
H2SO4
Na2SO4(aq) + 2H2O(l)
Moles of
NaOH
Mole Ratio
Between H2SO4
and NaOH
Volume of
Solution of NaOH
Molarity of
NaOH
Chapter 6/34
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)
Chapter 6/35
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?
Chapter 6/36
Titration
buret
Erlenmeyer
flask
standard solution
(known concentration)
unknown concentration solution
An indicator is added which changes
color once the reaction is complete
Chapter 6/37
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
Molarity of
NaOH
Moles of
NaOH
NaCl(aq) + 2H2O(l)
Moles of
HCl
Mole Ratio
Between NaOH
and HCl
Volume of
Solution of HCl
Molarity of
HCl
Chapter 6/39
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
Chapter 6/40
Percent Composition and
Empirical Formulas
Percent Composition: Expressed by identifying the
elements present and giving the mass percent of each.
Empirical Formula: It tells the smallest whole-number
ratios of 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
Chapter 6/41
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
Molar
masses
Moles
Mole ratios
Subscripts
Relative
mole ratios
Chapter 6/42
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
Chapter 6/43
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
Chapter 6/44
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
Chapter 6/45