Chapter 4: The Mole
• Atomic mass provides a means to count
atoms by measuring the mass of a sample
• The periodic table on the inside cover of the
text gives atomic masses of the elements
• The mass of an atom is called its atomic
mass
• When using atomic masses, retain a
sufficient number of significant figures so
the atomic mass data contributes only
slightly to the uncertainty of the result
• The mass of the formula unit is called the
formula mass
• Formula masses are calculated the same
way as molecular masses
– For example the formula mass of calcium
oxide, CaO, is the mass of calcium (40.08) plus
the mass of oxygen (15.999) = 56.08
• One mole of a substance contains the same
number of formula units as the number of
atoms in exactly 12 g of carbon-12
• Counting formula units by moles is no
different than counting eggs by the dozen
(12 eggs) or pens by the gross (144 pens)
• Avogadro’s number is huge because atoms
and molecules are so small: a huge number
of them are needed to make a lab-sized
sample
• Avogadro’s number links moles and atoms,
or moles and molecules and provides an
easy way to link mass and atoms or
molecules
• The molecular mass allows counting of
molecules by mass
• The molecular mass is the sum of atomic
masses of the atoms in the compounds
formula
– For example the molar mass of water, H2O, is
twice the mass of hydrogen (1.008) plus the
mass of oxygen (15.999) = 18.015
• Strictly speaking, ionic compounds do not
have a “molecular mass” because they don’t
contain molecules
• One mole of a substance has a mass in
grams numerically equal to its formula mass
• The mass of one mole of a substance is also
called its molar mass
• One mole of any substance contains the
same number of formula units
• This number is called Avogadro’s number
or constant
1 mol formula units = 6.02 x 1023 formula units
• Using water (molar mass 18.015) as an
example:
1 mole H2O Ù 6.022 x 1023 molecules H2O
1 mole H2O Ù 18.015 g H2O
18.015 g H2O Ù 6.022 x 1023 molecules H2O
• Within chemical compounds, moles of
atoms always combine in the same ratio as
the individual atoms themselves so:
1 mole H2O Ù 2 mole H
1 mole H2O Ù 1 mole O
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• Stoichiometry is the study of the mass
relationships in chemical compounds and
reactions
• A common use for stoichiometry is to relate
the masses of reactants needed to make a
compound
• These calculations can be solved using the
factor-label method and equivalence
relations relating molecular masses and/or
formula masses
• Example: How many grams of iron are in a
15.0 g sample of iron(III) oxide?
• The usual form for describing the relative
masses of the elements in a compound is a
list of percentages by mass
• This is called the percentage composition
or percentage composition by mass
• The percentage by mass is the number of
grams of the element in 100 g of the
compound and can be calculated using:
• Example: A sample was analyzed and found
to contain 0.1417 g nitrogen and 0.4045 g
oxygen. What is the percentage composition
of this compound?
% element =
mass of element
mass of whole sample
× 100%
• Hydrogen peroxide consists of molecules
with the formula H2O2
• This is called the molecular formula
• The simplest formula for hydrogen peroxide
is HO and is called the empirical formula
• It is possible to calculate the empirical
formula for a compound from mass data
• The goal is to produce the simplest whole
number mole ratio between atoms
ANALYSIS: 15.0 g Fe2O3 Ù ? g Fe
LINKS: 1 mol Fe2O3 Ù 2 mol Fe
1 mol Fe2O3 Ù 159.7 g Fe2O3
1 mol Fe Ù 55.85 g Fe
SOLUTION:
1 mol Fe 2 O 3
2 mol Fe
15.0 g Fe 2 O 3 × 159
.9 g Fe 2 O 3 × 1 mol Fe 2 O 3
.85 g Fe
× 551 mol
Fe = 10.5 g Fe
ANALYSIS: Find sample mass and calculate %
LINKS: whole sample = 0.5462 g
SOLUTION:
%N=
0.1417 g N
0.5462 g sample
× 100% = 25.94 %
%O=
0.4045 g O
0.5462 g sample
× 100% = 74.06 %
• Example: A 2.012 g sample of a compound
contains 0.522 g of nitrogen and 1.490 g of
oxygen. Calculate its empirical formula
ANALYSIS: We need the simplest whole number
mole ratio between nitrogen and oxygen
SOLUTION:
1 mol N
0 . 522 g N × 14.01
g N = 0.0373 mol N
1 mol O
1 .490 g O × 15.999
g O = 0.0931 mol O
N 0.0373 O 0.0931 = N 1.00 O 2.50
0.0373
0.0373
N 1.00 × 2 O 2.50 × 2 = N 2.00 O 5.00 = N 2 O 5
2
• Empirical formulas may also be calculated
indirectly
• When a compound made only from carbon,
hydrogen, and oxygen burns completely in
pure oxygen only carbon dioxide and water
are produced
• This is called combustion
• Empirical formulas may be calculated from
the analysis of combustion information
12.011 g C
7.406 g CO 2 × 44.009
g CO 2 = 2.022 g C
2.0158 g H
4.512 g H 2 O × 18.015
= 0.5049 g H
g H 2O
total mass of C and H = 2.473 g
mass O = 5.217 g - 2.527g = 2.690 g
1 mol C
C : 2.022 g C × 12.011
g C = 0.1683 mol C
1 mol H
H : 0.5049 g H × 1.008
= 0.5009 mol H
gH
1 mol O
O : 2.690 g O × 15.999
= 0.1681 mol O
gO
C 0.1683 H 0.5009 O 0.1681 = C1.001 H 2.980 O1.000 = CH 3O
0.1681
0.1681
• Example: The combustion of a 5.217 g
sample of a compound of C, H, and O gave
7.406 g CO2 and 4.512 g of H2O. Calculate
the empirical formula of the compound.
ANALYSIS: This is a multi-step problem. The
mass of oxygen is obtained by difference:
g O = 5.217 g sample – ( g C + g H )
The masses of the elements may then by used to
calculate the empirical formula of the
compound
SOLUTION:
• The formula for ionic compounds is the
same as the empirical formula
• For molecules, the molecular formula and
empirical are usually different
• If the experimental molecular mass is
available, the empirical formula can be
converted into the molecular
• The molecular formula will be a common
multiplier times all the coefficients in the
empirical formula
0.1681
• Example: The empirical formula of
hydrazine is NH2, and its molecular mass is
32.0. What is its molecular formula?
ANALYSIS: The molecular mass (32.0) is some
simple multiple of the mass calculated from the
empirical formula (16.03)
SOLUTION:
32.0
multiplier = 16.03
= 2.00
The correct formula is then
N 1× 2.00 H 2× 2.00 = N 2 H 4
• The coefficients of a balanced chemical
equation provide the mole-to-mole ratios for
the substances involved in the reaction
• Whenever a problem asks you to convert
between different substances, the
calculation usually involves a mole-to-mole
relationship
• How to detect unbalanced equations will be
covered shortly
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• Example: If 0.575 mole of CO2 is produced
by the combustion of propane, C3H8, how
many moles of oxygen are consumed? The
balanced equation is:
C3H8 + 5 O2 Æ 3 CO2 + 4 H2O
ANALYSIS: Relating two compounds usually
requires a mole-to-mole ratio
SOLUTION:
• In many problems you will also need to
perform one or more mole-to-mass
conversions
• These types of stoichiometry problems are
summarized with the flowchart:
mol O 2
0.575 mol C 3 H 8 × 1 5mol
C 3 H 8 = 2.88 mol O 2
• Example: How many grams of Al2O3 are
produced when 41.5 g Al react?
2Al(s) + Fe2O3(s) Æ Al2O3(s) + 2 Fe(l)
ANALYSIS:
41.5 g Al Ù ? g Al2O3
SOLUTION:
grams Al → moles Al → moles Al 2 O 3 → grams Al 2 O 3
1 mol Al 2 O 3
102.0 g Al 2 O 3
1 mol Al
41.5 g Al × 26.98
g Al × 2 mol Al × 1 mol Al 2 O 3 = 78.45 g Al 2 O 3
•
Guidelines for Balancing Equations:
1) Balance elements other than H and O first
2) Balance as a group any polyatomic ions that
appears unchanged on both sides of the arrow
3) Balance separately those elements that appear
somewhere by themselves
•
As a general rule you should use the
smallest whole-number coefficients when
writing balanced chemical equations
•
•
•
Chemical equations provide quantitative
descriptions of chemical reactions
Conservation of mass is the basis for
balancing equations
To balance an equation:
1) Write the unbalanced equation
2) Adjust the coefficients to get equal numbers
of each kind of atom on both sides of the
arrow
• All reactions eventually use up a reactant
and stop
• The reactant that is consumed first is called
the limiting reactant because it limits the
amount of product that can form
• Any reagent that is not completely
consumed during the reactions is said to be
in excess and is called an excess reactant
• The computed amount of product is always
based on the limiting reagent
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• Example: How many grams of NO can form
when 30.0 g NH3 and 40.0 g O2 react
according to:
4 NH3 + 5 O2 Æ 4 NO + 6 H2O
ANALYSIS: This is a limiting reactant problem
SOLUTION:
NH3
.01 g NO
mol NO
30.0 g NH3 × 171 mol
× 44mol
× 301 mol
= 52.9 g NO
.03 g NH3
NH3
NO
O2
30.01 g NO
4 mol NO
40.0 g O 2 × 321 mol
.00 g O 2 × 5 mol O 2 × 1 mol NO = 30.01 g NO
O 2 is the limiting reagent and 30.01 g NO form
percentage yield =
actual yield
theoretica l yield
• The amount of product isolated from a
chemical reactions is almost always less
than the calculated, or maximum, amount
• The actual yield is the amount of the
desired product isolated
• The theoretical yield is the amount that
would be recovered if no loss occurred (the
calculated, maximum amount)
• The percentage yield is the actual yield as
a percentage of the theoretical yield
× 100 %
• When working with percentage yield:
– Remember they involve a measured (actual
yield) and calculated (theoretical yield) quantity
– The calculation may be done in either grams or
moles
– The result can never be a number larger than
100%
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