Introductory Chemistry:
A Foundation
FIFTH EDITION
by Steven S. Zumdahl
University of Illinois
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1
Chemical Composition
Chapter 8
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2
These seashells from Sanibel
Island, Florida, contain the
mineral calcium carbonate.
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by Houghton
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Rebman/Photo
Researchers, Inc.
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3
The Enzo Ferrari is built of
composite materials.
Source: APby
Photo
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4
8.1 Counting by Weighing
• Aim-To understand the concept
of average mass and explore
how counting can be done by
weighing.
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5
The Jelly Bean Analogy
• If you worked in a
candy store and
someone came in
and asked for 1000
jellybeans, no more
and no less, how
would you give the
customer what they
wanted?
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6
JellyBean Analogy
• Suppose each jellybean has a
mass of exactly 5 grams?
• You could simply weigh out 5000
grams of jellybeans and you
would have 1000 of them.
Counting out 1000 would take a
long time.
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Jellybean Analogy
• In reality, jellybeans do not all
weigh exactly the same.
• You could weigh out 10 separate
beans and find their average mass.
• Now you could simply multiply the
average mass of the beans by the
desired number and find how much
1000 beans would weigh.
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8
8.2 Atomic Masses:Counting Atoms
by Weighing
• Aim-To understand atomic
mass and its experimental
determination.
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9
Atomic Masses
• Balanced equation tells us the relative
numbers of molecules of reactants and
products
C + O2  CO2
1 atom of C reacts with 1 molecule of O2
to make 1 molecule of CO2
• If I want to know how many O2 molecules I
will need or how many CO2 molecules I can
make, I will need to know how many C atoms
are in the sample of carbon I am starting with
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10
Atomic Masses
• Dalton used the percentages of elements in
compounds and the chemical formulas to
deduce the relative masses of atoms
• Unit is the amu.
– Atomic Mass Unit
– 1 amu = 1.66 x 10-24g
• We define the masses of atoms in terms of
atomic mass units
– 1 Carbon atom = 12.01 amu,
– 1 Oxygen atom = 16.00 amu
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11
Atomic Mass
• Atomic mass is defined as the weighted
average mass of all the isotopes of a
particular element.
– Includes a correction to allow for the relative
abundance of each element.
• Atomic mass is found by looking on your
periodic table, (number to the upper right
corner, in red)
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12
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13
What is the mass of a molecule of
Oxygen?
• Oxygen is diatomic, so the mass of a
molecule (O2) would apparently be more
than the mass of one single atom of oxygen.
• The mass of O2 would be simply the mass
of 2 atoms of oxygen, added together, or
15.9994 amu + 15.9994 amu = 31.9988 amu
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14
Formula Mass
• Formula mass is the mass of 1 formula unit
of an ionic compound.
• Formula mass is found by adding the
masses of all of the atoms that make up 1
formula unit of the compound.
Example: CaCO3
F.M. = 1-Ca
+1-C
+3-O
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40.078 amu
12.0107 amu
(3x15.9994) amu)
100.087 amu
15
Molecular Mass
• Molecular mass is the mass of 1 molecule
of a covalently bonded compound.
• Calculated in the same manner as the
formula mass mentioned earlier.
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16
Atomic Masses
• Atomic/formula/molecular masses allow us
to convert weights into numbers of particles
If a sample of carbon weighs 3.00 x 1020 amu, how
many atoms of carbon are in the sample?
1 C atom
3.00 x 10 amu x
 2.50 x 1019 C atoms
12.01 amu
Since our equation tells us that 1 C atom reacts
with 1 O2 molecule, if I have 2.50 x 1019 C
atoms, I will need 2.50 x 1019 molecules of O2
20
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Example #1
Calculate the Mass (in amu) of 75 atoms of Al
•
•
Determine the mass of 1 Al atom
1 atom of Al = 26.98 amu
Use the relationship as a conversion factor
26.98 amu
75 Al atoms x
 2024 amu
1 Al atom
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18
8.3 The Mole
• Aims-To understand the mole concept and
Avogadro’s number. To learn to convert
among moles, mass, and number of atoms
in a given sample.
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19
Chemical Packages - Moles
• We use a package for atoms and molecules
called a mole
• A mole is the number of particles equal to the
number of Carbon atoms in 12 g of C-12
• One mole = 6.022 x 1023 units
• One mole of particles= 1 molar mass
(atomic/formula mass expressed in grams)
• 1 mole of C atoms weighs 12.01 g and has 6.02 x 1023 atoms
• The number of particles in 1 mole is called
Avogadro’s Number
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20
Molar Mass
• The molar mass is the mass in grams of one
mole of a compound
• The relative weights of atoms/molecules can be
calculated from atomic masses
water = H2O = 2(1.008 amu) + 16.00 amu
= 18.02 amu
• 1 mole of H2O will weigh 18.02 g, therefore
the molar mass of H2O is 18.02 g
• 1 mole of H2O will contain 16.00 g of oxygen
and 2.02 g of hydrogen
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21
A 1-mol sample of graphite
(a form of carbon) weighs 12.01
g.
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22
Figure 8.1: All these samples of pure
elements contain the same number (a
mole) of atoms: 6.022 x 1023 atoms.
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23
Example 8.6 from Text
• Calcium carbonate, CaCO3, is the principal
mineral found in limestone, marble, chalk,
and pearls. Calculate the molar mass of
calcium carbonate.
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24
Example 8.6 cont.
• Calcium carbonate is composed of 1 ion of
Ca2+ and one ion of CO32-. One mol of
CaCO3 contains one mol of Ca2+ ions and
one mol of CO32- ions. We calculate the
molar mass by summing the masses of the
components.
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Example 8.6 cont.
• Mass of one mol Ca2+ = 40.08 grams
• Mass of one mol CO32- (contains 1 mol C
and 3 mol O)
– 1 mol C = 12.01 grams
– 3 mol O = 3 x 16.00 grams
– One mol CO32- = 60.01 grams
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Example 8.6 cont.
• 40.08 grams + 60.01 grams = 100.09 grams
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27
Figure 8.2: One-mole samples of iron
(nails), iodine crystals, liquid mercury,
and powdered sulfur.
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Glenn Izett/U.S.byGeological
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Example #2
• How many atoms of boron are contained in
156.00 amu’s of boron?
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30
Example #3
Calculate the equivalent number
of moles of 26.2 grams of gold.
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Example #4
• How many atoms are there in 2.71 X 10-4
mol of platinum?
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32
A silicon chip of the type used in
electronic equipment.
by Houghton
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33
Example 8.4 from text
• A silicon chip used in an integrated circuit
of a microcomputer has a mass of 5.68
milligrams. How many silicon (Si) atoms
are present in this chip? The average
atomic mass of silicon is 28.09 amu.
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Example 8.4 cont.
• Convert mg to grams
• Convert grams to mol Si
• Convert mol Si to atoms of Si
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35
Figure 8.3: Various numbers of methane
molecules showing their constituent atoms
(cont’d).
36
Figure 8.3: Various numbers of methane
molecules showing their constituent atoms
(cont’d).
37
Black
walnuts
with and
without
their
green
hulls.
8.5 Percent Composition
• Aim-To learn to find the mass percent of an
element in a given compound
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39
•
Percent Composition
Percentage of each element in a compound
– By mass
• Can be determined from
 the formula of the compound or
 the experimental mass analysis of the
compound
• The percentages may not always total to 100%
due to rounding
part
Percentage 
 100%
whole
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40
Example #4
Determine the Percent Composition
from the Formula C2H5OH
 Determine the mass of each element in 1
mole of the compound
2 moles C = 2(12.01 g) = 24.02 g
6 moles H = 6(1.008 g) = 6.048 g
1 mol O = 1(16.00 g) = 16.00 g
 Determine the molar mass of the
compound by adding the masses of the
elements
1 mole C2H5OH = 46.07 g
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Example #4
Determine the Percent Composition
from the Formula C2H5OH
 Divide the mass of each element by the molar
mass of the compound and multiply by 100%
24.02g
 100%  52.14%C
46.07g
6.048g
 100%  13.13%H
46.07g
16.00g
 100%  34.73%O
46.07g
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42
8.6 Formulas of Compounds
• Aim-To understand the meaning of
empirical formulas of compounds.
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43
Empirical Formulas
• The simplest, whole-number ratio of atoms in a
molecule is called the Empirical Formula
– can be determined from percent composition or
combining masses
• The Molecular Formula is a multiple of the
Empirical Formula
100g
%A
mass A (g)
100g
%B
mass B (g)
MMA
MMB
moles A
moles A
moles B
moles B
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44
Figure
8.4:
The
glucose
molecul
e.
Example #5
Determine the Empirical Formula of
Benzopyrene, C20H12
 Find the greatest common factor (GCF) of the
subscripts
factors of 20 = (10 x 2), (5 x 4)
factors of 12 = (6 x 2), (4 x 3)
GCF = 4
 Divide each subscript by the GCF to get the
empirical formula
C20H12 = (C5H3)4
Empirical Formula = C5H3
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46
8.7 Calculation of Empirical
Formulas
• Aim-To learn to calculate empirical
formulas
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47
Steps for Determining the Empirical
Formula of a Compound
• 1. Obtain the mass of each element present (in
grams)
• 2. Determine the number of moles of each type of
atom present
• 3. Divide the number of moles of each element by
the smallest number of moles to convert the
smallest number to 1.
• 4. If the numbers from 3 were not whole numbers,
multiply the numbers by the smallest integer that
will convert all of them to whole numbers. This
set of numbers represents the subscripts.
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Example #6
Determine the Empirical Formula of
Acetic Anhydride if its Percent Composition is
47% Carbon, 47% Oxygen and 6.0% Hydrogen
 Convert the percentages to grams by
assuming you have 100 g of the compound
– Step can be skipped if given masses
47gC
100g 
 47gC
100g
47gO
100g 
 47gO
100g
6.0gH
100g 
 6.0gH
100g
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49
Example #6
Determine the Empirical Formula of
Acetic Anhydride if its Percent Composition is
47% Carbon, 47% Oxygen and 6.0% Hydrogen
 Convert the grams to moles
1 mol C
47g C 
 3.9 mol C
12.01g
1 mol H
6.0 g H 
 6.0 mol H
1.008g
1 mol O
47 g O 
 2.9 mol O
16.00g
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50
Example #6
Determine the Empirical Formula of
Acetic Anhydride if its Percent Composition is
47% Carbon, 47% Oxygen and 6.0% Hydrogen
 Divide each by the smallest number of moles
3.9 mol C  2.9  1.3
6.0 mol H  2.9  2
2.9 mol O  2.9  1
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51
Example #6
Determine the Empirical Formula of
Acetic Anhydride if its Percent Composition is
47% Carbon, 47% Oxygen and 6.0% Hydrogen
 If any of the ratios is not a whole number, multiply
all the ratios by a factor to make it a whole number
– If ratio is ?.5 then multiply by 2; if ?.33 or ?.67 then
multiply by 3; if ?.25 or ?.75 then multiply by 4
Multiply all the
Ratios by 3
Because C is 1.3
3.9 mol C  2.9  1.3 x 3  4
6.0 mol H  2.9  2 x 3  6
2.9 mol O  2.9  1 x 3  3
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Example #6
Determine the Empirical Formula of
Acetic Anhydride if its Percent Composition is
47% Carbon, 47% Oxygen and 6.0% Hydrogen
° Use the ratios as the subscripts in the
empirical formula
3.9 mol C  2.9  1.3 x 3  4
6.0 mol H  2.9  2 x 3  6
2.9 mol O  2.9  1 x 3  3
C4H6O3
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53
8.8 Calculation of Molecular
Formulas
• Aim-To learn to calculate the molecular
formula of a compound, given its empirical
formula and molar mass
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54
Molecular Formulas
• The molecular formula is a multiple of the
empirical formula
• To determine the molecular formula you
need to know the empirical formula and the
molar mass of the compound
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55
Example #7
Determine the Molecular Formula of Benzopyrene
if it has a molar mass of 252 g and an
empirical formula of C5H3
 Determine the empirical formula
• May need to calculate it as previous
C5H3
 Determine the molar mass of the empirical
formula
5 C = 60.05 g, 3 H = 3.024 g
C5H3 = 63.07 g
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56
Example #7
Determine the Molecular Formula of Benzopyrene
if it has a molar mass of 252 g and an
empirical formula of C5H3
 Divide the given molar mass of the
compound by the molar mass of the
empirical formula
– Round to the nearest whole number
252 g
4
63.07 g
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57
Example #7
Determine the Molecular Formula of Benzopyrene
if it has a molar mass of 252 g and an
empirical formula of C5H3
 Multiply the empirical formula by the
calculated factor to give the molecular
formula
(C5H3)4 = C20H12
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58
8.5: The
structure
of P4O10
as a
"ballandstick"
model.