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Chapter 3: Atoms: The
Building Blocks of Matter
CHEMISTRY 1-2
MR. CHUMBLEY
Section 1: The Atom: From
Philosophical Idea to Scientific
Theory
The Origin of the Atom
 Humans have thought about the existence of matter
for a very long time
 In ancient Greece, two competing idea for how
matter exists rose to prominence
 The word “atom” comes from the ancient Greek
words, “a-” and “-tomos”
 This translates to “uncuttables”
Democritus’ Particle Theory of Matter
 Around 400 BC, a Greek philosopher named
Democritus is credited for creating the first model of
the atom
 While Democritus did not come up with the idea
himself, he made it a fuller idea
 According to Democritus, “the universe is composed
of two elements: the atoms and the void in which
they exist and move.”
Aristotle’s Continuous Theory of Matter
 After Democritus’s time, other Greek philosophers
came to support a different idea
 According to Aristotle, matter was continuous and
could be broken down indefinitely
 Aristotle’s idea were more widely accepted as the way
in which matter existed, and his opinion was
accepted for nearly 2000 years
Understanding how Matter Behaves
 By the late 1700’s, most chemists knew and agreed
about elements
 However, they had difficulty explaining how
different substances could combine with one another
to form new ones
 This is the origin of modern chemistry, which
focused on quantitative analysis of chemical
reactions
Law of Conservation of Mass
 With the help of improved balances and careful
measuring, the first law was discovered
 The law of conservation of mass states that
matter is neither created nor destroyed during
ordinary chemical reactions or physical changes
Law of Definite Proportions
 The second discovery was that pure substances have
a fixed proportion, regardless of how it is made
 The law of definite proportions states that a
chemical compound contains the same elements in
exactly the same proportions by mass regardless of
the size of the sample or the source of the compound
Law of Multiple Proportions
 Early chemists also knew that the same elements
could combine to create different compounds
 The law of multiple proportions states that if
two or more different compounds are composed of
the same two elements, then the ratio of the masses
of the second element combined with a certain mass
of the first element is always a ratio of small whole
numbers
Dalton’s Atomic Theory
 In 1808 John Dalton proposed an explanation that
included all three laws
 Dalton’s atomic theory has five points:





All matter is composed of extremely small particles called atoms.
Atoms of an element are identical in size, mass and other properties;
atoms of different elements differ in size, mass, and other properties.
Atoms cannot be subdivided, created, or destroyed.
Atoms of different elements combine in simple whole-number ratios
to form chemical compounds.
In chemical reactions, atoms are combined, separated, or
rearranged.
Democritus to Dalton
 By using the measurable property of mass,
Democritus’s idea was made into Dalton’s scientific
theory
 While we have since learned that not all points of
Dalton’s theory hold true, his basic theory has not
been discarded, merely modified and improved
Section 3: Counting Atoms
Atomic Number
 All atoms of an element have the same number of
protons
 The atomic number of an element is the number
of protons of each atom of that element
 Atomic number is noted by the letter Z
 In a neutral atom, the number of electrons is equal to
the number of protons
Isotopes
 Atoms of the same element can have different number of
neutrons
 Isotopes are atoms of the same element that have
different masses
 Therefore isotopes are atoms with the same number of
protons, but different numbers of neutrons
 Nuclide is a general term for a specific isotope of an
element
Mass Number
 Since both protons and neutrons contribute to the
mass of the atom, we can find the mass by adding the
number of protons to the number of neutrons
 The mass number is the total number of protons
and neutrons that make up the nucleus of an isotope
 Mass number is noted by the letter A
Identifying Isotopes
 Isotopes can be identified using hyphen notation
that combines the name of the element and the mass
number
 For example:
 uranium-238
 carbon-12
 hydrogen-3
Identifying Isotopes
 Isotopes can be identified using the nuclear symbol,
which combines the atom’s chemical symbol (X), mass
number (A), and atomic number (Z) like so:
𝐴
𝑍𝑋
 For example:
238
 92𝑈
12
 6𝐶
3
 1𝐻
Sample Problem A
How many protons, electrons, and neutrons are there in
an atom of chlorine-37?
 Step 1: Analyze

Given: name and mass number of chlorine-37

Unknown: numbers of protons, electrons, and neutrons
 Step 2: Plan

Atomic number = number of protons = number of electrons

Mass number = number of protons + number of neutrons
Sample Problem A
How many protons, electrons, and neutrons are there in
an atom of chlorine-37?
 Step 3: Solve

The mass number of chlorine-37 is 37. Using the periodic table
shows the atomic number of chlorine is 17.
Atomic number = number of protons = number of electrons
= 17 protons and electrons
Number of neutrons = mass number – atomic number
= 37 – 17 = 20 neutrons
 An atom of chlorine-37 is made up of 17 electrons, 17
protons, and 20 neutrons
Homework
 Chapter 3, Section 3
 Practice (p.76) #1-3
 Chapter Review (p.86) #8, 11
Atomic Mass
 When the mass of a single atom is expressed in
kilograms, the value is very small
 For most calculations it is much easier to use a relative
atomic mass
 For this, an arbitrary measure has been assigned to
determine the mass of individual isotopes
 One unified atomic mass unit, or 1 u, is exactly 1/12
the mass of a carbon-12 atom
Atomic Mass
 However, the mass of isotopes is not simply the mass
number expressed in atomic mass units
 Careful calculation of the masses of different isotopes has
shown that differ slightly from the mass number
 The difference is due to a few factors



Protons and neutrons deviate slightly from 1 u
The mass of electrons is included in the atomic mass
A small amount of mass is changed into energy when proton and
neutrons combine to form the nucleus
Average Atomic Mass
 Since most elements occur naturally as isotopes, the
mass of that element is determined by the weighted
average of those isotopes
 Average atomic mass is the weighted average of
the atomic masses of the naturally occurring isotopes
of an element
 The weighting of the mass values is determined by
the percentage that isotope has in natural abundance
Average Atomic Mass
 For example: copper (Cu) is
found in two naturally
occurring isotopes: copper-63
and copper-65
 The mass number, percentage
natural abundance, and
atomic mass in AMU is shown
to the right
 The average atomic mass is
then found by adding the
values when the mass of each
isotope is multiplied by its
natural abundance
Isotope
Mass
#
Abundance
Atomic
Mass (u)
Copper-63
63
69.15%
62.929601
Copper-65
65
30.85%
64.927794
Average atomic mass calculation:
+
0.6915 × 62.929 601 u
0.3085 × 64.927 794 u
63.55 u
Homework
 Use Figure 3.5 on p. 78
 Calculate the average atomic mass of carbon, oxygen,
and uranium
The Mole
 Establishing a relative atomic mass makes it possible to
know how many atoms of an element are present in a
sample of that element with a known mass
 A mole is the amount of a substance that contains as
many particles as there are atoms in exactly 12g of
carbon-12
 The mole is abbreviated as mol
 The mole is the base unit of amount in SI
Avogadro’s Number
 Avogadro’s number is the number of particles in
exactly one mole of a pure substance
 This number has been experimentally determined to
be 6.022 141 79 × 1023
 We will define Avogadro’s number as 6.022 × 1023
 This number is named after 19th century scientist,
Amedeo Avogadro, who did not actually determine
this number
Molar Mass
 Since one mole of a substance contains particles, we can
calculate the mass of one mole of that substance
 The molar mass is the mass of one mole of a pure
substance
 This is a constant value and a property of the material
 Molar mass is measured in g/mol
 The value of the molar mass is equivalent in grams to the
average atomic mass in atomic mass units
Mass and Amount of an Element
Mass of
an
element
in grams
=
×
𝐦𝐨𝐥𝐚𝐫 𝐦𝐚𝐬𝐬 𝐨𝐟 𝐞𝐥𝐞𝐦𝐞𝐧𝐭
×
𝟏 𝐦𝐨𝐥
𝟏 𝐦𝐨𝐥
=
𝐦𝐨𝐥𝐚𝐫 𝐦𝐚𝐬𝐬 𝐨𝐟 𝐞𝐥𝐞𝐦𝐞𝐧𝐭
Amount
of an
element
in moles
=
×
𝟏 𝐦𝐨𝐥
×
𝟔. 𝟎𝟐𝟐 × 𝟏𝟎𝟐𝟑 𝐚𝐭𝐨𝐦𝐬
𝟔. 𝟎𝟐𝟐 × 𝟏𝟎𝟐𝟑 𝐚𝐭𝐨𝐦𝐬
=
𝟏 𝐦𝐨𝐥
Number
of atoms
of an
element
Sample Problem B (p. 80)
What is the mass in grams of 3.50 mol of the element
copper, Cu?
 Step 1: Analyze

Given: 3.50 mol of Cu

Unknown: mass of Cu in grams
 Step 2: Plan
 Amount of Cu in moles  mass of Cu in grams

The mass of an element in grams can be found by multiplying the amount
of moles by the molar mass
grams Cu
moles Cu ×
= grams Cu
moles Cu
Sample Problem B (p. 80)
What is the mass in grams of 3.50 mol of the element
copper, Cu?
 Step 3: Solve

The molar mass of copper from the periodic table is 63.55 g/mol
63.55 g Cu
3.50 mol Cu ×
= 222 g Cu
1 mol Cu
 3.50 mol of copper has a mass of 222 g
Sample Problem C (p. 81)
A chemist produces 11.9 g of aluminum, Al. How may
moles of aluminum were produced?
 Step 1: Analyze

Given: 11.9 g Al

Unknown: amount of Al in moles
 Step 2: Plan
 mass of Al in grams  amount of Al in moles

The amount of moles of an element can be found by dividing the mass in
grams by the molar mass
moles Al
grams Al ×
= moles Al
grams Al
Sample Problem C (p. 81)
A chemist produces 11.9 g of aluminum, Al. How may
moles of aluminum were produced?
 Step 3: Solve

The molar mass of aluminum from the periodic table is 26.98 g/mol
1 mol Al
11.9 g Al ×
= 0.441 mol Al
26.98 g Al
 11.9 g of aluminum has 0.441 mols of Al
Homework
 P. 81
 Practice Set B: #1-4
 Practice Set C: #1-3
Conversions with Avogadro’s Number
 Similar to how the mass of a substance can be
converted to moles, and vice-versa, the number of
atoms or particles can be determined using
Avogadro’s number
Amount
of an
element
in
moles
=
=
𝟏 𝐦𝐨𝐥
×
𝟔. 𝟎𝟐𝟐 × 𝟏𝟎𝟐𝟑 𝐚𝐭𝐨𝐦𝐬
𝟔. 𝟎𝟐𝟐 × 𝟏𝟎𝟐𝟑 𝐚𝐭𝐨𝐦𝐬
×
𝟏 𝐦𝐨𝐥
Number
of
atoms
of an
element
Sample Problem D (p. 82)
How many moles of silver, AG, are in 3.01 × 1023 atoms of
silver?
 Step 1: Analyze

Given: 3.01 × 1023 atoms of Ag

Unknown: amount of Ag in moles
 Step 2: Plan
 Number of atoms of Ag amount of Ag in moles

The number of atoms is converted to amount in moles by dividing by
Avogadro’s number
moles Ag
Ag atoms ×
= moles Ag
Avogafro′ s number of Ag atoms
Sample Problem D (p. 82)
How many moles of silver, AG, are in 3.01 × 1023 atoms of
silver?
 Step 3: Solve

Avogadro’s number is 6.022 × 1023 .
23
3.01 × 10
1 mol Ag
Ag atoms ×
= 0.500 mol Ag
23
6.022 × 10 Ag atoms
 3.01 × 1023 atoms of Ag is 0.500 moles of Ag
Sample Problem E (p. 82-83)
What is the mass in grams of 1.20 × 1023 atoms of copper,
Cu?
 Step 1: Analyze

Given: 1.20 × 1023 atoms of Cu

Unknown: mass of Cu in grams
 Step 2: Plan
 Number of atoms of Cu amount of Cu in moles  mass of Cu in grams

The number of atoms is converted to amount in moles by dividing by Avogadro’s
number. The amount in moles is then multiplied by molar mass to find the mass in
grams
Cu atoms ×
moles Cu
grams Cu
×
= grams Cu
Avogadro′ s number of Cu atoms moles Cu
Sample Problem E (p. 82-83)
What is the mass in grams of 1.20 × 1023 atoms of copper,
Cu?
 Step 3: Solve

Avogadro’s number is 6.022 × 1023 .

The molar mass of Cu is 63.55 g/mol.
1.20 ×
1023 Cu
1 mol Cu
63.55 Cu
atoms ×
×
= 1.27 × 10−14 g Cu
23
6.022 × 10 Cu atoms 1 mol Cu
 1.20 × 1023 atoms of Cu has a mass of 1.27 × 10-14
grams
Homework
 P. 82
 Practice Set D: #1-3
 P. 83
 Practice Set E: #1-3
Conversions with Avogadro’s Number
 Similar to how the mass of a substance can be
converted to moles, and vice-versa, the number of
atoms or particles can be determined using
Avogadro’s number
Amount
of an
element
in
moles
=
=
𝟏 𝐦𝐨𝐥
×
𝟔. 𝟎𝟐𝟐 × 𝟏𝟎𝟐𝟑 𝐚𝐭𝐨𝐦𝐬
𝟔. 𝟎𝟐𝟐 × 𝟏𝟎𝟐𝟑 𝐚𝐭𝐨𝐦𝐬
×
𝟏 𝐦𝐨𝐥
Number
of
atoms
of an
element
Sample Problem D (p. 82)
How many moles of silver, AG, are in 3.01 × 1023 atoms of
silver?
 Step 1: Analyze

Given: 3.01 × 1023 atoms of Ag

Unknown: amount of Ag in moles
 Step 2: Plan
 Number of atoms of Ag amount of Ag in moles

The number of atoms is converted to amount in moles by dividing by
Avogadro’s number
moles Ag
Ag atoms ×
= moles Ag
Avogafro′ s number of Ag atoms
Sample Problem D (p. 82)
How many moles of silver, AG, are in 3.01 × 1023 atoms of
silver?
 Step 3: Solve

Avogadro’s number is 6.022 × 1023 .
23
3.01 × 10
1 mol Ag
Ag atoms ×
= 0.500 mol Ag
23
6.022 × 10 Ag atoms
 3.01 × 1023 atoms of Ag is 0.500 moles of Ag
Sample Problem E (p. 82-83)
What is the mass in grams of 1.20 × 1023 atoms of copper,
Cu?
 Step 1: Analyze

Given: 1.20 × 1023 atoms of Cu

Unknown: mass of Cu in grams
 Step 2: Plan
 Number of atoms of Cu amount of Cu in moles  mass of Cu in grams

The number of atoms is converted to amount in moles by dividing by Avogadro’s
number. The amount in moles is then multiplied by molar mass to find the mass in
grams
Cu atoms ×
moles Cu
grams Cu
×
= grams Cu
Avogadro′ s number of Cu atoms moles Cu
Sample Problem E (p. 82-83)
What is the mass in grams of 1.20 × 1023 atoms of copper,
Cu?
 Step 3: Solve

Avogadro’s number is 6.022 × 1023 .

The molar mass of Cu is 63.55 g/mol.
1.20 ×
1023 Cu
1 mol Cu
63.55 Cu
atoms ×
×
= 1.27 × 10−14 g Cu
23
6.022 × 10 Cu atoms 1 mol Cu
 1.20 × 1023 atoms of Cu has a mass of 1.27 × 10-14
grams
Homework
 P. 82
 Practice Set D: #1-3
 P. 83
 Practice Set E: #1-3