Ch02 lecture

© 2016 Pearson Education, Inc.Instructor’s Resources Materials (Download Only) for Principles of Chemistry: A Molecular Approach, 3/e
Nivaldo J. Tro
Catherine E. MacGowan
Armstrong State University
Lecture Presentation
Chapter 2
Atoms and
Elements

Nivaldo J. Tro
Imaging and Moving Atoms
• Gerd Binnig and Heinrich Rohrer were experimenting with an electrical
current passing over a flat metal surface and discovered that this “tunneling”
current allowed them to scan the surface on an atomic scale.
• They were essentially taking pictures of atoms on the surface.
• Their discovery led to the development of electron scanning tunneling
microscopy (STM).

Nivaldo J. Tro
Atomic Theory of Matter and Its Laws
• Law of the conservation
of matter
• In a chemical reaction
matter is neither created
nor destroyed.
- Total mass of used
reactants = total mass
of products produced.
- Total number of
reactant atoms = total
number of product
atoms.

Nivaldo J. Tro
• Dalton’s atomic theory of matter
– Atoms are small, discreet, indivisible pieces of matter.
– All elements are made up particles called atoms.
– An element’s atoms are identical in size, mass, and
chemical properties.
• Scientists did not know about isotopes.
– Isotopes are elemental atoms that differ in
their mass due to different number of
neutrons..
– Molecules (compounds) are formed when two or more
elements are combined.
– Molecules are simple whole-number ratios of the
combined elements.

Nivaldo J. Tro
• Dalton’s Atomic Theory notes that
– “Atoms of different elements can combine to
form molecules in simple whole-number
combination of atoms.”
• Law of definite proportions
• Law of multiple proportions

Nivaldo J. Tro
Law of Definite Proportions
• For a given compound, the elements always combine in the same
proportion.
– For example
• A sodium chloride molecule (NaCl) is always a 1:1 ratio of
one sodium atom to one chlorine atom.
– A 100.0 g sample of NaCl contains 39.3 g Na and 60.7 g Cl.
Mass Cl = 60.7 g = 1.54
Mass Na 39.3 g
– A 58.44.0 g sample of NaCl contains 22.99 g Na and 35.44 g Cl.
Mass Cl = 35.44 g = 1.54
Mass Na 22.99 g

Nivaldo J. Tro
Law of Multiple Proportions
• Two elements, A and X, can form different compounds by combining
in different proportions.
• These combinations can be represented as a ratio.
– For example
• A molecule of carbon dioxide (CO2) has a ratio of one C atom
to every two atoms of oxygen or 1:2.
• A molecule of hydrogen peroxide (CO) has a ratio of one C
atom to one atom of oxygen or 1:1

Nivaldo J. Tro

Nivaldo J. Tro
Discovery of the Electron and Its Properties
• Discovery of electron
– Thomson’s charge to mass experiment used
cathode rays.
• Thomson believed that these particles were the ultimate
building blocks of matter.
• These cathode ray particles became known as electrons.
– The Millikan oil drop experiment
• showed that the particle had the same amount of charge as
the hydrogen ion as Thomson observed in his experiments.

Nivaldo J. Tro
Discovery of the Atom’s Electron
• Thomson’s charge to mass experiment investigated the effect on a
cathode ray when placing an electric field around a tube.
1. Charged matter is attracted to an electric field.
2. Light’s path is not deflected by an electric field.
• Cathode rays are made of tiny particles.
– These particles had a negative charge because the beam always
deflected toward the (+) plate.
• The amount of deflection was related to the charge and mass of the
particles.
– The charge–mass ratio of these particles was
(−)1.76 × 108 C/g.

Nivaldo J. Tro
Charge to Mass Ratio of Electron
Experiment

Nivaldo J. Tro
Millikan’ Oil Drop Experiment
• Investigation led to determining
the charge of the electron.
– Charge on an electron
(–)1.60 × 10–19 C
charge × (mass/charge) = mass
–1.60 × 10–19 C x
(1 gram/1.76 × 108 C)
= 9.10 × 10–28 grams
• Thomson’s experiments
determined that the mass of one
electron would be 2000 times
lighter than a hydrogen atom.

Nivaldo J. Tro
Atomic Structure of the Electron
• Electrons are particles found in all atoms.
– One of the fundamental pieces of matter
• The electron has a charge of −1.60 × 10–19 C.
• The electron has a mass of 9.1 × 10−28 g.

Nivaldo J. Tro
J.J. Thomas: The Plum Pudding Model of
the Atom
• J.J. Thomas (plum pudding model)
– The atom is composed of a positive cloud of matter
in which electrons are embedded.
• Explains the positive (+) and negative (–)
charged behavior of matter

Nivaldo J. Tro
Rutherford’s Gold Foil Experiment
• Gold foil experiment
– could not explain
Thomson’s plum
pudding atom model.
– led to the discovery
of the atom’s
nucleus.

Nivaldo J. Tro
Rutherford’s Gold Foil Experiment:
Discovery of the Nucleus
• From the gold foil experiment the following conclusions
were proposed:
– The atom contains a tiny dense center called the
nucleus.
– The nucleus has essentially the entire mass of the atom.
• The electrons weigh so little that they give practically no mass to
the atom.
– The nucleus is positively charged.
• The amount of positive charge balances the negative charge of
the electrons.
• The electrons are dispersed in the empty space of the atom
surrounding the nucleus.

Nivaldo J. Tro
Atomic Structure: Historical Perspective
• Rutherford’s model (solar system)
– stated that the atom is mostly empty space with a dense
center of mass (nucleus) and circling electrons.
– proposed that the nucleus had a particle that had the same
amount of charge as an electron but the opposite sign.
• These particles are called protons.
– charge = +1.60 × 1019 C
– mass = 1.67262 × 10−24 g
– Since protons and electrons have the same amount of
charge, for the atom to be neutral, there must be equal
numbers of protons and electrons.

Nivaldo J. Tro
Number of Protons in an Atom Determines
the Element’s Identity
• The number of protons located in an atom’s nucleus determines the
element’s identity.
– The number of protons in the nucleus of an atom is called the
atomic number.
• Each element has a unique chemical name and symbol.
– The symbol can either be one or two letters.
• O (oxygen) or Fe (iron)
• The elements are arranged
on the periodic table in order
of their atomic numbers.

Nivaldo J. Tro
Isotopes: Same Element but with Different
Masses
• Isotopes are elements whose atoms differ in mass due to
varying numbers of neutrons.
– They are the same element because they have the same
number of protons (atomic number).
– They are chemically identical.
• Isotopes are identified by their mass numbers.
– Protons + neutrons = mass number
• Isotopic symbol
13
Al
26.981
Atomic number, Z
Atom symbol
Atomic weight

Nivaldo J. Tro
How many protons, electrons, and neutrons
are in the following atoms?
Protons Electrons Neutrons
1. 32
S
16
2. 65
Cu
29
3. U-240

Nivaldo J. Tro
How many protons, electrons, and neutrons
are in the following atoms?
Protons Electrons Neutrons
1. 32
S 16 16 16
16
2. 65
Cu 29 29 36
29
3. U – 240 92 92 148
Note: Neutral atoms will have the same number of protons as electrons.

Nivaldo J. Tro
Problem:
Complete the following table:
Protons Neutrons Electrons
Atomic
Number
Mass
Number
Atomic
Symbol
6 7
42 96
55 133

Nivaldo J. Tro
Answer:
Protons Neutrons Electrons
Atomic
Number
Mass
Number
Atomic
Symbol
6 7 6 6 13
42 54 42 42 96
13 14 13 13 27
55 78 55 55 133

Nivaldo J. Tro
Isotopes and Atomic Mass (weight)
• What is the connection between an element’s isotopes and its
atomic mass?
– The observed mass of an element is a weighted average of
the weights of all the naturally occurring atoms.
• Average mass = atomic weight (amu)
• What information is needed to determine the atomic mass of
an element?
– The natural abundance of each of the element’s isotopes
• The percentage of an element that is one isotope is called the
isotope’s natural abundance.
– Example problem
• Boron has two isotopes.
– Boron is 19.9% 10B and 80.1% 11B.
• Boron atomic weight
= 0.199 (10.0 amu) + 0.801 (11.0 amu) = 10.8 amu

Nivaldo J. Tro
Problem:
Calculating an element’s atomic mass
The element silver (Ag) has an atomic mass of 107.868 amu.
It has two isotopes—Ag-109 (108.905 amu) with an
abundance of 48.1600% and Ag-107. Determine the amu for
Ag-107.

Nivaldo J. Tro
Answer:
Calculating an element’s atomic mass
The element silver (Ag) has an atomic mass of 107.868 amu. It has two
isotopes—Ag-109 (108.905 amu) with an abundance of 48.1600% and
Ag-107. Determine the amu for Ag-107.
Problem Strategy:
1. Set up an algebraic equation.
2. Solve for x, which is the amu for Ag-107.
Answer:
1. Set up an algebraic equation.
(0.481600) 108.905 amu + x (0.518400) = 107.868 amu
2. Solve for x, which is the amu for Ag-107.
52.4490 amu + 0.518400x = 107.868 amu
0.518400 x = 55.4180 amu
x = 106.905 amu

Nivaldo J. Tro
Ions are Charged Atoms
• IONS are atoms or groups of atoms with a positive (+) or negative (–)
charge.
• Taking away an electron from an atom gives a cation with a
positive charge.
– More protons in nucleus versus electrons surrounding nucleus
– Metals tend to form cations to achieve the “full shell” look.
• Adding an electron to an atom gives an anion with a negative
charge.
– Fewer protons in the nucleus versus electrons surrounding
nucleus
– Nonmetals tend to form anions to achieve the “full shell” look.

Nivaldo J. Tro
Cations
• A cation forms when an atom
loses one or more electrons from
its outer (valence) shell (energy
level).
• Cations are positively charged
because the atom has more
protons (+) than electrons (–).
– Mg atom has 12 protons and
12 electrons.
– Mg+2 ion has 12 protons and
10 electrons.
• Metal elements tend to form
cations.
• Example
– Mg  Mg2+ + 2 e–
Anions
• An anion forms when an atom
gains one or more electrons into
its outer (valence) shell (energy
level).
• Anions are negatively charged
because the atom has less
protons (+) than electrons (–).
– F atom has 9 protons and
9 electrons.
– F– ion has 9 protons and 10
electrons.
• Nonmetal elements tend to form
anions.
• Example
– F + e–  F–

Nivaldo J. Tro
Periodic Table: Historic perspective
• Dmitri Mendeleev (1834–1907)
developed the modern periodic table.
– Argued that element properties
are periodic functions of their
atomic weights
• Periodic law
– When the elements are arranged
in order of increasing atomic
mass, certain sets of properties
recur periodically.
– Elements having similar physical
and chemical properties fall within
a column.
• We now know that element properties
are periodic functions of their atomic
numbers.

Nivaldo J. Tro
Periodic Table: Organization
• The elements are arranged from left to right in increasing
atomic number (number of protons an element has).
• Rows in the periodic table are referred to as periods.
• Columns within the periodic table are sometimes referred
to as families.
– Because the elements within the column have similar
physical and chemical properties
• The elements, their names, and symbols are given on the
periodic table.

Nivaldo J. Tro
Areas of the Periodic Table
• Major Areas
– Metals
• Main group metals
• Transition metals
• Inner transition metals
– Lanthanides
– Actinides
– Metalloids
– Non metals

Nivaldo J. Tro
Periodic Table: Major Divisions

Nivaldo J. Tro
Periodic Table: Metals
• Characteristics
– Solid at room temperature (except Hg)
– Reflective surface
• Shiny
– Conduct heat and electrical current
– Malleable
• Can be shaped
– Ductile
– Lose electrons and form cations
– About 75% of the elements in the periodic table are metals.

Nivaldo J. Tro
Periodic Table: Metalloids
• Characteristics
– Can exhibit the properties of metals
and/or nonmetals
– Known as semiconductors
• Poor conductors of heat
– Solids at room temperature

Nivaldo J. Tro
Periodic Table: Nonmetals
• Characteristics
– Can be found in all three states (gas, liquid, and solid)
of matter
– Poor conductors of heat and electricity
– Solids are brittle.
– Gain electrons to become anions
– Except for H, found mostly in the upper right on the
periodic table

Nivaldo J. Tro
Periodic Table: Families

Nivaldo J. Tro
Families: Grouping of the Periodic Table
• Elements within a column are considered
“families.”
– They have similar chemical and physical properties.
– They all have the same number of electrons in the
outermost shell (energy level).
• Families of the periodic table
– Alkali metals (column 1/Group number 1A)
– Alkaline earth metals (column 2/Group number 2A)
– Halogens (column 17/Group number 7A)
– Noble/inert gases (column 18/Group number 8A)

Nivaldo J. Tro
Ions and the Periodic Table
• Metals tend to form cations.
– Lose electrons in their outermost energy level to
“obtain the nearest noble gas electron configuration”
• Nonmetals tend to form anions.
– Gain electrons in their outermost energy level to
“obtain the nearest noble gas electron configuration”
• Metalloids tend to form cations but can also form anions.

Nivaldo J. Tro
Periodic Table: Ion Formation

Nivaldo J. Tro
What is a Mole?
• Chemistry is quantitative in nature—its unit is the mole.
• The mole as a unit versus “dozen” as a unit.
– The unit “dozen” is associated with twelve units.
– The unit mole is associated with 6.02 × 1023 units.
• There is Avogadro’s number of units in every mole.
– 6.02 × 1023 units is known as Avogadro’s number.
– 1 mole = 6.02 × 1023 units
• One mole of Cu atoms has
– an atomic mass of 63.55 grams, which is
– 6.022 × 1023 Cu atoms, which is
– approximately 22 pennies.

Nivaldo J. Tro
Problem:
How large is a mole?
You have inherited ¼ mole of pennies. If you spent
1 × 1012 dollars every second how many decades would it
take for you to spend your inheritance?

Nivaldo J. Tro
Answer:
How large is a mole?
You have inherited ¼ mole of pennies. If you spent
1 × 1012 dollars every second how many decades would it
take for you to spend your inheritance?
¼ mol × ( 6.02 x 1023 cents/1 mol) × ($1/100 cents)
= 1.51 × 1021 dollars
1.51 × 1021 dollars × (1 sec/1 × 1012 dollars) × (1 min/60 sec)
× (1 hr /60 min) × (1 day/24 hrs) × (1 yr/365 days) ×
(1 decade/10 yrs) = 4.79 decades.
Every mole contains 6.02 × 1023 units. That is a lot of
pieces!

Nivaldo J. Tro
Mass to Mole Conversions
• To go from mass to mole
Mass(g) × (1 mol/atomic mass of element) = mole of element
• Example
– 24.00 grams C × (1 mole carbon/12.011 grams) = 2.00 moles of
carbon atoms
Mass of element (g) Mole of element

Nivaldo J. Tro
Mole to Mass Conversions
• To go from mole to mass
Mole of element × (atomic mass (element)/1 mole) = mass (g)
• Example
– 8.00 mole He × (4.00 grams/1 mole He)
= 32.0 grams He atoms
Mass of element (g)Mole of element

Nivaldo J. Tro
Mass to Number of Atoms Conversions
• To go from mass to number of atoms, you must go through the mole.
Mass (g) × (1 mol/atomic mass) × (6.02 × 1023 atoms/1 mole) = # atoms
• Example:
– 45.99 g Na × (1 mol Na/22.99 g) × (6.02 × 1023 Na atoms/1 mole)
= 1.204 × 1024 Na atoms
Mole of
element
Mass of
element (g)
Number of atoms
of the element

Nivaldo J. Tro
Atoms to Mass Conversions
• To go from atoms to mass you must go through the mole.
(# of atoms) × (1 mole/6.02 × 1023 atoms) × (atomic mass/1 mole) = mass (g)
• Example:
– 8 atoms Am × (1 mole/6.02 × 1023 Am atoms) × (243 g/1 mol Am) =
3.23 × 10–21 grams of Am
Mole of
element
Mass of
element (g)
Number of atoms
of the element

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