General Chemistry Principles and Modern Applications 8ed

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Chapter 2: Atoms and the Atomic Theory
Philip Dutton
University of Windsor, Canada
Prentice-Hall © 2002
General Chemistry
Principles and Modern Applications
Petrucci • Harwood • Herring
8th
Edition

2. Prentice-Hall ©General Chemistry:Slide 2 of 25
Contents
• Early chemical discoveries
• Electrons and the Nuclear Atom
• Chemical Elements
• Atomic Masses
• The Mole

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Early Discoveries
Lavoisier 1774 Law of conservation of mass
Proust 1799 Law of constant composition
Dalton 1803-1888 Atomic Theory

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Dalton’s Atomic Theory
 Each element is composed of small particles called atoms.
 Atoms are neither created nor destroyed in chemical reactions.
 All atoms of a given element are identical
 Compounds are formed when atoms of more than one element
combine

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Consequences of Dalton’s theory
 In forming carbon monoxide, 1.33 g
of oxygen combines with 1.0 g of
carbon.
 In the formation of hydrogen
peroxide 2.66 g of oxygen combines
with 1.0 g of hydrogen.
 Law of Definite Proportions: combinations of elements
are in ratios of small whole numbers.

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Behavior of charges

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Cathode ray tube

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Properties of cathode rays
Electron m/e = -5.6857 x 10-9
g coulomb-1

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Charge on the electron
 From 1906-1914 Robert Millikan showed ionized oil drops
can be balanced against the pull of gravity by an electric field.
The charge is an integral multiple of the electronic charge, e.

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Radioactivity
Radioactivity is the spontaneous emission of radiation
from a substance.
 X-rays and γ-rays are high-energy light.
 α-particles are a stream of helium nuclei, He2+.
 β-particles are a stream of high speed electrons
that originate in the nucleus.

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The nuclear atom
Geiger and Rutherford
1909

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The α-particle experiment
 Most of the mass and all of the
positive charge is concentrated in a
small region called the nucleus .
 There are as many electrons outside
the nucleus as there are units of
positive charge on the nucleus

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The nuclear atom
Rutherford
protons 1919
James Chadwick
neutrons 1932

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Atomic Diameter 10-8
cm Nuclear diameter 10-13
cm
Nuclear Structure
Particle Mass Charge
kg amu Coulombs (e)
Electron 9.109 x 10-31
0.000548 –1.602 x 10-19
–1
Proton 1.673 x 10-27
1.00073 +1.602 x 10-19
+1
Neutron 1.675 x 10-27
1.00087 0 0
1 Å

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Scale of Atoms
Useful units:
 1 amu (atomic mass unit) = 1.66054 x 10-24 kg
 1 pm (picometer) = 1 x 10-12 m
 1 Å (Angstrom) = 1 x 10-10 m = 100 pm = 1 x 10-8
cm
The heaviest atom has a mass of only 4.8 x 10-22 g
and a diameter of only 5 x 10-10
m.
Biggest atom is 240 amu and is 50 Å across.
Typical C-C bond length 154 pm (1.54 Å)
Molecular models are 1 Å /inch or about 0.4 Å /cm

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Isotopes, atomic numbers and mass numbers
To represent a particular atom we use the symbolism:
A= mass number Z = atomic number

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Measuring atomic masses

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The Periodic tableAlkali Metals
Alkaline Earths
Transition Metals
Halogens
Noble Gases
Lanthanides and Actinides
Main Group
Main Group

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The Periodic Table
• Read atomic masses.
• Read the ions formed by main group elements.
• Read the electron configuration.
• Learn trends in physical and chemical properties.
We will discuss these in detail in Chapter 10.

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The Mole
• Physically counting atoms is impossible.
• We must be able to relate measured mass to
numbers of atoms.
– buying nails by the pound.
– using atoms by the gram

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Avogadro’s number
The mole is an amount of substance that
contains the same number of elementary
entities as there are carbon-12 atoms in
exactly 12 g of carbon-12.
NA = 6.02214199 x 1023
mol-1

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Molar Mass
• The molar mass, M, is the mass of one mole
of a substance.
M(g/mol 12
C) = A(g/atom 12
C) x NA (atoms 12
C /mol 12
C)

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Combining Several Factors in a Calculation—Molar Mass, the
Avogadro Constant, Percent Abundance.
Potassium-40 is one of the few naturally occurring radioactive
isotopes of elements of low atomic number. Its percent natural
abundance among K isotopes is 0.012%. How many 40K
atoms do you ingest by drinking one cup of whole milk
containing 371 mg of K?
Want atoms of 40
K, need atoms of K,
Want atoms of K, need moles of K,
Want moles of K, need mass and M(K).
Example 2-9

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Convert strategy to plan
mK(mg) x (1g/1000mg)  mK (g) x 1/MK (mol/g)  nK(mol)
Convert mass of K(mg K) into moles of K (mol K)
Convert moles of K into atoms of 40
K
nK(mol) x NA  atoms K x 0.012%  atoms 40
K
nK = (371 mg K) x (10-3
g/mg) x (1 mol K) / (39.10 g K)
= 9.49 x 10-3
mol K
and plan into action
atoms 40
K = (9.49 x 10-3
mol K) x (6.022 x 1023
atoms K/mol K)
x (1.2 x 10-4 40
K/K)
= 6.9 x 1017 40
K atoms

25. Prentice-Hall ©General Chemistry:Slide 25 of 25
Chapter 2 Questions
3, 4, 11, 22, 33,
51, 55, 63, 83.