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Prentice-Hall © 2002General Chemistry: Chapter 2Slide 1 of 25
Chapter 2: Atoms and the Atomic Theory
Philip DuttonUniversity of Windsor, Canada
Prentice-Hall © 2002
General ChemistryPrinciples and Modern Applications
Petrucci • Harwood • Herring
8th Edition
Prentice-Hall © 2002General Chemistry: Chapter 2Slide 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.
Prentice-Hall © 2002General Chemistry: Chapter 2Slide 10 of 25
Radioactivity
Radioactivity is the spontaneous emission of radiation from a substance.
X-rays and -rays are high-energy light.
-particles are a stream of helium nuclei, He2+.
-particles are a stream of high speed electrons
that originate in the nucleus.
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The nuclear atom
Geiger and Rutherford1909
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The -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
Rutherfordprotons 1919
James Chadwickneutrons 1932
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Atomic Diameter 10-8 cm Nuclear diameter 10-13 cm
Nuclear Structure
Particle Mass Chargekg amu Coulombs (e)
Electron 9.109 x 10-31 0.000548 –1.602 x 10-19 –1Proton 1.673 x 10-27 1.00073 +1.602 x 10-19 +1Neutron 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
Prentice-Hall © 2002General Chemistry: Chapter 2Slide 19 of 25
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 12C) = A (g/atom 12C) x NA (atoms 12C /mol 12C)
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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 40K, 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 40K
nK(mol) x NA atoms K x 0.012% atoms 40K
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 40K = (9.49 x 10-3 mol K) x (6.022 x 1023 atoms K/mol K)
x (1.2 x 10-4 40K/K)= 6.9 x 1017 40K atoms
Prentice-Hall © 2002General Chemistry: Chapter 2Slide 25 of 25
Chapter 2 Questions
3, 4, 11, 22, 33, 51, 55, 63, 83.