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Atomic Theories
Atomic timeline
• Your poster should contain:
– The date the theory was developed
– A picture of the ‘model’
– The name of the main contributor
– A short summary of the theory
Atomic Timeline
450 BC
1896
1803
1911
1910
450 BC - Democritus
• Said that matter is composed of tiny,
indivisible particles called “atomos”
• Atom (2006): Smallest part of an element
that still has the property of that element
I was close
enough. You little
punks living in
your AD don’t
know how good
you have it.
1803 – John Dalton
• The theory that atoms are the fundamental
building blocks of matter reemerged in the
early 19th century
• John Dalton: School teacher
• Based on observations of
how elements combine
•
1803
–
John
Dalton
1. Elements are composed of extremely small
particles summarized
called atoms what was already said
• Basically
•in2.his
AllTop
atoms
of an
element are identical
in mass
Five
Descriptions
of Matter
& properties
• 3. Atoms of one element are different from
atoms of another element
Haha, that’s like
• 4. Atoms of an element
are not changed into
my top ten list that
atoms of another element;
I do nor are they created
everynight..except
or destroyed in chemical
reactions
it’s the top 5 – and
about matter.
I
• 5. Compounds are its
formed
when
atoms
get it.
combine; a given compound
always has the
same relative number & kind of atoms
Daltons Table of Elements
Dalton’s Postulates…
• Law of Constant Composition (was known at the time)
– In a compound, the relative #’s and kinds of atoms stay the
same
5. Compounds are formed when atoms combine; a given compound always has the same
relative number & kind of atoms
• Law of Conservation of Mass (was known at the time)
– The total mass of materials present after a chemical reaction is
the same as the total mass present before
Massreactants = MassProducts
4. Atoms of an element are not changed into atoms of another element; nor are they
destroyed in chemical reactions
created or
• Law of Multiple Proportions (Dalton predicted!)
– If two elements A and B combine to form more than one
compound, they do so in simple whole number ratios
J.J. Thomson – The Cathode Ray Tube
• Discovered negatively charged part of an
atom
• This was the first Sub Atomic particle
• Was able to use his cathode ray tube to
determine Charge:Mass ratio
J.J. Thomson’s Model
•Electrons were small mass, so they must
be small size compared to the atom
•Explain where the
electrons were located in his
Plum Pudding Model
•Uniform positive sphere
with electrons embedded
Millikan’s Oil Drop Experiment
• Robert Millikan
– Determined the charge of an electron
– Used this & Thomson’s charge:mass ratio to
determine the mass
– Mass e- = 9.10 x 10-28 g
Rutherford’s Gold Foil
Rutherford (and his gold foil)
• Rutherford
– Gold Foil Experiment
– Atom is mostly empty space (how empty?)
– Nucleus: Contains all of the mass (neutrons and
protons)
– Electrons: Account for most
of the volume
Questions
• Why do things have different mass?
• Why does this happen?
Modern Atomic Theory
2 Problems with Dalton’s
Theory
1. Atoms are divisible
2. Atoms of an element can
be different
Subatomic Particles
• Proton: Discovered by Rutherford in 1919
– Charge: 1.602 x 10-19 C
• Neutron: Discovered by James Chadwick in
1932
– Charge: 0
1 amu = -19
1.66054 x 10-24 g
• Electron: (J.J.) Charge: -1.602 x 10
C
Identifying Atoms
• Atomic number:
– # of protons in the nucleus
– “Address” of the atom – identifies what element it
is
• Atomic Weight:
– Average mass of the atom
• Mass Number
Why don’t
we count
electrons?
– The number of protons + neutrons in the nucleus
– Whole #
1836 Electrons = mass of 1 proton
Using these numbers…
• Atomic # = # of protons = # of electrons
• Mass # = protons + neutrons
• Neutrons = Mass # - protons
Example’s
Determine the number of P+, No, and E- in…
1) Mg
Atomic # = 12 12protons, 12electrons
Mass # = 24
12 neutrons
2) Tc
Atomic # = 43 43protons, 43electrons
Mass # = 99
56 neutrons
3) Mn
Atomic # = 25 25protons, 25electrons
Mass # = 55
30 neutrons
4) Sn
Atomic # = 50 50protons, 50electrons
Mass # = 119
69 neutrons
Chemical Symbols
a.
1.
2.
3.
4.
Cannot determine number of electrons without
additional information.
The atom has 30 electrons.
The atoms has 15 electrons.
The atom has no electrons unless it is charged.
a.
1.
2.
3.
4.
Cannot determine number of electrons without
additional information.
The atom has 30 electrons.
The atoms has 15 electrons.
The atom has no electrons unless it is charged.
b.
1.
2.
3.
4.
The protons reside in the nucleus of the atom.
The protons are evenly distributed throughout the
atom.
The protons are dispersed with the electrons
around the nucleus.
The protons reside in a shell just outside the
nucleus.
b.
1.
2.
3.
4.
The protons reside in the nucleus of the atom.
The protons are evenly distributed throughout the
atom.
The protons are dispersed with the electrons
around the nucleus.
The protons reside in a shell just outside the
nucleus.
Modern Atomic Theory
2 Problems with Dalton’s
Theory
1. Atoms are divisible
2. Atoms of an element can
be different
Isotopes
• Atoms of the same element with different #’s of
neutrons
11
C
6
12
C
6
13
C
6
14
C
6
D:\Chapter_02\Present\eMedia_Library\HydrogenI
sotopes\HydrogenIsotopes.html
Average Atomic Mass
• Isotopes in Banana’s
19
K
39.098
This number represents the average
of all the naturally occurring isotopes
of Potassium
Relative Abundance
• To calculate average mass, we must know
how much each isotope occurs
• Calculating Average:
(mass isotope 1 x %) + (mass isotope 2 x %) +….
A hypothetical element has two isotopes.
One of the isotopes has an abundance of
75.0% and a relative mass of 12.0 amu, while
the other has an abundance of 25.0% and a
relative mass of 14.0 amu. The atomic mass
of this hypothetical element is
1. 12.5 amu.
2. 13.0 amu.
3. 13.5 amu.
4. 13.8 amu.
Correct Answer:
1. 12.5 amu.
2. 13.0 amu.
3. 13.5 amu.
4. 13.8 amu.
Atomic mass = (0.750)(12.0 amu) + (0.250)(14.0 amu)
Atomic mass = 9.0 amu + 3.5 amu = 12.5 amu
SAMPLE EXERCISE 2.4 Calculating the Atomic Weight of an Element from
Isotopic Abundances
Naturally occurring chlorine is 75.78% 35Cl, which has an atomic mass of 34.969 amu, and
24.22% 37Cl, which has an atomic mass of 36.966 amu. Calculate the average atomic mass (that
is, the atomic weight) of chlorine.
Solution The average atomic mass is found by multiplying the abundance of each isotope
by its atomic mass and summing these products. Because 75.78% = 0.7578 and 24.22% =
0.2422, we have
This answer makes sense: The average atomic mass of Cl is between the masses of the
two isotopes and is closer to the value of 35Cl, which is the more abundant isotope.
PRACTICE EXERCISE
Three isotopes of silicon occur in nature: 28Si (92.23%), which has an atomic mass of
27.97693 amu; 29Si (4.68%), which has an atomic mass of 28.97649 amu; and 30Si
(3.09%), which has an atomic mass of 29.97377 amu. Calculate the atomic weight of
silicon.
Answer: 28.09 amu
Ions
• Atoms of the same element with different #’s
of electrons
SAMPLE EXERCISE 2.3 Writing Symbols for Atoms
Magnesium has three isotopes, with mass numbers 24, 25, and 26. (a) Write
the complete chemical symbol (superscript and subscript) for each of them.
(b) How many neutrons are in an atom of each isotope?
Solution (a) Magnesium has atomic number 12, and so all atoms of magnesium
contain 12 protons and
12 electrons. The three isotopes are therefore represented by
(b) The number of neutrons in each isotope is the mass number minus the
number of protons. The numbers of neutrons in an atom of each isotope are
therefore 12, 13, and 14, respectively.
PRACTICE EXERCISE
Give the complete chemical symbol for the atom that contains
82 protons, 82 electrons, and 126 neutrons.
How many neutrons are there in an
atom of 14C?
1.
2.
3.
4.
6
8
12
14
Correct Answer:
1. 6
2. 8
3. 12
4. 14
Mass number 14
Atomic number 6
C
The difference between the mass number and the
atomic number is the number of neutrons (14  6) = 8.
An isotope of Cr3+ ion containing 24
protons and 27 neutrons would
contain ___ electrons.
1.
2.
3.
4.
21
24
27
3
Correct Answer:
1.
2.
3.
4.
21
24
27
3
The +3 charge
indicates there
are 3 more
protons than
electrons:
24  3 = 21