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Topic 2:
Atoms and The Atomic Theory
Contents
• Early chemical discoveries
• Electrons and the nuclear atom
• Chemical elements
• Atomic masses
• The Avogadro constant & the Mole concept
Early Discoveries
Democritus (B.C 470-400)
Democritus is a philosoph lived in the
ancient times of Greece. He brought up the
modernest theory concerned with the nature
of the physical world in that period of time.
According to Democritus;
• Every thing existing in nature is composed
of atoms.
• Atoms are very tiny particles which can
not be perceived by our senses.
• All of them are made of the same matter,
but they all have different sizes,shapes and
weights.
• Atoms are those particles,which can not
be split into parts,can not be annihilated and
created
Early Discoveries
Lavoisier 1774 Law of Conservation of Mass
Proust 1799
Law of Constant Composition
Dalton 1803-1808
Dalton’s Atomic Theory
Law of Conservation of Mass
Lavoisier, known as the establisher of Chemistry
The mass of substances formed by a chemical
reaction is the same as the mass of substances
entering into the reaction.
Reactants
Silvernitrate
and
potasium
chromate
Products
Silverchromate
and
potasiumnitrate
2 AgNO3 (aq) + K2CrO4 (aq)  Ag2CrO4 (s) + 2 KNO3 (aq)
Law of Constant Composition
All samples of a compound have the
same composition- the same proportions
by mass of the constituent elements.
Proust 1799
This is called Law of Constant Composition
Sample A
10.0000 g
1.1190 g H
8.8810 g O
Composition
% H=11.19
% O=88.81
Sample B
27.0000 g
3.0213 g H
23.9787 g O
Water consists of %11,19 of
hydrogen and %88,81 of oxygen
Dalton’s Atom Theory
•
John Dalton
(1766-1844)
•
•
•
John Dalton’s Theory involves three
assumptions:
All atoms of an element are alike in mass
and other properties, but the atoms of
one element are different from those of
all other elements.
Each chemical element is composed of
minute, indestructible particles called
atoms. Atoms can neither be created nor
destroyed during a chemical change.
In each of their compounds different
elements combine in a simple numerical
ratio.For example one atom of A to one of
B(AB), or one atom A to two of B(AB 2
EFFECT OF ELECTRICITY and MAGNETISM ON
THE ATOMIC STRUCTURE
.
Behaviour of electrically charged particles
All objects of matter are made up of electrically
charged particles.
Q  Q  Neutral
 
Q  Q
Q  Q




 Positively charged


 Negatively charged
• Electrically charged particles
deflect in a magnetic field.
Q1 Q2
F k 2
r
k:Coulomb
constant
EFFECT OF ELECTRICITY and
MAGNETISM ON THE ATOMIC
STRUCTURE
• Objects with like charges repel one another.
•Objects with unlike charges attract one another.
The force, (F),of attraction or repulsion is
directly proportional to the magnitude or
quantity of charges (Q1 ve Q2) and inversely
proportional to the square of the distance
between them (r2). This force is called Coulomb
force.
Behaviour of electrically charged
particles
Electrically
charged
particles are deflected from
the straight line path into
the
curved
path
perpendicular to the field
Negatively
charged
particles are deflected in
one direction, whereas
positively charged particles
are
deflected
in
the
opposite direction, when
they travel at right angles
through a magnetic field.
Assumption is made that the
influence of the magnet as
represented lines of force
run from the north pole to the
south pole of the magnet.
The Discovery of Electrons
CRT,the abbreviation for cathode ray tube has become an
everyday expression. The CRT is the heart of the familiar
computer monitor or TV set. The first cathode ray tube was
made by Michael Faraday (1791-1867) .
Faraday discovered cathode rays, a type of radiation
emitted by the negative terminal cathode that
crossed the evacuated tube to the positive terminal
or anode.
Later scientists found that cathode rays travel in
straight lines and have properties that are
independent of the cathode material.
The Discovery of Electrons
Cathode rays are invisible and they can only be
detected by the light emitted by materials they
strike. Fluorescence is the term used to describe
the emission of light by a material when it is struck
by energetic radiation.
In 1897, Thomson established the ratio of mass to
electric charge (m/e). He concluded that cathode
rays are negatively charged fundamental particles
of matter found in all atoms. Cathode rays
subsequently became electrons, the term used
firstly by George Stoney in 1874
Figure 2.1 Determining the mass/charge ratio for cathode rays
The properties of Cathode rays
In 1897, Thomson established the ratio of mass to electric
charge (m/e). He concluded that cathode rays are negatively
charged fundamental particles of matter found in all atoms.
Cathode rays subsequently became electrons, the term
used firstly by George Stoney in 1874
Cathode beam rays
are deflected by the
electrical and
magnetic field.
The ratio m/e = -5.6857 x 10-9 g/coulomb
Millikan’s Oil Drop Experiment
•Ions are produced by energetic X rays.Some of
these ions become attached to oil droplets,giving
them a charge.
•The fall of a droplet between the condenser
plates is speeded up or slowed down, depending
on the magnitude and sign of the charge on the
droplet
Millikan’s Oil Drop Experiment
Millikan
concluded
that
the
magnitude of the charge, q on a
droplet is an integral multiple of the
electronic charge,e ; q= ne (where
n=1,2,3…..)
q= ne
Robert Millikan
(1868-1953)
• Millikan determined the electronic charge e
through a series of oil drop experiments
• By combining this value with an accurate value of the
mass to charge ratio for an electron we get the mass
of an electron .
e = 1,6022 x 10-19 C
me = 9,1094 x 10-28 g
Figure 2.3 Millikan’s oil drop
experiment
Thomson’s Atom Model
He atoms and ions
according to Thomson’s
atom model
Thomson’s Atom Model
After the discovery of electrons Thomson
brought up an atom model.
The positive charge needed to counterbalance
the negative charges of electrons in a neutral
atom is in the form of a nebulous cloud.
Electrons float in this diffused cloud of
positive charge (Plum pudding model).
X-Rays and Radioactivity
• 1895 Wilhelm Roentgen noticed
that when cathode ray tubes
were operating certain materials
outside the tubes would glow or
fluoresce.
Because
of
the
unknown nature of this radiation,
he coined the term X-rays.
• Antoine Becquerel discovered radioactivity as a result
of series experiments with the uranium containing
material. Radioactivity is the spontaneous emission of
energy from unstable atoms. Radioactive decay is the
process by which an atomic nucleus loses its energy
by emitting ionizing particles
X-Rays and Radioactivity
•Ernest
Rutherford,
4
2  identified two types of
2
radiation from radioactive
materials, alpha and beta
0
rays,
Paul
Villard
1
discovered gamma rays.
He
e
• Marie and Pierre Curie
discovered
plenty
of
radioactive elements
. The most important
ones are Ra and Po.
X-Rays and Radioactivity
Alpha particles: carry two fundamental units
of positive charge and have the same mass
as helium atoms. They have +2 charges.
Beta particles: are negatively charged
particles produced by changes occuring
within the nuclei of radioactive atoms and
have the same properties as electrons.
Gama rays are defined as the electromagnetic
radiation of extremely high penetrating power.
When the radiation is passed through the electric field,
gamma rays are undeflected. Alpha particles and beta
particles are deflected in the opposite directions of
each other.
Three types of radiation
The Nuclear Atom
The experiment of thin foils of gold
In 1909 Geiger and Ernest Marsden bombarded very thin
foils of gold with alpha particles, here is what they
observed:
The Nuclear Atom
The scattering of α particles by metal foil
The deflection of alpha particles according to
the Thomson’s model
The Nuclear Atom
1.The majority of α particles penetrated the foil
undeflected
2.Some of α particles are slightly scattered as
they encounter electrons
3.A few(about one in 20000) suffer rather
serious deflections as they penetrate the foil
4.A similar number can not pass through the
foil at all but bounce back in the direction from
which they have come.
These results can not be explained by
Thomson’s model.
The Nuclear Atom
Rutherford’s Atom Model
1. Most of the mass and all of the
positive charge of an atom are
centered in a very small region
called the nucleus.
2. The magnitude of positive charge
is different for different other
atoms and is approximately one
half the atomic weight of the
element.
3. There exist as many electrons
outside the nucleus as there are
units of positive charge on the
nucleus. The atom as a whole is
electrically neutral.
The Nuclear Atom
Rutherford discovered in 1919 the particles called
protons, in studies involving the scattering of alpha
particles by nitrogen atoms in air.
In 1932, James Chadwick discovered the neutrons
originated from the nuclei of atoms .
They are
uncharged particles.
The sum of the protons and neutrons
located in the nucleus of an atom is defined
as nucleon.
Rutherford’s
atom model
Fundamental Particles of Atom
Electron (e) unique atom unit has negative
charge.
Proton (p) unique atom unit has positive
charge .
Neutron (n) is uncharged.
~
m p  mn  me
Atomic mass unit: 1/12 of the mass of the atom
known as carbon-12, its abbreviation is amu .
Fundamental Particles of Atom
The number of protons in the nucleus of an atom
is called atomic number (Z), the total number of
protons and neutrons in its nucleus is called
mass number (A) .
In a neutral
atom
Q= p-e
Z  p pe
A pn
n  A Z
Atomic Scale
The mass of the heaviest atom is only 4.32 x 10-22 g
and the diameter of this atom is 5 Å’.
The units frequently used in the atomic scale:
1 amu (atomic mass unit) = 1.66054 x 10-24 kg
1 pm (pikometer) = 1 x 10-12 m
1 Å (Angstrom) = 1 x 10-10 m = 100 pm = 1 x
10-8 cm
Tipical C-C bond length= 154 pm (1.54 Å).
Chemical Elements
All the atoms of an element have the same
atomic number (Z). At present the elements
with the atomic numbers ranging from Z=1 to
Z=109 are known.
Each element has a name and a distinctive
symbol.
For most elements the chemical symbol is an
abbreviation of the English name and consists
of one or two letters.
Some elements known since ancient times
have symbols based on their Latein names,
such as Fe for iron(ferrum )
Chemical Elements
Element
Carbon
Oxygen
Nitrogen
Sulphur
Neon
Silicon
Symbol
C
O
N
S
Ne
Si
Element
Ferrium
Plumbum
Natrium
Kalium
Symbol
Fe
Pb
Na
K
Elements beyond uranium (Z=92) do not occur
naturally and must be synthesized in particle
accelerators .
Chemical Elements
Number of protons +
number of neutrons
Number of protons
A
Z
X

Charge
(p-e)
Symbol of
Element
Dalton put up the argument that all of the
atoms of each element had the same
atomic mass but we know at present not
every atom of one element may have the
same atomic mass.
Chemical Elements
Example:
20
10
Ne
21
10
Ne
22
10
All of the Ne atoms
have 10 protons in
their
nuclei.However the
number of neutrons
in their structures
varies from 11 to
12.
Ne
Isotopes:
Two or more atoms having the
same atomic number (Z) but
different mass numbers (A) are
called isotopes
20
10
Percent Natural Abundance
Ne
%90,9
21
10
Ne
%0,3
22
10
Ne
%8,8
Chemical Elements
I.
20
10
22
II.
10
Ne

Ne
2
ion
An element that has either lost or gained electrons
is called an ion and carries a net electric charge.
The number of protons never changes when an
atom becomes an ion. The first neon ion has 10p,
10n, 9e . (+1 charged)
The second one has 10p, 10n, 8e . (+2 charged)
Chemical Elements
Example:
Indicate the number of protons,electrons and
neutrons in
Z  p  17 A  p  n  35  17  n  n  18 p  e  17
Example:
Write an appropriate symbol for the
species consisting of 29 protons, 34
neutrons and 27 electrons
Isotopic Masses
We can not determine the mass of an individual
atom just by adding up the masses of its
fundamental particles.
When an atomic nucleus is created from
protons and neutrons, a small quantity of the
original mass is converted to energy.
This is the nuclear binding energy that holds
the protons and neutrons together and we can
not predict exactly how much this will be.
Isotopic Masses
How can we determine atomic masses?
We arbitrarily choose one atom and assign
it a certain mass. By international
agreement this standard is an atom of the
isotope C-12 which is assigned a mass of
12 amu. The masses of other atoms relative
to this value are determined. To do this we
use a mass spectrometer .
Mass spectrometry
A gaseous sample is ionized by bombardment
with electrons in the lower part of the
apparatus. The positive ions thus formed are
subjected to an electrical force by the
electrically charged velocity selector plates
and a magnetic force by a magnetic field
perpendicular to the page. Only ions with a
particular velocity
pass through and are
deflected into circular paths by the magnetic
field . Ions with different masses strike the
detector(here
the phtographic plate) in
different regions.
Mass spectrometry
Mass spectrometry
With mass spectral data,the ratio of the mass
Example: of 16O to 12C is found to be 1,33291. What is
the mass of an 16O atom?
The ratio of the masses is 16O / 12C =1,33291 ,the mass of an
16O is 1,33921 times the mass 12C ;
16O
(mass of oxygen) = 1,33291 x 12,00000 amu
=15,9949 amu
16O is found
By
mass
spectrometry
an
atom
of
Example:
to have 1,06632 times the mass of 15N. What
is the mass of an 15N, expressed in atomic
mass units?
16O mass = 1,06632 x 15N mass
15N
mass = 15,9949 amu / 1,06632 =15,00009 amu
Atomic masses
In a table of atomic masses the value listed for
carbon is 12,011, yet the atomic standard mass is
exactly 12. What is the reason for this difference?
The atomic mass standard is based on a sample
of carbon containing only atoms of 12C isotope,
whereas naturally occuring carbon contains some
13C and 14C isotopes as well. The existence of
these three isotopes causes the observed atomic
mass to be greater than 12
Atomic masses
Atomic mass (weight) of an element is the average
of the isotopic masses, weighted according to the
naturally occuring abundances of the isotopes.
We can calculate it by using the formula given
below:
Atomic
mass of
an
element
Fractional
abundance
Mass of
= of isotope x
isotope(1)
(1)
Fractional
abundance
+ of isotope x Mass of
isotope (2)
(2)
+…
Atomic masses
12C
isotope: 12,00000 amu; percent natural abundance: 98,892 %
13C isotope: 13,00335 amu; percent natural abundance: 1,108 %
Atomic mass of naturally occuring C is calculated as:
At. mass
Fractional
Fractional
of
x Mass of
abundance
abundance
naturally =
x Mass of +
13C
12
13
of C
12C
of C
occuring
carbon
At. mass
of
=
naturally
occuring
carbon
0,98892 12,00000  0,01108 13,00335
Atomic mass of
C
= 12,011 amu
The Avogadro Constant and
the concept of the mole
• Since atoms are very tiny particles and there are as
many as uncountable atoms used in Chemistry, it is
impossible to measure their masses individually. For this
reason a certain kind of system has been improved. The
SI quantity describes it as mole.
• A Mole is an amount of substance that contains the
same number of elementary
entities(atoms,molecules,other particles as there are
12C in in exactly 12 g of carbon. An italian scientist
namely Avogadro provided that the number of
elementary entities are defined as the Avogadro
constant,
1 mol = 6,02214 x 1023 atoms,molecules
The Avogadro Constant and
the Concept of the Mole
12,00g 12C = 1mol 12C atom = 6,022x1023
12,00g sample of 12C
6,022 x 1023
12C
12C
atoms
1,993 x 10-23g = the mass of a unique
atoms
1
12C
atom = 1,993 x 10-23g
1
12C
atom = 12amu
1amu = 1,661 x 10-24g
1,661 x 10-24g
12C
atom
1g
6,022 x 1023
6,022 x 1023 amu
1 mol
Oxygen: 16 amu
Oxygen= 16g/mol
The atomic mass ( the mass of one atom) and the
molecular weight (the mass of 6,022 x 1023 atoms)
are quantitatively the same but the units of them are
different :
1 oxygen atom = 16 amu
1 mol oxygen = 16 g
F-19
Mg-24,26,25
Cl-35,37
Pb-207,206,208,204
The Avogadro Constant and the
Concept of the Mole
Example: Potasyum-40 (40K) is one of the few naturally
occuring isotopes of elements of low atomic number. Its
percent natural abundance is 0.012% . How many 40K atoms
do you ingest by drinking one cup of whole milk containing
371 mg K ? K= 39,1
mK(mg) x (1g/1000mg)  mK (g) x 1/MK (mol/g)  nK(mol)
nK = (371 mg K) x (10-3 g/mg) x (1 mol K) / (39,10 g K) = 9,49 x 10-3 mol K
nK(mol) x NA  atoms K x 0.012%  atoms 40K
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