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TOPIC 2: Structure of Matter
OUTLINE
2.0 STRUCTURE OF
MATTER
2.1 Early models of atom
2.2 Modern models of atom
2.3 Structure of atom
OUTLINE
2.1 Early models of atom
2.1.1 Democritus
2.1.2 John Dalton
2.1.3 J.J.Thompson
2.1.4 Ernest Rutherford
2.1.1 Democritus (460 B.C. – 370 B.C.)
Greek
philosopher
(“atomos”)IDEAS
1.All matter composed of
atoms in void
2.Atoms were indivisible &
indestructible
3.Properties of atom differ in
shape, arrangement and sizes
IDEAS
4.Shape of atom
determined
the substance properties
*Eg:
-Fire atoms had sharp points
-Wine atoms were spheres
-Clay atoms were jagged
2.1.2 John Dalton (1766 – 1844)
2.1.3 J.J Thomson – (1856 -1940)
2.1.3 J.J Thomson – (1856 -1940)
Sir William Crookes - 1875
From
experiments he
was able to prove
the ray was
matter
When he connected
the plates to the + and
-
terminals of a high
voltage battery he
observed a
mysterious glowing
blue ray
He removed
almost all
the air from the
tube and sealed
it
Constructed a glass
tube
with two metal
plates inside it
Extra Note
2.1.3 J.J Thomson – (1856 -1940)
2.1.3 J.J Thomson – (1856 -1940)
2.1.3 J.J Thomson – (1856 -1940)
Result’s
Experiment
The ray composed of negatively charged
matter
The ray attracted to + plate and
repelled by - plate
100 years earlier, Dalton proposed
atoms were neutral, solid spheres
Thomson’s experiment disproved
Dalton’s theory
IDEAS
Proposed that matter was made of
atoms with negatively charged
particles embedded in a positive
cloud
He called these negative particles
“electrons”.
The negative particles and positive
cloud neutralized each other so the
atom had an overall neutral charge.
2.1.3 J.J Thomson – (1856 -1940)
Plum Pudding Model
2.1.4 ERNEST RUTHERFORD
(1871-1937)
Rutherford used
this alpha particle
to investigate the
structure
Rutherford and
Geiger in the
Cavendish Lab of
the atom.
This particle was
called an
alpha particle
(α).
One type of
radioactivity is
when
an atom throws out
a positively
charged particle
from the nucleus
4. A screen
around the gold
to detect
where the alpha
particles were
traveling.
2. He fired these positive particles at a
thin piece of gold (dense metal).
3. This produced a beam of alpha particles
traveling in a straight line
1. He encased
uranium in lead
(which absorbs
alpha particles)
2.1.4 ERNEST RUTHERFORD
(1871-1937)
Gold Foil Experiment
Uranium is a radioactive
element that gives off
positive particles
(alpha particles).
Rutherford used these
positive particles to
investigate the makeup of
the atom.
URANIUM
2.1.4 ERNEST RUTHERFORD
(1871-1937)
•Rutherford shot alpha particles at a thin sheet of gold to
observe what happened when the positive α particles
passes through the gold atoms.
•If Thompson’s model was correct, the alpha particles
should pass through the diffused positive cloud with ease.
2.1.4 ERNEST RUTHERFORD
(1871-1937)
Gold Foil experiment
He concluded that the
atom had a dense,
positive central nucleus
composed of + charged
protons
He proposed his
Planetary Model of the Atom.
His model created positively
charged protons located in the
nucleus and placed
electrons in orbit around the
nucleus - like
planets around the sun.
2.1.4 ERNEST RUTHERFORD
(1871-1937)
• Based on his experimental evidence:
–The atom is mostly empty space
–All the positive charge, and almost all the
mass is concentrated in a small area in the
center. He called this a “nucleus”
–The electrons distributed around the nucleus,
and occupy most of the volume
–His model was called a “nuclear model”
2.1.4 ERNEST RUTHERFORD
(1871-1937)
• VIDEO 1
• VIDEO 2
SUMMARY
· Dalton's model of the atom - solid, tiny, indivisible
particles.
· Thomson's model - often describe as the "plum
pudding" model - electrons are scattered throughout the
atom.
· Rutherford's model - includes the solid nucleus in
the center of the atom.
OUTLINE
2.2 Modern models of atom
2.2.1 Niels Bohr
2.2.2 James Chadwid
2.2.1 NIELS BOHR
Modification of Rutherford Model
The Bohr model of the atom is based
on the quantum mechanics.
2.2.1 NIELS BOHR
1
• Electrons orbit the nucleus (in orbits) that have a set size and
energy (7 Energy level)
2
• The energy of the orbit is related to its size. The height energy is
found in the largest orbit
3
• The arrangement of these electrons in certain energy levels
determines the chemical and physical properties of the elements
4
• Radiation is absorbed or emitted when an electron moves from
one orbit to another. And the energy change E = E2 – E1
IDEAS
VIDEO 3
2.2.2 JAMES CHADWICK
1
• In 1932 James Chadwick discovered a 3rd
subatomic particle in the atom
2
• He called this particle a neutron because it
was neutral
3
• He proposed that neutrons resided in the
nucleus of the atom along with protons
VIDEO 4
2.1 Early models of atom
2.1.1 Democritus
2.1.2 John dalton
2.1.3 J.J.Thompson
2.1.4 Ernest Rutherford
OUTLINE
2.3 Structure of atom
2.3.1 Defining the atom
2.3.2 Distinguishing among atoms
OUTLINE
2.3.1 Defining the atom
2.3.1.1 Electron
2.3.1.2 Nucleus
2.3.1.3 Structure nucleus
2.3.1 Defining the atom
Neutrons have no lectrical
charge
1
• Atomic radius: 40 -270 picometer
2
• Smallest part of an element
3
• 3 subatomic particles: protons, neutrons, electrons
4
• Protons = Positive (+) charge
• Neutrons = Neutral
• Electron = Negative (-) charge
2.3.1.1 Electron
ELECTRON
Discovery
J J Thomson
-Cathode ray
tube
- Cathode
rays
composed of
negatively
charged
particles
(electrons)
Mass
(0.005
amu)
Determine the
atomic
number (Z)
(the atom’s
identity)
Location
Outside
nucleus
(Orbital)
Charge
Negative
charge
2.3.1.2 Nucleus
NUCLEUS
Discovery
Ernest
Rutherford
Determine
the Atom's
mass
Location
Central part
of an atom
Made up
Proton and
neuton
2.3.1.2 Nucleus
2.3.1.3 Structure of Nucleus
NEUTRONS
Mass = 1 amu,
charge = 0
Neutrons act as a
type of “insulation”
between the protons,
preventing them
from repelling each
other
PROTONS
Mass = 1 amu,
charge = +1
Determines the
atom's identity
(atomic number)
OUTLINE
2.3.2 Distinguishing among atoms
2.3.2.1 Atomic Number
2.3.2.2 Mass Number
2.3.2.3 Atomic Mass
2.3.2.4 Isotopes
2.3.2.5 Electronic Configuration
2.3.2.1 Atomic Number (Z)
5. Atoms are neutral,
# proton = # electron
1. Atoms are
identified by their
atomic number
4. The atomic number of an
element never changes
6. Eg: Hydrogen atoms,
contain one proton and
have an atomic number 1
2. Number of protons
determine an element's
atomic number
3. Periodic Table is
in order of increasing
atomic number
Element # of protons Atomic # (Z)
Carbon 6 6
Phosphorus 15 15
Gold 79 79
2.3.2.1 Atomic Number
Example…….
Mass number is the number of protons and
neutrons in the nucleus
p+ + n0 = Mass #
2.3.2.2 Mass number
Nuclide p+ n0 e- Mass #
Oxygen 8 10 8 18
Arsenic 33 42 33 75
Phosphorus 15 16 15 31
2.3.2.2 Mass number
Example…..
1. An element's or isotope's atomic
number tells how many protons are in
its atoms.
1. An element's or isotope's mass
number tells how many protons and
neutrons in its atoms.
SUMMARY
1. An element's or isotope's atomic
number tells how many protons are in
its atoms.
2. An element's or isotope's mass
number tells how many protons and
neutrons in its atoms.
SUMMARY
X
Mass
number
Atomic
number
Subscript →
Superscript →
SUMMARY
Br
80
35
 Find each of these:
a) number of protons
b) number of neutrons
c) number of electrons
d) Atomic number
e) Mass Number
Lv
298
116
QUIZ……
QUIZ…..
 If an element has an atomic number
of 34 and a mass number of 78, what
is the:
a) number of protons
b) number of neutrons
c) number of electrons
d) complete symbol
QUIZ……
Measuring Atomic Mass
• Instead of grams (because the numbers would
be too small), the unit we use is the Atomic
Mass Unit (amu)
• It is defined as one-twelfth the mass of a
Carbon-12 atom.
– Carbon-12 chosen because of its isotope purity.
• Each isotope has its own atomic mass, thus
we determine the average from percent
bundance.
2.3.2.3 Atomic mass
 Eg: How heavy is an atom of oxygen?
 It depends, because there are different kinds
of oxygen atoms.
 We are more concerned with the average
atomic mass.
2.3.2.3 Atomic mass
To calculate the average:
• Multiply the atomic mass of each isotope
by it’s abundance (expressed as a
decimal), then add the results.
A = [(mass of isotope) (%abundance) ] + [(mass of isotope) %abundance)] +
100
2.3.2.3 Atomic mass
Atomic Masses
Isotope Symbol Composition of
the nucleus
% in nature
Carbon-12 12C 6 protons
6 neutrons
98.89%
Carbon-13 13C 6 protons
7 neutrons
1.11%
Carbon-14 14C 6 protons
8 neutrons
<0.01%
Atomic mass is the average of all the naturally occurring
isotopes of that element.
Carbon = 12.011
2.3.2.3 Atomic mass
The natural abundance for boron isotopes is:
19.9% 10B (10.013 amu) and 80.1% 11B (11.009amu). Calculate the
atomic weight of boron.
Atomic Mass = [19.9)(10.013)] + [(80.1)(11.009)]
100%
= [199] + [882]
100%
= 10.81 amu
so, the atomic weight of B = 10.81 amu
PROBLEM 1
Verify that the atomic mass of magnesium of 24.31 amu, given the
following: 24Mg= 23.985042 amu, 78.99%
25Mg= 24.985837 amu, 10.00%
26Mg= 25.982593 amu, 11.01%
Atomic mass = [(0.7899)(23.985042)] + [(0.1)(24.985837)] +
[(0.1101)(25.982593)]
= 18.946 + 2.499 + 2.861
= 24.306 amu
PROBLEM 2
Solution:
PROBLEM 3
The atomic mass of lithium is 6.94 amu, the naturally occurring isotopes
are 6Li = 6.015121 amu, and 7Li = 7.016003 amu. Determine the
percent abundance of each isotope.
6.94 = [(% 6Li)(6.015121)] + [(%7Li)(7.016003)]
Since I don’t know what the percentage are, I will have to use variables.
100% of Lithium is determined by these two naturally occurring
isotopes.
We will let 6Li = x and 7 Li = 1-x; we use 1 – x instead of 100
– x because the small number is easier to work with.
(in other words we reduced 100% to decimal form 1.00)
Solution:
PROBLEM 4PROBLEM 3
6.94 = [(% 6Li)(6.015121)] + [(%7Li)(7.016003)]
6.94 = [(x)(6.015121)] +[(1-x)(7.016003)]
6.94 = 6.015121x + 7.016003 – 7.016003x
6.94 -7.016003 = (6.015121x - 7.016003x)
-0.076003 = -1.000882 x
Solve for x:
x = 0.075936
6Li = x and 7 Li = 1-x
Therefore, 6Li = 0.075936 x 100% = 7.59%
7Li = (1- 0.075936 ) x 100% =
Copper exists as two isotopes: 63Cu (62.9298 amu) and 65Cu
(64.9278 amu). What are the percent abundances of the isotopes?
Atomic mass for Cu = 63.546 amu 63Cu % = 1-x 65Cu % = x
63.546 = [(1-x)(62.9298)] + [(x)(64.9278)]
63.546 = 62.9298 - 62.9298x + 64.9278x
1.3818 = 1.998x
x = 1.3818 / 1.998
x = 0.6916
65Cu = 0.6916 x 100% =69.16%
63Cu = 1 – x = 1 – 0.6916 = 0.3084 x 100% =30.84%
PROBLEM 4
Solution:
PROBLEM 5
Lithium has an elemental atomic mass of 6.941 and has two
naturally occurring isotopes, 6Li and 7Li. Their masses are
6.0151 and 7.0160 respectively. What are the natural
abundances (to 2 decimal places in percentage) of the isotopes
of Lithium?
Answer:
6Li: 7.49%, 7Li: 92.51%
• Frederick Soddy (1877-1956)
• Proposed the idea of isotopes in 1912
• Elements with the same number of
protons and a different number of
neutrons (thus have different masses)
• different masses
2.3.2.4 Isotopes
2.3.2.4 Isotopes
EXAMPLES:
ATOMIC STRUCTURE
Electrons are arranged in Energy Levels or Shells
around the nucleus of an atom.
• first shell = a maximum of 2 electrons
• second shell = a maximum of 8 electrons
• third shell = a maximum of 8 electrons
2.3.2.5 Electronic Configuration
ATOMIC STRUCTURE
There are two ways to represent the atomic structure of an
element or compound;
1. Electronic Configuration
2. Dot & Cross Diagrams
2.3.2.5 Electronic Configuration2.3.2.5 Electronic Configuration
ELECTRONIC CONFIGURATION
With electronic configuration, elements are represented numerically by the
number of electrons in their shells and number of shells. For example;
N
Nitrogen
7
14
2 in 1st shell
5 in 2nd shell
Configuration = 2 , 5
2 + 5 = 7
2.3.2.5 Electronic Configuration
1. Electronic Configuration
ELECTRONIC CONFIGURATION
Write the electronic configuration for the following elements;
Ca O
Cl Si
Na
20
40
11
23
8
17
16
35
14
28
B 11
5
a) b) c)
d) e) f)
2.3.2.5 Electronic Configuration
2. DOT & CROSS DIAGRAMS
With Dot & Cross diagrams elements and compounds are represented
by Dots or Crosses to show electrons, and circles to show the shells.
For example;
Nitrogen
N XX X
X
XX
X
N
7
14
2.3.2.5 Electronic Configuration
Draw the Dot & Cross diagrams for the following elements;
O Cl
8 17
16 35
a) b)
2.3.2.5 Electronic Configuration
SUMMARY
1. The Atomic Number of an atom = number of
protons in the nucleus.
2. The Atomic Mass of an atom = number of
Protons + Neutrons in the nucleus.
3. The number of Protons = Number of Electrons.
4. Electrons orbit the nucleus in shells.
5. Each shell can only carry a set number of electrons.
VIDEO 5
LINK….
https://www.nde-
ed.org/EducationResources/HighSchool/Radiography/hs_rad_index.htm
Topic 2: Structure of matter
Topic 2: Structure of matter

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Topic 2: Structure of matter

  • 1. TOPIC 2: Structure of Matter
  • 2. OUTLINE 2.0 STRUCTURE OF MATTER 2.1 Early models of atom 2.2 Modern models of atom 2.3 Structure of atom
  • 3. OUTLINE 2.1 Early models of atom 2.1.1 Democritus 2.1.2 John Dalton 2.1.3 J.J.Thompson 2.1.4 Ernest Rutherford
  • 4.
  • 5. 2.1.1 Democritus (460 B.C. – 370 B.C.) Greek philosopher (“atomos”)IDEAS 1.All matter composed of atoms in void 2.Atoms were indivisible & indestructible 3.Properties of atom differ in shape, arrangement and sizes IDEAS 4.Shape of atom determined the substance properties *Eg: -Fire atoms had sharp points -Wine atoms were spheres -Clay atoms were jagged
  • 6. 2.1.2 John Dalton (1766 – 1844)
  • 7. 2.1.3 J.J Thomson – (1856 -1940)
  • 8. 2.1.3 J.J Thomson – (1856 -1940)
  • 9. Sir William Crookes - 1875 From experiments he was able to prove the ray was matter When he connected the plates to the + and - terminals of a high voltage battery he observed a mysterious glowing blue ray He removed almost all the air from the tube and sealed it Constructed a glass tube with two metal plates inside it Extra Note
  • 10. 2.1.3 J.J Thomson – (1856 -1940)
  • 11. 2.1.3 J.J Thomson – (1856 -1940)
  • 12. 2.1.3 J.J Thomson – (1856 -1940) Result’s Experiment The ray composed of negatively charged matter The ray attracted to + plate and repelled by - plate 100 years earlier, Dalton proposed atoms were neutral, solid spheres Thomson’s experiment disproved Dalton’s theory
  • 13. IDEAS Proposed that matter was made of atoms with negatively charged particles embedded in a positive cloud He called these negative particles “electrons”. The negative particles and positive cloud neutralized each other so the atom had an overall neutral charge. 2.1.3 J.J Thomson – (1856 -1940) Plum Pudding Model
  • 14. 2.1.4 ERNEST RUTHERFORD (1871-1937) Rutherford used this alpha particle to investigate the structure Rutherford and Geiger in the Cavendish Lab of the atom. This particle was called an alpha particle (α). One type of radioactivity is when an atom throws out a positively charged particle from the nucleus
  • 15. 4. A screen around the gold to detect where the alpha particles were traveling. 2. He fired these positive particles at a thin piece of gold (dense metal). 3. This produced a beam of alpha particles traveling in a straight line 1. He encased uranium in lead (which absorbs alpha particles) 2.1.4 ERNEST RUTHERFORD (1871-1937) Gold Foil Experiment
  • 16. Uranium is a radioactive element that gives off positive particles (alpha particles). Rutherford used these positive particles to investigate the makeup of the atom. URANIUM 2.1.4 ERNEST RUTHERFORD (1871-1937)
  • 17. •Rutherford shot alpha particles at a thin sheet of gold to observe what happened when the positive α particles passes through the gold atoms. •If Thompson’s model was correct, the alpha particles should pass through the diffused positive cloud with ease. 2.1.4 ERNEST RUTHERFORD (1871-1937) Gold Foil experiment
  • 18. He concluded that the atom had a dense, positive central nucleus composed of + charged protons He proposed his Planetary Model of the Atom. His model created positively charged protons located in the nucleus and placed electrons in orbit around the nucleus - like planets around the sun. 2.1.4 ERNEST RUTHERFORD (1871-1937)
  • 19.
  • 20. • Based on his experimental evidence: –The atom is mostly empty space –All the positive charge, and almost all the mass is concentrated in a small area in the center. He called this a “nucleus” –The electrons distributed around the nucleus, and occupy most of the volume –His model was called a “nuclear model” 2.1.4 ERNEST RUTHERFORD (1871-1937)
  • 21. • VIDEO 1 • VIDEO 2
  • 22. SUMMARY · Dalton's model of the atom - solid, tiny, indivisible particles. · Thomson's model - often describe as the "plum pudding" model - electrons are scattered throughout the atom. · Rutherford's model - includes the solid nucleus in the center of the atom.
  • 23. OUTLINE 2.2 Modern models of atom 2.2.1 Niels Bohr 2.2.2 James Chadwid
  • 24. 2.2.1 NIELS BOHR Modification of Rutherford Model The Bohr model of the atom is based on the quantum mechanics.
  • 25. 2.2.1 NIELS BOHR 1 • Electrons orbit the nucleus (in orbits) that have a set size and energy (7 Energy level) 2 • The energy of the orbit is related to its size. The height energy is found in the largest orbit 3 • The arrangement of these electrons in certain energy levels determines the chemical and physical properties of the elements 4 • Radiation is absorbed or emitted when an electron moves from one orbit to another. And the energy change E = E2 – E1 IDEAS
  • 27. 2.2.2 JAMES CHADWICK 1 • In 1932 James Chadwick discovered a 3rd subatomic particle in the atom 2 • He called this particle a neutron because it was neutral 3 • He proposed that neutrons resided in the nucleus of the atom along with protons
  • 29. 2.1 Early models of atom 2.1.1 Democritus 2.1.2 John dalton 2.1.3 J.J.Thompson 2.1.4 Ernest Rutherford OUTLINE 2.3 Structure of atom 2.3.1 Defining the atom 2.3.2 Distinguishing among atoms
  • 30. OUTLINE 2.3.1 Defining the atom 2.3.1.1 Electron 2.3.1.2 Nucleus 2.3.1.3 Structure nucleus
  • 31. 2.3.1 Defining the atom Neutrons have no lectrical charge 1 • Atomic radius: 40 -270 picometer 2 • Smallest part of an element 3 • 3 subatomic particles: protons, neutrons, electrons 4 • Protons = Positive (+) charge • Neutrons = Neutral • Electron = Negative (-) charge
  • 32. 2.3.1.1 Electron ELECTRON Discovery J J Thomson -Cathode ray tube - Cathode rays composed of negatively charged particles (electrons) Mass (0.005 amu) Determine the atomic number (Z) (the atom’s identity) Location Outside nucleus (Orbital) Charge Negative charge
  • 34. 2.3.1.2 Nucleus 2.3.1.3 Structure of Nucleus NEUTRONS Mass = 1 amu, charge = 0 Neutrons act as a type of “insulation” between the protons, preventing them from repelling each other PROTONS Mass = 1 amu, charge = +1 Determines the atom's identity (atomic number)
  • 35.
  • 36. OUTLINE 2.3.2 Distinguishing among atoms 2.3.2.1 Atomic Number 2.3.2.2 Mass Number 2.3.2.3 Atomic Mass 2.3.2.4 Isotopes 2.3.2.5 Electronic Configuration
  • 37. 2.3.2.1 Atomic Number (Z) 5. Atoms are neutral, # proton = # electron 1. Atoms are identified by their atomic number 4. The atomic number of an element never changes 6. Eg: Hydrogen atoms, contain one proton and have an atomic number 1 2. Number of protons determine an element's atomic number 3. Periodic Table is in order of increasing atomic number
  • 38.
  • 39. Element # of protons Atomic # (Z) Carbon 6 6 Phosphorus 15 15 Gold 79 79 2.3.2.1 Atomic Number Example…….
  • 40. Mass number is the number of protons and neutrons in the nucleus p+ + n0 = Mass # 2.3.2.2 Mass number
  • 41. Nuclide p+ n0 e- Mass # Oxygen 8 10 8 18 Arsenic 33 42 33 75 Phosphorus 15 16 15 31 2.3.2.2 Mass number Example…..
  • 42. 1. An element's or isotope's atomic number tells how many protons are in its atoms. 1. An element's or isotope's mass number tells how many protons and neutrons in its atoms. SUMMARY
  • 43. 1. An element's or isotope's atomic number tells how many protons are in its atoms. 2. An element's or isotope's mass number tells how many protons and neutrons in its atoms. SUMMARY
  • 45.  Find each of these: a) number of protons b) number of neutrons c) number of electrons d) Atomic number e) Mass Number Lv 298 116 QUIZ……
  • 46. QUIZ…..  If an element has an atomic number of 34 and a mass number of 78, what is the: a) number of protons b) number of neutrons c) number of electrons d) complete symbol QUIZ……
  • 47. Measuring Atomic Mass • Instead of grams (because the numbers would be too small), the unit we use is the Atomic Mass Unit (amu) • It is defined as one-twelfth the mass of a Carbon-12 atom. – Carbon-12 chosen because of its isotope purity. • Each isotope has its own atomic mass, thus we determine the average from percent bundance. 2.3.2.3 Atomic mass
  • 48.  Eg: How heavy is an atom of oxygen?  It depends, because there are different kinds of oxygen atoms.  We are more concerned with the average atomic mass. 2.3.2.3 Atomic mass
  • 49. To calculate the average: • Multiply the atomic mass of each isotope by it’s abundance (expressed as a decimal), then add the results. A = [(mass of isotope) (%abundance) ] + [(mass of isotope) %abundance)] + 100 2.3.2.3 Atomic mass
  • 50. Atomic Masses Isotope Symbol Composition of the nucleus % in nature Carbon-12 12C 6 protons 6 neutrons 98.89% Carbon-13 13C 6 protons 7 neutrons 1.11% Carbon-14 14C 6 protons 8 neutrons <0.01% Atomic mass is the average of all the naturally occurring isotopes of that element. Carbon = 12.011 2.3.2.3 Atomic mass
  • 51. The natural abundance for boron isotopes is: 19.9% 10B (10.013 amu) and 80.1% 11B (11.009amu). Calculate the atomic weight of boron. Atomic Mass = [19.9)(10.013)] + [(80.1)(11.009)] 100% = [199] + [882] 100% = 10.81 amu so, the atomic weight of B = 10.81 amu PROBLEM 1
  • 52. Verify that the atomic mass of magnesium of 24.31 amu, given the following: 24Mg= 23.985042 amu, 78.99% 25Mg= 24.985837 amu, 10.00% 26Mg= 25.982593 amu, 11.01% Atomic mass = [(0.7899)(23.985042)] + [(0.1)(24.985837)] + [(0.1101)(25.982593)] = 18.946 + 2.499 + 2.861 = 24.306 amu PROBLEM 2 Solution:
  • 53. PROBLEM 3 The atomic mass of lithium is 6.94 amu, the naturally occurring isotopes are 6Li = 6.015121 amu, and 7Li = 7.016003 amu. Determine the percent abundance of each isotope. 6.94 = [(% 6Li)(6.015121)] + [(%7Li)(7.016003)] Since I don’t know what the percentage are, I will have to use variables. 100% of Lithium is determined by these two naturally occurring isotopes. We will let 6Li = x and 7 Li = 1-x; we use 1 – x instead of 100 – x because the small number is easier to work with. (in other words we reduced 100% to decimal form 1.00) Solution:
  • 54. PROBLEM 4PROBLEM 3 6.94 = [(% 6Li)(6.015121)] + [(%7Li)(7.016003)] 6.94 = [(x)(6.015121)] +[(1-x)(7.016003)] 6.94 = 6.015121x + 7.016003 – 7.016003x 6.94 -7.016003 = (6.015121x - 7.016003x) -0.076003 = -1.000882 x Solve for x: x = 0.075936 6Li = x and 7 Li = 1-x Therefore, 6Li = 0.075936 x 100% = 7.59% 7Li = (1- 0.075936 ) x 100% =
  • 55. Copper exists as two isotopes: 63Cu (62.9298 amu) and 65Cu (64.9278 amu). What are the percent abundances of the isotopes? Atomic mass for Cu = 63.546 amu 63Cu % = 1-x 65Cu % = x 63.546 = [(1-x)(62.9298)] + [(x)(64.9278)] 63.546 = 62.9298 - 62.9298x + 64.9278x 1.3818 = 1.998x x = 1.3818 / 1.998 x = 0.6916 65Cu = 0.6916 x 100% =69.16% 63Cu = 1 – x = 1 – 0.6916 = 0.3084 x 100% =30.84% PROBLEM 4 Solution:
  • 56.
  • 57. PROBLEM 5 Lithium has an elemental atomic mass of 6.941 and has two naturally occurring isotopes, 6Li and 7Li. Their masses are 6.0151 and 7.0160 respectively. What are the natural abundances (to 2 decimal places in percentage) of the isotopes of Lithium? Answer: 6Li: 7.49%, 7Li: 92.51%
  • 58. • Frederick Soddy (1877-1956) • Proposed the idea of isotopes in 1912 • Elements with the same number of protons and a different number of neutrons (thus have different masses) • different masses 2.3.2.4 Isotopes
  • 60. ATOMIC STRUCTURE Electrons are arranged in Energy Levels or Shells around the nucleus of an atom. • first shell = a maximum of 2 electrons • second shell = a maximum of 8 electrons • third shell = a maximum of 8 electrons 2.3.2.5 Electronic Configuration
  • 61. ATOMIC STRUCTURE There are two ways to represent the atomic structure of an element or compound; 1. Electronic Configuration 2. Dot & Cross Diagrams 2.3.2.5 Electronic Configuration2.3.2.5 Electronic Configuration
  • 62. ELECTRONIC CONFIGURATION With electronic configuration, elements are represented numerically by the number of electrons in their shells and number of shells. For example; N Nitrogen 7 14 2 in 1st shell 5 in 2nd shell Configuration = 2 , 5 2 + 5 = 7 2.3.2.5 Electronic Configuration 1. Electronic Configuration
  • 63. ELECTRONIC CONFIGURATION Write the electronic configuration for the following elements; Ca O Cl Si Na 20 40 11 23 8 17 16 35 14 28 B 11 5 a) b) c) d) e) f) 2.3.2.5 Electronic Configuration
  • 64. 2. DOT & CROSS DIAGRAMS With Dot & Cross diagrams elements and compounds are represented by Dots or Crosses to show electrons, and circles to show the shells. For example; Nitrogen N XX X X XX X N 7 14 2.3.2.5 Electronic Configuration
  • 65. Draw the Dot & Cross diagrams for the following elements; O Cl 8 17 16 35 a) b) 2.3.2.5 Electronic Configuration
  • 66. SUMMARY 1. The Atomic Number of an atom = number of protons in the nucleus. 2. The Atomic Mass of an atom = number of Protons + Neutrons in the nucleus. 3. The number of Protons = Number of Electrons. 4. Electrons orbit the nucleus in shells. 5. Each shell can only carry a set number of electrons.