1
1.1
1.2
1.3
1.4
1.5
1.6
1
Atomic Structure
and Relative Masses
The Atomic Nature of Matter
The Experimental Evidence of Atomic Structure
Sub-atomic Particles
Atomic Number, Mass Number and Isotopes
Mass Spectrometer
Relative Isotopic, Atomic and Molecular Masses
1.1
The Atomic
Nature of Matter
2
1.1 The atomic nature of matter (SB p.2)
What is “atom”?
Atomos = indivisible
Atomism(原子論)
The Greek philosopher Democritus
(~460 B.C. – 370 B.C.)
3
1.1 The atomic nature of matter (SB p.2)
Atomos = indivisible
These are
iron atoms!!
Continuous
division
Iron
Continuous
division
4
1.1 The atomic nature of matter (SB p.2)
Atomos = indivisible
管子<內業篇>
靈氣在心，一來一逝，
其細無內，其大無外
5
1.1 The atomic nature of matter (SB p.2)
Dalton’s atomic theory
1803 AD John Dalton
6
1.1 The atomic nature of matter (SB p.2)
Main points of Dalton’s atomic theory
1. All elements are made up of atoms.
2. Atoms cannot be created, divided into
smaller particles, nor destroyed in the
chemical process.
A chemical reaction simply changes the
way atoms are grouped together.
7
1.1 The atomic nature of matter (SB p.2)
Main points of Dalton’s atomic theory
3. Atoms of the same element are identical. They
have the same mass and chemical properties.
4. Atoms of different elements are different.
They have different masses and chemical
properties.
5. When atoms of different elements combine to
form a compound, they do so in a simple whole
number ratio to each other.
Check Point 1-1
8
1.2
The Experimental
Evidence of
Atomic Structure
9
1.2 The experimental evidence of atomic structure (SB p.3)
Steps to Thomson’s Atomic Model
• 1876
Goldstein
Discovery of cathode rays from
discharge tube experiment.
10
1.2 The experimental evidence of atomic structure (SB p.3)
Discovery of Cathode Rays
• A beam of rays came out from the
cathode and hit the anode
• Goldstein called the beam cathode rays
11
1.2 The experimental evidence of atomic structure (SB p.3)
Steps to Thomson’s Atomic Model
• 1876
Goldstein
Discovery of cathode rays from
discharge tube experiment.
• 1895
Crookes
Cathode rays are negatively charged
particles which travelled in straight
line.  electrons
12
1.2 The experimental evidence of atomic structure (SB p.3)
Deflected in
the electric
field
13
Deflected in
the magnetic
field
1.2 The experimental evidence of atomic structure (SB p.3)
The beam was composed of negatively
charged fast-moving particles.
14
1.2 The experimental evidence of atomic structure (SB p.3)
Measurement of the m/e ratio of ‘electron’
1897
J J Thomson (1856-1940)
15
1.2 The experimental evidence of atomic structure (SB p.3)
Measure the mass to
charge ratio (m/e) of
the particles produced
The particles
were constituents
of all atoms!!
16
Independent of the
nature of the gas inside
the discharge tube
Thomson called the
particles ‘electrons’.
1.2 The experimental evidence of atomic structure (SB p.3)
Thomson’s atomic model
+
+
+
17
+
+
+
Atom
• An atom was a positively
charged sphere of low
density
• The positively charged
sphere is balanced
electrically by negatively
charged electrons
Electron
1.2 The experimental evidence of atomic structure (SB p.3)
How are the particles distributed in
an atom?
+
+
+
+
Positive
charge
18
+
• Most of the mass of the
atom was carried by the
electrons (>1000 e-)
• An atom was a positively
+
charged sphere of low
density with negatively
charged electrons
embedded in it like a
plum pudding
Electron
1.2 The experimental evidence of atomic structure (SB p.3)
How are the particles distributed in
an atom?
+
+
+
+
+
Like a raisin bun (提子飽)
+
Positive
charge
19
Electron
1.2 The experimental evidence of atomic structure (SB p.3)
How are the particles distributed in
an atom?
Experimental evidence : Powerful projectiles such as -particles passes
straight through a thin gold foil.
Analogy : -particle vs a thin gold foil
 15-inch canon ball vs a piece of paper
20
1.2 The experimental evidence of atomic structure (SB p.3)
Steps to Rutherford’s Atomic Model
• Nobel laureates, Physics, 1903
Becquerel
21
Pierre Curie
Marie Curie
1.2 The experimental evidence of atomic structure (SB p.3)
Steps to Rutherford’s Atomic Model
• 1896
Becquerel
1st discovery of radioactive substance.
(an uranium salt)
22
1.2 The experimental evidence of atomic structure (SB p.3)
Steps to Rutherford’s Atomic Model
• 1898
Pierre & Marie Curie
Radioactive polonium and radium were
isolated
1g from 500 Kg
pitchblende
23
1.2 The experimental evidence of atomic structure (SB p.3)
The Curie Family
• Pierre & Marie Curie
Nobel laureate, Physics, 1903
• Marie Curie
Nobel laureate, Chemistry, 1911
• Federic Joliet & Irene Joliet-Curie
Nobel laureate, Chemistry, 1935
24
1.2 The experimental evidence of atomic structure (SB p.3)
Steps to Rutherford’s Atomic Model
• 1899
Rutherford
(Nobel laureate, Physics, 1908)
Discovery of  and  radiations.
 radiation  He2+
 radiation  e
25
1.2 The experimental evidence of atomic structure (SB p.3)
Rutherford’s scattering experiment
26
1.2 The experimental evidence of atomic structure (SB p.3)
• A thin gold foil was bombarded with a beam of
fast-moving -particles (+ve charged)
Observation:
• most -particles
passed through the
foil without
deflection
• very few -particles
were scattered or
rebounded back
27
It was quite the most incredible event that
has ever happened to me in my life.
It was almost as incredible as if you fired a
15-inch shell at a piece of tissue paper and it
came back and hit you.
28
1.2 The experimental evidence of atomic structure (SB p.3)
Interpretation of the experimental results
• Nucleus is positively charged because it
repels the positively charged alpha particles.
29
1.2 The experimental evidence of atomic structure (SB p.3)
Interpretation of the experimental results
• Nucleus occupies a very small space (10-12 of
size of atom) because very few  particles
are deflected.
30
1.2 The experimental evidence of atomic structure (SB p.3)
Interpretation of the experimental results
• The radius of an atom is about 20,000 times
that of the nucleus. Thus, if we imagine a
large football stadium as being the whole
atom, then the nucleus would be about the
size of a peanut.
31
1.2 The experimental evidence of atomic structure (SB p.3)
Interpretation of the experimental results
• Nucleus is relatively massive and highly
charged because of the large deflection.
32
1.2 The experimental evidence of atomic structure (SB p.3)
Interpretation of the experimental results
• Number of positive charges in each nucleus can
be calculated from experimental results
 Presence of protons in nucleus
33
1.2 The experimental evidence of atomic structure (SB p.3)
Steps to Chadwick’s Atomic Model
• 1919
F. W. Aston
(Nobel laureate, Chemistry, 1922)
Isotopes of Neon were discovered using
mass spectrometry
20
10
34
Ne
22
10
Ne
1.2 The experimental evidence of atomic structure (SB p.3)
Steps to Chadwick’s Atomic Model
• 1920
Rutherford
Postulated the presence of neutrons in
the nucleus
35
1.2 The experimental evidence of atomic structure (SB p.3)
Steps to Chadwick’s Atomic Model
• James Chadwick
(Nobel laureate, Physics, 1935)
Discovery of the neutron
36
1.2 The experimental evidence of atomic structure (SB p.3)
Chadwick’s Experiments
37
1.2 The experimental evidence of atomic structure (SB p.3)
Steps to Chadwick’s Atomic Model
Interpretation : -
9
4
38
Be
+
4
2
He
12
6
C
+
1
0
n
1.2 The experimental evidence of atomic structure (SB p.3)
Chadwick’s atomic model
Electron
Neutron
39
Proton
Check Point 1-2
1.3
40
Sub-atomic
Particles
1.3 Sub-atomic particles (SB p.6)
Sub-atomic particles
3 kinds of sub-atomic particles:
• Protons
• Neutrons
• Electrons
41
Inside the condensed
nucleus
Moving around the
nucleus
1.3 Sub-atomic particles (SB p.6)
A carbon-12 atom
42
1.3 Sub-atomic particles (SB p.6)
Characteristics of sub-atomic particles
Sub-atomic
particle
Symbol
Proton
p or 1 H
1
Neutron
n or 1 n
0
Electron
e- or 0e
-1
Location in
atom
Nucleus
Nucleus
Surrounding the
nucleus
Actual charge
(C)
1.6  10-9
0
1.6 x 10-9
Relative charge
+1
0
-1
Actual mass (g)
1.7  10-24
1.7  10-24
9.1  10-28
1
1
0
Approximate
relative mass
(a.m.u.)
43
1.3 Sub-atomic particles (SB p.6)
1 a.m.u.
= 1/12 of the mass of a C-12 atom
One C-12 atom has 6 p, 6n and 6e

mass of e can be ignored
mass of a C-12 atom  6p + 6n
 mass of p  mass of n
mass of a C-12 atom  6p + 6n  12p  12n

44
mass of p  mass of n  1 a.m.u.
1.3 Sub-atomic particles (SB p.6)
Express the masses of the following isotopes
in a.m.u..
12
6
C
12
45
13
6
C
~13
14
6
C
~14
1.4
Atomic Number,
Mass Number
and Isotopes
46
1.4 Atomic number, mass number and isotopes (SB p.7)
Atomic number
The atomic number (Z) of an element is
the number of protons contained in the
nucleus of the atom.

Atoms are electrically neutral
Atomic
number
47
=
Number of
protons
=
Number of
electrons
1.4 Atomic number, mass number and isotopes (SB p.8)
Mass number
The mass number (A) of an atom is the sum
of the number of protons and neutrons in
the nucleus.
Mass
number
=
Number of
protons
+
Number of
neutrons
Number of neutrons = Mass number – Atomic number
48
1.4 Atomic number, mass number and isotopes (SB p.8)
Isotopes
Isotopes are atoms of the same element with
the same number of protons but
different numbers of neutrons. Or
Isotopes are atoms of the same element with
the same atomic number but
different mass numbers
49
1.4 Atomic number, mass number and isotopes (SB p.8)
Notation for an isotope
Mass number
A
Z
Atomic number
50
X
Symbol of
the
element
1.4 Atomic number, mass number and isotopes (SB p.8)
51
Atomic
number
Mass
number
Number
of
protons
No. of
electron
s
5
10
5
5
5
10
5
B
8
17
8
8
9
14
28
14
14
14
17
8
28
14
O
Si
10
22
10
10
12
22
10
34
78
34
34
44
30
66
30
30
36
78
34
66
30
No. of
Notation
neutrons
Ne
Se
Zn
A boron isotope has a relative mass of ~10 a.m.u.
Give the isotopic notation.
10
5
52
B
1.4 Atomic number, mass number and isotopes (SB p.8)
Discovery of isotopes by mass spectrometry
What is the difference in mass between
the two isotopes of hydrogen ?
1
1
H
1 a.m.u.
= 1.7  10-24 g
2
1
H
No balance is accurate enough
to distinguish this difference
= 0.0000000000000000000000017 g
53
1.4 Atomic number, mass number and isotopes (SB p.8)
What is the difference in mass between
the two isotopes of hydrogen ?
What is the relative abundances of
the two isotopes of hydrogen ?
1
1
H
99.8%
2
1
H
0.02%
Both tasks can be accomplished with a
mass spectrometer !!
54
1.5 Mass spectrometer (SB p.10)
Mass spectrometer
Extremely
accurate
Resolution :
 1024 g
55
1.5 Mass spectrometer (SB p.10)
Mass spectrometer
Highly precise
Results of
measurement are
reproducible
56
1.5 Mass spectrometer (SB p.10)
Mass spectrometer
Highly
sensitive
Sample size :
as small as 1 g
57
1.5 Mass spectrometer (SB p.10)
Mass spectrometer
Highly
efficient
Analysis can be
accomplished in a
couple of minutes.
58
1.5 Mass spectrometer (SB p.10)
+

The sample (element or compound) is
vaporized
59
1.5 Mass spectrometer (SB p.10)
+

Positive ions are produced from the vapour
X(g) + e  X+(g) + 2e
60
1.5 Mass spectrometer (SB p.10)
+

X(g) + e  X+(g) + 2e
Atom
Molecule
61
Simple ion
Molecular/polyatomic ion
1.5 Mass spectrometer (SB p.10)
+

+ve ions accelerated by a known and fixed
electric field
62
1.5 Mass spectrometer (SB p.10)
+

+ve ions are then deflected by a known and
variable magnetic field
63
1.5 Mass spectrometer (SB p.10)
+

The ions are detected
64
1.5 Mass spectrometer (SB p.10)
+

The mass spectrum is traced out by the
recorder
65
1.5 Mass spectrometer (SB p.10)
Mass spectrum of Rb:
x-axis :-
For singly charged ions, e = 1
m/e = m
= isotopic mass (relative to C-12)
 mass number (whole number)
66
Relative isotopic mass
The relative isotopic mass of a particular
isotope of an element is the relative mass of
one atom of that isotope on the 12C = 12.0000
scale.
67
1.5 Mass spectrometer (SB p.10)
Mass spectrum of Rb:
Y-axis :Relative abundance,
Ion intensity, or
Detector current
68
Relative atomic mass
The relative atomic mass of an element is
the weighted average of the relative
isotopic masses of the natural isotopes on
the 12C = 12.0000 scale.
69
1.9 Relative isotopic, atomic and molecular masses (SB p.22)
Q.1
72.12%
Relative atomic mass of Rb
= 85  72.12% + 87  27.88%
= 85.56
27.88%
70
The mass spectrum of lead is given below. Given that the
relative atomic mass of lead is 207.242, calculate the
relative abundance of the peak at m/e of 208.
Let x be the relative abundance of the peak at m/e of 208
207.242  204 
x = 52.3
71
1.5
23.6
22.6
x
 206 
 207 
 208 
47.7  x
47.7  x
47.7  x
47.7  x
1.9 Relative isotopic, atomic and molecular masses (SB p.22)
Q.2(a)
Relative atomic mass of Pb
0.2
2.4
2.2
5.2
 (204)(
)  (206)(
)  (207)(
)  (208)(
)
10
10
10
10
= 207.2
Pb
Q.2(b) 103 
206
104 
208
72
2
Pb
2
1.9 Relative isotopic, atomic and molecular masses (SB p.22)
Q.3(a)(i)/(ii)
Rn
222

Rn
220
The lighter ions(220Rn+)
with a smaller m/e ratio
are defected more
73

1.9 Relative isotopic, atomic and molecular masses (SB p.22)
3.(b)
 the strength of the magnetic field or
 the strength of the electric field would
bring the ions from Y onto the detector.
In practice, the strength of the electric
field is fixed while that of the magnetic
field is increased gradually to bring ions of
increasing m/e ratios onto the detector.
74
1.9 Relative isotopic, atomic and molecular masses (SB p.22)
3.(c)
Rn2+ would be deflected more than the
ions at X and Y. (Rn2+ has a smaller m/e)
If magnetic field strength and electric
field strength are fixed,
m/e   deflection 
75
1.9 Relative isotopic, atomic and molecular masses (SB p.22)
m/e
Ionic
species
N

O
 [16O16O]2
,
14
4.0
14
16
0.8
16
0.3
20
20
28
29
76
Relative
abundance
14
14
2
, [ N N]
Ne , Ar
 40
100
[ N N]
0.76
[ N N]
14
14
14
15
2


1.9 Relative isotopic, atomic and molecular masses (SB p.22)
m/e
Ionic
species
32
23
[ O O]
33
0.02
[ O O]
0.09
[ O O]
34
77
Relative
abundance
40
2.0
44
0.10
16
16
16

17
17

17

Ar
40

[ O C O]
16
12
16

1.9 Relative isotopic, atomic and molecular masses (SB p.22)
Relative molecular mass
The relative molecular mass is the relative
mass of a molecule on the carbon-12 scale.
Relative molecular mass can be determined
by mass spectrometer directly.
78
1.5 Mass spectrometer (SB p.10)
Mass spectrum of Cl2:
The peaks with higher m/e ratio
correspond to molecular ions
Fragmentation of molecules always
occurs during the ionization process.
Cl2(g)  Cl(g) + Cl(g)
79
1.5 Mass spectrometer (SB p.10)
Mass spectrum of Cl2:
The scale has been enlarged
for these two peaks.
80
1.5 Mass spectrometer (SB p.10)
Complete the following table
81
m/e ratio
Corresponding
ion
35
35Cl+
37
37Cl+
70
[35Cl-35Cl]+
72
[35Cl-37Cl]+
74
[37Cl-37Cl]+
1.9 Relative isotopic, atomic and molecular masses (SB p.22)
What is the relative atomic mass of Cl?
The relative abundances of
Cl-35 and Cl-37 are 75.77
and 24.23 respectively
75.77
24.23
RAM  35 
 37 
100
100
= 35.48
82
1.9 Relative isotopic, atomic and molecular masses (SB p.22)
What is the relative molecular mass of Cl2 ?
Method 1
RMM  2  35.48  70.96
Method 2
27
18
3
RMM  70 
 72 
 74 
48
48
48
= 71
83
1.9 Relative isotopic, atomic and molecular masses (SB p.22)
What is the RMM of CH3Cl?
Molecular
ions
123
2
40
RMM  50 
 51 
 52 
165
165
165
= 50.50
84
Complete the following table
m/e
35
35Cl+
37
37Cl+
50
51
52
85
Corresponding ion
[12C1H335Cl]+
[13C1H335Cl]+ , [12C2H1H235Cl]+
[12C1H337Cl]+
The mass spectrum of dichloromethane is given below.
Calculate the relative molecular mass of dichloromethane.
RMM  84 
94
3
59
2.2
13
2.5
0.8
 85 
 86 
 87 
 88 
 89 
 90 
174.5
174.5
174.5
174.5
174.5
174.5
174.5
= 85.128
86
The END
87
1.1 The atomic nature of matter (SB p.3)
Back
(a) What does the word “atom” literally mean? (a) Indivisible
(b) Which point of Dalton’s atomic theory is based on the law
of conservation of mass proposed by Lavoisier in 1774
which states that matter is neither created nor destroyed
(b) Atoms can neither be
in the course of a chemical reaction?
created nor destroyed.
(c) Which point of Dalton’s atomic theory is based on the law
of constant proportion proposed by Proust in 1799 which
states that all pure samples of the same chemical
compound contain the same elements combined together
in the same proportions by mass?
(c) Atoms of different elements combine to form a
compound. The numbers of various atoms combined
88
bear
a simple whole number ratio to each other.
Answer
1.2 The Experimental evidence of atomic structure (SB p.4)
Back
(a) Atoms were found to be divisible. What names were
(a) Electron, proton
given to the particles found inside the atoms?
and neutron
(b) Give the most important point of the following
experiments:
(i) E. Goldstein’s gas discharge tube experiment;
(ii) J. J. Thomson’s cathode ray tube experiment;
(iii) E. Rutherford’s gold foil scattering experiment.
(b) (i) Discovery of cathode rays
(ii) Discovery of electrons
(iii) Discovery of nucleus in atoms
89
Answer
1.3 Sub-atomic particles (SB p.6)
The identity of an element is determined by the
number of which sub-atomic particle?
Answer
The identity of an element is determined by
the number of protons in its atomic nucleus.
Back
90
1.3 Sub-atomic Particles (SB p.7)
Back
(a) Which part of the atom accounts for almost all the mass of
that atom? (a) Nucleus
(b) The mass of which sub-atomic particle is often assumed
to be zero? (b) Electron
Answer
91
1.3 Sub-atomic particles (SB p.7)
Are there any sub-atomic particles other than protons,
neutrons and electrons?
Answer
Other than the three common types of subatomic particles (proton, neutron and
electron), there are also some sub-atomic
particles called positron (anti-electron) and
quark.
Back
92
1.3 Sub-atomic particles (SB p.7)
If bromine has two isotopes, 79Br and 81Br, how many
physically distinguishable combinations of Br atoms are
there in Br2?
79Br—79Br
79Br—81Br
81Br—81Br
They have different molecular
masses and thus have different
density
Back
93
1.4 Atomic number, mass number and isotopes (SB p.8)
Back
Write the symbol for the atom that has an atomic number of
11 and a mass number of 23. How many protons, neutrons
and electrons does this atom have?
23
11
94
Na, 11 protons, 12 neutrons, 11 electrons.
Answer
1.5 Mass spectrometer (SB p.12)
Back
Label the different parts of the mass spectrometer.
A – Vaporization chamber
B – Ionization chamber
C – Accelerating electric field
D – Deflecting magnetic field
E – Ion detector
95
Answer
1.5 Mass spectrometer (SB p.12)
Back
The mass spectrum of neon is given below. Determine the
relative atomic mass of neon.
Relative atomic mass of neon
(20  114)  (21 0.2)  (22  11.2)
=
(114  0.2  11.2)
= 20.18
Answer
96
1.6 Relative isotopic, atomic and molecular masses (SB p.14)
(a) The mass spectrum of lead is given below. Given that the
relative atomic mass of lead is 207.242, calculate the
relative abundance of the peak at m/e of 208.
Let x be the relative abundance
of the peak at m/e of 208.
(204  1.5 + 206  23.6 + 207 
22.6 + 208x)  (1.5 + 23.6 + 22.6
+ x) = 207.242
x = 52.3
The relative abundance of the
peak at m/e of 208 is 52.3.
97
Answer
1.6 Relative isotopic, atomic and molecular masses (SB p.14)
Back
(b) The mass spectrum of dichloromethane is given below.
Calculate the relative molecular mass of dichloromethane.
The relative molecular mass of
dichloromethane
= (84  94 + 85  3.0 + 86  59 +
87  2.2 + 88  13 + 89  2.5 +
90  0.8)  (94 + 3.0 + 59 + 2.2 +
13 + 2.5 + 0.8)
= 85.128
The relative molecular mass of
dichloromethane is 85.128.
Answer
98