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Atomic Structure Atomic Structure and Relative Massesand Relative Masses
1.11.1 The Atomic Nature of MatterThe Atomic Nature of Matter
1.21.2 The Experimental Evidence of Atomic StructureThe Experimental Evidence of Atomic Structure
1.31.3 Sub-atomic ParticlesSub-atomic Particles
1.41.4 Atomic Number, Mass Number and IsotopesAtomic Number, Mass Number and Isotopes
1.51.5 Mass SpectrometerMass Spectrometer
1.61.6 Relative Isotopic, Atomic and Molecular MassesRelative Isotopic, Atomic and Molecular Masses
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1.1.11 The Atomic The Atomic
Nature of Nature of MatterMatter
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What is “atom”?What is “atom”?
1.1 The atomic nature of matter (SB p.2)
The Greek philosopher Democritus (~460 B.C. – 370 B.C.)
Atomos = indivisibleAtomism(原子論 )
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Iron
Continuous division
Continuous division
These are iron atoms!!
Atomos = indivisible
1.1 The atomic nature of matter (SB p.2)
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Atomos = indivisible管子 <內業篇 >靈氣在心，一來一逝，其細無內，其大無外
1.1 The atomic nature of matter (SB p.2)
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Dalton’s atomic Dalton’s atomic theorytheory1803 AD John Dalton
1.1 The atomic nature of matter (SB p.2)
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Main points of Dalton’s atomic Main points of Dalton’s atomic theorytheory1. 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.
1.1 The atomic nature of matter (SB p.2)
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Main points of Dalton’s atomic Main points of Dalton’s atomic theorytheory
5. When atoms of different elements combine to form a compound, they do so in a simple whole number ratio to each other.
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.
Check Point 1-1Check Point 1-1
1.1 The atomic nature of matter (SB p.2)
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The The Experimental Experimental Evidence of Evidence of
Atomic Atomic StructureStructure
1.1.22
10
1.2 The experimental evidence of atomic structure (SB p.3)Steps to Thomson’s Atomic Steps to Thomson’s Atomic
ModelModel
•1876 Goldstein
Discovery of cathode rays from discharge tube experiment.
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Discovery of Cathode RaysDiscovery of Cathode Rays
• A beam of rays came out from the cathode and hit the anode
• Goldstein called the beam cathode rays
1.2 The experimental evidence of atomic structure (SB p.3)
12
Steps to Thomson’s Atomic Steps to Thomson’s Atomic ModelModel
•1876 Goldstein
Discovery of cathode rays from discharge tube experiment.
•1895 Crookes
Cathode rays are negatively charged particles which travelled in straight line. electrons
1.2 The experimental evidence of atomic structure (SB p.3)
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Deflected in the electric
field
Deflected in the magnetic
field
1.2 The experimental evidence of atomic structure (SB p.3)
14
The beam was composed of negatively charged fast-moving particles.
1.2 The experimental evidence of atomic structure (SB p.3)
15
Measurement of the m/e ratio of Measurement of the m/e ratio of ‘electron’‘electron’
J J Thomson (1856-1940)
1897
1.2 The experimental evidence of atomic structure (SB p.3)
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Measure the mass to charge ratio (m/e) of the particles produced
Independent of the nature of the gas inside the discharge tube
The particles were constituents of all atoms!!
Thomson called the particles ‘electrons’.
1.2 The experimental evidence of atomic structure (SB p.3)
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Thomson’s atomic Thomson’s atomic modelmodel
Atom
1.2 The experimental evidence of atomic structure (SB p.3)
• An atom was a positively charged sphere of low density+ +
+
+ +
+
• The positively charged sphere is balanced electrically by negatively charged electrons
Electron
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How are the particles distributed How are the particles distributed in an atom?in an atom?
+ +
+
+ +
+
Positive charge
• Most of the mass of the atom was carried by the electrons (>1000 e-)
Electron
1.2 The experimental evidence of atomic structure (SB p.3)
• An atom was a positively charged sphere of low density with negatively charged electrons embedded in it like a plum pudding
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How are the particles distributed How are the particles distributed in an atom?in an atom?
+ +
+
+ +
+
Positive charge
Like a raisin bun ( 提子飽 )
Electron
1.2 The experimental evidence of atomic structure (SB p.3)
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How are the particles distributed How are the particles distributed in an atom?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
1.2 The experimental evidence of atomic structure (SB p.3)
21
Steps to Rutherford’s Atomic Steps to Rutherford’s Atomic ModelModel
•Nobel laureates, Physics, 1903
Becquerel Marie Curie
Pierre Curie
1.2 The experimental evidence of atomic structure (SB p.3)
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Steps to Rutherford’s Atomic Steps to Rutherford’s Atomic ModelModel
•1896 Becquerel
1st discovery of radioactive substance.
(an uranium salt)
1.2 The experimental evidence of atomic structure (SB p.3)
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Steps to Rutherford’s Atomic Steps to Rutherford’s Atomic ModelModel•1898 Pierre & Marie Curie
Radioactive polonium and radium were isolated
1g from 500 Kg pitchblende
1.2 The experimental evidence of atomic structure (SB p.3)
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The Curie FamilyThe Curie Family
•Pierre & Marie Curie
Nobel laureate, Physics, 1903•Marie Curie
Nobel laureate, Chemistry, 1911•Federic Joliet & Irene Joliet-Curie
Nobel laureate, Chemistry, 1935
1.2 The experimental evidence of atomic structure (SB p.3)
25
Steps to Rutherford’s Atomic Steps to Rutherford’s Atomic ModelModel
•1899 Rutherford
(Nobel laureate, Physics, 1908)
Discovery of and radiations.
radiation He2+
radiation e
1.2 The experimental evidence of atomic structure (SB p.3)
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Rutherford’s scattering Rutherford’s scattering experimentexperiment
1.2 The experimental evidence of atomic structure (SB p.3)
27
• 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
1.2 The experimental evidence of atomic structure (SB p.3)
28
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.
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Interpretation of the experimental resultsInterpretation of the experimental results
• Nucleus is positively charged because it repels the positively charged alpha particles.
1.2 The experimental evidence of atomic structure (SB p.3)
30
Interpretation of the experimental resultsInterpretation of the experimental results
• Nucleus occupies a very small space (10-
12 of size of atom) because very few particles are deflected.
1.2 The experimental evidence of atomic structure (SB p.3)
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Interpretation of the experimental resultsInterpretation 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.
1.2 The experimental evidence of atomic structure (SB p.3)
32
Interpretation of the experimental resultsInterpretation of the experimental results
• Nucleus is relatively massive and highly charged because of the large deflection.
1.2 The experimental evidence of atomic structure (SB p.3)
33
Interpretation of the experimental resultsInterpretation of the experimental results
1.2 The experimental evidence of atomic structure (SB p.3)
Presence of protons in nucleus
• Number of positive charges in each nucleus can be calculated from experimental results
34
Steps to Chadwick’s Atomic Steps to Chadwick’s Atomic ModelModel
Ne2010 Ne22
10
• 1919 F. W. Aston
(Nobel laureate, Chemistry, 1922)
1.2 The experimental evidence of atomic structure (SB p.3)
Isotopes of Neon were discovered using mass spectrometry
35
Steps to Chadwick’s Atomic Steps to Chadwick’s Atomic ModelModel
• 1920 Rutherford
Postulated the presence of neutrons in the nucleus
1.2 The experimental evidence of atomic structure (SB p.3)
36
• James Chadwick
(Nobel laureate, Physics, 1935)
Steps to Chadwick’s Atomic Steps to Chadwick’s Atomic ModelModel
1.2 The experimental evidence of atomic structure (SB p.3)
Discovery of the neutron
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Chadwick’s ExperimentsChadwick’s Experiments
1.2 The experimental evidence of atomic structure (SB p.3)
38
Steps to Chadwick’s Atomic Steps to Chadwick’s Atomic ModelModel
Be94 He4
2 C126
n10+ +
Interpretation : -
1.2 The experimental evidence of atomic structure (SB p.3)
39
Chadwick’s atomic Chadwick’s atomic modelmodel
ProtonElectron
Neutron Check Point Check Point 1-21-2
1.2 The experimental evidence of atomic structure (SB p.3)
40
Sub-atomic Sub-atomic ParticlesParticles
1.1.33
41
Sub-atomic particlesSub-atomic particles
1.3 Sub-atomic particles (SB p.6)
3 kinds of sub-atomic particles:
• Protons
• Neutrons
• Electrons
Inside the condensed nucleus
Moving around the nucleus
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A carbon-12 atomA carbon-12 atom
1.3 Sub-atomic particles (SB p.6)
43
Characteristics of sub-atomic Characteristics of sub-atomic particlesparticles
Sub-atomic particle
Proton Neutron Electron
Symbol p or n or e- or
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
Approximate relative mass (a.m.u.)
1 1 0
H11 n1
0e0
-1
1.3 Sub-atomic particles (SB p.6)
44
1 a.m.u. = 1/12 of the mass of a C-12 atom
mass of a C-12 atom 6p + 6n
mass of p mass of n 1 a.m.u.
One C-12 atom has 6 p, 6n and 6e
mass of p mass of n
mass of e can be ignored
mass of a C-12 atom 6p + 6n 12p 12n
1.3 Sub-atomic particles (SB p.6)
45
Express the masses of the following isotopes in a.m.u..
C146C13
6C126
12
1.3 Sub-atomic particles (SB p.6)
~13 ~14
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Atomic Atomic Number, Number,
Mass Mass Number and Number and
IsotopesIsotopes
1.1.44
47
Atomic numberAtomic number1.4 Atomic number, mass number and isotopes (SB p.7)
The atomic number (Z) of an element is the number of protons contained in the nucleus of the atom.
Atomic number =
Number of protons
Number of electrons=
Atoms are electrically neutral
48
Mass numberMass number
1.4 Atomic number, mass number and isotopes (SB p.8)
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
49
IsotopesIsotopes
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
1.4 Atomic number, mass number and isotopes (SB p.8)
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XAZ
Notation for an isotope
Symbol of the
element
Mass number
Atomic number
1.4 Atomic number, mass number and isotopes (SB p.8)
51
Atomic numbe
r
Mass number
Number of
protons
No. of electron
s
No. of neutron
s
Notation
5 10
8 8 9
28 14 14
10 12
78 44
66 30
5
5 5
8 17
14 14
22 10 10
34 34 34
30 30 36
B105
O178
Si2814
Ne2210
Se7834
Zn6630
1.4 Atomic number, mass number and isotopes (SB p.8)
52
A boron isotope has a relative mass of ~10 a.m.u.
Give the isotopic notation.
B105
53
Discovery of isotopes by mass Discovery of isotopes by mass spectrometryspectrometry
What is the difference in mass between the two isotopes of hydrogen ?
H11 H2
1
1 a.m.u.
= 1.7 10-24 g
= 0.0000000000000000000000017 g
No balance is accurate enough to distinguish this difference
1.4 Atomic number, mass number and isotopes (SB p.8)
54
What is the relative abundances of the two isotopes of hydrogen ?
H11 H2
1
99.8% 0.02%
Both tasks can be accomplished with a mass spectrometer !!
What is the difference in mass between the two isotopes of hydrogen ?
1.4 Atomic number, mass number and isotopes (SB p.8)
55
1.5 Mass spectrometer (SB p.10)
Mass spectrometerMass spectrometer
Extremely accurate
Resolution :
1024 g
56
1.5 Mass spectrometer (SB p.10)
Mass spectrometerMass spectrometer
Highly preciseResults of measurement are reproducible
57
1.5 Mass spectrometer (SB p.10)
Mass spectrometerMass spectrometer
Highly sensitive
Sample size : as small as 1 g
58
1.5 Mass spectrometer (SB p.10)
Mass spectrometerMass spectrometer
Highly efficient
Analysis can be accomplished in a couple of minutes.
59
The sample (element or compound) is vaporized
+
1.5 Mass spectrometer (SB p.10)
60
Positive ions are produced from the vapour
X(g) + e X+(g) + 2e
+
1.5 Mass spectrometer (SB p.10)
61
X(g) + e X+(g) + 2e
+
1.5 Mass spectrometer (SB p.10)
Atom Simple ionMolecule Molecular/polyatomic ion
62
+ve ions accelerated by a known and fixed electric field
+
1.5 Mass spectrometer (SB p.10)
63
+ve ions are then deflected by a known and variable magnetic field
+
1.5 Mass spectrometer (SB p.10)
64
The ions are detected
+
1.5 Mass spectrometer (SB p.10)
65
The mass spectrum is traced out by the recorder
+
1.5 Mass spectrometer (SB p.10)
66
Mass spectrum of Rb: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)
1.5 Mass spectrometer (SB p.10)
67
Relative isotopic massRelative 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.
68
Mass spectrum of Rb:Mass spectrum of Rb:
Y-axis :-
Relative abundance,
Ion intensity, or
Detector current
1.5 Mass spectrometer (SB p.10)
69
Relative atomic massRelative 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.
70
Q.172.12%
27.88%
Relative atomic mass of Rb
= 85 72.12% + 87 27.88%
= 85.56
1.9 Relative isotopic, atomic and molecular masses (SB p.22)
71
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.
x47.7x
208x47.7
22.6207
x47.723.6
206x47.7
1.5204 207.242
Let x be the relative abundance of the peak at m/e of 208
x = 52.3
72
Q.2(a)
Relative atomic mass of Pb
)105.2
(208)()102.2
(207)()102.4
(206)()100.2
(204)(
= 207.2
Q.2(b) 103
2206Pb
104
2208Pb
1.9 Relative isotopic, atomic and molecular masses (SB p.22)
73
Q.3(a)(i)/(ii)
Rn222
Rn220
The lighter ions(220Rn+) with a smaller m/e ratio are defected more
1.9 Relative isotopic, atomic and molecular masses (SB p.22)
74
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.
3.(b)
1.9 Relative isotopic, atomic and molecular masses (SB p.22)
75
Rn2+ would be deflected more than the ions at X and Y. (Rn2+ has a smaller m/e)
3.(c)
1.9 Relative isotopic, atomic and molecular masses (SB p.22)
If magnetic field strength and electric field strength are fixed,
m/e deflection
76
m/eRelative
abundance
Ionic species
14 4.0
16 0.8
20 0.3
28 100
29 0.76
2Ar40
N]N[ 1414
N]N[ 1514
1.9 Relative isotopic, atomic and molecular masses (SB p.22)
Ne20,
21414 N]N[
21616 O]O[
N14,
O16,
77
m/eRelative
abundance
Ionic species
32 23
33 0.02
34 0.09
40 2.0
44 0.10
O]O[ 1616
O]O[ 1716
O]O[ 1717
Ar40
O]CO[ 161216
1.9 Relative isotopic, atomic and molecular masses (SB p.22)
78
Relative molecular Relative molecular massmass
The relative molecular mass is the relative mass of a molecule on the carbon-12 scale.
1.9 Relative isotopic, atomic and molecular masses (SB p.22)
Relative molecular mass can be determined by mass spectrometer directly.
79
Mass spectrum of Mass spectrum of ClCl22:: 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)
1.5 Mass spectrometer (SB p.10)
80
Mass spectrum of Mass spectrum of ClCl22::
The scale has been enlarged for these two peaks.
1.5 Mass spectrometer (SB p.10)
81
Complete the following tableComplete the following table
m/e ratio
Corresponding ion
35
37
70
72
74
1.5 Mass spectrometer (SB p.10)
35Cl+
37Cl+
[35Cl-35Cl]+
[35Cl-37Cl]+
[37Cl-37Cl]+
82
What is the relative atomic mass of Cl?What is the relative atomic mass of Cl?The relative abundances
of Cl-35 and Cl-37 are 75.77 and 24.23
respectively
1.9 Relative isotopic, atomic and molecular masses (SB p.22)
10024.23
37100
75.7735 RAM
= 35.48
83
What is the relative molecular mass of What is the relative molecular mass of ClCl22 ? ?
1.9 Relative isotopic, atomic and molecular masses (SB p.22)
Method 1 70.96 35.482 RMM
Method 2
483
744818
724827
70 RMM
= 71
84
What is the RMM of CHWhat is the RMM of CH33Cl?Cl?
1.9 Relative isotopic, atomic and molecular masses (SB p.22)
16540
521652
51165123
50 RMM
= 50.50
Molecular ions
85
Complete the following tableComplete the following table
m/e Corresponding ion
35
37
50
51
52
35Cl+
37Cl+
[12C1H335Cl]+
[12C1H337Cl]+
[13C1H335Cl]+ , [12C2H1H2
35Cl]+
86
The mass spectrum of dichloromethane is given below. Calculate the relative molecular mass of dichloromethane.
174.50.8
90174.52.5
89174.5
1388
174.52.2
87174.5
5986
174.53
85174.5
9484 RMM
= 85.128
87
The END
88
(a)What does the word “atom” literally mean?
(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 in the course of a chemical reaction?
(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?
1.1 The atomic nature of matter (SB p.3)
(a) Indivisible
(b) Atoms can neither be created nor destroyed.
(c) Atoms of different elements combine to form a compound. The numbers of various atoms combined bear a simple whole number ratio to each other.
Back
Answer
89
(a)Atoms were found to be divisible. What names were given to the particles found inside the atoms?
(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.
1.2 The Experimental evidence of atomic structure (SB p.4)
(a) Electron, proton and neutron
(b) (i) Discovery of cathode rays
(ii) Discovery of electrons
(iii) Discovery of nucleus in atoms
Back
Answer
90
The identity of an element is determined by the number of which sub-atomic particle?
Back
1.3 Sub-atomic particles (SB p.6)
The identity of an element is determined by the number of protons in its atomic nucleus.
Answer
91
(a)Which part of the atom accounts for almost all the mass of that atom?
(b) The mass of which sub-atomic particle is often assumed to be zero?
1.3 Sub-atomic Particles (SB p.7)
(a) Nucleus
(b) Electron
Back
Answer
92
Are there any sub-atomic particles other than protons, neutrons and electrons?
Back
1.3 Sub-atomic particles (SB p.7)
Other than the three common types of sub-atomic particles (proton, neutron and electron), there are also some sub-atomic particles called positron (anti-electron) and quark.
Answer
93
If bromine has two isotopes, 79Br and 81Br, how many physically distinguishable combinations of Br
atoms are there in Br2?
Back
1.3 Sub-atomic particles (SB p.7)
79Br—79Br
79Br—81Br
81Br—81Br
They have different molecular masses and thus have
different density
94
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?
1.4 Atomic number, mass number and isotopes (SB p.8)
Back
, 11 protons, 12 neutrons, 11 electrons.Na23
11
Answer
95
Label the different parts of the mass spectrometer.
1.5 Mass spectrometer (SB p.12)
Back
A – Vaporization chamber
B – Ionization chamber
C – Accelerating electric field
D – Deflecting magnetic field
E – Ion detector
Answer
96
The mass spectrum of neon is given below. Determine the relative atomic mass of neon.
1.5 Mass spectrometer (SB p.12)
Back
Relative atomic mass of neon
=
= 20.18
)2.112.0114()2.1122()2.021()11420(
Answer
97
(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.
1.6 Relative isotopic, atomic and molecular masses (SB p.14)
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.
Answer
98
(b) The mass spectrum of dichloromethane is given below. Calculate the relative molecular mass of dichloromethane.
Back
1.6 Relative isotopic, atomic and molecular masses (SB p.14)
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
