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Transcript
Atoms and the Atomic Theory
All the matters can be broken down into elements. Is matter
continuously divisible into ever smaller and smaller pieces, or is
there an ultimate limit? What is an element made of?
Greeks
•Aristotle- Continuous Theory of Matter
•Democritus- Discontinous Theory of Matter
•Atomos- “indivisible”
Early Chemical Discoveries and Atomic Theory
Three important fundamental laws in chemistry
1) Law of conservation of mass
2) Law of constant composition
3) Law of multiple proportions
1
Law of Conservation of Mass
1774 Antoine Lavoisier –showed heating the red power HgO causes it to
decompose into the silvery liquid mercury and the colorless gas oxygen.
2HgO  2Hg +O2 then show that oxygen is the key substance involved
in combustion.
Furthermore, results of combustion reactions
•Total mass of products = total mass of reactants
(tin + air+ sealed glassed vessel)  (tin oxide + remaining air + glass
vessel)
Law of Conservation of Mass
~ The total mass of substances present after a chemical reaction is
the same as the total mass of substances before the reaction.
Matter is neither created nor destroyed in a chemical reaction.
2
E.g. A 0.382g sample of magnesium reacts with 2.652g of
nitrogen gas. The sole product is magnesium nitride. After
reaction, the mass of unreacted nitrogen is 2.505g. What mass
of magnesium nitride is produced?
Mass before reaction =0.382g Mg + 2.652g N2 gas = 3.034g
Mass after reaction = ?g Mg3N2 gas + 2.505 N2 gas
3.034g – 2.505g = 0.529g
3
Law of Constant Composition
1799 Joseph Proust –
Law of Constant Composition (Definite Proportion)
~ All samples of a compound have the same composition- the same
proportion by mass of the constituent elements. This means that the
relative amount of each element in a particular compound is always the
same, regardless of the source of the compound or how the compound is
prepared.
E. g. Water is made up of two elements H and O. The two sample of water
below have the same proportions of the two elements, expressed as
percentages by mass. Every sample of water contains 1 part hydrogen and
8 parts oxygen by mass.
_________________________________________
Sample A
10.000g
1.119g H
8.881g O
Composition
%H = 11.19
%O = 88.81
Sample B
27.000g
3.031g H
23.979g O
4
Dalton’s Atomic Theory
How can the Law of conservation of mass and Law of
constant composition be explain? Why do element
behave as they do?
1803-1808 John Dalton : proposed a new theory of matter.
1. Each chemical element is composed of minute,
indestructible particles called atoms.
2. All atoms of a given element are identical to one
another in mass and other properties, but the atoms of
one element are different from the atoms of other
elements.
5
Dalton’s Atomic Theory
3. Atoms of an element are not changed into
atoms of a different element by chemical
reactions; atoms are neither created nor
destroyed in chemical reactions. Chemical
compounds are formed when atoms combine
with each other.
– If atoms of an element are indestructible, then the
same remains unchanged. This explains the law of
conservation of mass.
6
Dalton’s Atomic Theory
4. In each of their compounds, different elements
combine in a simple numerical ratio: e.g. one atom
of A to one of B (AB) or one atom of A to two of B
(AB2).
– If all atoms of an element are alike in mass (assumption
2) and if atoms unite in fixed numerical ratio
(assumption 3), the percent composition of a compound
must have a unique value, regardless of the origin of
the sample analyzed. This explains the law of constant
composition.
7
Law of Multiple Proportions
Dalton’s theory leads to a prediction- the law of
multiple proportions.
~ If two elements form more than a single compound,
the masses of one element combined with a fixed
mass of the second are in the ratio of small whole
numbers.
Same elements to combine in different ratios to give
different substances.
8
E.g. Oxygen and carbon can combine either in a 1: 1.333 mass ratio to
make a substance or in a 1: 2.667 mass ratio to make a substance.
first
1 g carbon per 1.333 g oxygen C:O mass ratio = 1: 1.333
second 1 g carbon per 2.667 g oxygen C:O mass ratio = 1: 2.667
comparison C:O mass ratio in second sample = (1 g C)/(2.667g O) = 2
of C:O ratios C:O mass ratio in first sample
(1 g C)/(1.333g O)
• Compare two substances clearly the second substance contains exactly
twice as much oxygen as the first for a given number of carbon. If the
first oxide has the molecular formula CO then the second oxide will be
CO2.
9
E.g.There are two compounds, both contain nitrogen and
hydrogen. Compound A contains 1.50g of N and 0.216g
H. Compound B contains 2.00g of N and 0.144g H. If the
formula of compound B is N2H2, what is the formula of
compound A?
N:H ratio in A = 1.50: 0.216g = 1.00: 0.144
N:H ratio in B = 2:00: 0.144 = 1.00: 0.0720
H in A is (0.144/0.0720 = 2 ) twice as much in B
IF B is N2H2 then A is N2H4
10
Atomic Mass Ratio
Dalton’s theory enables us to set up a scale of relative atomic masses.
He cannot measure the exact mass of atoms but relative mass.
E.g. Consider calcium sulfide, which consists of 55.6% calcium by
mass and 44.4% sulfur by mass. Suppose there is one calcium atom
for each sulfur atom in calcium sulfide. Because we know that the
mass of a calcium atom relative to that of a sulfur atom must be the
same as the mass % in calcium, we know that the ratio of the mass of
a calcium atom to that of a sulfur atom is
mass of Ca atom = 55.6 = 1.25
mass of a S atom 44.4
or mass of a Ca atom = 1.25 x mass of a sulfur atom
By continuing in this manner with other compounds, it is possible to
build up a table of relative atomic masses. We define a quantity called
atomic mass ratio, which is the ratio of the mass of a given atom to the
mass of some particular reference atom.
11
The Structure of Atoms
What is an atom made of ?
Discovery of subatomic particle
The Discovery of Electrons
*1897 J.J. Thomson- cathode ray experiment
Thomson’s experiment involved the use of
cathode-ray tube. When a sufficiently high voltage
is applied across the electrode, an electric current
flows through the tube from negatively charged
electrode ( the cathode) to the positively charged
electrode (the anode).
12
Thomson’s Experiment
-
Voltage source
+
Vacuum tube
Metal Disks
13
Thomson’s Experiment
Voltage source
+
 By adding an electric field
14
Thomson’s Experiment
Voltage source
+
 By adding an electric field he found that
the moving pieces were negative
15
Thomson’s Model
• Found the electron
• Couldn’t find positive
(for a while)
• Said the atom was like
plum pudding
• A bunch of positive
stuff, with the electrons
able to be removed
• established the ratio of
mass to electric charge
for cathode ray
m/e = -5.6857x10-9
g/coulomb.
Spherical
cloud of
positive
charge
Electrons
16
Millikan’s Oil-Drop Experiment: Mass of Electron
1909 Robert Millikan determined the electronic charge through a
series of oil-drop experiments. The currently accepted value of the
charge of the e is
–1.6022x10-19C.
Substituting into Thomson’s mass to charge ratio then gives the
mass of electron as 1/1836(= 9.1094x10-28g).
17
X-Ray and Radioactivity
Ernest Rutherford identified two type of radiation from radioactive
materials, alpha () and beta (). -particles (He2+)carry two
fundamental units of positive charge and have essentially the same
mass as He atoms. -particles are negatively charged particles
produced by changes occurring within the nuclei of radioactive atoms
and have the same properties as electrons.
A third form of radiation, that is not affected by an electric field was
discovered in 1900 by Paul Villard. This radiation, called -ray, is not
made up of particles; it is electromagnetic radiation of extremely high
penetrating power.
Properties of the three radioactive emissions discovered
Original name
Modern name
Mass (amu) Charge
-ray
-particle
4.00
+2
-ray
-particle (electron) 5.49x10-4
-1
-ray
-ray
0
0_______
18
1909 Ernest Rutherford Scattering Experiment
~ used  particle to study the inner structure of atoms.
~directed a beam of -particles at a thin gold foil
Lead
block
Uranium
Florescent
Screen
Gold Foil
19
Rutherford Expected
• The alpha particles to pass through
without changing direction very much
• WHY?
• The positive charges were spread out
evenly. Alone they were not enough to
stop the alpha particles
20
What he expected
21
Because
22
Rutherford thought the mass
was evenly distributed in the
atom
23
Rutherford thought the
mass was evenly
distributed in the atom
particles should pass
through the low + density
model.
24
What he got
The majority of -particles penetrated the foil undeflected.
 Some  particles experienced slightly deflections.
 A few (about one in every 20,000) suffered rather serious
deflections as they penetrated the foil.
 A similar number did not pass through the foil at all, but
bounced back in the direction from which they had come.
25
How he explained it
• Atom is mostly empty
• Small dense,
positive piece
at center
• Alpha particles
are deflected by
it if they get close
enough
+
26
+
27
The Nuclear Atom: Protons and Neutrons
1911 Rutherford explained his results by proposing a model of the atom
known as the nuclear atom and having these features.
1. Most of the mass and all of the positive charge of an atom are
centered in a very small region called the nucleus. The atom is mostly
empty space.
2. The magnitude of the positive charge is different for different atoms
and is approximately one-half the atomic weight of the element.
3. There are as many electrons outside the nucleus as there are units of
positive charge on the nucleus. The atom as a whole is electrically
neutral.
Rutherford’s nuclear atom suggested the existence of positively charged
fundamental particles of matter in the nuclei of atoms- called protons.
He predicted the existence in the nucleus of electrically neutral
particles.
* 1932 James Chadwick
~ verified that there is another type of particles in atom called neutron.
28
The Structure of Atoms
Therefore
•Modern picture of an atom, then, consist of three types of particleselectrons, protons and neutron.
Electric Charge
Particle SI (C )
Electron -1.602x10-19
Proton +1.602x10-19
Neutron
0
Mass
Atomic
SI (g)
-1
9.109x10-28
+1
1.673x10-24
0
1.675x10-24
amu
5.49x10-4
1.0073
1.0087
Located
outside nucleus
in nucleus
in nucleus
29
Size of an atom
Atoms are small ~10-10 meters
Hydrogen atom, 32 pm radius
Nucleus tiny compared to atom
IF the atom was the size of a stadium, the nucleus
would be the size of a marble.
• Radius of the nucleus near 10-15m.
• Density near 1014 g/cm
•
•
•
•
30
Conclusion:
• Matter is composed, on a tiny scale, of particles
called atoms. Atoms are in turn made up of
minuscule nuclei surrounded by a cloud of particles
called electrons. Nuclei are composed of particles
called protons and neutrons, which are themselves
made up of even smaller particles called quarks.
Quarks are believed to be fundamental, meaning
that they cannot be broken up into smaller particles.
31
Chemical Elements
Atomic number
What is that makes one atom different from another?
Elements differ from one another according to the number
of protons in their nucleus
• atomic number (Z) = Number of proton in atom’s
nucleus
• mass number (A) = # of protons (Z) + # of neutrons (N)
E.g.
Mass #
Atomic #
12
C
6
Symbol of element
32
Isotopes
• Contrary to what Dalton thought, we know that atoms of
an element do not necessarily all have the same mass.
• Isotope- atoms of the same element containing different
numbers of neutrons and therefore having different
masses.
33
Isotopes of Hydrogen
34
Mass Spectrometer-
The most accurate
means of determining atomic and molecular weights.
•
35
Mass Spectrum of Elemental Carbon
This small
peak
represents
the relative
abundance
of C13 in
nature.
When 12C and 13C are analyzed in a mass spectrometer, the ratio of their
masses is found to be : Mass13C = 1.0836129
Mass12C
Since the atomic mass unit is defined such that the mass of 12C is exactly
12 amu, then on this same scale,
Mass13C = (1.0836129)(12amu) = 13.003355 amu
36
Average Atomic Mass
When considering atomic masses from the P-Table,
recall that reported values are actually weighted
averages of all the naturally occurring isotopes.
Average atomic mass = (% of each isotope)(atomic mass of each isotope)
100
Boron has two isotopes 10B and 11B. They have
the abundance 18.7% and 81.3% respectively.
Determine the average atomic mass for Boron.
37
Computing Average Mass from Mass Spectrometer
Data = Atomic Weight
When natural copper is
vaporized and injected into a
mass spectrometer, the
results shown below are
obtained. Use these data to
compute the average mass
of copper. The mass values
for 63Cu and 65Cu are 62.93
amu and 64.93 amu
respectively.
38
Isotopes and Average Atomic Mass Questions
1. Do either of the following pairs represent isotopes
of one another?
a.
40K
19
and 40Ar18
b. 90Sr38 and 94Sr38
2. The nobel gas Neon, has three isotopes of
masses, 22, 21and 20. If the isotopes have the
abundance of 8.01%, 1.99% and 90.00%
respectively, what is the average atomic mass of
neon atoms?
3. A naturally occurring sample of an element
consists of two isotopes, one of mass 85 and
one of mass 87. The abundance of these
isotopes is 71% and 29%. Calculate the
average mass of an atom of this element.
4. If 69Ga and 71Ga occur in the %’s 62.1 and 37.9,
calculate the average atomic mass of gallium 39
Mass Spectrum of Chlorine Molecule
(35Cl-35Cl)+, (35Cl-37Cl)+, or (37Cl-37Cl)+
40
Ions
Ion= an electrically charged particle obtained from an atom or
chemically bonded group of atoms lose or gain electrons. The
charge on an ion is equal to the # of protons minus the # of
electrons. An atom that gains extra electrons becomes a
negatively charged ion, called an anion. An atom that loses
electrons becomes positively charged ion, called a cation.
number p + number n
number p
A +?
E
Z
number p - number e
E.g. Determine numbers of electrons in Mg2+ cation
and the S2- anion?
Mg2+
number e =?
S2-
number e =?
41
Introduction to Periodic Table
With discovery of many elements
1869 Mendeleev and Meyer
~ independently proposed periodic table organized
the elements
In modern periodic table, The periodic table of the
elements is organized into 18 groups and 7 periods.
Elements are represented by one or two-letter
symbols and are arranged according to atomic
number.
* a horizontal row of elements- a period
* a vertical row of elements- a group or family
42
Periodic Table of Elements
Group
1
2
Period
1A
1
2
1
H
He
4
5
6
7
8
9
10
11
12
2A
3
4
Li
11
12
Na
20
K
Ca
Sr
16
17
18
5A
5
6
7
6A
7A
8
9
4.003
6B
7B
21
22
23
24
25
Y
88.91
Ti
V
Cr
Mn
--- --8B -- ---
26
Fe
27
Co
28
Ni
1B
2B
29
30
Cu
47.88 50.94 52.00 54.94 55.85 58.93 58.69 63.55
40
Zr
41
Nb
42
Mo
91.22 92.91 95.94
43
Tc
(98)
44
Ru
45
Rh
46
Pd
47
Ag
101.1 102.9 106.4 107.9
Zn
Al
C
N
O
10
14
15
16
Si
P
26.98 28.09 30.97
31
Ga
32
Ge
33
As
65.39 69.72 72.61 74.92
48
Cd
49
In
50
Sn
51
Sb
112.4 114.8 118.7 121.8
17
18
S
Cl
Ar
32.07
35.45
39.95
34
35
36
Se
Br
Kr
78.96 79.90 83.80
52
53
I
Xe
127.6
126.9
131.3
Te
54
55
56
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
Cs
Ba
Lu
Hf
Ta
W
Re
Os
Ir
Pt
Au
Hg
Tl
Pb
Bi
Po
At
Rn
(209)
(210)
(222)
116
117
118
132.9 137.3
87
7
5B
39
85.47 87.62
6
4B
44.96
38
Rb
3B
Sc
39.10 40.08
37
4A
13
Mg
19
5
15
3A
B
22.99 24.30
4
14
F
Ne
10.81 12.01 14.01 16.00 19.00 20.18
Be
6.941 9.012
3
13
8A
1.008
2
3
103
104
Rf
Db
Sg
Bh
Hs
Mt
226.0
227.0
(216)
(262)
(266)
(264)
(269)
(268)
*Lanthanoids
**Actinoids
138.91 178.5 180.9 183.8 186.2 190.2 192.2 195.1 197.0
88
Fr
(223)
*
Ra **
*
Lr
57
La
89
** Ac
58
Ce
90
Th
105
59
106
60
Pr
Nd
91
92
Pa
U
107
61
Pm
93
Np
108
62
Sm
94
Pu
109
63
Eu
95
Am
110
Ds
64
Gd
96
Cm
111
Rg
65
Tb
97
Bk
200.6 204.4 207.2 209.0
112
Uub
66
Dy
98
Cf
113
Uut
67
Ho
99
Es
114
Uuq
68
Er
100
Fm
115
Uup
69
Tm
101
Md
Uuh
Uus
70
Yb
102
No
43
Uuo
It is customary also to divide the elements into broad categories
known as
Metals: Except mercury (liquid), metals are solid s at room
temperature. They are generally malleable, ductile, good
conductors of heat and electricity, and have a lustrous or shiny
appearance.
Nonmetals: generally have opposite properties of metals; e.g.
poor conductors of heat and electricity.
Metalloid (semimetal): is an element having both metallic and
nonmetallic properties.
Or into three groups
Main group elements are those in groups 1, 2 and 13-18. when
form ions, group 1, 2 lose the same # e as their group #; group 13
lose group #-10; group 14-18 gain 18-group #.
Transition elements: from group 3 to 12, and because all of them
are metals, they are also called the transition metals. The # of
electrons lost in TM is not related to their group #.
44
Inner transition metals which include Lanthanides and Actinindes.
Nuclear Chemistry
• Nuclear reactions involve
changes that originate in the
nucleus of the atom.
• Chemical changes involve
changes in the electron cloud.
• Uses:
–
60Co-
gamma ray emitter- ionizing
radiation for treatment of cancerous
tumors.
– 201Thallium stress test of heart
muscle
– Radiocarbon dating 14C ½ life 5730
years
– Nuclear power ~ 20% of US
electricity production
45
Radioactivity
• Recall that all atoms of the same element have
the same number of protons. The number of
neutrons in the atoms nucleus, however, may be
different from one atom to the next= Isotopes.
Uranium- 234
Uranium-235
Uranium-238
92 protons
92 protons
92 protons
142 neutrons
143 neutrons
146 neutrons
Trace
0.7%
99.3%
• Different isotopes have different abundances
• Different isotopes have different stabilities
46
Patterns of Nuclear Stability
As the atomic number
increases, the neutron to
proton ratio of the stable
nuclei increases. The
stable nuclei are located
in the shaded area of the
graph known as the belt
of stability. The majority
of radioactive nuclei
occur outside this belt.
47
Nuclear Equations
• Radionuclides are unstable nuclei that
emit particles and electromagnetic
radiation to transform into a stable
nucleus.
238
92
U
234
90
Th
+
4
2
He
48
Nuclear Equations
• Mass numbers and atomic numbers
must be balanced in all nuclear
equations.
49
What product is formed when
thorium-232 undergoes alpha
decay?
50
Types of Radioactive Decay
Alpha decay- nucleus
emits 2 protons and 2
neutrons (He nucleus)
Beta decay- a neutron
in the nucleus decays
into a proton and an
electron, the electron
is emitted
Gamma- high energy, short
wavelength electromagnetic
radiation- accompanies other
51
radioactive emissions.
Types of Radioactive Decay
Electron Capture- capture by
the nucleus of an electron
from the electron cloud
surrounding the nucleus.
Positron- particle with the
same mass as an electron,
but an opposite charge
collides with an electron
and produces gamma
radiation.
52
Penetrating Power of Radioactive
Decay
53
Radioactive Decay Particles
Particle
Nuclear
Equation
Example
Alpha= 2
Nucleus  4He
226Ra
Beta =neutron
1n
 1p + 0e
131I131Xe
53
54
+ 0e 1n
81Rb
+ 0e  81Kr
protons and 2
neutrons
converts to proton
and a high energy
electron
Electron
Capture=
2
1p
1
-1
1
0
-1
0
88
37
222Rn +4He
86
-1
2
+ 0e
-1
36
electron captured
by nucleus
54
Radioactive Decay Particles
Particle
Positron=proton
Nuclear
Equation
1p  1n + 0e
converted to a
neutron and an
electron
1
Gamma=
Not shown in
equations, but
almost always
accompanies other
decay.
electromagnetic
radiation
0
1
Example
11C
6
11B +0e
5
Remember a positron has the same mass as an electron, but the
opposite charge
1
55
Radioactive Decay Particles
Particle
Effect
Alpha
Decrease atomic mass by ___ and
atomic number by _____.
Atomic number _______________.
Beta
Electron
Capture
Positron
Atomic number _______________.
Atomic number _______________.
Gamma
56
Half Life- the time required for half of any
given quantity of a substance to react / decay.
(independent of initial quantity of atoms)
Half Life Simulation
Number of Th232 atoms in a
sample initially
containing 1
million atoms as
a function of
time. Th-232 has
a half-life of 14
billion years.
57
Half Life Problems
Example:
An isotope of cesium (cesium-137) has a
half-life of 30 years. If 1.0 mg of cesium137 disintegrates over a period of 90
years, how many mg of cesium-137 would
remain?
58
Half Life Problems
1. A 2.5 gram sample of an isotope of strontium-90 was formed in a 1960 explosion of an
atomic bomb at Johnson Island in the Pacific Test Site. The half-life of strontium-90 is
28 years. In what year will only 0.625 grams of this strontium-90 remain?
2. Actinium-226 has a half-life of 29 hours. If 100 mg of actinium-226 disintegrates over a
period of 58 hours, how many mg of actinium-226 will remain?
3. Thallium-201 has a half-life of 73 hours. If 4.0 mg of thallium-201 disintegrates over a
period of 6.0 days and 2 hours, how many mg of thallium-201 will remain?
4. Sodium-25 was to be used in an experiment, but it took 3.0 minutes to get the sodium
from the reactor to the laboratory. If 5.0 mg of sodium-25 was removed from the
reactor, how many mg of sodium-25 were placed in the reaction vessel 3.0 minutes
later if the half-life of sodium-25 is 60 seconds?
5. Selenium-83 has a half-life of 25.0 minutes. How many minutes would it take for a 10.0
mg sample to decay and have only 1.25 mg of it remain?
59
Uranium-238: an example of an unstable nucleus
decaying to form other unstable nuclei
Uranium-238 is
radioactive,
undergoing alpha
decay. But, the
daughter nuclide is
also radioactive,
undergoing beta
decay, to produce yet
another radioactive
nuclide, which
decays. The atom
goes through a rather
involved sequence of
radioactive decays
(both alpha and beta),
until a stable isotope
(lead-206) is reached.
60
Fission Reaction
Collision of a neutron with a U-235 nucleus can
cause the nucleus to split, creating two smaller
nuclides and three free neutrons. The three
neutrons may travel outward from the fission,
colliding with nearby U-235 nuclei, causing them to
split as well. Each split (fission) is accompanied by 61
Fission Chain Reaction
Mousetrap Chain Reaction
Collision of a neutron with a
U-235 nucleus can cause the
nucleus to split, creating two
smaller nuclides and three
free neutrons. The three
neutrons may travel outward
from the fission, colliding with
nearby U-235 nuclei, causing
them to split as well. Each
split (fission) is accompanied
by a large quantity of energy.
If sufficient neutrons are
present, we may achieve a
chain reaction. If only one
neutron were produced with
each fission, no chain
reaction would occur,
because some neutrons
62
would be lost through the
surface of the uranium
Fission Reaction
63
Fusion Reaction
Tremendous energy
needed to overcome the
repulsion between nuclei.
Heat required for this
reaction is on the order of
40,000,000 K. The energy
from an atomic bomb
could generate this heat
(hydrogen or
thermonuclear weapon).
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