Download II Atomic Theory

III Atomic Theory
• Democritus
– All matter consists of very small indivisible
particles “atomes” [Greek for indivisible]
• Dalton’s Atomic Theory (beginning of
the modern era of chemistry
– Postulates
• Elements composed of small particles called
atoms. All atoms of a given element are
identical (size, mass, chemical properties) and
can not be changed into another element
• Compounds made up of more than one
element. Ratio of the number of atoms of the
element is an integer or simple fraction
• Chem reaction involves only separation,
combination, or rearrangement of atoms. It
does not result in creation or destruction
• Dalton’s New Theory supported the mass
laws
• Mass Laws
– Law of Conversation of Mass
• Matter can not be created or destroyed
– Proust’s Law of Definite Proportion [Constant
Composition]
• Different samples of the same compound always have
the same elements in the same ratio
– Law of Multiple Proportions
• If two elements can combine to form more than one
compound, the masses of one element with a fixed mass
of the second are in a ratio of small whole numbers
It is found that 14 grams of nitrogen, can
react with 16 grams of oxygen, 32
grams of oxygen and 40 grams of
oxygen. This is evidence of of the law
of?
A) Constant composition
B) Conservation of mass
C) Multiple proportions
It is found that 14 grams of nitrogen, can
react with 16 grams of oxygen, 32
grams of oxygen and 40 grams of
oxygen. This is evidence of of the law
of?
– Law of Multiple Proportions
•
If two elements can combine to form more
than one compound, the masses of one
element with a fixed mass of the second are
in a ratio of small whole numbers
• Discovery of Particles Smaller than the Atom
• Crookes Cathode Rays
anode
cathode
Gas filled tube
– When current applied, the tube begins to glow.
Flow goes from cathode to anode. Called
CATHODE RAYS
– Same beam occurs no matter what the gas.
Therefore, cathode rays are part of every atom.
– Beams deflected in a magnetic field, so, they have
negative charge
– Term electron coined by Stanley in 1891. Cathode
rays are beams of electrons
• Mass to charge ratio (1897) Thomson
– Thomson used a Crookes tube and placed
it in an adjustable magnetic field. He was
able to determine the mass to charge (m/e)
ratio of the cathode rays. By comparing
the ratio to the smallest mass to charge
ratio in solution discovered that the mass
of the cathode ray had to be 1/1000 the
mass of hydrogen atom. Therefore,
contrary to Dalton’s hypothesis, there were
particles smaller than the atom.
– Found (m/e) = - 5.686 x 10-12 kg/C
• C = Coulomb, unit of charge
• Millikan Oil Drop experiment (1909)
– Measured the speed of charged drops of
oil with and without an electric field.
– Was able to calculate charge on the
particles.
– Found, all charges were an integral
multiple of the same number.
– Determined, this number had to be the
charge of an electron.
– e = - 1.602 x 10-19 C
If you use Thomson’s m/e ratio and
Millikan’s charge on the electron (e) you
can calculate the mass of the electron.
(m/e)(e) =
(-5.686 x 10-12kg/C)(- 1.602 x 10-9 C)
= 9.109 x 10 -31kg/electron
Current accepted value
= 9.1093897 x 10-31 kg/electron
• X - Rays
– Roentgen, working with the Crookes tube
noticed that other material in the room
fluoresced when the tube was turned off.
– Decided that the radiation must be coming
from Crookes tube but not regular light. It
was some unknown radiation
– Called X - ray.
• Protons
anode
cathode
– Goldstein used cathode ray tube with a perforated
cathode
– Noticed, cathode rays went to the anode, but +
ions formed also
– Mass of the + ions varied depending on the gas in
the tube.
– Lightest particle produced when hydrogen gas
was used.
– All other masses were multiples of the mass form
when H was used. Therefore, the hydrogen ion
had to be a fundamental particle in all matter
– Called it the proton
• Neutrons
– Mass of the + ions formed varied
depending on the gas.
– But, total mass could not be accounted for
only by the mass of the protons.
– Therefore, there must be other particles
with about the same mass as the proton
but neutral
• Found by Chadwick (1932), called
neutrons.
•
A)
B)
C)
D)
E)
His experiment determined the charge
on an electron.
Crookes
Thomson
Roentgen
Stanley
Millikan
•
•
His experiment determined the charge
on an electron.
Millikan Oil Drop experiment (1909)
– Measured the speed of charged drops of
oil with and without an electric field.
– Was able to calculate charge on the
particles.
• Models of the Atom developed after the
discovery of electrons
• Thomson Model (~ 1900)
– “plum pudding” model.
– Cloud of positive charge with nuggets of
negative charge embedded in it so that it is
neutral
• Rutherford Model (1910)
– Devised an experiment to test Thomson model
– Shot a beam of particles (He2+; He nucleus) at
a thin metal foil
– If Thomson is correct, particles should pass
through
Metal foil
particles
expected
– Concluded: 1) mass of atom and positive charge
is in the center of the atom called the nucleus 2)
electrons moved at random about the nucleus
Atomic Structure
• Fundamental Particles of Atom
– Proton
– Neutron
– Electron
p+
no
e-
nucleons
• Atom is neutral because the number of
p+ = number of e-.
• Nuclide
– Representation of any specific atomic
species
A
Z
X
X = element
A = atomic mass (p + n)
Z = atomic number (p)
# neutrons(N) = A - Z
# electron in a neutral species
= # protons = Z

28
14
Si
p+ = ?
no = ?
e=?
14
14
14
For the nuclide 158 O -2
1. The number of protons is?

For the nuclide 158 O -2
1. The number of protons is? 8
2. The number of neutrons is?

For the nuclide 158 O -2
1. The number of protons is? 8
2. The number of neutrons is? 7
3. The number of electrons is?

For the nuclide 158 O -2
1. The number of protons is? 8
2. The number of neutrons is? 7
3. The number of electrons is? 10


• Isotopes
– Same element (same Z), but have different
mass, therefore have different number of
neutrons
12
C
13
C
14
C
– The value of z = 6 is understood


• Atomic Mass (Atomic Weight) (AW)
– Mass of atom measured relative to the
atomic standard.
– Current standard is 126 C
• Unit of atomic mass is atomic mass unit
(u) or (amu)
• 1 u defined as 1/12 the mass of 1 atom
of 126 C 
• So 1 atom of 12C has a mass of 12 u
exactly!
• All other atomic masses based on this

specific isotope
• On periodic table each element has it
own “box”
Z
X
mass
X = element
Z = atomic number
Mass = mass of
naturally occurring
material
• u is an extremely small quantity
– Difficult to measure
– So: defined an amount that can be measured
• Mole
– SI unit for amount of a substance
– Defined as: the amount of a substance that
contains the same number of entities as
there are atoms is exactly 12 grams of 12C.
– Therefore: 1 mole 12C = 12.00 grams exactly
– Defined conversion so it is exact
• The quantity of 12C atoms that have a mass
of 12.00 grams (i.e. contained in 1 mole of
12C) is 6.022 x 1023 atoms.
• Because of the definition of a mole, then, 1
mole of anything contains 6.022 x 1023
particles. Called Avogadro’s Number
• Because of the way the mole is defined,
Avogadro’s number can actually be used for
two different conversions.
– The atom to mole conversion
• 6.022 x 1023 atoms/mole
– A mass conversion
• 6.022 x 1023 u/gram
• Mass of Elements (Weighted Average)
• Most elements exist naturally as more
than one isotope. So, “mass” of the
element is actually the average of the
mass of each isotope.
• It is, however, not a direct average, but
a weighted average. That is, we have
to account for the abundance of each
isotope.
• Example: lithium exists and 6Li and 7Li.
Isotope
mass
6Li
%
abundance
7.42
7Li
92.58
7.01600
6.01512
Average (%1 /100)(mass 1 )(%2 /100)(mass 2 )
Average (0.0742)(6.01512)  (0.9258)(7.01600)
Average (0.4463219)  (6.495412)
Average 6.9417339
Average 6.942
• How are isotope mass and abundance
determined?
• By using mass spectrometer
• Lightest mass most strongly deflected, so
have greatest arc and are determined first.
• Find the number of protons, neutrons and
electrons in:(n.b. periodic table on back cover
of notes)
A) 24Mg+2P = . N = . E =
• Find the number of protons, neutrons and
electrons in:(n.b. periodic table on back cover
of notes)
A) 24Mg+2P = 12 . N = 12 . E = 10
• Find the number of protons, neutrons and
electrons in:(n.b. periodic table inside back
cover of notes)
A) 24Mg+2 P = 12 . N = 12 . E = 10
B) 15N -3
P=
.N=
. E=
• Find the number of protons, neutrons and
electrons in:(n.b. periodic table inside back
cover of notes)
A) 24Mg+2 P = 12 . N = 12 . E = 10
B) 15N -3
P = 7 . N = 8 . E =10
• Find the number of protons, neutrons and
electrons in:(n.b. periodic table inside back
cover of notes)
A) 24Mg+2 P = 12 . N = 12 . E = 10
B) 15N -3
C)
65Zn
P = 7 . N = 8 . E =10
P=
. N=
. E=
• Find the number of protons, neutrons and
electrons in:(n.b. periodic table inside back
cover of notes)
A) 24Mg+2 P = 12 . N = 12 . E = 10
B) 15N -3
C)
65Zn
P = 7 . N = 8 . E =10
P = 30 . N = 35 . E = 30
Boron has two isotopes, 10B and 11B. If a
the mass of boron is 10.81, the mass
of 10B = 10.0129, and 11B = 11.0093,
what is the % abundance of 10B?
A) 80.%
B) 20.%
C) 15.%
D) 85.%
Boron has two isotopes, 10B and 11B. If a
the mass of boron is 10.81, the mass
of 10B = 10.0129, and 11B = 11.0093,
what is the % abundance of 10B?
Average (%1 /100)(mass 1 )(%2 /100)(mass 2 )
Let x = fraction 10B, so 1-x = fraction 11B
10.81(x)(10.0129) (1- x)(11.0093)
10.8110.0129x 11.0093 -11.0093 x
0.9964x = 0.1993
x = 0.20 = 10B
10B = 20.%
1 - x = 0.80 = 11B
11B = 80. %
• Determining the mass of a compound
– A compound is a collection of elements
– Made up of specific atoms in a specific
ratio e.g. water = H2O = 2 H + 1 O;
methane = CH4 = 1 C + 4 H
– So, mass of the compound is the sum of
the mass of each element
• Mass of the element called Atomic Mass
(Atomic Weight) AW
• Mass of the compound called Molar
Mass (Molecular Weight) M or MW
• Examples
• Mass of water: H2O
Mass = 2(AW H) + AW O
= 2(1.01) + (16.00)
= 18.02
• Units of mass
– If you are talking about an atom or a collection of
atoms (molecule),
Unit = u/ atom or u/ molecule
– If you are talking about a mole of atoms or
molecules
Unit = grams/ mole
What is the mass of 1 mole of Na2CO3,
given: Na = 22.99; C = 12.01; O = 16.00
(give numeric answer)
What is the mass of 1 mole of Na2CO3,
given: Na = 22.99; C = 12.01; O = 16.00
(give numeric answer)
2(22.99) + 12.01 + 3(16.00) =
45.98
+ 12.01 + 48.00 =
105.99