Download Isotopes PPT

Isotopes
• Atoms of the same element with different
mass numbers.
• Nuclear symbol:
Mass #
12
Atomic #
6
• Hyphen notation: carbon-12
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C
Isotopes
Neutron
+
Electrons
Nucleus
+
+
+
+
+
Nucleus
Proton
Proton
Nucleus
Carbon-12
Neutrons 6
Protons
6
Electrons 6
+
+
+
+
Neutron
Electrons
+
+
Carbon-14
Neutrons 8
Protons
6
Electrons 6
Nucleus
6Li
7Li
3 p+
3 n0
3 p+
4 n0
2e– 1e–
2e– 1e–
Neutron
Neutron
Electrons
Electrons
+
Nucleus
+
+
Nucleus
+
Nucleus
Lithium-6
Neutrons 3
Protons
3
Electrons 3
Proton
+
+
Nucleus
Lithium-7
Neutrons 4
Protons
3
Electrons 3
Proton
17
Cl
Isotopes
37
• Chlorine-37
– atomic #:
17
– mass #:
37
– # of protons:
17
– # of electrons: 17
– # of neutrons: 20
Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem
37
17
Cl
Relative Atomic Mass
•
12C
atom = 1.992 × 10-23 g
• atomic mass unit (amu)
• 1 amu = 1/12 the mass of a 12C atom
• 1 p = 1.007276 amu
1 n = 1.008665 amu
1 e- = 0.0005486 amu
+
Electrons
Nucleus
+
Neutron
+
+
+
+
Nucleus
Carbon-12
Neutrons 6
Protons
6
Electrons 6
Proton
Average Atomic Mass
• weighted average of all isotopes
• on the Periodic Table
• round to 2 decimal places
Avg.
(mass)(%) + (mass)(%)
Atomic =
100
Mass
Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem
Average Atomic Mass
• EX: Calculate the avg. atomic mass of oxygen if its
abundance in nature is 99.76% 16O, 0.04% 17O, and
0.20% 18O.
Avg.
(16)(99.76) + (17)(0.04) + (18)(0.20)
16.00
Atomic =
=
amu
100
Mass
Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem
Average Atomic Mass
• EX: Find chlorine’s average atomic mass
if approximately 8 of every 10 atoms are
chlorine-35 and 2 are chlorine-37.
Avg.
(35)(8) + (37)(2)
Atomic =
= 35.40 amu
10
Mass
Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem
17
100
Mass spectrum of chlorine. Elemental chlorine (Cl2) contains
only two isotopes: 34.97 amu (75.53%) and 36.97 (24.47%)
90
80
Cl-35
70
Abundance
AAM = (34.97 amu)(0.7553) + (36.97 amu)(0.2447)
60
AAM =
(26.412841 amu)
AAM =
+
(9.046559 amu)
35.4594 amu
50
40
30
Cl-37
20
10
0
34
36
35
Mass
37
Cl
35.4594
Mass Spectrophotometer
magnetic field
heaviest
ions
stream
of ions of
different
masses
electron
beam
gas
Dorin, Demmin, Gabel, Chemistry The Study of Matter 3rd Edition, page 138
lightest
ions
Weighing atoms
gas sample
enters here
.
ions accelerate
towards charged
slit
magnetic field
deflects lightest ions
most
filament current
ionizes the gas
The first mass spectrograph was
built in 1919 by F. W. Aston, who
received the 1922 Nobel Prize for
this accomplishment
ions separated by mass
expose film
• mass spectrometry is used to experimentally determine isotopic masses
and abundances
• interpreting mass spectra
• average atomic weights
- computed from isotopic masses and abundances
- significant figures of tabulated atomic weights gives some idea
of natural variation in isotopic abundances
Copyright © 1997-2005 by Fred Senese
Mass Spectrometry
198
200
202
Photographic plate
196
-
+
Stream of positive ions
Hill, Petrucci, General Chemistry An Integrated Approach 1999, page 320
199
201
204
Mass spectrum of mercury vapor
Mass Spectrum for Mercury
(The photographic record has been converted to a scale of relative number of atoms)
The percent natural abundances
for mercury isotopes are:
Hg-196
Hg-198
Hg-199
Hg-200
Hg-201
Hg-202
Hg-204
Relative number of atoms
30
25
198
0.146%
10.02%
16.84%
23.13%
13.22%
29.80%
6.85%
196
200
199
15
10
5
197
198
201
204
Mass spectrum of mercury vapor
20
196
202
199
200
Mass number
201
202
203
204
80
Hg
200.59
The percent natural abundances
for mercury isotopes are:
A
B
C
D
E
F
G
Hg-196
Hg-198
Hg-199
Hg-200
Hg-201
Hg-202
Hg-204
0.146%
10.02%
16.84%
23.13%
13.22%
29.80%
6.85%
(% "A")(mass "A") + (% "B")(mass "B") + (% "C")(mass "C") + (% "D")(mass "D") + (% "E")(mass "E") + (% F)(mass F) + (% G)(mass G) = AAM
(0.00146)(196) + (0.1002)(198) + (0.1684)(199) + (0.2313)(200) + (0.1322)(201) + (0.2980)(202) + (0.0685)(204) = x
0.28616 + 19.8396 + 33.5116 + 46.2600 + 26.5722 + 60.1960 + 13.974 = x
x = 200.63956 amu
92
Separation of Isotopes
U
238
Natural uranium, atomic weight = 238.029 g/mol
Density is 19 g/cm3. Melting point 1000oC.
Two main isotopes:
238
92
235
92
U
U
99.3%
0.7%
(238 amu) x (0.993) + (235 amu) x (0.007)
236.334 amu + 1.645 amu
Because isotopes are chemically identical
(same electronic structure), they cannot be
separated by chemistry.
So Physics separates them by diffusion or
centrifuge (mass spectrograph is too slow)…
237.979 amu
17
Cl
35.453
• Assume you have only two atoms of chlorine.
• One atom has a mass of 35 amu (Cl-35)
• The other atom has a mass of 36 amu (Cl-36)
• What is the average mass of these two isotopes?
35.5 amu
• Looking at the average atomic mass printed on the
periodic table...approximately what percentage is Cl-35
and Cl-36?
55% Cl-35 and 45% Cl-36 is a good approximation
17
Cl
35.453
Using our estimated % abundance data
55% Cl-35 and 45% Cl-36
calculate an average atomic mass for chlorine.
Average Atomic Mass = (% abundance of isotope "A")(mass "A") + (% "B")(mass "B")
AAM = (% abundance of isotope Cl-35)(mass Cl-35) + (% abundance of Cl-36)(mass Cl-36)
AAM = (0.55)(35 amu) + (0.45)(36 amu)
AAM = (19.25 amu) + (16.2 amu)
AAM = 35.45 amu
Isotopes
Dalton was wrong.
Atoms of the same element can have
different numbers of neutrons
different mass numbers
called isotopes
C-12
California WEB
vs.
C-14
Naming Isotopes
• Put the mass number after the name of
the element
• carbon- 12
• carbon -14
• uranium-235
California WEB
Using a periodic table and what you know about atomic
number, mass, isotopes, and electrons, fill in the chart:
Element
Symbol
Atomic
Number
Atomic
Mass
# of
protons
# of
neutron
# of
electron
8
8
8
39
Potassium
+1
Br
45
30
35
-1
30
Atomic Number = Number of Protons
Number of Protons + Number of Neutrons = Atomic Mass
Atom (no charge) : Protons = Electrons
Ion (cation) : Protons > Electrons
charge
Ion (anion) : Electrons > Protons
Using a periodic table and what you know about atomic
number, mass, isotopes, and electrons, fill in the chart:
ANSWER KEY
Element
Symbol
Atomic
Number
Atomic
Mass
# of
protons
# of
neutron
# of
electron
charge
Oxygen
O
8
16
8
8
8
0
Potassium
K
19
39
19
20
18
+1
Bromine
Br
35
80
35
45
36
-1
Zinc
Zn
30
35
30
65
30
0
Atomic Number = Number of Protons
Number of Protons + Number of Neutrons = Atomic Mass
Atom (no charge) : Protons = Electrons
Ion (cation) : Protons > Electrons
Ion (anion) : Electrons > Protons
Atomic Mass
•
•
•
•
•
How heavy is an atom of oxygen?
There are different kinds of oxygen atoms.
More concerned with average atomic mass.
Based on abundance of each element in nature.
Don’t use grams because the numbers would be
too small
Measuring Atomic Mass
• Unit is the Atomic Mass Unit (amu)
• One twelfth the mass of a carbon-12 atom.
• Each isotope has its own atomic mass we need
the average from percent abundance.
(1 amu)
(1 amu)
(1 amu)
(1 amu)
carbon atom
(1 amu)
(1 amu)
(1 amu)
(1 amu)
(12 amu)
(1 amu)
(1 amu)
(1 amu)
(1 amu)
Mass spectrums reflect the abundance of
naturally occurring isotopes.
Natural Abundance of Common Elements
Hydrogen
1H
= 99.985%
2H
= 0.015%
Carbon
12C
= 98.90%
13C
= 1.10%
Nitrogen
14N
= 99.63%
15N
= 0.37%
Oxygen
16O
= 99.762%
17O
= 0.038%
Sulfur
32S
= 95.02%
33S
= 0.75%
34S
= 4.21%
36S
= 0.02%
Chlorine
35Cl
= 75.77%
37Cl
= 24.23%
Bromine
79Br
= 50.69%
81Br
= 49.31%
18O
= 0.200%
For example….Methane
For carbon 1 in approximately 90
atoms are carbon-13
The rest are carbon-12 the isotope that
is 98.9% abundant.
So, for approximately 90 methane
molecules…1 carbon is carbon-13
C-13
Where’s
Waldo?
Where’s
Waldo?
Calculating averages
• You have five rocks, four with a mass of 50 g, and
one with a mass of 60 g. What is the average
mass of the rocks?
• Total mass = (4 x 50) + (1 x 60) = 260 g
• Average mass = (4 x 50) + (1 x 60) = 260 g
5
5
• Average mass = 4 x 50 + 1 x 60 = 260 g
5
5
5
California WEB
Calculating averages
• Average mass = 4 x 50 + 1 x 60 = 260 g
5 5
5
• Average mass = .8 x 50 + .2 x 60
• 80% of the rocks were 50 grams
• 20% of the rocks were 60 grams
• Average = % as decimal x mass +
% as decimal x mass +
% as decimal x mass +
California WEB
Isotopes
• Because of the existence of isotopes, the mass of a
collection of atoms has an average value.
• Average mass = ATOMIC WEIGHT
• Boron is 20% B-10 and 80% B-11.
That is, B-11 is 80 percent abundant on earth.
• For boron atomic weight
= 0.20 (10 amu) + 0.80 (11 amu) = 10.8 amu
Periodic Table
• Dmitri Mendeleev developed the
modern periodic table.
• Argued that element properties are
periodic functions of their atomic
weights.
• We now know that element
properties are periodic
functions of their
ATOMIC NUMBERS.
Atomic Mass
Magnesium has three isotopes.
78.99% magnesium 24 with
a mass of 23.9850 amu,
10.00% magnesium 25 with
a mass of 24.9858 amu, and
the rest magnesium 26 with
a mass of 25.9826 amu.
What is the atomic mass of
magnesium?
If not told otherwise,
the mass of the isotope is
the mass number in amu.
California WEB
Isotope
Percent
Abundance
Mg-24
78.99
23.9850 18.94575
Mg-25
10.00
24.9585
2.49585
Mg-26
11.01
25.9826
2.86068
Mass
24.304 amu
Atomic Mass
Calculate the atomic mass of copper if copper has two isotopes.
69.1% has a mass of 62.93 amu and the rest has a mass of
Percent
64.93 amu.
Isotope
Mass
Abundance
Cu-63
69.1
62.93
43.48463
Cu-65
30.9
64.93
20.06337
63.548
Average atomic mass (AAM)  (% " A" )(mass " A" )  (% " B" )(mass " B" )  ...
A.A.M.  (0.691)(62.93 amu)  (0.309)(64.93 amu)
A.A.M.  43.48463 amu  20.06337 amu
A.A.M.  63.548 amu for Copper
29
Cu
63.548
Protons
Neutrons
Electrons
Mass
number
Cu-65
A
B
29
C
A.
B.
C.
Argon
D
E
F
40
D.
E.
F.
Ba2+
56
G
H
I
G.
H.
I.
Given the average atomic mass of an element is 118.21 amu and it has
three isotopes (“A”, “B”, and “C”):
isotope “A” has a mass of 117.93 amu and is 87.14% abundant
isotope “B” has a mass of 120.12 amu and is 12.36% abundant
Find the mass of isotope “C”.
Show work for credit.
Extra Credit: What is a cation?
Protons
Neutrons
Electrons
Mass
number
Cu-65
A = 29
B = 36
29
C = 65
Argon
D = 18 E = 22
F = 18
40
Ba2+
56
G = 81 H = 54 I = 137
Given the average atomic mass of an element is 118.21 amu and it has
three isotopes (“A”, “B”, and “C”):
isotope “A” has a mass of 117.93 amu and is 87.14% abundant
isotope “B” has a mass of 120.12 amu and is 12.36% abundant
Find the mass of isotope “C”.
Show work for credit.
119.7932 amu
Extra Credit: What is a cation?
A positively charged atom. An atom that has lost a(n) electron(s).
Given the average atomic mass of an element is 118.21 amu and it has
three isotopes (“A”, “B”, and “C”):
isotope “A” has a mass of 117.93 amu and is 87.14% abundant
isotope “B” has a mass of 120.12 amu and is 12.36% abundant
Find the mass of isotope “C”.
Show work for credit.
Average Atomic Mass  (% " A" )(mass " A" )  (% " B" )(mass " B" )  (% " C" )(mass " C" )
118.21 amu  (0.8714)(117.93 amu)  (0.1236)(120.12 amu)  (0.005)(X amu)
118.21 amu  102.764202 amu  14.846832 amu  (0.005)(X amu)
0.598966  0.005 X amu
0.598966  0.005 X amu
0.005
0.005
X  119.7932 amu