Download Isotopes

Isotopes
• Atoms of the same element with different
mass numbers.
• Nuclear symbol:
Mass #
Atomic #
12
6
• Hyphen notation: carbon-12
Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem
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
17
Cl
Isotopes
37
• Chlorine-37
– atomic #:
17
– mass #:
37
– # of protons:
17
– # of electrons:
17
– # of neutrons:
20
37
17
Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem
Cl
Average Atomic Mass
• weighted average of all isotopes
• on the Periodic Table
• round to 2 decimal places
Avg.
Atomic
Mass
(mass)(% )  (mass )(% )

100
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 
100
amu
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.
Atomic
Mass
(35)(8)  (37)(2)

 35.40 amu
10
Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem
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
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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
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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.
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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
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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 +
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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
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.
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Isotope
Percent
Abundance
Mass
Mg-24
78.99
23.9850
18.94575
Mg-25
10.00
24.9585
2.49585
Mg-26
11.01
25.9826
2.86068
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
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  14.846832  (0.005)(X amu)
0.598966  0.005 X amu
0.598966  0.005 X amu
0.005
0.005
X  119.7932 amu