isotopes atoms of a given element that differ in the number of neutrons …and consequently in mass
TRANSCRIPT
Isotopesatoms of a given element that
differ in the number of neutrons
…and consequently in mass.
Working with Atomic Notation
AzX
Atomic Number
Chemical Symbol
Atomic Mass(of Isotope)
Example #1
126C
Atomic Number
Chemical Symbol
Atomic Mass(of Isotope)
Example #2
136C
Atomic Number
Chemical Symbol
Atomic Mass(of Isotope)
Example #3
146C
Atomic NumberAtomic Number
Chemical Symbol
Atomic Mass(of Isotope)
Some isotopes of carbon
126C 13
6C 146C
10
NeNeNeon
20.1797
3
LiLiLithium
6.941
47
AgAgSilver
107.8682
11
NaNaSodium
22.98977
Various ways of identifying Isotopes
• Using atomic notation,Example: 12
6 C or simply 12 C
•Using the mass notation.
Example: Carbon-12 or C-12 (read “carbon twelve” or “C twelve”)
Some IsotopesIsotopes of Carbon
mass
notation
atomic
notation
# of p+ # of e- # of no
C-11 11C 6 6
C-12 12C 6 6
C-13 13C 6 6
C-14 14C 6 6
Another example of isotopes
11H 2
1H 31H
or
H-1 H-2 H-3
The IsotopesIsotopes of Hydrogen
mass
notation
atomic
notation
# of p+ # of e- # of no
Hydrogen-1 1H 1 1
Hydrogen-2 2H 1 1
Hydrogen-3 3H 1 1
p+ =
no =
e- =
p+ =
no =
e- =
p+ =
no =
e- =
Figure #1 Figure #2 Figure #3
Isotopesatoms of a given element that
differ in the number of neutrons …and consequently in mass.
Why are masses on the periodic table usually expressed as decimal
numbers?
• masses on the table are masses on the table are weightedweighted averagesaverages of all known isotopes of of all known isotopes of the element of interestthe element of interest
Keep in mind:It is not possible to determine how many different isotopes exist by looking at the periodic table.
It is not possible to determine the frequency of various nuclides by looking at the periodic table.
The following does not occur in nature!
11H occurrence 33.3%
21H occurrence 33.3%
31H occurrence 33.3%
The following does occur in nature!
11H occurrence 99.98%
21H occurrence 0.0156%
31H occurrence 0.0044%
another way of looking at it:
11H occurrence 9,998
21H occurrence 1.56
31H occurrence 0.44
Imagine having 10,000 H atoms
That means the weighted average is:
11H 1 x 0.9998 = 0.9998
21H 2 x 0.00156 = 0.00312
31H 3 x 0.00004 = 0.00012
Weighted Average (0.9998 + 0.00312 + 0.00012) 1.01
Zn has 5 naturally occurring isotopes
6430Zn occurrence
6630Zn occurrence
6730Zn occurrence
6830Zn occurrence
7030Zn occurrence
Zn has 5 naturally occurring isotopes64
30Zn occurrence 49 %
6630Zn occurrence ~28 %
6730Zn occurrence ~4 %
6830Zn occurrence ~18 %
7030Zn occurrence ~1 %
Zn has 5 naturally occurring isotopes
6430Zn 49 %
6630Zn ~28 %
6730Zn ~4 %
6830Zn ~18 %
7030Zn ~1 %
65.39
(parenthesis) on the Periodic Table indicate the
most stable isotopestable means “longest living”
Parenthesis also suggest the element of
interest is radioactive.
Review problem #1Represent the following using atomic notation.
92
UUUranium
238.0289
Review problem #2 Represent the following using mass notation.
10
NeNeNeon
20.1797
Review problem #3If the atom described below had 2 naturally occurring isotopes, which of the 2 would have a greater frequency of occurrence? Express your answer in atomic and mass notation.
3
LiLiLithium
6.941
Review problem #4How many total subatomic particles are in the following “neutral” atoms of Fe-55 and Fe-57?
