AVERAGE ATOMIC MASS
Why is the mass of each element (on the periodic table) not an integer?
 Not all atoms in an element have the same mass
 Atoms of any element are composed of several isotopes
RECALL: Isotopes are atoms of an element that have the same number of protons but
different numbers of neutrons.
 Isotopes with different mass numbers exist in a fixed ratio in a sample of an element
 The relative mass of each isotope is measured using atomic mass units (u or amu) which are a
relative measure defined by the mass of carbon-12 (ie. 1 atom of C-12 has a mass of 12 u)
 Why are relative atomic masses not integers?
 Mass of prots & neutrons close to but not exactly 1.
 Mass of e-s (though almost relatively 0) do contribute to the overall mass.
 mass of a proton  mass of a neutron  1 u
The mass of each element on the periodic table has been determined using a naturally
occurring sample of that element.
 The percent abundance of each isotope can be determined by mass spectroscopy
 The mass is an average of the masses of all the element’s isotopes usually found in nature and
is called the average atomic mass.
Calculating Average atomic mass
This calculation involves more than simply averaging the masses of each isotope.
 Isotopic abundance - the relative amount in which each isotope is present in an element in
nature (usually expressed as a percent or a decimal).
 Since each isotope is not present in equal amounts, each isotopes percent isotopic abundance
is used together with the weighted measure of the mass of a specific isotope to calculate the
average atomic mass of the element
AVERAGE ATOMIC MASS [A and B are isotope masses]
Average
Relative
Isotopic
Relative
Isotopic
atomic = atomic mass X abundance of + atomic mass X abundance of + ...
mass of isotope A
isotope A
of isotope B
isotope B
-orAAM = mass A x % abundance + mass B x % abundance
100
100
Example 1:
A sample of carbon has two isotopes C-12 and C-13, with
C-12 comprising 98.89 % of the sample and C-13 comprising 1.11 % . Find the average
atomic mass.
A.A.M. =
(0.9889)(12) + (0.0111)(13)
11.8668
+ 0.1443
12.0111 a.m.u.
Therefore the average atomic mass of carbon is 12.01 u.
Example #2: The element boron consists of two isotopes, 105B and 115B. Their masses, based on
the carbon scale, are 10.01 and 11.01, respectively. The abundance of 105B is 20.0%.What is the
atomic abundance of and the abundance of 115B?
Solution:
The percentages of multiple isotopes must add up to 100%.
Since boron only has two isotopes, the abundance of one must be 100.0 - the abundance of the
other.
abundance of 115B = 100.0 - abundance of 105B
abundance of 115B = 100.0 - 20.0
abundance of 115B = 80.0
Answer:
The atomic abundance of 115B is 80%
Example #3: Copper is made up of two isotopes, Cu-63 (62.9296 amu) and Cu-65 (64.9278 amu).
Given copper's atomic weight of 63.546, what is the percent abundance of each isotope?
Solution:
Let x represent the abundance of the Cu-63 isotope
1) Write the following equation:
(62.9296) (x) + (64.9278) (1 - x) = 63.546
Once again, notice that 'x' and 'one minus x' add up to one.
2) Distribute equation and combine like terms to solve for x:
x = 0.6915
and therefore 1-x= 0.3085
Answer: Therefore the % abundance of Cu-63 isotope is 69.15% and Cu-65 is 30.85%
Note that this calculation technique works only with two isotopes. If you have three or more, there
are too many variables and not enough equations.