Isotopes and Atomic Mass
SCH 3U
Matter, Chemical Trends and Chemical Bonding
B3.2,3.1
l
What are Isotopes?
Two atoms are isotopes if they have the
same number of protons, but they have
different numbers of neutrons.
This means that:
Isotopes are atoms of the same element.
Isotopes have different atomic masses.
Isotopes have different number of neutrons in
their nuclei.
Comparing isotopes of Magnesium
 Magnesium has 3 isotopes, here is how they
compare:
Magnesium-24
Magesium-25
Magnesium-26
12 protons
12 protons
12 protons
12 neutrons
12 electrons
Isotope mass
24u
13 neutrons
12 electrons
Isotope mass
25u
14 neutrons
12 electrons
Isotope mass
26u
Comparing the isotopes of Magnesium
Similarities:
Same number of protons.
Same number of electrons.
Same appearance and chemical properties.
Differences:
Different number of neutrons.
Different atomic masses.
Isotopic Abundance (% Abundance)
A sample of magnesium is a mixture of the
three isotopes of magnesium.
Each isotope is a fraction of the mixture,
and has its own isotopic abundance
(expressed as a percentage of the whole).
The isotopic abundance is fixed so that
every sample of the element (in the
universe) has the same proportions of the
isotopes.
Isotopic Abundance
A sample of magnesium is a mixture of
three isotopes, present as:
79% Mg-24
10% Mg-25
11% Mg-26
Average Atomic Mass and Isotope
Abundance
 The Average Atomic Mass seen on the Periodic
Table is a weighted average of all of the isotope
masses.
 The weighted average takes into account the
isotope masses and their percent abundances.
 In a weighted average calculation, the isotope
with the greatest % abundance has the biggest
influence on the average atomic mass.
Average Atomic Mass
The average atomic mass is a weighted
average of all the isotope masses for a
particular element.
When you calculate average atomic mass,
you need three pieces of information:
The number of isotopes
The masses of each isotope
The % abundance of each isotope
Average Atomic Mass
The average atomic mass for carbon on
the Periodic table is 12.01u.
This means:
 Carbon has more than one isotope;
 One of carbon’s isotopes has a mass of 12,
another is greater than 12;
 The most abundant isotope is Carbon-12.
Calculating Average Atomic Mass
Use the equation:
AAM (u) = % ab1 x mass1 + % ab2 x mass2 + … % abn x massn
Where:
AAM = Average Atomic Mass
%ab1 = Percent abundance of isotope 1
mass1 = Mass of isotope 1
Sample Problem (p 167)
Silver has two naturally occuring isotopes,
Ag-107 (m=106.9u) and Ag-109
(m=108.9u) with isotopic abundances of
51.8% and 48.2% respectively. Calculate
the average atomic mass of silver.
AAM (u) = % ab1 x mass1 + % ab2 x mass2
AAM (u) = (0.518)x 106.9u + (0.482u) x 108.9u
AAM (u) = 55.4u + 52.5u
AAM (u) = 107.9u
Sample Problem (p 169)
 Boron exists as two naturally occuring isotopes:
Boron-10 (10.01u) and Boron-11(11.01).
Calculate the relative abundance of each
isotope, if the average atomic mass of boron is
10.81u.
To solve,
The percent abundance of all the isotopes should
add up to 1.
The % abundance of boron-10 is (x), and the %
abundance of boron-11 is (1-x).
Sample Problem (p 169 – continued)
Solution:
AAM (u) = % ab1 x mass1 + % ab2 x mass2
10.81 = x(10.01) + (1-x) (11.01)
10.81 = 10.01x + 11.01 – 11.01x
11.01x-10.01x = 11.01 -10.81
x = 0.2000
The abundance of Boron-10 is 20.0%
The abundance of Boron-11 is (1-x) or 80.0%