Overview: This tutorial discusses
isotopes and isotopic abundance. Nuclear binding energy and mass defect are
presented and the technique of mass spectrometry is introduced.
The current system of atomic masses was instituted in 1961 and is based on
the mass of 12C (read carbon twelve). By definition the atomic mass
of a single 12C atom is exactly 12 atomic mass units (denoted by
the abbreviation amu or u). The masses of all other elements are based on this
Note: These latter two processes only occur in nuclear reactions, not in normal
chemical reactions. Chemical reactions are processes in which the number of
electrons held or shared by an atom change. Nuclear reactions are processes
that involve changing the number of neutrons or protons held in the nucleus
of an atom.
How can we determine isotopic masses? It seems we should be able to add together
the masses of the constituent subatomic particles to determine the isotopic mass.
In the following example we will see how accurate this approach is.
Estimate the atomic mass of 7Li based on the masses of the constituent subatomic particles.
7Li: Mass Number: A = 7 = # of protons + # of neutrons
Atomic Number: Z = 3 = # of protons
number of neutrons = 4
number of electrons = 3 (this is a neutral atom).
Atomic mass = (#p+)(mp+) + (#n0)(mn0) + (# e-)(me-)
=(3)(1.00728 u) + (4)(1.00867 u) + (3)(5.5 x 10-4 u)
The experimentally determined value, measured by a technique called mass spectrometry, is 7.016005 u.
The difference in mass is: Dmass = 7.05817 u - 7.016005
u = 0.042165 u.
Approximately 0.6% of the mass is missing. This raises the question, what happened to this mass?
Therefore 3.79 x 109 kJ/mol is the amount of energy needed to break
the nucleus apart. This is much larger than the energy involved in normal chemical
reactions or processes. For instance, to remove an electron from an atom requires
only 5 x 102 kJ/mol. Hence it takes ~107 times more energy
to break apart the 7Li nucleus. This is a massive amount of energy.
The periodic table lists the atomic mass for each element. For instance, the
entry for copper (Cu) in the periodic table indicates an atomic mass of 63.546
u, but what does this really mean? In nature, Cu exists in two different isotopic
forms, 63Cu and 65Cu, and their natural abundances are
69% and 31%, respectively. We can use this data to solve for the elemental
i = an index identifying each isotope for the element
fi = fractional abundance of isotope i
mi = mass of isotope i
The elemental atomic mass is the atomic mass that appears in the periodic table.
It is nothing more than a weighted average of the isotopic masses of all the
naturally occurring isotopes.
We have been talking about isotopes for a while, but still have not formally
defined them. Isotopes are atoms of the same element that differ in the
number of neutrons in the nucleus and therefore they have different masses.
Nevertheless isotopes have practically identical properties in terms of chemical
What is the elemental atomic mass of naturally occurring Silicon? The naturally
occurring isotopes and their isotopic abundances are:
*Note that the natural abundances must add up to 100%!
The atomic mass of a specific atom or molecule is determined by using an experimental
technique called mass spectrometry. This technique separates the different isotopes
of atoms to allow determination of the percent abundance or isotopic composition
of the element in the given sample. Follow this link to learn the details of how a mass spectrometer works:
Each isotope appears as a peak in the mass spectrum. The intensity (height)
of each peak depends on the abundance of that isotope in the sample and the
unique location of the peak on the x-axis indicates the mass-to-charge ratio
(m/q) of the isotope.
Mass spectrometry is used in a diverse range of applications, such as accurate determinations of molecular masses, drug testing, determining the age of archaelogical artifacts (14C dating) and for studying the chemistry of DNA (see the Advanced Application below).
Consider the mass spectrum of silicon, shown below. The abundances are the
same as those in Example 2. As you can see, there are three isotopes. Each peak
represents one of the isotopes. The most abundant isotope has the highest peak
intensity and the least abundant isotope has the smallest intensity. Since the
peak intensities (heights) are proportional to the isotopic abundances, analysis
of the data allows for the determination of the relative abundances of each
isotope in the sample.
There are only two naturally occurring isotopes of Boron, 10B and
11B. If 10B has a mass of 10.013 u and 11B
has a mass of 11.006 u, what are the percent natural abundances of 10B
Now we have two unknowns and only one equation. We need another equation related
to the fractional abundances. We know that the fractional abundances must add
up to 1.00 (the percent abundances must add up to 100%), so:
Notice that there are no units associated with the fractional abundance. So
naturally occuring Boron is 19.64% 10B and 80.36% 11B.
This is all you need to know about mass spectrometry at this point, however,
if you would like to know more about how a mass spectrometer works go to: http://www.chemguide.co.uk/analysis/masspec/howitworks.html
Now you should have a good understanding of atomic, isotopic, and missing masses
along with nuclear binding energy. You should be comfortable performing calculations
to determine isotopic abundances, elemental atomic masses and determining nuclear
binding energies. Additionally, you have been introduced to the method of mass
spectrometry, which is used to determine atomic mass.