Atomic Mass

Overview: This tutorial discusses isotopes and isotopic abundance. Nuclear binding energy and mass defect are presented and the technique of mass spectrometry is introduced.

Skills:

• Estimation of atomic masses for isotopes
• Determination of missing mass and nuclear binding energy
• Calculation of isotopic abundance

New terms:

• Atomic Mass Unit (u or amu)
• Missing Mass
• Nuclear Binding Energy
• Isotope
• Elemental Atomic Mass
• Mass Spectrometry

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 standard.

Subatomic Particle Masses

Three processes that change the number of subatomic particles in an atom

1. Ion formation (Ionization): Changing the number of electrons in an atom.
   For example, starting with a neutral sodium atom: Na Na⁺ + e^-. The cation, Na⁺, is generated by removal of an electron.
   The anion, I^-, is formed by adding an electron to the neutral iodine atom: I + e^- I^-
   
   
2. Isotope Conversion: Changing the number of neutrons in the nucleus of an atom.
   As an example, ⁵⁶Fe + n^{0 57}Fe. ⁵⁶Fe and ⁵⁷Fe are two different isotopes of iron. They have different nuclear masses, but have the same nuclear charge (same number of protons) and essentially identical chemical reactivity.
   
   
3. Transmutation Changing the number of protons in the nucleus. This converts one element into another.
   ⁵⁵Mn + p^{+ 56}Fe⁺
   

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.

Example 1.

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⁺)(m_p+) + (#n⁰)(m_n0) + (# e^-)(m_e-)
=(3)(1.00728 u) + (4)(1.00867 u) + (3)(5.5 x 10^-4 u)
=7.05817 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?

Missing Mass and Nuclear Binding Energy

In atom: 4 n0, 3 p+, 3 e- mass = 7.016005 u			Separate: 4 n0, 3 p+, 3 e- mass = 7.05817 u

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.

Elemental Atomic Mass

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 atomic mass.


i = an index identifying each isotope for the element
f_i = fractional abundance of isotope i
m_i = 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 reactivity.

Example 2.

What is the elemental atomic mass of naturally occurring Silicon? The naturally occurring isotopes and their isotopic abundances are:

Silicon isotope	natural abundance	isotopic mass
28Si	92.21%	27.97693 u
29Si	4.70%	28.97649 u
30Si	3.09%	29.97379 u

*Note that the natural abundances must add up to 100%!

Mass Spectrometry

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: http://www.chemguide.co.uk/analysis/masspec/howitworks.html.

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.

Example 3.

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 and ¹¹B?

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

Summary

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.

Practice Quiz: Atomic Mass