How to Find the Most Common Isotope
- 1). Determine if your element is naturally occurring on Earth. Naturally occurring elements on the periodic table are listed with decimal places in their atomic masses. For example, iodine (I) has a standard atomic weight of 126.9045 atomic mass units (amu). This indicates that iodine is present on Earth naturally and could be composed of different isotopes. Alternatively, Technetium (Tc) does not exist on Earth naturally, but has been synthesized in labs. Its standard atomic weight is reported as 98 or (98) and this whole number presentation indicates it is produced from a synthetic process.
- 2). Compile available information about a naturally occurring element's isotopes. If a relative atomic mass is not available for an isotope, use atomic masses. The atomic mass of an isotope is the sum of the neutrons and protons in its nucleus, and is a whole number. The relative atomic mass is the actual mass of the isotope, which is slightly different from the atomic mass. Consider bromine (Br) as an example. Bromine has a standard atomic weight of 79.904 and and two isotopes, Br-79 and Br-81. Br-79 has a relative atomic mass of 78.9183371 and Br-81 has a relative atomic mass of 80.916 2906.
- 3). Creating an equation that sets the atomic mass equal to the weighted average of each isotope's specific mass.
Standard Atomic Weight = Abundance1 x Relative Atomic Mass1
+ Abundance2 + Relative Atomic Mass2 + ...
This relationship sums the product of each isotopes abundance and relative atomic mass.
Bromine will continue to serve as an example:
79.904 = (% Abundance Br-79) x 78.9183371 + (% Abundance Br-81) x 80.9162906 - 4). Determine the unknown abundances of isotopes present in a naturally occurring element by solving the equation for unknown abundances, noting that the sum of abundances equals 100 percent.
In the case of Br, there are only two isotopes, so:
100% = % Abundance Br-79 + % Abundance Br-81
We can solve for % Abundance Br-79 to get
% Abundance Br-79 = 100% - % Abundance Br-81
We can substitute this into the original equation:
79.904 = (% Abundance Br-79) x 78.9183371 + (% Abundance Br-81) x 80.9162906
becomes
79.904 = (100% - % Abundance Br-81) x 78.9183371 + (% Abundance Br-81) x 80.9162906
Solving this equation gives % Abundance Br-81 = 49.334%
Subtracting this value from 100%, we find % Abundance Br-81 = 50.066% - 5). Designate the isotope with this largest percent abundance as the most common isotope.
Using bromine as an example, Br-81 is the most common isotope.
