How to Calculate Percent Abundances

The nuclei of atoms contain only protons and neutrons, and each of these has, by definition, a mass of approximately 1 atomic mass unit (amu). The atomic weight of each element – which does not include the weights of electrons, which are considered negligible – should therefore be a whole number. A quick perusal of the periodic table, however, shows that the atomic weights of most elements contain a decimal fraction. This is because the listed weight of each element is an average of all the naturally occurring isotopes of that element. A quick calculation can determine the percentage abundance of each isotope of an element, provided you know the atomic weights of the isotopes. Because scientists have accurately measured the weights of these isotopes, they know the weights vary slightly from integral numbers. Unless a high degree of accuracy is needed, you can ignore these slight fractional differences when calculating abundance percentages.

TL;DR (Too Long; Didn't Read)

You can calculate the percent abundance of isotopes in a sample of an element with more than one isotope as long as the abundances of two or fewer are unknown.

What Is an Isotope?

The elements are listed in the periodic table according to the number of protons in their nuclei. Nuclei also contain neutrons, however, and depending on the element, there may be none, one, two, three or more neutrons in the nucleus. Hydrogen (H), for example, has three isotopes. The nucleus of 1H is nothing but a proton, but the nucleus of deuterium (2H) contains a neutron and that of tritium (3H) contains two neutrons. Six isotopes of calcium (Ca) occur in nature, and for tin (Sn), the number is 10. Isotopes may be unstable, and some are radioactive. None of the elements that occur after Uranium (U), which is 92nd in the periodic table, has more than one natural isotope.

Elements With Two Isotopes

If an element has two isotopes, you can readily set up an equation to determine the relative abundance of each isotope based on the weight of each isotope (W1 and W2) and the weight of the element (We) listed in the periodic table. If you denote the abundance of isotope 1 by x, the equation is:

W1 • x + W2 • (1 - x) = We

since the weights of both isotopes must add to give the weight of the element. Once you find (x), multiply it by 100 to get a percentage.

For example, nitrogen has two isotopes, 14N and 15N, and the periodic table lists the atomic weight of nitrogen as 14.007. Setting up the equation with this data, you get: 14x + 15(1 - x) = 14.007, and solving for (x), you find the abundance of 14N to be 0.993, or 99.3 percent, which means the abundance of 15N is 0.7 percent.

Elements With More Than Two Isotopes

When you have a sample of an element that has more than two isotopes, you can find the abundances of two of them if you know the abundances of the others.

As an example, consider this problem:

The average atomic weight of oxygen (O) is 15.9994 amu. It has three naturally occurring isotopes, 16O, 17O and 18O, and 0.037 percent of oxygen is made up of 17O. If the atomic weights are 16O = 15.995 amu, 17O = 16.999 amu and 18O = 17.999 amu, what are the abundances of the other two isotopes?

To find the answer, convert percentages to decimal fractions and note that the abundance of the other two isotopes is (1 - 0.00037) = 0.99963.

  1. Define a Variable

  2. Set one of the unknown abundances – say that of 16O – to be (x). The other unknown abundance, that of 18O, is then 0.99963 - x.

  3. Set up an Average Atomic Weight Equation

  4. (atomic weight of 16O) • (fractional abundance of 16O) + (atomic weight of 17O) • (fractional abundance of 17O) + (atomic weight of 18O) • (fractional abundance of 18O) = 15.9994

    (15.995) • (x) + (16.999) • (0.00037) + (17.999) • (0.99963 - x) = 15.9994

  5. Expand and Collect Numerical Values on the Right Side

  6. 15.995x - 17.999x = 15.9994 - (16.999) • (0.00037) - (17.999) (0.99963)

  7. Solve for x

  8. x = 0.9976

    Having defined (x) to be the abundance of 16O, the abundance of 18O is then (0.99963 - x) = (0.99963 - 0.9976) = 0.00203

    The abundances of the three isotopes are then:

    16O = 99.76%

    17O = 0.037%

    18O = 0.203%


About the Author

Chris Deziel holds a Bachelor's degree in physics and a Master's degree in Humanities, He has taught science, math and English at the university level, both in his native Canada and in Japan. He began writing online in 2010 with the goal of exploring scientific, cultural and practical topics, and at last count had reached over a hundred million readers through various sites.

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