Each element is a substance composed of atoms with an identical number of protons in their nuclei. For example, an atom of the element nitrogen always has seven protons. All elements except hydrogen also have neutrons in their nuclei, and the element's atomic weight is the sum of the weights of the protons and neutrons. "Isotope" refers to variant forms of elements with different neutron counts -- each variant, with its unique neutron count, is an isotope of the element. The periodic table of the elements lists the atomic weight of each element, which is the weighted average of the isotope weights based on the abundance of each. You could easily look up the percent abundance of each isotope in a chemistry book or on the Web, but you may have to calculate the percent abundance by hand, for example, to answer a question on a chemistry test at school. You can perform this calculation for only two unknown isotope abundances at a time.
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The general formula for relative abundance is (M1)(x) + (M2)(1-x) = Me, where Me is the atomic mass of the element from the periodic table, M1 is the mass of the isotope for which you know the abundance, x is the relative abundance of the known isotope, and M2 is the mass of the isotope of unknown abundance. Solve for x to get the relative abundance of the unknown isotope.
Identify Atomic Weights
Identify the element's atomic weight and the atomic count of protons and neutrons for each of the two isotopes. This is information that will be given to you on a test question. For example, nitrogen (N) has two stable isotopes: N14 has a weight, rounded to three decimal places, of 14.003 atomic mass units (amu), with seven neutrons and seven protons, whereas N15 weighs 15.000 amu, with eight neutrons and seven protons. The atomic weight of nitrogen is given as 14.007 amu.
Set Abundance Equal to x
Let x equal the percentage abundance of one of the two isotopes. The other isotope must then have an abundance of 100 percent minus x percent, which you express in decimal form as (1 - x). For nitrogen, you can set x equal to the abundance of N14 and (1 - x) as the abundance of N15.
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Write out Equation
Write out the equation for the element's atomic weight, which equals the weight of each isotope times its abundance. For nitrogen, the equation is thus 14.007 = 14.003x + 15.000(1 - x).
Solve for x
Solve for x using simple algebra. For nitrogen, simplify the equation to 14.003x + (15.000 - 15.000x) = 14.007 and solve for x. The solution is x = 0.996. In other words, the abundance of the N14 isotope is 99.6 percent, and the abundance of the N15 isotope is 0.4 percent, rounded to one decimal place.