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Average Atomic Weights and Tracking Migrating Songbirds

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    What do songbirds have to do with atomic weights? It may be surprising that studies fundamental to establishing the entire set of atomic weights in chemistry are also the foundation for understanding songbird migration.

    Figure \(\PageIndex{1}\) The black throated blue warbler[1], whose migration was traced by stable isotope techniques


    The exact definition of atomic weight is technical [2], but atomic weights are easy to calculate once we understand the isotopic composition of elements (see below).

    Most elements consist of more than one isotope, each with a different mass. If elements are to have a single atomic weight (as we see they do from stoichiometry), the atomic weight must be an average of the isotopic weights of the "typical mixture" of naturally occurring (not necessarily stable) isotopes. Slight variations in the abundances of isotopes from region to region, however, allow bird tracking studies. Below is a table showing the typical abundances of the naturally occurring isotopes of a few important elements, taken from the Table of the Nuclides:

    Table \(\PageIndex{1}\): Natural Abundance of Elements

    Isotope % nat. abundance isotopic mass
    1H 99.985 1.007825
    2H 0.015 2.0140
    12C 98.89 12 (definition)
    13C 1.11 13.00335
    14N 99.64 14.00307
    15N 0.36 15.00011
    16O 99.76 15.99491
    17O 0.04 16.999131
    18O 0.2 17.9991610
    28Si 92.23 27.97693
    29Si 4.67 28.97649
    30Si 3.10 29.97376

    The average we use to calculate atomic weight isn't a straight average, though; it's a weighted average, like the one used to calculate your Grade Point Average (GPA).

    Atomic Weight =

    \(\left(\dfrac{\%\text{ abundance isotope 1}}{100}\right)\times \left(\text{mass of isotope 1}\right)~ ~ ~ +\)

    \(\left(\dfrac{\%\text{ abundance isotope 2}}{100}\right)\times \left(\text{mass of isotope 2}\right)~ ~ ~ + ~ ~ ...\)

    Similar terms would be added for all the isotopes.

    A grade point average is also a weighted average: It reflects the grade in courses with more semester hours more than lighter classes. suppose you get an A (4.0) in a chemistry course with 4.0 credit hours, and an F (0.0) in a course with 1.0 semester hours. We multiply the grade by the fraction of time for each and add. Since we have a total of 5 hours, the fraction of time for chemistry is 4.0/5.0, and the result (3.2) is much closer to the course with more credit hours:

    GPA =

    \(\left(\dfrac{\text{Credit Hours Course 1}}{\text{total credit hours}}\right)\times \left(\text{Grade in Course 1}\right)~ ~ ~ +\)

    \(\left(\dfrac{\text{Credit Hours Course 2}}{\text{total credit hours}}\right)\times \left(\text{Grade in Course 2}\right)~ ~ ~ + ~ ~ ...\)

    \(\dfrac{4.0}{5.0} \times \text{4.0} + \dfrac{1.0}{5.0} \times \text{0.00} = \text{3.2}\)

    Compare this calculation to the calculation of the atomic weight of carbon, where once again, we have one component that represents a small fraction of the whole, and should have little impact on the weighted average:

    Example \(\PageIndex{1}\): The Atomic Weight of Carbon

    Since Carbon is central to the definition of atomic weight, and also to biological and ecological studies, let's use it as an example here and in the next section. Naturally occurring carbon is found by mass spectrometry to consist of the two isotopes in the table. Calculate the atomic weight of an average naturally occurring sample of carbon.

    Solution A sample of naturally occurring carbon contains 98.89% \({}_{\text{6}}^{\text{12}}\text{C}\) whose relative nuclidic mass is 12.0000 (by definition) and 1.11% \({}_{\text{6}}^{\text{13}}\text{C}\) whose relative nuclieic mass is 13.00335. The atomic weight of carbon would be

    \(\dfrac{\text{98.89}}{\text{100.00}} \times \text{12.00} + \dfrac{\text{1.11}}{\text{100.00}} \times \text{13.003} = \text{12.011}\)

    Thus the atomic weight of carbon is 12.011, as mentioned earlier in the discussion.

    Stable Isotope Ecological Studies

    An important corollary to the existence of isotopes should be emphasized at this point. When highly accurate results are obtained, atomic weights may vary slightly depending on where a sample of an element was obtained. This can occur because the percentages of different isotopes may depend on the source of the element. For example, when water evaporates from the ocean, the heavier water, 2H1HO (water with one deuterium atom) condenses and forms rain sooner than regular water, 1H1HO. Since most water vapor in the atmosphere comes from tropical regions, rain from those regions is higher in 2H1HO than rain from northern latitudes. This fact has been used to trace migrating butterflies, by their 2H:1H ratios. Tagging methods don't work nearly as well, because the depend on trapping the tagged individuals.

    Songbird Migration

    Similarly, migration pathways of songbirds are traced by 13C:12C ratios [4]. Other methods failed: Tagged birds have not been found; radio transmitters have failed, and DNA sampling did not work. The carbon isotope ratio in bird feathers is the same as the insects they eat, and the insects' ratio is the same as the plants they eat, and the ratio in plants depends on latitude. The ratio is called the isotopic signiture, and it varies by the type of plant (C3 or C4 and latitude. Black throated blue warblers appear to segregate at the wintering grounds (in the Greater Antilles) depending on their breeding latitude (in temperate N. America), as demonstrated by the \({}_{\text{6}}^{\text{12}}\text{C}\) to \({}_{\text{6}}^{\text{13}}\text{C}\) ratios in their feathers.

    Atomic Mass Standard

    After the possibility of variations in the isotopic composition of the elements was recognized, it was suggested that the scale of relative masses of the atoms (the atomic weights) should use as a reference the mass of an atom of a particular isotope of one of the elements. The standard that was eventually chosen was \({}_{\text{6}}^{\text{12}}\text{C}\), and it was assigned an atomic-weight value of exactly 12.000 000. Thus the atomic weights given in the Table of Atomic Weights are the ratios of weighted averages (calculated as in the Example) of the masses of atoms of all isotopes of each naturally occurring element to the mass of a single 126C atom. Since carbon consists of two isotopes, 98.99% 126C isotopic weight 12.000 and 1.11% 136C of isotopic weight 13.003, the average atomic weight of carbon is

    \(\dfrac{\text{98.89}}{\text{100.00}} \times \text{12.00} + \dfrac{\text{1.11}}{\text{100.00}} \times \text{13.003} = \text{12.011}\)

    If there are three or more naturally occurring isotopes, as in the case of oxygen, the same method applies:

    \(\dfrac{\text{92.23}}{\text{100.00}} \times \text{27.97693} + \dfrac{\text{4.67}}{\text{100.00}} \times \text{28.97649} + \dfrac{\text{3.10}}{\text{100.00}} \times \text{29.97376} = \text{28.0855}\)

    for example. Deviations from average isotopic composition are usually not large, and so the average atomic weights serve quite well for nearly all chemical calculations. In the study of nuclear reactions, however, one must be concerned about isotopic weights. This is discussed further in the section on Nuclear Chemistry.

    The SI definition of the mole also depends on the isotope \({}_{\text{6}}^{\text{12}}\text{C}\) and can now be stated. One mole is defined as the amount of substance of a system which contains as many elementary entities as there are atoms in exactly 0.012 kg of \({}_{\text{6}}^{\text{12}}\text{C}\) . The elementary entities may be atoms, molecules, ions, electrons, or other microscopic particles. This official definition of the mole makes possible a more accurate determination of the Avogadro constant than was reported earlier. The currently accepted value is NA = 6.02214179 × 1023 mol–1. This is accurate to 0.00000001 percent and contains five more significant figures than 6.022 × 1023 mol–1, the number used to define the mole previously. It is very seldom, however, that more than four significant digits are needed in the Avogadro constant. The value 6.022× 1023 mol–1 will certainly suffice for most calculations needed.

    From ChemPRIME: 4.13: Average Atomic Weights


    2. "The ratio of the [weighted] average mass of the atom[s] to the unified atomic mass unit [1/12 the mass of the isotope 12C, which is defined as 12.0000] IUPAC Gold Book,
    3. CRC Handbook of Chemistry and Physics, 83rd Edition
    4. Rubenstein, D.R. et. al.,Science, 295, 1062 (2002).

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