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3.4: Isotopic Abundances

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    278759
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    Learning Objectives

    Goals: 

    • Understand Isotopic Distribution of the atoms for various elements as they exist in their natural state
    • Recognize the meaning of the atomic mass on the periodic table

    By the end of this module, students should be able to:

    • Determine the average atomic mass from the natural isotopic distribution of the atoms of an element
    • Using the atomic mass on the periodic table for an element with two isotopes, and knowing two of the four data values of % composition and isotopic mass of each isotope, determine the other two values.

    Prior knowledge:

     

    Atomic Weight

    The atomic weight is the mass of an atom, typically expressed in atomic mass units (amu). For an isotope, it is the mass of the nucleus, that is the mass of the protons and neutrons, as the mass of the electrons are considered negligible. The atomic weight is often posted on the periodic table under the elment's symbol.  In their natural state only 21 elements exist as single isotopes (see section 2.2.2 of Gen Chem 1),  and most elements exist as a mixture of nuclei from multiple isotopes.

    Sodium is one of the elements that exists as a single isotope, 23Na.

    isotope_4.PNG isotope_3b.PNG

    Figure \(\PageIndex{1}\): Sodium only has one isotope, and note, the uncertainty of this value is expressed by the (2) in 22.989 769 28(2), indicating that the last value has an uncertainty of +/- .000 000 02. Adapted from 2018 IUPAC Technical Report - IPTEI for Education Community, (cc. 4.0), p. 1864.

     

    For  elements that have more than one isotope the atomic weight is the average weight based on the fractional abundance of each isotope, and this is the value given on the periodic table. Copper has two isotopes, 63Cu (69.15%, mass=62.9300 amu) and 65Cu (30.85%, mass = 64.928 amu), and so the respective mole fractions are 0.6915 and 0.3085, resulting in an average atomic weight of 63.55 amu, even though there is not a single atom that weighs 63.55 amu.

    isotope_2.PNG \[\underbrace{0.6915}_{fraction \; ^{63}Cu}\underbrace{(62.9300\, amu)}_{mass \; ^{63}Cu} + \underbrace{ 0.3085}_{fraction \; ^{65}Cu} \underbrace{(64.928 \,amu)}_{mass \; ^{65}Cu} = \underbrace{63.55\, amu}_{\text{average mass}} \\ note: \; 0.6915 + 0.3085 = 1\]

    Figure \(\PageIndex{2}\): Natural samples of copper contain two isotopes, and its atomic weight is to four significant digits is 63.55 amu, even though there is not a single atom of copper that weights 63.55 amu.

    Exercise \(\PageIndex{1}\)

    True or False: The atomic weight of copper is 63.456 amu, which means an atom of copper weights 63.546 amu.

    Answer

    False, 63.456 amu is the average atomic weight of copper, there are actually two isotopes, one weighing 62.930 amu and one weighing 64.928 amu.

     

    Exercise \(\PageIndex{2}\)

    The atomic weight of chorine is ______________and the atomic number of chlorine-35 is________________.

    1. 35, 17
    2. 17, 35
    3. 35.4527; 17
    4. 35.4527; 35
    Answer

    C) the atomic weight is the average of mass of all isotopes of chlorine atoms and found below the symbol on the periodic table. The atomic number is the number of protons in all chlorine atoms and is found on the top of the symbol in the periodic table.

    Lead has 4 stable isotopes as shown in Figure \(\PageIndex{3}\) . Lead is a fairly large element and has many radioisotopes, in fact it has 29 isotopes with a half life of less than one hour, seven between an hour and a year, and two with half lives over a year (202Pb = 5.25x104 years & 210Pb = 22.3 years), but these are insignificant to the thousandth position (note the four mole fractions add to one).

    isotope_6.PNG isotope_5.PNG

    Figure \(\PageIndex{3}\): Lead has 4 stable isotopes. Adapted from 2018 IUPAC Technical Report - IPTEI for Education Community, (cc. 4.0), p 2007.

    Exercise \(\PageIndex{3}\)

    Nitrogen has two isotopes, the first isotope has a mass of 14.004amu, while the second isotope has a mass of 15.000 amu, which isotope has the greatest natural abundance?  Tip, you need to look at the periodic table.
    1. 14N
    2. 15N
    3. There are equal amounts.
    4. Not enough information provided.
    5. none of the above
    Answer

    A) The average atomic weight is 14.007 amu which is much closer to 14 then 15, so 14 is the most abundant isotope.  In the next section you will use this kind of data to calculate fractional abundances.

    Isotopic Abundance Calculations

    For elements with more than one isotope there are a variety of problems you need to be able to solve. You not only need to be able to calculate the average mass from the isotopic abundance and masses, but go backwards, using the atomic weight on the periodic table as the average atomic mass. Rewriting eq. \(\PageIndex{1}\) in terms of algebraic variables gives us a feel for the types of problems (review section 2.9 Fractions and Percent). Consider an element with two isotopes A & B, each having a mass of mA or mb​​​​​​ and a mole fraction of XA or Xb​​​​​​.

    \[\underbrace{X_A}_{\text{fraction A }}\underbrace{(m_A)}_{\text{ mass A}} + \underbrace{X_B}_{\text{fraction B }}\underbrace{(m_B)}_{\text{ mass B}} = \underbrace{m_{ave}}_{\text{average mass}}\]

    In general the equation for an element with [n] isotopes is:

    \[\sum_{i=1}^{n}X_im_i=m_{ave}\]

    Note, the equation for two isotopes has 5 unknowns, and you can get mave from the periodic table, so you need to be given three of the other variables to solve the equation (one equation can only be solved for one unknown). But there is another equation we can use to reduce the number of unknowns. That is the sum of the fractions equals 1

    \[X_A + X_B=1\]

    or in general

    \[\sum_{i=1}^{n}X_i=1\]

    (this is the same equation as 2.9.4 when we looked at the fraction of salt in salt water).

     

    Example \(\PageIndex{1}\): Calculating Average Atomic Mass

    What is the average atomic mass of Neon, given that it has 3 isotopes with the follow percent abundances;

    20Ne = 19.992 amu (90.51%), 21Ne = 20.993 amu (0.27%), 22Ne = 21.991 amu.

    What we know: since you know what the element is, you can solve this without doing any math by using the periodic table, but you need to be able to do the math because it might be an unknown, and that is the only way you can Figure out the correct significant Figures.

    Since Ne-20 has the greatest percent abundance, it should have the most impact on your average. Therefore, we expect the average atomic mass to be closer to the mass of Ne-20 (about 19.992 amu). Click the following video tutor to see if we estimated correctly.

    Video \(\PageIndex{2}\): 4:53 minute YouTube solving above problem.

    Answer According to the correct number of significant Figures, we came up with 20.18 amu as the average atomic weight even though the average atomic weight from the periodic table is 20.179 amu. However, it is still a good check to make sure that you are on the right path.

    Check Yourself: We predicted earlier that our answer should be closer to the mass of Ne-20 (19.992 amu) instead of Ne-21 or Ne-22 because it has the greatest natural abundance, and thus, impacts the average more. We can see that the math does align with our logic!

    The following problem is more complicated.  This is mathematically the same problem as equation \(\PageIndex{2}) or 2.9.4 of the last chapter, and has 5 variables, but you have only been given two (fraction of 35Cl and the mass of 35Cl) and you are asked what is the mass of the other isotope.  So you can algebraically solve it for the other isotope (MB of equation \(\PageIndex{2}\)), but your algebraic solution would have two variables in it (fraction or percent of the other isotope and the average mass).  By now you should realize the value of the average is what is given on the periodic table, and so you can look that up, and you can solve for the fraction of the second isotope by noting the sum of the fractions equals 1 (\(eq \(\PageIndex{4}\)).  Try to follow this problem on your own, and then watch the video

    Example \(\PageIndex{2}\): Isotopic Mass Calculation

    Chlorine has two isotopes, with 75.53% being 35Cl with an isotopic mass of 34.969 amu, what is the mass of the other isotope?

    What we know: In this case, you have the average atomic mass (from the periodic table). You are trying to find the mass of the individual isotope. You also know that the individual isotopes have to add up to 100%.

    Answer: The correct answer is 36.9 amu.

     

    Test Yourself

    Homework: Section 3.4

    Graded Assignment: Section 3.4

    Query \(\PageIndex{1}\)

     


    3.4: Isotopic Abundances is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

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