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Wize High School Grade 11 Chemistry Textbook > Atomic Structure and Periodic Trends

Isotopes

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Isotopes and Atomic Mass

Isotopes and Radioisotopes

  • When two atoms have the same atomic number, but a different number of neutrons, we call these isotopes.
  • Since they have a different number of neutrons, isotopes will have different mass numbers

^{ }12C13CC−12C−13carbon−12carbon−1398.9%1.1%\begin{array}{cc} ^{12}C&^{13}C\\ C-12 & C-13\\ \text{carbon}-12&\text{carbon}-13\\ 98.9\%&1.1\% \end{array}

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  • Isotopes can be stable, meaning that they do not decay.
  • Radioisotopes are radioactive isotopes of an element. The nucleus of these isotopes is unstable and will decay with time, emitting radiation.
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Calculating the Atomic Mass

  • The atomic mass number is a weighted average based on the relative abundance of isotopes Isotopes have similar reactivity to one another, that's why we can form C bonds with either C-12 or C-13
  • To determine the average mass of an element, use this equation:
atomic mass=(mass of isotope 1)(abundance of isotope 1)+(mass of isotope 2)(abundance of isotope 2)+...\boxed{{\text {atomic mass}=(\text{mass of isotope 1}) (\text{abundance of isotope 1})+(\text{mass of isotope 2}) (\text{abundance of isotope 2})+...}}
  • Plug in the mass of each isotope in a.m.u. and plug in the relative abundance of each isotope as a decimal Example: we are told the relative abundance of C-12 is 98%. You would want to plug in
    0.98
    for this isotope's relative abundance.

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Example: Solving for the Atomic Mass

Chlorine can be found in nature as 35Cl (mass 34.969a.m.u, 75.78% abundance) and 37Cl (mass 36.966a.m.u, 24.22% abundance). What is the average atomic mass of Cl?


atomic mass=(mass 35Cℓ)(abundance 35Cℓ)+(mass 37Cℓ)(abundance 37Cℓ)\text{atomic mass}= (\text{mass } ^{35}C\ell)(\text{abundance } ^{35}C\ell)+(\text{mass } ^{37}C\ell)(\text{abundance } ^{37}C\ell)

atomic mass=(34.969a.m.u.)(0.7578)+(36.966a.m.u.)(0.2422)\text{atomic mass}=(34.969\text{a.m.u.})\left(0.7578\right)+(36.966\text{a.m.u.})\left(0.2422\right)

atomic mass=35.45a.m.u.\text{atomic mass}= 35.45\text{a.m.u.}

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Example: Finding the Number of Subatomic Particles in a Nucleus


a) How many protons are in an 53Cr nuclei?

24, The Atomic Number


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b) How many protons are in a 232Th nuclei?

90, The Atomic Number


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c) How many neutrons are in a 18O nuclei?

18 - Z = 18 - 8 = 10


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d) How many neutrons are in a 64Cu nuclei?

64 - Z = 64 - 29 = 35



Practice: Identifying Isotopes of Elements

An atom with 6 neutrons and a mass number of 13 is an isotope of which element?


Practice: Solving for the Weight of an Isotope

Naturally occurring potassium contains two stable isotopes. The lighter isotope, 39K ( 38.9637 amu) is the more abundant isotope, accounting for 93.26% of the nuclei. What is the weight of the heavier isotope, 41K?

Practice: Finding Abundance

The two stable isotopes of Boron are as follows:

Isotope 1: 510B^{10}_5B (10.013a.m.u)
Isotope 2: 511B^{11}_5B (11.009a.m.u.).

Calculate the percentage abundance of each isotope based on boron's average atomic mass. Express your answer in a percentage value, rounded to the nearest whole integer. Do not include any symbols.