Atomic Mass and Isotope Abundances (Notes)

Distinctions: atomic number, mass number, and atomic mass

• Atomic number (Z): the number of protons in the nucleus; a whole number (integer).
• Mass number (A): the total number of protons and neutrons in the nucleus; A = Z + N; a whole number (integer).
• Atomic mass: a measured quantity that is related to Z and A but is a decimal value for most elements; not simply Z or A.
• Key distinction: Z and A are integers, while atomic mass is usually a decimal value.

Atomic mass unit (amu)

• Definition: the atomic mass unit is defined as
  $1\ \mathrm{u} = \frac{1}{12} \ m(\mathrm{C}^{12})$
  i.e., one twelfth the mass of a carbon-12 atom.
• This is a conventional unit used to express atomic and molecular masses.

Atomic mass vs. isotopic composition

• Atomic mass is the weighted average mass of an element in nature.
• Elements exist as several isotopes; the atomic mass reflects the natural abundances of these isotopes.
• Only stable isotopes are used when calculating the standard atomic mass for an element; radioactive isotopes are typically excluded because they decay.
• In practice, you compute the atomic mass by taking the weighted average of the masses of the stable isotopes, using their fractional abundances.

Isotopes and natural abundance

• Isotopes: same element, different neutron numbers; different masses.
• Abundance: the fraction of each isotope present in nature (often reported as a percent).
• Convert percent abundance to fractional abundance (decimal) by dividing by 100.
• The sum of all fractional abundances for an element equals 1:
  $\sum<em>i f</em>i = 1\,.$
• If there is only one stable isotope, the atomic mass equals the mass of that isotope.
• If there are multiple stable isotopes, the standard atomic mass is $M = \sum<em>i f</em>i m_i$ where:
  • $f_i$ = fractional abundance of isotope $i$,
  • $m_i$ = mass of isotope $i$ (the isotopic mass).
• For a two-isotope case, this reduces to
  $M = f<em>1 m</em>1 + (1 - f<em>1) m</em>2.$

Worked examples from the transcript

Copper (Cu)

• Isotopes: Cu-63 and Cu-65
• Given:
  • Cu-63: abundance $69.17\% \Rightarrow f1 = 0.6917$, mass $m1 = 62.9296$ amu
  • Cu-65: abundance $30.83\% \Rightarrow f2 = 0.3083$, mass $m2 = 64.9278$ amu
• Atomic mass of copper (element in nature):
  $M_{\mathrm{Cu}} = (0.6917)(62.9296) + (0.3083)(64.9278) \ \approx 63.546\ \text{amu}.$
• Quick breakdown of the arithmetic (shown for clarity):
  • $62.9296 \times 0.6917 \approx 43.5288$ amu
  • $64.9278 \times 0.3083 \approx 20.0172$ amu
  • Sum $\approx 63.5460$ amu
• This matches the standard atomic mass of copper on the periodic table (approximately $63.546$ amu).

Bromine (Br)

• Two naturally occurring isotopes: Br-79 and Br-81

• Given:

  • Br-79: abundance $50.69\% \Rightarrow f{79} = 0.5069$, mass $m{79} = 78.918$ amu
  • Br-81: abundance $a{81}$ (unknown in the setup), $f{81} = 1 - 0.5069 = 0.4931$ (since sum of fractional abundances must be 1), mass $m_{81}$ is unknown in the setup.

• Known standard atomic mass for bromine: $M_{\mathrm{Br}} = 79.904$ amu.

• Equation to solve for $m{81}$: $M</em>{\mathrm{Br}} = f<em>{79} m</em>{79} + f<em>{81} m</em>{81}$
  $79.904 = (0.5069)(78.918) + (0.4931) m_{81}$

• Solve for $m{81}$: $m</em>{81} = \frac{79.904 - (0.5069)(78.918)}{0.4931} \approx 80.916\ \text{amu}.$

• Note on abundances: the sum of fractional abundances must equal 1, so if you know one abundance, the other is $1$ minus that value (or $100\%$ minus that percentage if using percent form).

• This demonstrates how you can determine the mass of a less common isotope from the overall atomic mass and the known isotope masses/abundances.

• The transcript also notes that while the example uses two isotopes, you can extend the same idea to more isotopes by including all stable isotopes in the sum:
  $M = \sum<em>i f</em>i m<em>i$ with $\sumi f_i = 1$.

Practical notes and exam-oriented ideas

• When calculating the atomic mass, always convert percent abundances to fractional abundances (divide by 100) before multiplying by isotope masses.
• If an element has only one stable isotope, its standard atomic mass equals the mass of that isotope (no averaging).
• Always use the stable isotopes for standard atomic mass; radioactive isotopes are generally excluded from the standard calculation.
• The atomic mass given in the periodic table is the weighted average mass of naturally occurring isotopes, expressed in amu.
• If asked to find an unknown isotope mass from a given standard atomic mass and known abundances, you can rearrange the core equation accordingly.
• A two-isotope problem can be solved with a single equation if the abundances sum to 1; a two-unknown system (two unknown abundances) requires an additional constraint (the sum-to-one condition) to solve.
• Common pitfalls to avoid:
  • Forgetting to convert percent to decimal.
  • Using mass number instead of isotopic mass.
  • Including radioactive isotopes when calculating standard atomic mass.
  • Not ensuring that fractional abundances sum to 1.

Summary formulas to remember

• Atomic mass unit definition:
  $1\ \mathrm{u} = \frac{1}{12} m(\mathrm{C}^{12})$
• Weighted average mass (standard atomic mass):
  $M = \sum<em>i f</em>i m_i$
• Fractional abundances from percent abundances:
  $f<em>i = \frac{p</em>i}{100}$
• Two-isotope special case:
  $M = f<em>1 m</em>1 + (1 - f<em>1) m</em>2$
• Relationship for a two-isotope system when one isotope’s mass is unknown:
  $m<em>2 = \frac{M - f</em>1 m<em>1}{f</em>2}$

Connections and relevance

• Understanding atomic mass is foundational for quantitative chemistry: calculating molar masses, stoichiometry, and predicting behavior in reactions.
• The concept ties directly to isotopes, nuclear structure, and the real-world isotopic composition of elements found in nature.
• The idea of weighted averages extends beyond chemistry to any context where mixtures of different components contribute to an overall measured value.