Average atomic mass Knowledge-Based Study

Study notes

1. Why Are Atomic Masses Not Whole Numbers?

   When examining the periodic table, you will notice that atomic masses are rarely whole numbers. For example, carbon is listed as 12.011 amu, and chlorine is listed as 35.45 amu. However, an individual atom must contain a whole number of protons and neutrons, meaning its mass number is always an integer.

   This discrepancy occurs because most elements exist naturally as a mixture of isotopes—atoms of the same chemical element that possess identical numbers of protons but differing numbers of neutrons. The atomic mass recorded on the periodic table represents the average atomic mass of all naturally occurring isotopes of that element.

   Measuring Isotopic Composition

   In any naturally occurring sample of an element, different isotopes are present in characteristic proportions known as their percent natural abundance.

   • Percent Abundance: The percentage of atoms of a specific isotope found in a natural sample of the element. The sum of the percent abundances of all isotopes of an element must always equal 100%.
   • Fractional Abundance: The percent abundance expressed in decimal form, obtained by dividing the percentage by 100:
     $\text{Fractional Abundance} = \frac{\text{Percent Abundance}}{100}$

   Because fractional abundance represents parts of a whole, the sum of all fractional abundances for an element equals exactly 1.

   Calculating the Weighted Average

   The average atomic mass is not a simple arithmetic mean. Instead, it is a weighted average, which accounts for both the mass and the relative prevalence of each isotope.

   To calculate the average atomic mass, multiply the mass of each isotope by its fractional abundance, then sum the results:

   $$\text{Average Atomic Mass} = (\text{fraction}_1 \times m_1) + (\text{fraction}_2 \times m_2) + \dots + (\text{fraction}_n \times m_n)$$

   Alternatively, using percent abundances directly:

   $$\text{Average Atomic Mass} = \frac{(\%_1 \times m_1) + (\%_2 \times m_2) + \dots + (\%_n \times m_n)}{100}$$

   Where $m_i$ represents the isotopic mass of isotope $i$, typically measured in atomic mass units (amu).

   Worked Example

   Naturally occurring chlorine consists of two stable isotopes:

   • Chlorine-35: Isotopic mass = 34.969 amu; Percent abundance = 75.77%
   • Chlorine-37: Isotopic mass = 36.966 amu; Percent abundance = 24.23%

   Calculation:

   1. Convert percent abundances into fractional abundances:
      $\text{Fraction of Cl-35} = \frac{75.77}{100} = 0.7577$
      $\text{Fraction of Cl-37} = \frac{24.23}{100} = 0.2423$
   2. Multiply each mass by its fractional abundance:
      $\text{Contribution of Cl-35} = 0.7577 \times 34.969\text{ amu} = 26.496\text{ amu}$
      $\text{Contribution of Cl-37} = 0.2423 \times 36.966\text{ amu} = 8.957\text{ amu}$
   3. Sum the contributions:
      $\text{Average Atomic Mass} = 26.496\text{ amu} + 8.957\text{ amu} = 35.453\text{ amu}$

   Key Takeaway: Notice that 35.45 amu is much closer to 35 amu than to 37 amu because Chlorine-35 is substantially more abundant in nature.