Isotopes and Average Atomic Mass

Isotopes

  • Definition: Isotopes are versions of an element that have different masses.
  • Key Characteristic: Isotopes of an element possess the same number of protons but differ in their number of neutrons.
  • Example: Hydrogen Isotopes:
    • Hydrogen-1 (Protium): Has 1 proton and 0 neutrons.
    • Hydrogen-2 (Deuterium): Has 1 proton and 1 neutron.
    • Hydrogen-3 (Tritium): Has 1 proton and 2 neutrons.
    • In all these isotopes, the number of protons remains constant (1), while the number of neutrons varies, leading to different atomic masses.

The Periodic Table and Average Atomic Mass

  • The atomic mass listed for an element on the periodic table is a weighted average of the masses of all its naturally occurring isotopes.
  • Weighted Average Principle: More common (abundant) isotopes exert a greater influence on the calculated average atomic mass. Consequently, the average atomic mass of an element will be numerically closest to the mass of its most common isotope.
  • Determining the Most Common Isotope: To identify the most common isotope among a set of an element's isotopes, compare their mass numbers to the average atomic mass provided on the periodic table. The isotope whose mass is closest to the average atomic mass is generally the most abundant.
    • Example: Magnesium (Mg): Given isotopes 24extMg^{24} ext{Mg}, 25extMg^{25} ext{Mg}, and 26extMg^{26} ext{Mg}. By looking up Magnesium's average atomic mass on the periodic table (approximately 24.305extamu24.305 ext{ amu}), we can infer that 24extMg^{24} ext{Mg} is the most common version as its mass is closest to the average.
    • Example: Carbon (a.a.m = 12.01extamu12.01 ext{ amu}):
      • Isotopes include Carbon-12, Carbon-13, and Carbon-14.
      • Based on the average atomic mass, Carbon-12 is the most common isotope.
    • Example: Chlorine (a.a.m = 35.45extamu35.45 ext{ amu}):
      • Isotopes include Chlorine-35 and Chlorine-37.
      • Based on the average atomic mass, Chlorine-35 is the most common isotope.
    • Example: Fluorine (a.a.m = 19.00extamu19.00 ext{ amu}):
      • Only one stable isotope, Fluorine-19.
      • When an element has only one stable isotope, the atomic mass listed on the periodic table is simply the mass of that specific isotope.

Calculating the Average Atomic Mass

  • The average atomic mass of an element is calculated using the following formula, which accounts for the mass and relative abundance of each isotope: Average Atomic Mass=f<em>1M</em>1+f<em>2M</em>2+…\text{Average Atomic Mass} = f<em>1M</em>1 + f<em>2M</em>2 + \dots
    • ff: Represents the fractional abundance of an isotope. This is obtained by dividing the percent abundance by 100100. For instance, a 76%76\% abundance becomes 0.760.76. Fractional abundances must sum to 11 (or 100%100\%).
    • MM: Represents the exact atomic mass of each specific isotope, typically measured in atomic mass units (amu).

Calculation Examples

Example 1: Chlorine
  • Given Information:
    • Chlorine-35: Exact mass of 34.97 amu34.97 \text{ amu}, 76%76\% abundant.
    • Chlorine-37: Exact mass of 36.97 amu36.97 \text{ amu}, 24%24\% abundant.
  • Calculation Steps:
    1. Convert percent abundances to fractional abundances:
      • Chlorine-35: 76%→0.7676\% \rightarrow 0.76
      • Chlorine-37: 24%→0.2424\% \rightarrow 0.24
    2. Apply the weighted average formula:
      Average Atomic Mass=(0.76)(34.97 amu)+(0.24)(36.97 amu)\text{Average Atomic Mass} = (0.76)(34.97 \text{ amu}) + (0.24)(36.97 \text{ amu})
    3. Perform the multiplications:
      Average Atomic Mass=26.58 amu+8.87 amu\text{Average Atomic Mass} = 26.58 \text{ amu} + 8.87 \text{ amu}
    4. Sum the products:
      Average Atomic Mass=35.45 amu\text{Average Atomic Mass} = 35.45 \text{ amu}
Example 2: Lithium
  • Given Information:
    • Lithium-6: Mass of 6.015 amu6.015 \text{ amu}, 7.59%7.59\% abundant.
    • Lithium-7: Mass of 7.016 amu7.016 \text{ amu}, 92.41%92.41\% abundant.
  • Calculation Steps:
    1. Convert percent abundances to fractional abundances:
      • Lithium-6: 7.59%→0.07597.59\% \rightarrow 0.0759
      • Lithium-7: 92.41%→0.924192.41\% \rightarrow 0.9241
    2. Apply the weighted average formula:
      Average Atomic Mass=(0.0759)(6.015 amu)+(0.9241)(7.016 amu)\text{Average Atomic Mass} = (0.0759)(6.015 \text{ amu}) + (0.9241)(7.016 \text{ amu})
    3. Perform the multiplications:
      Average Atomic Mass=0.457 amu+6.483 amu\text{Average Atomic Mass} = 0.457 \text{ amu} + 6.483 \text{ amu}
    4. Sum the products:
      Average Atomic Mass=6.940 amu\text{Average Atomic Mass} = 6.940 \text{ amu}
Example 3: Silicon (with three isotopes)
  • Given Information:
    • Silicon-28: Mass of 27.98 amu27.98 \text{ amu}, 92.2%92.2\% abundant.
    • Silicon-29: Mass of 28.98 amu28.98 \text{ amu}, 4.7%4.7\% abundant.
    • Silicon-30: Mass of 29.97 amu29.97 \text{ amu}, 3.1%3.1\% abundant.
  • Calculation Steps:
    1. Convert percent abundances to fractional abundances:
      • Silicon-28: 92.2%→0.92292.2\% \rightarrow 0.922
      • Silicon-29: 4.7%→0.0474.7\% \rightarrow 0.047
      • Silicon-30: 3.1%→0.0313.1\% \rightarrow 0.031
    2. Apply the weighted average formula for three isotopes:
      Average Atomic Mass=(0.922)(27.98 amu)+(0.047)(28.98 amu)+(0.031)(29.97 amu)\text{Average Atomic Mass} = (0.922)(27.98 \text{ amu}) + (0.047)(28.98 \text{ amu}) + (0.031)(29.97 \text{ amu})
    3. Perform the multiplications:
      Average Atomic Mass=25.80 amu+1.36 amu+0.93 amu\text{Average Atomic Mass} = 25.80 \text{ amu} + 1.36 \text{ amu} + 0.93 \text{ amu}
    4. Sum the products:
      Average Atomic Mass=28.09 amu\text{Average Atomic Mass} = 28.09 \text{ amu}