Relative atomic mass: averaging the isotopes

Comparing the masses of atoms

Chemists need to compare the masses of atoms. Without relative masses, we would not know in what ratio of masses to mix reactants for a chemical reaction.

In 1961, chemists adopted a scale based on carbon-12, whose atom is given a mass of exactly 12. The unit, one atomic mass unit (amu), is one twelfth of the mass of a carbon-12 atom.

The relative atomic mass of an element is the mass of its atom compared with this scale. On it, one hydrogen-1 atom is 1.007 amu and one sulphur-32 atom is 31.972 amu.

Most elements are a mix of isotopes. Each isotope's mass number is its relative isotopic mass (m), and its natural share is its isotopic abundance (p), in percent.

Relative atomic mass = (m₁p₁ + m₂p₂ + m₃p₃ + …) ÷ 100. It is a weighted average, so it is often not a whole number: chlorine's is 35.45.

Multiply each isotope's mass by its percentage, add them up, and divide by 100. Common isotopes count for more.

Key terms

Atomic mass unit (amu)

The unit for comparing the masses of atoms.

Relative atomic mass

Giving one mass for an element, compared with carbon-12.

Relative isotopic mass

The mass number of one isotope, used in the formula.

Isotopic abundance

How common an isotope is in nature, as a percentage.

Short questions with model answers

1. Q1. Sample Example 1 (p. 28): find the relative atomic mass of krypton from Table 2.2.

   • 80 × 2.0 + 82 × 12.0 + 83 × 12.0 + 84 × 57.0 + 85 × 17.0
   • 160 + 984 + 996 + 4788 + 1445 = 8373
   • 8373 ÷ 100 = 83.73 ≈ 83.7

   83.7, the book's answer. It needs 17.0 for krypton-85; with 57.0 the total would be 117.7.

2. Q2. Sample Example 2 (p. 28): chlorine is 35.45. Chlorine-37 is 24.23% and the light isotope is 75.77%. Find the light isotope's relative isotopic mass.

   • 35.45 = (m × 75.77 + 37 × 24.23) ÷ 100
   • 3545 = m × 75.77 + 896.51
   • m = (3545 − 896.51) ÷ 75.77 = 2648.49 ÷ 75.77 = 34.95

   34.95. Your book calls this the light isotope's relative atomic mass; it is its relative isotopic mass.

Common mistakes

• ✗ Krypton's relative atomic mass is (80 + 82 + 83 + 84 + 85) ÷ 5 = 82.8.

  ✓ Each isotope must be weighted by its abundance. Krypton-84 is 57.0%, so the answer, 83.7, sits near 84.

• ✗ Krypton's relative atomic mass is 8373.

  ✓ The abundances are percentages, so divide the total by 100: 83.73.

• ✗ A relative atomic mass must be a whole number, like a mass number.

  ✓ It is an average of several isotopes, so it is usually a decimal: chlorine 35.45, krypton 83.7.

MCQs

1. 1. What is the relative atomic mass of nitrogen if ¹⁴N is 99.64% and ¹⁵N is 0.35%?

   1. (a) 14.0210
   2. (b) 14.0021
   3. (c) 14.2100
   4. (d) 14.1200
   Show answer

   (b) Exercise MCQ (vi) (p. 30): (14 × 99.64 + 15 × 0.35) ÷ 100 = 1400.21 ÷ 100 = 14.0021.

2. 2. Rubidium has two isotopes, ⁸⁵Rb and ⁸⁷Rb. The light isotope is 72.2%. What is the abundance of the heavier isotope?

   1. (a) 15%
   2. (b) 27.8%
   3. (c) 37%
   4. (d) 72%
   Show answer

   (b) Exercise MCQ (x) (p. 30): the two abundances add up to 100%, so 100 − 72.2 = 27.8%.

3. 3. One atomic mass unit (amu) is defined as:

   1. (a) 1/12 of the mass of a carbon-12 atom
   2. (b) The mass of a hydrogen-1 atom
   3. (c) The mass of one proton
   4. (d) 1/16 of the mass of an oxygen atom
   Show answer

   (a) p. 27: the amu is one twelfth of the mass of a carbon-12 atom; hydrogen-1 is 1.007 amu, not exactly 1.

4. 4. Using Table 2.2, what is the relative atomic mass of krypton?

   1. (a) 84.0
   2. (b) 82.8
   3. (c) 117.7
   4. (d) 83.7
   Show answer

   (d) Sample Example 1 (p. 28): 8373 ÷ 100 = 83.7. 82.8 is the simple average, and 117.7 comes from the misprinted 57.0.

5. 5. Chlorine is 35.45, with chlorine-37 at 24.23%. What is the relative isotopic mass of the light isotope?

   1. (a) 35.45
   2. (b) 37
   3. (c) 34.95
   4. (d) 36
   Show answer

   (c) Sample Example 2 (p. 28): (3545 − 896.51) ÷ 75.77 = 34.95.

Quick revision

• Atomic masses are compared on the carbon-12 scale; 1 amu is one twelfth of a carbon-12 atom's mass.
• Relative atomic mass = (m₁p₁ + m₂p₂ + …) ÷ 100, a weighted average of the isotopes.
• The same formula run backwards finds an unknown isotopic mass, as for chlorine's light isotope (34.95).