Relative Masses of Atoms and Molecules

3.2 Relative masses of atoms and molecules

Relative Atomic Mass (Aᵣ)

• Definition:
  Relative atomic mass (Aᵣ) of an element is the weighted average mass of all naturally occurring isotopes of the element compared to 1/12th of the mass of a carbon-12 atom.
• Standard reference:
  • The carbon-12 isotope is assigned a mass of exactly 12 atomic mass units (u).
  • 1 atomic mass unit (1 u) = 1/12th the mass of one atom of carbon-12.
• Mathematical expression:Aᵣ = (average mass of one atom of the element) ÷ (1/12 × mass of one atom of carbon-12)
• Important points:
  • Aᵣ is dimensionless (no units).
  • It is not the same as mass number (A).
    • Mass number = number of protons + neutrons in a particular isotope.
    • Aᵣ = weighted average considering all isotopes and their abundances.
  • The weighted average means more abundant isotopes influence the Aᵣ more than less abundant ones.
• Example calculation:
  Chlorine exists as:
  • ⁷³.⁰⁰% ³⁵Cl (mass = 35.0 u)
  • ²⁷.⁰⁰% ³⁷Cl (mass = 37.0 u)

  Weighted mean:

  Aᵣ = (0.7300 × 35.0) + (0.2700 × 37.0)
  Aᵣ = 25.55 + 9.99
  Aᵣ = 35.54

• Uses of Aᵣ:
  • Converting between mass of an atom and number of moles.
  • Determining molar mass of elements and compounds.
  • Balancing chemical equations with mole ratios.

Isotopes and their effect on Aᵣ

• Isotopes = atoms of the same element with same number of protons but different numbers of neutrons.
• Isotopes have different mass numbers but identical chemical properties.
• The natural abundance of each isotope determines the Aᵣ value in the periodic table.

Example: Carbon has:

• ¹²C (98.9% abundance) → mass 12 u (by definition)
• ¹³C (1.1% abundance) → mass 13.003 u

Aᵣ(C) = (0.989 × 12.000) + (0.011 × 13.003)
Aᵣ(C) ≈ 12.011

Relative Molecular Mass (Mᵣ)

• Definition:
  Relative molecular mass (Mᵣ) is the sum of the relative atomic masses (Aᵣ values) of all the atoms in a molecule.
• Formula:Mᵣ = Σ Aᵣ (of all atoms in the molecular formula)
• Important points:
  • Used for covalent molecules.
  • For ionic compounds, the same calculation is performed, but the term used is relative formula mass (Mᵣ) instead of molecular mass.
• Example:
  Water (H₂O):
  • H: Aᵣ = 1.008, 2 atoms → 2 × 1.008 = 2.016
  • O: Aᵣ = 16.00, 1 atom → 16.00
    Mᵣ = 2.016 + 16.00 = 18.016

Relative Formula Mass (Mᵣ) for Ionic Compounds

• Definition:
  The sum of the Aᵣ values of all the ions in the empirical formula of an ionic compound.
• Reason for using “formula mass”:
  Ionic compounds do not exist as discrete molecules; instead, they exist as giant ionic lattices.
• Example:
  Sodium chloride (NaCl):
  • Na: Aᵣ = 22.99
  • Cl: Aᵣ = 35.45
    Mᵣ = 22.99 + 35.45 = 58.44

Calculating Mᵣ from chemical formula

Step-by-step process:

1. Write down the chemical formula of the substance.
2. Identify the number of each type of atom.
3. Multiply each Aᵣ by the number of atoms present.
4. Add all values to get Mᵣ.

Example 1: Glucose (C₆H₁₂O₆)
C: 12.01 × 6 = 72.06
H: 1.008 × 12 = 12.096
O: 16.00 × 6 = 96.00
Mᵣ = 72.06 + 12.096 + 96.00 = 180.156

Example 2: Calcium carbonate (CaCO₃)
Ca: 40.08 × 1 = 40.08
C: 12.01 × 1 = 12.01
O: 16.00 × 3 = 48.00
Mᵣ = 40.08 + 12.01 + 48.00 = 100.09

Mass spectrometry in determining Aᵣ

• Mass spectrometer is used to:
  • Identify isotopes present in an element.
  • Determine relative abundances.
  • Calculate Aᵣ accurately.

Key stages in mass spectrometry:

1. Ionisation of atoms/molecules.
2. Acceleration through electric field.
3. Deflection by magnetic field (proportional to mass/charge ratio).
4. Detection and analysis.

Example data analysis:
Element X has:

• Isotope ¹⁰X: abundance 20.0% → mass 10.012 u
• Isotope ¹¹X: abundance 80.0% → mass 11.009 u

Aᵣ = (0.200 × 10.012) + (0.800 × 11.009)
Aᵣ = 2.0024 + 8.8072
Aᵣ = 10.8096

Applications of Aᵣ and Mᵣ

• Calculating the number of moles:n = m ÷ Mᵣ
• Converting between mass and particle count:number of particles = n × Nₐ
  (where Nₐ = Avogadro constant = 6.022 × 10²³ mol⁻¹)
• Determining empirical and molecular formula from experimental data.

Worked example: molecular formula from mass data

Problem: A compound contains 40.00% C, 6.71% H, and 53.29% O by mass. Its Mᵣ is 60.05. Find molecular formula.

1. Convert % to moles:
   C: 40.00 ÷ 12.01 = 3.33 mol
   H: 6.71 ÷ 1.008 = 6.66 mol
   O: 53.29 ÷ 16.00 = 3.33 mol
2. Divide by smallest:
   C: 3.33 ÷ 3.33 = 1
   H: 6.66 ÷ 3.33 = 2
   O: 3.33 ÷ 3.33 = 1→ Empirical formula = CH₂O
3. Calculate empirical mass:
   C: 12.01 × 1 = 12.01
   H: 1.008 × 2 = 2.016
   O: 16.00 × 1 = 16.00
   Empirical mass = 30.026
4. Compare with molecular mass:
   Mᵣ / empirical mass = 60.05 ÷ 30.026 ≈ 2→ Molecular formula = C₂H₄O₂

Distinguishing between Aᵣ and Mᵣ

Key points to remember

• Aᵣ is always a weighted average; it can be a decimal.
• Mᵣ is always a sum of atomic masses; for ionic compounds, it’s called relative formula mass.
• Carbon-12 is the standard reference.
• Mass spectrometry is the most accurate method for determining Aᵣ.
• For ionic compounds, the lattice is considered; no discrete molecules exist.
• Calculations often involve converting between % composition, empirical formula, and Mᵣ.