1.3 Pure Substances

Particulate Diagram Interpreter

Eight pure substances drawn to real bond angles and ionic radii, five as discrete molecules and three as lattices with one formula unit outlined. Count the atoms in the box and the empirical formula falls out; change the sample size and the percent by mass refuses to move.

Formula UnitsEmpirical FormulaDefinite ProportionsPercent by Mass
Topic 1.3

Elemental Composition of Pure Substances

Explain the quantitative relationship between the elemental composition by mass and the empirical formula of a pure substance.

Pure substances come in two structural flavors: some are built from discrete molecules, others from atoms or ions held in fixed proportions described by a formula unit. Either way, the composition is fixed.

That fixedness is the law of definite proportions: every pure sample of a given compound has the same ratio of constituent masses. Water from a glacier and water from a lab synthesis are both 11.2% H and 88.8% O by mass.

The empirical formula is the lowest whole-number ratio of atoms in the compound. Getting there from mass data is a fixed four-step routine:

  1. Assume 100 g of compound, which turns each mass percent directly into grams.
  2. Convert each mass to moles by dividing by that element's molar mass.
  3. Divide every mole count by the smallest one.
  4. Clear any fractions by multiplying all subscripts by 2, 3, or 4 (0.5 → ×2, 0.33 → ×3, 0.25 → ×4).

The molecular formula is a whole-number multiple of the empirical formula. Find n = (molar mass) ÷ (empirical formula mass) and multiply every subscript by n. Without a molar mass you can only report the empirical formula — and for an ionic compound the empirical formula is the formula, since no discrete molecule exists.

Key points

  • Law of definite proportions: a pure compound has one fixed mass ratio no matter its source.
  • Empirical formula = smallest whole-number atom ratio; molecular formula = empirical × n.
  • Combustion analysis is the classic route to empirical formulas: all C ends up in CO₂ and all H in H₂O; oxygen is found by mass difference.

Equations

  • not on the sheet
    • atoms of X per formula unit
    • molar mass of element X
  • not on the sheetMust be a whole number; if it comes out at 2.98, it is 3.
    • multiplier from empirical to molecular formula

Common mistakes

  • 0.33 does not round to 0. Ratios like 1 : 1.33 mean multiply everything by 3 to get 3 : 4.
  • Divide by the smallest mole count, not the smallest mass.
  • Ionic compounds have only empirical formulas. There is no “molecular formula of NaCl”.
  • In combustion analysis, oxygen is not measured directly. Get its mass by subtracting the C and H masses from the sample mass.

Worked example

A compound is 40.0% C, 6.71% H, and 53.3% O by mass and has a molar mass of 180 g/mol. Determine its empirical and molecular formulas.

Assume 100 g: 40.0 g C, 6.71 g H, 53.3 g O.

Moles:
C: 40.0 ÷ 12.01 = 3.331 mol
H: 6.71 ÷ 1.008 = 6.657 mol
O: 53.3 ÷ 16.00 = 3.331 mol

Divide by smallest (3.331): C 1.00, H 2.00, O 1.00 → empirical formula CH₂O.

Molecular formula: empirical mass = 30.03 g/mol; n = 180 ÷ 30.03 = 5.99 ≈ 6.
Molecular formula = C₆H₁₂O₆ (glucose).

Full notes for topic 1.3 →