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General Chemistry I · Stoichiometry

Empirical Formula

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On this page 7 sections
  1. In 30 seconds
  2. Why this matters
  3. The college version
  4. Eli explains
  5. Key takeaway
  6. Study tools
  7. Sources & references

In 30 seconds

The empirical formula is the simplest whole-number ratio of atoms in a compound. It is found by converting the mass (or percent) of each element to moles, then dividing every mole quantity by the smallest one to get a ratio. If that ratio contains a fraction, you multiply all subscripts by the smallest factor that turns every value into a whole number. The empirical formula gives the ratio, not necessarily the actual molecule — glucose (C₆H₁₂O₆) and formaldehyde (CH₂O) share the empirical formula CH₂O.

Why this matters

The empirical formula is the first concrete result when an unknown is analyzed — a forensic lab, a drug-discovery team, or a materials scientist often knows only an elemental breakdown and must deduce the simplest ratio before measuring the molar mass to get the true molecular formula. Getting the ratio right is the foundation of identifying any new compound.

The college version

Key Ideas

Empirical vs. molecular formula

  • Empirical formula: simplest whole-number ratio (CH₂O).
  • Molecular formula: the actual number of atoms per molecule (C₆H₁₂O₆).
  • The molecular formula is a whole-number multiple of the empirical formula.

The "assume 100 g" shortcut

  • Because percent composition is intensive, treating the sample as 100 g makes each percent read directly as a mass in grams (40.0% C → 40.0 g C).
  • This turns a percentage into concrete grams you can convert to moles.

From ratio to whole numbers

  • After dividing by the smallest number of moles, if you get 1 : 2 : 1 you are done.
  • If you get a fraction like 1.5, 1.33, or 1.25, multiply every subscript by 2, 3, or 4, respectively, to clear it.

Equations and Variables

  • Moles of element: n = m / M (mass ÷ molar mass).
  • Ratio step: divide each n by the smallest n.
  • Fraction-clearing multipliers:
    • 1.5 (or 0.5, 2.5) → multiply by 2
    • 1.33 (⅓ multiples, e.g., 0.33, 1.67) → multiply by 3
    • 1.25 (¼ multiples, e.g., 0.25, 0.75) → multiply by 4

How It Works (Problem-Solving Method)

  1. Assume a 100 g sample (if given percents) and write each element's mass in grams.
  2. Convert each mass to moles using molar mass.
  3. Divide every mole value by the smallest mole value.
  4. Inspect the ratio. If whole numbers, they are the subscripts.
  5. If a fraction remains, multiply all subscripts by the smallest integer that clears it.
  6. Write the empirical formula with the resulting whole-number subscripts.

Worked Example

A compound is 40.0% C, 6.71% H, and 53.3% O by mass. Find its empirical formula.

  1. Assume 100 g: 40.0 g C, 6.71 g H, 53.3 g O.
  2. Convert to moles:
    • C: 40.0 g ÷ 12.01 g/mol = 3.33 mol
    • H: 6.71 g ÷ 1.008 g/mol = 6.66 mol
    • O: 53.3 g ÷ 16.00 g/mol = 3.33 mol
  3. Divide by the smallest (3.33):
    • C: 3.33 ÷ 3.33 = 1.00
    • H: 6.66 ÷ 3.33 = 2.00
    • O: 3.33 ÷ 3.33 = 1.00
  4. Ratio 1 : 2 : 1 → empirical formula CH₂O.

Fractional example: Suppose a compound gives a 1 : 1.5 ratio for two elements. Multiply both by 2 → 2 : 3. An empirical formula with a subscript of 1.5 is never written; it becomes 2 and 3.

Common Confusions

  • "Empirical and molecular formulas are the same." — They match only when the ratio can't be reduced (H₂O, NaCl). Glucose has empirical CH₂O but molecular C₆H₁₂O₆.
  • "Divide by the largest number of moles." — You divide by the smallest so the smallest becomes 1.
  • "Just round 1.5 to 1 or 2." — No; 1.5 must be multiplied by 2 (never rounded off — it changes the ratio).
  • "Percent composition gives atoms directly." — Percents are masses; you must convert to moles first, since ratios are atom-based, not mass-based.
Eli, the EliExplains learning guide

Eli explains

The same idea, in plain words

Explain it like I’m 10

Finding an empirical formula is like simplifying a recipe. If a cake recipe lists "6 cups flour, 12 cups sugar, 6 eggs," the simplest version is "1 flour : 2 sugar : 1 egg" — you divided everything by the smallest number, 6. Chemists do the same with moles of each element. If the recipe comes out "1.5 cups," you can't leave a half in a ratio, so you double everything to make it whole. The catch: the simplified recipe tells you the ratio, not whether the actual cake used 6, 12, and 6 or 12, 24, and 12 — that bigger answer is the molecular formula.

Key takeaways

  • Empirical formula = simplest whole-number atom ratio.
  • "Assume 100 g" converts percents directly to grams.
  • Always divide by the smallest number of moles.
  • 1.5 → ×2; 1.33 → ×3; 1.25 → ×4 (multiply all subscripts).
  • Empirical formula may equal the molecular formula (H₂O) or be a divisor of it (CH₂O vs C₆H₁₂O₆).
  • Assume 100 g → convert percents to grams.
  • Grams → moles for each element.
  • Divide all by the smallest mole count.
  • Clear fractions by multiplying all subscripts (1.5→×2, 1.33→×3, 1.25→×4).

Keep learning

Ready to build on this? Continue to the next lesson.

Practice General Chemistry I

This lesson has no separate scored set. Practice draws from the subject’s question bank.

Study tools & related lessonsYou’ll learn to · Related

You’ll learn to

  • Define empirical formula and distinguish it from molecular formula.
  • Determine an empirical formula from percent composition or from measured masses.
  • Handle fractional mole ratios (1.5, 1.33, 1.25) by multiplying to whole numbers.
  • Explain the "assume 100 g" shortcut and why it works.

Sources & references

  1. OpenStax, "3.2 Determining Empirical and Molecular Formulas." *Chemistry 2e*.
  2. Chemistry LibreTexts, "10.12: Determining Empirical Formulas."
  3. Chemistry LibreTexts, "10.10: Percent Composition."

This lesson was adapted from the open educational references above; their licenses and attributions are preserved. See Copyright & Licensing.

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