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1 Atomic Structure and Properties 2 Compound Structure and Properties 3 Properties of Substances and Mixtures 4 Chemical Reactions 5 Kinetics 6 Thermochemistry 7 Equilibrium 8 Acids and Bases 9 Thermodynamics and Electrochemistry

1 Atomic Structure

1.1 Moles and Molar Mass 1.2 Mass Spectra of Elements 1.3 Elemental Composition of Pure Substances 1.4 Composition of Mixtures 1.5 Atomic Structure and Electron Configuration 1.6 Photoelectron Spectroscopy 1.7 Periodic Trends 1.8 Valence Electrons and Ionic Compounds

Empirical and Molecular Formula

Learning Objective 1.3.A Explain the quantitative relationship between the elemental composition by mass and the empirical formula of a pure substance.

Quick Notes

  • Empirical formula = simplest whole number ratio of atoms in a compound.
  • Molecular formula = actual number of atoms of each element in a molecule.
  • The law of definite proportions states that a pure compound always has the same ratio of elements by mass.
  • To determine an empirical formula from mass data:
    • Convert mass (or %) of each element to moles.
    • Divide each mole value by the smallest to get a ratio.
    • Multiply to get whole numbers if necessary.
  • Key definitions:
    • Empirical formula: lowest whole number ratio.
    • Molecular formula: actual number of atoms (may be a multiple of empirical formula).

Full Notes

A chemical formula tells us the types and number of atoms in a substance. Some substances are made up of individual molecules (like H2O or C6H12O6), while others consist of ions or atoms in fixed proportions, like NaCl or Fe2O3.

Empirical vs Molecular Formula

The empirical formula gives the simplest whole number ratio of elements in a compound.

The molecular formula shows the actual number of atoms of each element in one molecule of the compound.

ExampleHydrogen peroxide has the molecular formula H2O2, but its empirical formula is HO. Glucose has the molecular formula C6H12O6, and its empirical formula is CH2O.

Law of Definite Proportions

According to this law, every pure compound has the same elements in the same fixed mass ratio. So even if we have different samples of a compound, the mass ratios will always be the same.

Example Water (H2O) always contains 2 grams of hydrogen for every 16 grams of oxygen — a ratio of 1:8 by mass.

Determining the Empirical Formula from Mass Data

To find the empirical formula from mass or percentage composition, follow these steps:

Step 1: Convert each element’s mass (or percentage) to moles
If given %, treat the total as 100 g
Use: moles = mass ÷ molar mass
Divide mole amounts by the smallest value
Multiply up to get only whole numbers if needed (i.e. x everything by 2)

Worked Example

Find the empirical formula for a compound with a composition by mass of C 52.2%, H 13.0% and O 34.8%.

  1. Assume 100 g total
    C = 52.2 g → 52.2 ÷ 12.01 = 4.35 mol
    H = 13.0 g → 13.0 ÷ 1.008 = 12.9 mol
    O = 34.8 g → 34.8 ÷ 16.00 = 2.18 mol
  2. Divide by smallest mole value (2.18)
    C = 4.35 ÷ 2.18 = 2.00
    H = 12.9 ÷ 2.18 = 5.92 ≈ 6.00
    O = 2.18 ÷ 2.18 = 1.00
  3. Write empirical formula
    C2H6O
AP Chemistry worked example showing C 52.2%, H 13.0%, O 34.8% converted to moles and ratios to give empirical formula C2H6O.

Finding the Molecular Formula

If the molar mass of the compound is known, you can find the molecular formula:

Worked Example

The empirical formula of a compound is CH2O and its molar mass is 180 g/mol. Find the molecular formula.

  1. Calculate the empirical formula mass
    (1 × 12.0) + (2 × 1.0) + (1 × 16.0) = 30.0 g/mol
  2. Find n
    n = 180 ÷ 30.0 = 6
  3. Multiply empirical formula by n
    6 × CH2O = C6H12O6
AP Chemistry worked example showing CH2O empirical mass of 30.0, n = 180/30 = 6, giving molecular formula C6H12O6.

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Matt’s exam tip

Always check that your final empirical formula makes chemical sense. It should use whole numbers and reflect what’s possible based on the elements involved. If your ratio gives 1.5 or 1.33, multiply all ratios to clear the decimal.

Summary