Learning Objectives
1 objectiveBy the end of this note, you should be able to:
- 1.3.AExplain the quantitative relationship between the elemental composition by mass and the empirical formula of a pure substance.
Molecules, Formula Units, and Fixed Composition
The way atoms combine in a pure substance determines whether it is described by a molecular formula or a formula unit.
Some pure substances consist of discrete molecules — individual groups of covalently bonded atoms that move as a unit. Water (H₂O), carbon dioxide (CO₂), and glucose (C₆H₁₂O₆) are all molecular compounds. Each molecule has a definite number of atoms of each element, and that number is reflected in the molecular formula.
Other pure substances do not exist as separate molecules. Ionic compounds such as sodium chloride (NaCl) consist of cations and anions held together by electrostatic attraction in an extended lattice. The formula NaCl is a formula unit — it reports the simplest ratio of ions, not a discrete pair traveling together. Metallic elements (e.g., Fe, Cu) and covalent network solids (e.g., SiO₂) are similarly described by formula units rather than molecular formulas because their bonding extends throughout the entire structure.
| Substance Type | Structural Unit | Example | Formula Represents |
|---|---|---|---|
| Molecular compound | Discrete molecule | H₂O | Actual atoms per molecule |
| Ionic compound | Extended ionic lattice | NaCl | Simplest cation-to-anion ratio |
| Covalent network solid | Extended covalent lattice | SiO₂ | Simplest atom ratio in lattice |
| Metallic element | Extended metallic lattice | Fe | Single atom of the element |
Whether a substance is molecular or ionic, its formula always communicates the fixed proportions of atoms or ions present. This fixed composition is what makes the substance "pure" in the chemical sense — every sample has the same elemental makeup.
MisconceptionStudents sometimes assume NaCl means one Na⁺ ion is bonded to one specific Cl⁻ ion as a pair. In reality, NaCl describes the 1 : 1 ratio throughout the entire lattice — no individual "NaCl molecule" exists.
The Law of Definite Proportions
The law of definite proportions states that any pure sample of a given compound always contains the same elements in the same ratio by mass, regardless of the sample's size or source.
Because every molecule or formula unit of a compound contains a fixed number of each type of atom, and each type of atom has a characteristic average atomic mass, the mass ratio of the elements cannot vary. For example, every sample of pure water is 88.89% oxygen and 11.11% hydrogen by mass.
This law allows chemists to work backward: measuring the mass percent of each element in an unknown compound provides enough information to determine the compound's empirical formula. If a compound's mass percentages deviate from a known value, the sample is either impure or is a different compound entirely.
Examiner InsightAP free-response questions frequently give mass-percent data and ask students to derive an empirical formula. The law of definite proportions is the conceptual justification for why this procedure works.
Empirical Formulas from Mass Composition
The empirical formula of a compound gives the lowest whole-number ratio of atoms of each element present.
To determine an empirical formula from percent composition, convert mass data to moles and then simplify the mole ratio:
- Assume a convenient sample mass — typically 100.00 g so that mass percentages convert directly to grams.
- Convert grams of each element to moles by dividing by the element's molar mass.
- Divide every mole value by the smallest mole value to obtain a preliminary ratio.
- Multiply all values by the smallest whole number that converts every ratio to a whole number (if any ratio is not already a whole integer).
The molecular formula is always a whole-number multiple of the empirical formula. For example, glucose has the empirical formula CH₂O and the molecular formula C₆H₁₂O₆ — the molecular formula is exactly 6 × the empirical formula.
| Compound | Molecular Formula | Empirical Formula | Multiplier |
|---|---|---|---|
| Glucose | C₆H₁₂O₆ | CH₂O | 6 |
| Hydrogen peroxide | H₂O₂ | HO | 2 |
| Acetic acid | C₂H₄O₂ | CH₂O | 2 |
| Carbon dioxide | CO₂ | CO₂ | 1 |
MisconceptionStudents sometimes stop after step 3 even when the ratios are not whole numbers (e.g., 1 : 1.5). A ratio of 1 : 1.5 must be multiplied by 2 to give 2 : 3. Always check that the final subscripts are integers.
Exam TipCommon non-integer ratios to watch for — 0.33 → ×3, 0.5 → ×2, 0.25 → ×4, 0.67 → ×3.
Worked Example — Determining an Empirical Formula
A compound is found to contain 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. Determine its empirical formula.
Step 1 — Assume 100.00 g sample:
$${m}_{\text{C}}=40.0\text{ g}$$
$${m}_{\text{H}}=6.7\text{ g}$$
$${m}_{\text{O}}=53.3\text{ g}$$
Step 2 — Convert to moles:
Equation used
$$n=\frac{m}{M}$$
Where: n = moles (mol), m = mass (g), M = molar mass (g mol⁻¹).
Carbon:
$${n}_{\text{C}}=\frac{40.0}{12.01}$$
$${n}_{\text{C}}=3.33\text{ mol}$$
Hydrogen:
$${n}_{\text{H}}=\frac{6.7}{1.008}$$
$${n}_{\text{H}}=6.65\text{ mol}$$
Oxygen:
$${n}_{\text{O}}=\frac{53.3}{16.00}$$
$${n}_{\text{O}}=3.33\text{ mol}$$
Step 3 — Divide by the smallest mole value (3.33 mol):
$$\text{C}:\frac{3.33}{3.33}=1.00$$
$$\text{H}:\frac{6.65}{3.33}=2.00$$
$$\text{O}:\frac{3.33}{3.33}=1.00$$
Step 4 — Ratios are already whole numbers.
$$\text{Empirical formula}={\text{CH}}_{2}\text{O}$$
Practice Problem:
A compound contains 72.4% iron and 27.6% oxygen by mass. Determine its empirical formula.
Answer
Step 1 — Assume 100.00 g sample:
$${m}_{\text{Fe}}=72.4\text{ g}$$
$${m}_{\text{O}}=27.6\text{ g}$$
Step 2 — Convert to moles:
Equation used
$$n=\frac{m}{M}$$
Iron:
$${n}_{\text{Fe}}=\frac{72.4}{55.85}$$
$${n}_{\text{Fe}}=1.296\text{ mol}$$
Oxygen:
$${n}_{\text{O}}=\frac{27.6}{16.00}$$
$${n}_{\text{O}}=1.725\text{ mol}$$
Step 3 — Divide by the smallest mole value (1.296 mol):
$$\text{Fe}:\frac{1.296}{1.296}=1.00$$
$$\text{O}:\frac{1.725}{1.296}=1.33$$
Step 4 — Multiply all ratios by 3 to obtain whole numbers:
$$\text{Fe}:1.00\times 3=3$$
$$\text{O}:1.33\times 3=4$$
$$\text{Empirical formula}={\text{Fe}}_{3}{\text{O}}_{4}$$
QUICK RECAP
Key Points
- Molecular compounds consist of discrete molecules; ionic compounds are described by formula units.
- A formula unit gives the simplest ratio of ions in an ionic lattice.
- The law of definite proportions: fixed mass ratio in any pure sample.
- Mass percent composition is the same regardless of sample size or source.
- Empirical formula = lowest whole-number mole ratio of elements.
- Molecular formula = whole-number multiple of the empirical formula.
- Assume 100.00 g to convert percent composition directly to grams.
- Divide grams by molar mass to get moles of each element.
- Divide all mole values by the smallest to get the ratio.
- Multiply by a whole number if ratios are not integers (×2, ×3, ×4).
- Different compounds can share the same empirical formula (e.g., CH₂O).
- Deviations from expected mass ratios indicate impurity or a different compound.
CAN I…? PROGRESS CHECK
Self-Assessment
- Distinguish between a molecular formula and a formula unit for different types of pure substances.
- State the law of definite proportions and explain why it applies to all pure compounds.
- Convert percent composition by mass data into an empirical formula using a stepwise mole-ratio method.
- Identify when mole ratios require multiplication to reach whole-number subscripts.
- Explain why multiple compounds can share the same empirical formula.
- Determine an empirical formula from mass data for a compound containing three or more elements.