Elemental Composition of Pure Substances
Use fixed elemental composition to find mass fractions and empirical formulas.
AP Chemistry Topic 1.3 · CED printed p. 32 / PDF p. 39. This teaching, practice and feedback were added for the AP edition; original author material remains separately traceable.
Before you start
- Mole conversion
- Formula subscripts
A pure compound repeats one composition
Imagine cutting a pure sample into many smaller samples. Each piece still has the same ratio of elements by mass, because it has the same composition. That is the law of definite proportions. Molecules such as H2O are separate covalent units; an ionic compound such as MgCl2 is an extended solid described by formula units. Both have a fixed elemental ratio. A mixture can change composition, so do not force a single empirical formula onto a mixture.
For a known formula, calculate an element’s mass fraction by dividing its mass contribution in one mole by the compound’s molar mass. In CO2, oxygen contributes 32.00 g to 44.01 g per mole, so its mass fraction is 32.00/44.01. This ratio concerns mass. The O:C atom ratio remains 2:1, a different ratio.
Turn mass ratios into atom ratios
An empirical formula is the simplest whole-number atom ratio. To find it from masses or mass percentages, convert each element’s mass to moles using that element’s molar mass. Divide all amounts by the smallest. If a ratio is near 1.5 or 1.333, multiply the whole set by 2 or 3; do not round a chemically meaningful fraction away. If given percentages, choosing a 100 g sample is a bookkeeping convenience. It does not mean the actual sample was 100 g.
A molecular formula can be a multiple of the empirical formula. C2H4 and C3H6 both simplify to CH2; elemental analysis alone cannot choose between them. An independently measured molecular molar mass can identify the integer multiplier. This is a useful consistency check rather than an excuse to guess a familiar formula.
Let's work through it
Work it through
A pure binary iron oxide sample contains 2.80 g Fe and 1.20 g O. Use Fe = 55.85 and O = 16.00 g mol−1.
- Convert separately: nFe = 2.80/55.85 = 0.05013 mol; nO = 1.20/16.00 = 0.07500 mol.
- Divide by 0.05013: Fe:O ≈ 1:1.496. This is close to 1:1.5 at the measurement precision.
- Multiply both by 2: Fe:O = 2:3, so the empirical formula is Fe2O3.
- Check the mass ratio predicted by the formula: 111.70:48.00 ≈ 2.327:1, close to measured 2.80:1.20 ≈ 2.333:1. Small differences reflect rounded input masses.
Try it from memory
Pause & retrieve
Try it before you peek.
Write your answer and the reason from memory. Then compare with the worked answer. Your self-assessment is a study decision, not an automatically graded score.
Question 1
Why must you convert element masses to moles before finding an empirical formula?
See every recall question with its answer
These references preserve each prompt and its answer for direct evidence links. Try the interactive questions above before opening the worked reasoning.
Retrieval 1
Why must you convert element masses to moles before finding an empirical formula?
Worked answer and reason
Moles are proportional to atom counts; grams are weighted by each atom’s mass.
Chemical subscripts count atoms, not grams.
Retrieval 2
What does a 1:1.5 mole ratio become, and why not write a subscript 1.5?
Worked answer and reason
Multiply both terms by 2, giving 2:3. Empirical-formula subscripts use the smallest whole-number atom ratio.
Multiplication preserves the ratio rather than changing the measured composition.
Catch the tempting mistake
- Equal masses of two elements imply a 1:1 atom ratio.
- The lighter element contributes more atoms per gram.
- An empirical formula always gives the molecular formula.
- It gives only the lowest ratio; additional molar-mass evidence is needed for molecular size.
Put it to work
Give the chemical reason as well as the calculation. Worked solutions are on a separate page.
- Unit 1 problem 5 · ap application · data
- Unit 1 problem 6 · ap synthesis · multiple
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For the teacher
- Look for mass-to-mole conversion before ratio simplification.
- Ask students to predict whether a larger pure sample changes mass percentage.
Source objectives and scope
Source edition: ap-chemistry-ced-fall2024-local-b5dfe867. See this topic in the course outline →
- 1.3.A
- Explain the quantitative relationship between the elemental composition by mass and the empirical formula of a pure substance.
- 1.3.A.1
- Some pure substances are composed of individual molecules, while others consist of atoms or ions held together in fixed proportions as described by a formula unit.
- 1.3.A.2
- According to the law of definite proportions, the ratio of the masses of the constituent elements in any pure sample of that compound is always the same.
- 1.3.A.3
- The chemical formula that lists the lowest whole number ratio of atoms of the elements in a compound is the empirical formula.
Science references for this teaching
- AP Chemistry Course and Exam Description, effective Fall 2024 (supplied source) — Exact topic-numbered learning objectives, essential knowledge, science practices and AP exclusions; supplied PDF SHA256 b5dfe8677ef3d88c613865d2e2a3e8d6125d652e2b24c71ef1e8ce4e011094f0.
Glossary and deeper reading · 7 refreshers
These editorial refreshers supplement the reading. Definitions and examples are added guidance; original source wording and recorded answers remain in their original context.
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Mole
A mole is an amount of substance containing exactly 6.02214076 × 10²³ specified entities. The entities can be atoms, molecules, ions, or formula units, so the counted entity must be named. For example, one mole of oxygen molecules contains two moles of oxygen atoms.
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Formula unit
A formula unit specifies the simplest whole-number ratio of ions in an ionic compound. It is a counting unit for an extended solid rather than an isolated molecule. For example, one calcium chloride formula unit represents one calcium ion and two chloride ions.
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Mixture
A mixture contains two or more substances whose proportions can vary. Its components retain their chemical identities until a chemical reaction changes them. For example, water and ethanol can form mixtures with many different compositions.
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Mass fraction
Mass fraction is a component's mass divided by the total mass of the sample. Multiplying that fraction by 100 gives its mass percentage. For example, 5 grams of a solute in 100 grams of solution corresponds to a mass fraction of 0.05 and a mass percentage of 5 percent.
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Molar mass
Molar mass is the mass of one mole of a specified substance. It is commonly expressed in grams per mole and connects a sample's mass to its amount of substance. For example, water has a molar mass of approximately 18.02 grams per mole.
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Empirical formula
An empirical formula gives the simplest whole-number ratio of atoms in a compound. It is obtained by comparing amounts in moles of each element rather than comparing their masses directly. For example, both ethene and propene have the empirical formula CH₂ despite having different molecular formulas.
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Molecular formula
A molecular formula gives the actual number of each kind of atom in a molecule. It can be a whole-number multiple of the empirical formula. For example, glucose has molecular formula C₆H₁₂O₆ and empirical formula CH₂O.