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Composition of Mixtures

Distinguish pure-substance composition from a mixture and justify a mass-balance model.

AP Chemistry Topic 1.4 · CED printed p. 33 / PDF p. 40. This teaching, practice and feedback were added for the AP edition; original author material remains separately traceable.

Before you start

Composition can vary without creating a new formula

A pure substance has one chemical identity: one kind of atom, molecule or formula unit. A mixture contains two or more substances in proportions that can vary. You might have many identical water molecules in pure water, or water and ethanol molecules in a mixture. More than one element does not automatically mean “mixture”; every pure water molecule already contains two elements.

Elemental analysis reports how much of each element is present across the whole sample. Compare this to the mass fraction expected for a pure candidate. If an impurity lacks the measured element, all of that element must come from the candidate, giving a clean route to its mass. If both substances contain that element, you must account for both contributions.

Build a mass balance before solving

For a two-substance mixture of masses mA and mB, total sample mass is mA + mB. If the fraction of an element X in each pure substance is fA and fB, then measured X mass is fAmA + fBmB. Those two relationships let you solve for composition. The answer must lie between zero and the total mass, and the mixture’s elemental fraction must lie between the fractions of its two components.

A purity percentage needs an explicit impurity model. “Less chlorine than expected” may indicate a chlorine-free impurity if that is the known situation, but elemental data alone do not identify an arbitrary unknown impurity. The AP skill is to connect the measured quantity to a justified composition model and recognize what cannot be determined.

Let's work through it

Work it through

A 2.00 g sample is known to contain only NaCl and chlorine-free sand. Elemental analysis finds 0.910 g Cl. NaCl has M = 58.44 g mol−1; chlorine contributes 35.45 g mol−1.

  1. Find the chlorine mass fraction in pure NaCl: f = 35.45/58.44 = 0.6066.
  2. Because sand supplies no Cl, 0.910 g = f × mNaCl. Thus mNaCl = 0.910/0.6066 = 1.50 g.
  3. Subtract from total: sand mass = 2.00 − 1.50017 = 0.50 g. Purity by NaCl mass is (1.50017/2.00) × 100 = 75.0%.
  4. State the assumption: every measured chlorine atom belongs to NaCl. Without that assumption the same Cl mass could fit another composition.

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 of 20 self-assessed

Question 1

Is pure CO2 a mixture because it contains carbon and oxygen?

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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

Is pure CO2 a mixture because it contains carbon and oxygen?

Worked answer and reason

No. It contains one kind of molecule with a fixed formula.

Mixture classification counts substances, not element types.

Retrieval 2

A candidate substance is 40% X by mass and its only impurity contains no X. A 10.0 g sample has 3.00 g X. How much candidate is present?

Worked answer and reason

3.00/0.40 = 7.50 g, assuming all X comes from the candidate.

The element mass is a known fraction of the candidate mass.

Catch the tempting mistake

Any formula found from elemental analysis proves a sample is pure.
A mixture can imitate a bulk elemental ratio; purity is a model supported by additional evidence.
Purity equals measured element mass divided by sample mass.
That gives the element’s mass fraction. Divide by the element fraction in the pure candidate to obtain candidate purity.

Put it to work

Give the chemical reason as well as the calculation. Worked solutions are on a separate page.

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For the teacher
  • Require the impurity assumption in the written solution.
  • Use impossible fractions to prompt students to inspect a model rather than accept algebra mechanically.

Unit planning and formative guidance →

Source objectives and scope

Source edition: ap-chemistry-ced-fall2024-local-b5dfe867. See this topic in the course outline →

1.4.A
Explain the quantitative relationship between the elemental composition by mass and the composition of substances in a mixture.
1.4.A.1
Pure substances contain atoms, molecules, or formula units of a single type. Mixtures contain atoms, molecules, or formula units of two or more types, whose relative proportions can vary.
1.4.A.2
Elemental analysis can be used to determine the relative numbers of atoms in a substance and to determine its purity.
Science references for this teaching
Glossary and deeper reading · 4 refreshers

These editorial refreshers supplement the reading. Definitions and examples are added guidance; original source wording and recorded answers remain in their original context.

  • 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.

  • Pure substance

    A pure substance has one chemical identity and a fixed composition. It may be an element or a compound, so containing more than one element does not make it a mixture. For example, pure water contains hydrogen and oxygen in a fixed ratio.

  • 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.

    Read more: Moles and Molar Mass →

  • 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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