Composition of Mixtures: Study Guide (Chemistry 1.4)

Pure substances and mixtures at the particle level

In Topic 1.3 you saw that a pure compound always has the same composition. Water is always 11.19% hydrogen by mass, whether it comes from a tap in Lahore or a glacier in Iceland. Real samples are often not pure, though. A bag of salt may contain some sand. A rock may hold two different iron oxides. This topic asks a practical question: if you measure how much of one element a sample contains, what can you say about what the sample is made of?

Start with the particles.

• A pure substance contains particles of one type only. These may be atoms (in an element such as Fe), molecules (such as H₂O) or formula units (such as NaCl). Every particle is identical, so the proportions of the elements are fixed by the formula.
• A mixture contains particles of two or more types. The relative amounts of the components are not fixed. You can mix 1 g of salt with 10 g of sand or with 2 g of sand, and both are still “salt and sand”.

Notice that the compound in the middle box contains two elements, but it is still pure: every unit is identical. “Pure” means one type of particle, not one element.

Mass percent: fixed for a compound, variable for a mixture

The mass percent of an element is the mass of that element divided by the mass of the sample, times 100. For a pure compound you can calculate it from the formula alone:

mass percent of X = (number of X atoms in the formula × atomic mass of X) ÷ molar mass × 100

Pure compound	Working	Mass percent of Cl
NaCl	35.45 ÷ 58.44 × 100	60.66%
KCl	35.45 ÷ 74.55 × 100	47.55%

Every pure sample of NaCl is 60.66% chlorine. This is the law of definite proportions from Topic 1.3.

A mixture of NaCl and KCl is different. Its chlorine content depends on how much of each salt is present. If the mixture is all NaCl, it is 60.66% Cl. If it is all KCl, it is 47.55% Cl. Any real mixture of the two must lie between those limits.

One trap: the mixture’s mass percent is not simply the average of 60.66% and 47.55%. That average (54.11%) is right only for a mixture with equal masses of the two salts. In general the mixture’s value is a weighted average, weighted by the mass of each component.

Is the sample pure? Reading elemental analysis data

Elemental analysis means measuring how much of one or more elements a sample contains. One common approach is to convert the element into a compound you can collect and weigh, for example by precipitating it or by burning the sample and weighing the products. You do not need the lab details for this topic; you need to interpret the numbers.

Elemental analysis lets you answer two kinds of question.

1. What are the relative numbers of atoms? Convert each element’s mass to moles and compare. For a pure compound you get the whole-number ratio in its formula (this is how you found empirical formulas in Topic 1.3).
2. How pure is the sample? Compare the measured mass percent with the value the formula predicts.

Here is fictional data for samples taken from two bottles, each labelled “calcium carbonate, CaCO₃”. Pure CaCO₃ is 40.04% calcium by mass.

Sample	Bottle 1: % Ca by mass	Bottle 2: % Ca by mass
First	40.0	31.2
Second	40.0	35.6
Third	40.1	28.4

Bottle 1 gives the same value every time, and it matches the formula. That is what a pure substance does. Bottle 2 gives values that are all below 40.04% and that vary from sample to sample. Both facts point to a mixture: CaCO₃ with something that contains less calcium (or none), unevenly spread through the bottle.

The identification step here uses skill 5.A: before you calculate anything, decide which quantities in the data you need. For a purity question you need the sample mass, the mass of the analysed element, and the mass fraction of that element in the pure compound.

The purity calculation

If the impurity contains none of the analysed element, all of that element came from the compound you are interested in. Then:

1. mass of compound = mass of element ÷ (mass fraction of element in the compound)
2. purity (mass percent of compound in the sample) = mass of compound ÷ mass of sample × 100

You can do step 1 in moles instead: moles of element → moles of compound (using the formula) → mass of compound. Both routes give the same answer.

Worked example 1: purity of a contaminated salt

Question. A 2.50 g sample of potassium chloride, KCl, is contaminated with sand. Sand contains no chlorine. Analysis shows the sample contains 1.02 g of chlorine. What is the percent by mass of KCl in the sample?

1. Choose the quantities. You need the sample mass (2.50 g), the chlorine mass (1.02 g) and the mass fraction of Cl in KCl.
2. Mass fraction of Cl in KCl: 35.45 ÷ 74.55 = 0.47552.
3. Mass of KCl: all the chlorine is in the KCl, so m(KCl) = 1.02 g ÷ 0.47552 = 2.1450 g. Mole route: n(Cl) = 1.02 g ÷ 35.45 g mol⁻¹ = 0.028773 mol. One Cl per KCl, so n(KCl) = 0.028773 mol and m(KCl) = 0.028773 mol × 74.55 g mol⁻¹ = 2.1450 g. Same result.
4. Purity: 2.1450 g ÷ 2.50 g × 100 = 85.80%.

Answer. The sample is 85.8% KCl by mass (3 significant figures, matching 2.50 g and 1.02 g). The remaining 0.355 g is sand.

Check. The sample is 1.02 ÷ 2.50 × 100 = 40.8% chlorine. That is below 47.55% (pure KCl), as it must be when an impurity with no chlorine dilutes the salt. A purity above 100% would have told you the impurity also contains chlorine, or that the data are wrong.

Worked example 2: a mixture where both parts contain the element

Question. A 4.00 g mixture of sodium chloride, NaCl, and potassium chloride, KCl, contains 2.10 g of chlorine. Find (a) the mass of each salt and (b) the ratio of sodium atoms to potassium atoms in the mixture.

Now the chlorine comes from both salts, so one division is not enough. You need two equations. Let a = moles of NaCl and b = moles of KCl.

1. Chlorine equation (moles). Each formula unit has one Cl, so a + b = n(Cl) = 2.10 g ÷ 35.45 g mol⁻¹ = 0.059238 mol.
2. Mass equation. 58.44a + 74.55b = 4.00 g.
3. Solve. From step 1, a = 0.059238 − b. Substitute: 58.44(0.059238 − b) + 74.55b = 4.00 3.4619 + 16.11b = 4.00, so b = 0.033402 mol and a = 0.025836 mol.
4. Masses. m(NaCl) = 0.025836 × 58.44 = 1.510 g; m(KCl) = 0.033402 × 74.55 = 2.490 g. These add to 4.00 g, which checks the algebra.
5. Atom ratio. Each NaCl holds one Na atom (as Na⁺) and each KCl one K atom, so Na : K = 0.025836 : 0.033402 = 0.773 : 1.

Answer. (a) 1.51 g NaCl and 2.49 g KCl (37.7% and 62.3% by mass). (b) Na : K = 0.773 : 1 (or about 1 : 1.29).

Interpretation. The chlorine mass percent is 2.10 ÷ 4.00 × 100 = 52.5%, which sits between 47.55% and 60.66% (Figure 2). It is closer to the KCl value, so it makes sense that KCl is the larger part. The ratio of metal atoms is not a whole-number ratio, which is normal for a mixture: you could have made it with any proportions.

You can also solve with mass fractions: if x is the mass of NaCl, then 0.60661x + 0.47552(4.00 − x) = 2.10, giving x = 1.510 g. Use whichever route you find clearer.

Common misconceptions

• “A substance made of two elements is a mixture.” No. A compound such as NaCl contains two elements but only one type of formula unit, so it is pure. A mixture has two or more types of particle.
• “A uniform-looking sample must be pure.” Salt water looks the same everywhere but contains water molecules and ions from the salt. Appearance does not decide purity; particle make-up does.
• “The mass percent of a mixture is the simple average of the components’ values.” Only for equal masses. In general it is a weighted average and can be anywhere between the two pure values.
• “Divide the element mass by the sample mass and that is the purity.” That gives the mass percent of the element in the sample (40.8% Cl in Worked example 1), not the percent of the compound (85.8% KCl). You must divide by the element’s mass fraction in the compound as well.
• Assuming the impurity contains none of the element without checking. The one-step purity method works only if that is true. If both components contain the element, set up two equations as in Worked example 2.
• “A non-whole-number atom ratio means I made an arithmetic error.” For a pure compound, yes, check your working. For a mixture, a non-whole-number ratio is expected and is itself evidence that the sample is a mixture.
• Rounding mass fractions early. Rounding 0.47552 to 0.48 changes the purity in Worked example 1 from 85.8% to 85.0%. Keep full values until the end.

Where this leads

The idea that a mixture’s composition can vary returns in Unit 3, where you describe solutions and gas mixtures with concentration, mole fraction and partial pressure (Topics 3.4 and 3.7). Purity calculations reappear whenever you use stoichiometry with real, imperfect samples (Topic 4.5). Next you move from what samples are made of to what atoms are made of: Topic 1.5, Atomic Structure and Electron Configuration. If the empirical formula steps felt shaky, revisit Topic 1.3, Elemental Composition of Pure Substances.

Now try the practice questions, then use the revision notes and the checklist to consolidate.