Quantitative chemistry

1. Mass does not disappear

When wood burns, ash is left behind. That does not mean most of the matter has vanished: gases have moved into the air. In every chemical reaction, atoms are rearranged. No atoms are created or destroyed.

The law of conservation of mass says that the total mass of the reactants equals the total mass of the products. Count all substances, including gases entering or leaving the apparatus.

2. Read a formula before balancing an equation

A small number within a formula tells you how many atoms of that element occur in one molecule or formula unit. H₂O contains two H atoms and one O atom. A number in front multiplies the whole formula: 2H₂O represents four H atoms and two O atoms.

Balance equations by changing the numbers in front. Never change a correct formula to make the counts match: H₂O₂ is hydrogen peroxide, a different substance from water.

Use the smallest whole-number coefficients. In 2Mg + O₂ → 2MgO, the atom counts are Mg: 2 each side; O: 2 each side.

3. Relative formula mass: add the atoms

The relative atomic mass, Aᵣ, is the average mass of an atom compared with one-twelfth of the mass of a carbon-12 atom. This average takes account of the element’s isotopes. The relative formula mass, Mᵣ, adds the Aᵣ values of every atom in a formula. Both are relative values and have no unit.

In a balanced equation, include the coefficient when comparing total relative masses. For 2Mg + O₂ → 2MgO: left = 2 × 24 + 32 = 80; right = 2 × 40 = 80.

4. What percentage of a compound is one element?

Fertilisers contain nitrogen, but not every gram of fertiliser is nitrogen. Percentage by mass tells you how much of the compound’s mass comes from a particular element.

Percentage by mass = (total Aᵣ contribution of the element ÷ Mᵣ of compound) × 100

If the element occurs more than once, count every atom. In NH₄NO₃ there are two nitrogen atoms. With Aᵣ values N 14, H 1 and O 16, Mᵣ = 80, so nitrogen percentage = 28 ÷ 80 × 100 = 35%.

5. Why can a balance reading change?

An open system can exchange substances with its surroundings. If a gas escapes, less matter remains on the balance. If oxygen from the air reacts with a metal, extra matter joins the sample and its mass rises.

Example 1: Mg reacts with O₂ to form MgO. The solid gains mass because it contains oxygen atoms from the air.

Example 2: CaCO₃ → CaO + CO₂. Heating calcium carbonate produces carbon dioxide, which escapes from an open container. The solid loses mass, but solid + escaped gas still have the original total mass.

For a mass-against-time graph, read both axes and units. A decreasing line may show gas leaving. A plateau means the measured mass has stopped changing; check the experiment before deciding whether the reaction has finished.

6. Measurements come with uncertainty

Real measurements are not perfectly exact. A balance reading to 0.01 g cannot show every smaller change. Repeating a measurement helps reveal its spread, but a consistently incorrect balance can give repeated, incorrect results.

Mean = sum of readings ÷ number of readings. Range = highest − lowest. One estimate of uncertainty from repeated readings is half the range. State which method you used; it does not account for every possible error.

Plot individual readings or use a suitable chart to show their distribution. A smaller spread usually indicates more repeatable measurements. Investigate an unusual result: record the reason if a known method error justifies exclusion. Do not remove a result simply because it does not fit your prediction.

Improve a mass experiment by using a suitable balance, checking its zero, keeping all material on the balance, avoiding spills and allowing hot apparatus to cool before weighing. Repeat the procedure and compare like-for-like measurements.

7. Concentration in grams per dm³

A solute is the substance dissolved. A solvent does the dissolving. Together they form a solution. Concentration in g/dm³ tells you how many grams of solute are in each cubic decimetre of solution.

1 dm³ = 1000 cm³ = 1 litre. To change cm³ to dm³, divide by 1000. For example, 250 cm³ = 0.250 dm³.

Mass of solute (g) = concentration (g/dm³) × volume of solution (dm³).

The volume is the final solution volume, not simply the volume of water poured in. To make a stated volume, dissolve the weighed solute in some water, transfer it without losses, then add water until the final solution reaches the required volume and mix.

Concentration = mass ÷ volume. Rearrange to volume = mass ÷ concentration. With g and g/dm³, the volume is in dm³.

8. Higher only: a mole is a counting unit

Higher only. A mole is a fixed number of particles, rather like a dozen is 12. One mole contains 6.02 × 10²³ stated particles. This number is the Avogadro constant. Always say what you are counting: atoms, molecules, ions or electrons.

One mole of carbon atoms and one mole of carbon dioxide molecules contain the same number of those particles. A mole of CO₂ molecules contains three times as many atoms in total, because each molecule contains three atoms.

The mass of one mole in grams is numerically equal to Mᵣ (or Aᵣ for individual atoms). Thus one mole of CO₂ has mass 44 g.

Amount (mol) = mass (g) ÷ mass of one mole (g/mol).
Mass (g) = amount (mol) × mass of one mole (g/mol).

Keep extra digits during a calculation; round at the end to the requested significant figures. 0.250 has three significant figures; its leading zero is not significant.

9. Higher only: use the equation as a recipe

Higher only. The coefficients in a balanced equation give a mole ratio, not a mass ratio. In 2Mg + O₂ → 2MgO, two moles of Mg react with one mole of O₂ to form two moles of MgO.

1. Write the balanced equation.
2. Convert the known mass into moles.
3. Use the equation’s mole ratio to find the required amount.
4. Convert that amount into a mass.

Check the result: the product has gained 4 g oxygen, so it is reasonable that its mass is greater than the starting metal mass.

10. Higher only: work back from masses

Higher only. Experimental masses can reveal the balancing numbers. Convert each mass to moles, divide all amounts by the smallest, then express the ratio as the simplest whole numbers.

Do not use the gram values directly as coefficients. Different substances have different masses per mole. After calculating a ratio, check that each atom is balanced.

11. Higher only: which reactant runs out?

Higher only. A reaction can only continue while all necessary reactants remain. The limiting reactant is used up first and sets the maximum amount of product. Another reactant may be in excess, meaning some remains unused.

To compare given masses, convert both to moles before applying the ratio. Adding more of the excess reactant will not increase product after the limiting reactant has been used up.

12. Higher only: explain changes in concentration

Higher only. For a fixed solution volume, increasing the dissolved mass increases concentration. For a fixed dissolved mass, increasing solution volume decreases concentration because the same solute is spread through more solution.

There is no separate required practical in this topic. Its measurement and calculation skills support practicals throughout chemistry. This Trilogy worksheet does not require percentage yield, atom economy, molar gas volumes or mol/dm³ titration calculations.

Explain a mass investigation

Use the illustration to explain where the missing 1.4 g went in the open system. Then plan two improvements that would make the comparison more reliable. This is a measurement-skills activity, not an additional required practical.

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One final thought

Use a scientific idea and a specific example in your reflection.