Calculation problems

Chemistry problems in the open (written) part of the exam nearly always consist of a few simple steps: equation, moles, ratio, moles again. The difficulty is not in the formulae but in organising a lot of information: what reacted with what, what was left over, why the mass changed. This topic shows the general method and the most common problem types — mixtures, mass change, limiting and excess reagents, yield, samples split into portions, and finding formulae — using official test items. It brings together "Moles and calculations" and "Solutions".

§1. General method and key laws

It is convenient to solve any calculation problem in this order:

1. Write and balance all the equations.
2. Convert the given quantities into moles: n = m/M = V/22.4 = N/Nₐ = C·V.
3. Name the unknowns (x, y) — usually the amounts of the components of a mixture.
4. Set up equations from the data (mass, volume, mole ratio, change in mass).
5. Solve them, convert to the quantity asked for and check the answer.

The mole is the "currency" of every calculation: proportions taken directly between masses or volumes lead to mistakes, because the coefficients in an equation give mole ratios. Atom balancing saves time in multi-step problems: for example, the amount of iron stays the same through all its compounds, so you can relate the starting and final substances without writing the intermediate reactions.

§2. Mixture problems

If a mixture has two unknowns, you need two independent equations: usually one for mass and one for gas volume or reagent used. If only one component reacts (copper, for example, does not react with hydrochloric acid), the problem reduces to a single equation.

Worked example

7.8 g of a mixture of Mg and Al dissolved in excess HCl, giving 8.96 L of H₂ (STP). Find the composition of the mixture.

1. Mg → H₂ (1 : 1), 2Al → 3H₂; n(H₂) = 8.96/22.4 = 0.4 mol.
2. 24x + 27y = 7.8; x + 1.5y = 0.4.
3. x = 0.4 − 1.5y → 9.6 − 36y + 27y = 7.8 → y = 0.2, x = 0.1.
4. Mg 0.1·24 = 2.4 g, Al 0.2·27 = 5.4 g.

§3. Change-in-mass problems

When a plate is dipped in a solution or a substance is heated or roasted, ask: "why did the mass change?". On a plate, subtract the mass of metal dissolved and add the mass of metal deposited. On heating, subtract the gas released and add any oxygen taken up from the air. The quickest route is to find the change in mass per mole of reaction and divide the total change by it.

Worked example

97.6 g of a mixture of Cd(NO₃)₂ and Fe was heated in air and its mass did not change. Find the mass of the salt.

1. 2Cd(NO₃)₂ → 2CdO + 4NO₂ + O₂: 1 mol of salt (236 g) → 128 g CdO, a loss of 108 g.
2. 4Fe + 3O₂ → 2Fe₂O₃: 1 mol of Fe (56 g) → 80 g Fe₂O₃, a gain of 24 g.
3. 108x = 24y → y = 4.5x; 236x + 56·4.5x = 488x = 97.6 → x = 0.2.
4. Mass of salt 0.2·236 = 47.2 g.

§4. Limiting and excess reagents; yield

If the amounts of both reactants are given, first decide which is limiting: the product is calculated from the limiting reagent, while some of the excess reagent is left over. Divide the amount of each reactant by its coefficient and compare — the smaller value belongs to the limiting reagent.

The theoretical amount comes from the equation; the actual amount is always less than or equal to it.

Worked example

A mixture of 10 g of hydrogen and 64 g of oxygen was exploded. What mass of water forms, and which gas is left over?

1. n(H₂) = 10/2 = 5 mol, n(O₂) = 64/32 = 2 mol; 2H₂ + O₂ → 2H₂O.
2. Compare 5/2 = 2.5 with 2/1 = 2: oxygen is limiting.
3. n(H₂O) = 2·2 = 4 mol, m = 4·18 = 72 g; H₂ left over: 5 − 4 = 1 mol (2 g).

§5. Several reactions and samples split into portions

When a sample is divided into equal portions, a separate experiment is done on each. First find the amounts in one portion, and if the question asks about the whole sample, remember to multiply at the end.

Worked example

1.6 mol of a mixture of Cu, C and S was split into two equal portions. Burning portion 1 in excess O₂ gave 0.7 mol of gas; dissolving portion 2 in conc. H₂SO₄ gave 136.8 g of gas. How many moles of CO₂ were produced in the two experiments together?

1. Each portion is 0.8 mol. On burning, Cu becomes solid CuO, so the gases are only CO₂ and SO₂: n(C) + n(S) = 0.7 → n(Cu) = 0.1.
2. With conc. H₂SO₄: Cu → 1 SO₂; C → CO₂ + 2SO₂; S → 3SO₂.
3. n(C) = a, n(S) = 0.7 − a: n(SO₂) = 0.1 + 2a + 3(0.7 − a) = 2.2 − a.
4. 44a + 64(2.2 − a) = 136.8 → 140.8 − 20a = 136.8 → a = 0.2.
5. Each portion gives 0.2 mol CO₂, so 0.4 mol in total.

§6. Finding formulae, and organic problems

• From combustion products: n(C) = n(CO₂), n(H) = 2n(H₂O), oxygen from the mass difference.
• From the formula of a homologous series: for CₙH₂ₙ₊₂, 14n + 2 = M.
• From a functional-group reaction: H₂ released with Na, the silver mirror test, alkali used.

Worked example

A and B are saturated monocarboxylic acids (A comes before B in the series). 48 g of the mixture reacts completely with 160 g of 28 % KOH; after the silver mirror reaction the products reacted with 200 g of 24 % NaOH. Find the mass of A.

1. m(KOH) = 44.8 g = 0.8 mol → 0.8 mol of acids in total.
2. A gives a silver mirror, so it is methanoic acid HCOOH; it is oxidised to carbonic acid (CO₂ + H₂O).
3. n(NaOH) = 48/40 = 1.2 mol. Carbonic acid needs 2 mol NaOH, B needs 1 mol: 2x + (0.8 − x) = 1.2 → x = 0.4.
4. m(HCOOH) = 0.4·46 = 18.4 g; check: B 29.6/0.4 = 74 → propanoic acid.

Check yourself

1. A mixture of Fe and Cu (12 g) gave 2.24 L of H₂ (STP) with dilute HCl. What is the mass of copper?

2. 20 g of CaCO₃ was heated, leaving 15.6 g. What percentage decomposed?

3. Burning 0.1 mol of a hydrocarbon gave 0.3 mol CO₂ and 0.4 mol H₂O. What is its formula?

4. 50 g of 8 % NaOH was added to 50 g of 9.8 % H₂SO₄. Is the solution acidic, neutral or alkaline?

5. 5.6 g of Fe reacted with excess CuSO₄ solution. How does the plate’s mass change?

6. 2 mol of SO₂ and 1 mol of O₂ gave 1.6 mol of SO₃. What is the yield?

7. What mass of water forms from 4 g of hydrogen and 16 g of oxygen?

8. What mass of NaOH can be obtained from 2.3 g of sodium through all stages?

Common mistakes

• Starting calculations before balancing the equation.
• Not subtracting gas and precipitate from the mass of a solution.
• Forgetting the "half" in problems with a split sample.
• Ignoring the reducing (aldehyde) behaviour of methanoic acid.
• Calculating from the excess reagent without finding the limiting one.