Mole concept & stoichiometry — questions and answers
5 high-yield, syllabus-aligned questions on Mole concept & stoichiometry, covering 16 individually creditable mark points. Each one shows the answer that scores full marks under our guide, which phrase earns which mark, and a common incomplete answer — so you can see the difference rather than guess at it.
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Every question and solution is reviewed by examiners versed in WAEC and JAMB, each with more than thirty years of experience.
Mark allocations are TopMarks’ own, written against the published syllabus and Chief Examiners’ reports. They are a guide to how these answers are usually credited, not an official marking scheme.
Define2 marks
Define the mole.
TopMarks model answer
The mole is the SI unit of amount of substance. One mole contains exactly 6.02214076 x 10²³ specified elementary entities — atoms, molecules or ions — a number known as the Avogadro constant. Many textbooks and past papers state the same idea in its older form: the amount containing as many entities as there are atoms in 12 g of carbon-12.
Suggested marking guide
1Amount of substance, defined by a fixed number of entities — either the modern or the carbon-12 wording
2Names the Avogadro constant and its value
Why it scores
Two marks means two ideas: that the mole is an AMOUNT defined by a fixed NUMBER of entities, and the value of that number. Either wording earns the first — the modern “exactly 6.02214076 x 10²³ entities”, or the older “as many as there are atoms in 12 g of carbon-12”. Both describe the same quantity; the 2019 SI redefinition simply fixed the number rather than deriving it from a mass. A definition that says only “the molecular mass in grams” describes the molar mass, not the mole, and earns nothing.
A common incomplete answer
“A mole is the molecular mass of a substance expressed in grams.”
What it costs: Nothing under this guide. That is the definition of molar mass. The mole is an amount of substance defined by a NUMBER of particles, and neither the Avogadro constant nor the carbon-12 standard appears here.
Key ideas to include: amount of substance, elementary entities, carbon-12, Avogadro constant.
Calculate3 marks
Calculate the number of moles present in 10 g of calcium trioxocarbonate(IV), CaCO₃. [Ca = 40, C = 12, O = 16]
TopMarks model answer
Molar mass of CaCO₃ = 40 + 12 + (3 x 16) = 100 g mol⁻¹. Number of moles = mass ÷ molar mass = 10 ÷ 100 = 0.1 mol.
Suggested marking guide
1Correct molar mass of 100
2Correct relationship stated: moles = mass ÷ molar mass
3Correct answer with the unit
The working, line by line
1Molar mass of CaCO₃ = 40 + 12 + (3 x 16) = 100 g mol⁻¹ — Work the molar mass out in full and show it. This is a mark on its own, and you keep it even if everything after it goes wrong.
2Number of moles = mass ÷ molar mass — Write the relationship before you put numbers in. Examiners award this method mark separately from the arithmetic.
3= 10 ÷ 100 = 0.1 mol — The unit is mol. An answer of “0.1” with no unit loses a mark it did all the work to earn.
Why it scores
Three marks for a calculation that takes twenty seconds — and they are three SEPARATE marks: the molar mass, the relationship, the answer with its unit. A candidate who writes only “0.1” has done the chemistry correctly and collected one mark out of three.
A common incomplete answer
“10 ÷ 100 = 0.1”
What it costs: Two marks of three. The molar mass is visible in the working so that mark survives, but the relationship is never stated and the answer has no unit.
Key ideas to include: molar mass, mass, mole, mol.
Determine4 marks
A compound contains 40% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Determine its empirical formula. [C = 12, H = 1, O = 16]
TopMarks model answer
Divide each percentage by the relative atomic mass: C = 40 ÷ 12 = 3.33; H = 6.7 ÷ 1 = 6.7; O = 53.3 ÷ 16 = 3.33. Divide each result by the smallest, 3.33: C = 1; H = 2; O = 1. The empirical formula is CH₂O.
Suggested marking guide
1Divides each percentage by the relative atomic mass
2Correct mole ratios obtained
3Divides through by the smallest ratio
4Correct empirical formula
The working, line by line
1C: 40 ÷ 12 = 3.33 H: 6.7 ÷ 1 = 6.7 O: 53.3 ÷ 16 = 3.33 — Percentage ÷ relative atomic mass. Set it out in a table if you can — the examiner can then see each division.
2Divide each by the smallest (3.33): C = 1, H = 2, O = 1 — This step is a mark of its own. Never skip it, even when the ratio is obvious.
3Empirical formula = CH₂O — Empirical means SIMPLEST whole-number ratio — not the molecular formula. Do not multiply up unless the molar mass is given.
Why it scores
Three of the four marks here are for METHOD, not for the formula. The candidate who writes “CH₂O” alone — even though it is right — has shown no working and can only be given the answer mark. Setting the working out in three visible lines is the difference between 1 and 4.
A common incomplete answer
“The empirical formula is CH₂O.”
What it costs: Three marks of four. The answer is correct, but with no working there is nothing for the three method marks to attach to.
20.0 cm³ of 0.1 mol dm⁻³ hydrochloric acid exactly neutralised 25.0 cm³ of sodium hydroxide solution. Calculate the concentration of the sodium hydroxide solution in mol dm⁻³.
TopMarks model answer
Moles of HCl = concentration x volume ÷ 1000 = 0.1 x 20 ÷ 1000 = 0.002 mol. The equation HCl + NaOH → NaCl + H₂O shows a mole ratio of 1 : 1, so moles of NaOH = 0.002 mol. Concentration of NaOH = moles x 1000 ÷ volume = 0.002 x 1000 ÷ 25 = 0.08 mol dm⁻³.
Suggested marking guide
1Correct number of moles of acid
2Mole ratio taken from the balanced equation
3Correct conversion back to concentration
4Correct answer with the unit
The working, line by line
1Moles of HCl = C x V ÷ 1000 = 0.1 x 20 ÷ 1000 = 0.002 mol — Volumes in titrations are in cm³ but concentrations are per dm³, so you must divide by 1000. Forgetting this is the single commonest error in the topic.
2HCl + NaOH → NaCl + H₂O, mole ratio 1 : 1 — Always write the equation. The ratio mark is separate, and in a question with a 1 : 2 ratio this step is what saves you.
3Moles of NaOH = 0.002 mol — Carry the ratio across explicitly rather than assuming it.
4Concentration = 0.002 x 1000 ÷ 25 = 0.08 mol dm⁻³ — Convert back to per dm³ and state the unit.
Why it scores
The mole ratio mark is the one that separates candidates. In this question the ratio happens to be 1 : 1, so a student who never thinks about it still gets the right number — and still loses the mark, because the examiner cannot see that it was considered. Write the equation every single time.
A common incomplete answer
“Moles of HCl = 0.1 x 20 ÷ 1000 = 0.002 mol, so the concentration of NaOH is 0.002 mol dm⁻³.”
What it costs: Three marks of four. The moles of acid are right, but the equation and ratio are missing, and moles have been reported as though they were a concentration — the conversion back to per dm³ was never done.
Key ideas to include: concentration, mole ratio, balanced equation, mol dm⁻³, volume in cm³.
Explain3 marks
State the law of conservation of mass and explain how it is applied in balancing a chemical equation.
TopMarks model answer
The law of conservation of mass states that matter is neither created nor destroyed in the course of a chemical reaction, so the total mass of the reactants is equal to the total mass of the products. In balancing an equation, the number of atoms of each element must therefore be the same on both sides of the equation. This is achieved by placing coefficients in front of the formulae; the formulae themselves are never altered, because changing a subscript would change the substance.
Suggested marking guide
1Correct statement of the law
2Atoms of each element equal on both sides
3Balance using coefficients, never by changing formulae
Why it scores
The third mark is the practical one and the one candidates rarely earn: you balance by changing the BIG number in front, never the small number inside the formula. A student who balances H₂O by writing H₂O₂ has changed water into hydrogen peroxide and made the equation describe a different reaction.
A common incomplete answer
“Mass cannot be created or destroyed, so the equation must balance.”
What it costs: Two marks of three. The law is stated correctly, but nothing explains that the ATOMS of each element must be equal, and nothing says how balancing is actually achieved.
Key ideas to include: conservation of mass, reactants, products, coefficient, subscript.
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