AQA · Week 1 of the 13-week plan
Notes for every section, eight new exam-style questions with full worked solutions, and online recall cards for your review days. Nothing here is copied from the book.
What this chapter covers
- 2.1 The mole, the Avogadro constant and molar mass
- 2.2 Empirical and molecular formulae
- 2.3 Full and ionic equations
- 2.4 Reacting masses and gas volumes
- 2.5 Solutions, concentration and titrations
- 2.6 Errors and uncertainties
- 2.7 Percentage yield and atom economy
- 2.8 Hazards, risks and precautions
Learn it
Read one section at a time, then answer its check question before you open the answer.
2.1 The mole, the Avogadro constant and molar mass
Chemists count particles in moles. One mole of anything contains the same number of particles as there are atoms in exactly 12 g of carbon-12.
Avogadro constant, NA = 6.022 × 1023 mol−1: the number of particles in one mole.
Molar mass, M: the mass of one mole of a substance, in g mol−1. Numerically it equals the relative formula mass.
n = m ÷ M (moles = mass in g ÷ molar mass)
number of particles = n × NA
Always say which particle you are counting. One mole of H2O contains 6.022 × 1023 molecules but 3 × 6.022 × 1023 atoms. One mole of CaCl2 contains 2 mol of Cl− ions.
Relative formula mass, Mr, is the sum of the relative atomic masses in the formula. For ionic compounds some people call it relative formula mass rather than relative molecular mass; the calculation is the same.
Check yourself: How many moles of chloride ions are in 11.1 g of CaCl2 (M = 111.1 g mol−1)?
0.0999 mol CaCl2, so 0.200 mol Cl−
2.2 Empirical and molecular formulae
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2.3 Full and ionic equations
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2.4 Reacting masses and gas volumes
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2.5 Solutions, concentration and titrations
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2.6 Errors and uncertainties
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2.7 Percentage yield and atom economy
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2.8 Hazards, risks and precautions
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Extra worked questions
Cover the solution, answer on paper, then mark yourself against each step. Each step shows the mark it earns.
All 8 worked solutions are free for members. The first one is open to everyone.
Moles of calcium carbonate
(a) Calculate the mass of 0.250 mol of calcium carbonate, CaCO3. (b) Calculate the number of oxygen atoms in this sample. Ar: Ca 40.1, C 12.0, O 16.0. NA = 6.022 × 1023 mol−1.
Try it first, then show the worked solution
| (a) Molar mass | M = 40.1 + 12.0 + 3 × 16.0 = 100.1 g mol−1 | 1 |
|---|---|---|
| (a) Mass | m = 0.250 × 100.1 = 25.0 g | 1 |
| (b) Oxygen atoms | Each CaCO3 has 3 O atoms: 0.250 × 3 × 6.022 × 1023 = 4.52 × 1023 | 1 |
Answer: (a) 25.0 g (b) 4.52 × 1023 oxygen atoms
Examiner tip: Read the question for which particle it wants: formula units, atoms of one element, or ions.
From percentages to a molecular formula
A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Its relative molecular mass is 180. Find its empirical and molecular formulae. Ar: C 12.0, H 1.0, O 16.0.
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Volume of gas from a carbonate
5.00 g of calcium carbonate is heated until it fully decomposes: CaCO3(s) → CaO(s) + CO2(g). Calculate the volume of carbon dioxide formed, in dm3, at 100 kPa and 298 K. M(CaCO3) = 100.1 g mol−1; R = 8.31 J K−1 mol−1.
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Titrating sodium hydroxide
25.0 cm3 of sodium hydroxide solution is neutralised by a mean titre of 21.40 cm3 of 0.100 mol dm−3 sulfuric acid. 2NaOH + H2SO4 → Na2SO4 + 2H2O. Calculate the concentration of the sodium hydroxide.
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Uncertainty in a titration
A student records burette readings of 0.50 cm3 and 21.90 cm3. Each reading has an uncertainty of ±0.05 cm3. (a) Calculate the percentage uncertainty in the titre. (b) Suggest one change to the method that would reduce this percentage uncertainty without changing the apparatus.
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Water in hydrated magnesium sulfate
2.50 g of hydrated magnesium sulfate, MgSO4·xH2O, is heated to constant mass. 1.22 g of anhydrous MgSO4 remains. Find x. Ar: Mg 24.3, S 32.1, O 16.0, H 1.0.
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Yield and atom economy of an ester
Ethanoic acid reacts with ethanol: CH3COOH + C2H5OH ⇌ CH3COOC2H5 + H2O. 6.00 g of ethanoic acid (M = 60.0) with excess ethanol (M = 46.0) gives 5.28 g of ethyl ethanoate (M = 88.0). Calculate (a) the percentage yield and (b) the atom economy for ethyl ethanoate.
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Back titration of impure limestone
0.500 g of impure calcium carbonate is added to 50.0 cm3 of 0.200 mol dm−3 hydrochloric acid (an excess). When the reaction stops, the leftover acid needs 21.0 cm3 of 0.100 mol dm−3 sodium hydroxide. Calculate the percentage by mass of CaCO3 in the sample. CaCO3 + 2HCl → CaCl2 + CO2 + H2O. M(CaCO3) = 100.1.
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Recall cards
Answer out loud or on paper before you turn the card. Only a complete answer counts as "Got it".
These are 3 of 18 recall cards. Premium members get the full deck for every chapter, ready for each review day in their plan.
Your review days
Count from your Day 0, the day you finish learning this chapter. Your plan schedules these for you.
- +1 Recall cards (these online cards, or the cards in the book)
- +3 Recall cards and Q1 to Q4
- +7 Recall cards and Q5 to Q8
- +14 Recall cards and Q9 to Q12
- +30 Full chapter test, timed
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