Topic 3
30 questions: 15 multiple choice, 15 structured.
Extended candidates are examined on Core and Supplement content, so keep All.
Limestone is mainly calcium carbonate. Which formula represents calcium carbonate?
Answer: C
Calcium carbonate contains Ca2+ and CO32− ions in a 1:1 ratio, so the formula is CaCO3 (C). A wrongly doubles the calcium; B loses one oxygen from the carbonate ion; D wrongly doubles the carbonate ion and adds an unnecessary bracket.
What does the molecular formula of a compound show?
Answer: A
The molecular formula shows the number and type of every atom in one molecule (A). B describes the empirical formula; C describes an ionic formula (formula unit with ionic charges, not atoms in a molecule); D describes percentage composition by mass.
Magnesium ribbon burns in air to form magnesium oxide. Which symbol equation, including state symbols, correctly represents this reaction?
Answer: D
2Mg(s) + O2(g) → 2MgO(s) has 2 Mg and 2 O on each side, so it is balanced with correct state symbols (D). A gives the wrong formula for magnesium oxide (MgO2); B uses atomic oxygen, O(g), instead of the diatomic molecule O2(g); C is unbalanced (2 Mg atoms but only 1 MgO formed).
A compound has the molecular formula C4H10. What is its empirical formula?
Answer: B
The ratio C:H in C4H10 is 4:10, which simplifies (divide by 2) to 2:5, giving C2H5 (B). A is the molecular formula, not simplified to the smallest ratio; C (CH2, ratio 1:2) does not match 4:10; D (C2H10) only halves the carbon and leaves the hydrogen unchanged, which is not a valid simplification.
Aluminium ions, Al3+, and sulfate ions, SO42−, combine to form an ionic compound. What is the formula of this compound?
Answer: A
Charges must balance: 2 × (+3) from two Al3+ ions cancels 3 × (−2) from three SO42− ions, giving Al2(SO4)3 (A). B and C ignore charge balancing altogether. D swaps the subscripts incorrectly: 3 × (+3) + 2 × (−2) = +5, which does not balance to zero.
Aqueous silver nitrate is added to aqueous sodium chloride, forming a white precipitate. Which ionic equation, with state symbols, correctly represents this reaction?
Answer: C
The ionic equation shows only the species that actually change: Ag+(aq) + Cl−(aq) → AgCl(s) (C). A is the full (non-ionic) equation, which still includes the spectator ions Na+ and NO3−, so it is not an ionic equation. B shows silver and nitrate ions forming a precipitate, but silver nitrate is soluble and does not precipitate. D has the wrong formula and unbalanced charge (left-hand charge is +1, but the compound Ag2Cl is not a real, correctly charge-balanced formula).
Sodium reacts with water to produce sodium hydroxide and hydrogen gas.
Give the formula of each of the following:
sodium hydroxide
hydrogen gas
Write a balanced symbol equation for this reaction, including state symbols.
Sodium hydroxide is made of Na+ and OH− in a 1:1 ratio, so its formula is NaOH. Hydrogen gas exists as diatomic molecules, H2. For the equation, count atoms on each side: 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g) gives 2 Na, 4 H and 2 O on the left, and 2 Na, (2 + 2) = 4 H and 2 O on the right, so it is balanced. State symbols show sodium as a solid, water as a liquid, sodium hydroxide dissolved in the water as aqueous, and hydrogen escaping as a gas.
The table shows the number of each type of atom present in one molecule of three compounds, P, Q and R.
| Compound | Carbon atoms | Hydrogen atoms | Oxygen atoms |
|---|---|---|---|
| P | 1 | 4 | 0 |
| Q | 2 | 6 | 1 |
| R | 1 | 2 | 1 |
Deduce the molecular formula of compound P.
Deduce the molecular formula of compound Q.
Deduce the molecular formula of compound R.
State the meaning of the term molecular formula.
The molecular formula simply lists how many atoms of each element are in one molecule, in the order given in the table. P has 1 carbon and 4 hydrogen atoms and no oxygen, so its formula is CH4. Q has 2 carbon, 6 hydrogen and 1 oxygen atom, giving C2H6O. R has 1 carbon, 2 hydrogen and 1 oxygen atom, giving CH2O. This is exactly what "molecular formula" means: the actual number and type of atoms present in one molecule, as opposed to the simplest ratio (the empirical formula).
Calcium carbonate is a solid. When it is heated strongly in an open container, it decomposes into calcium oxide, which is also a solid, and carbon dioxide gas.
Write a word equation for this reaction.
Write a balanced symbol equation for this reaction, including state symbols.
Explain, using the equation, why the total mass of solid remaining decreases when calcium carbonate is heated in an open container.
Heating calcium carbonate breaks it down into calcium oxide and carbon dioxide: calcium carbonate → calcium oxide + carbon dioxide. As a symbol equation, CaCO3(s) → CaO(s) + CO2(g); checking atoms, there is 1 Ca, 1 C and 3 O on each side, so it is balanced. Because the container is open, the carbon dioxide gas produced is free to escape into the surrounding air. Only the solid calcium oxide remains in the container, and since some of the original mass has left as gas, the mass of solid remaining is less than the starting mass of calcium carbonate.
A hydrocarbon molecule contains 6 carbon atoms and 12 hydrogen atoms.
State the meaning of the term empirical formula.
Give the molecular formula of this hydrocarbon.
Deduce the empirical formula of this hydrocarbon.
A different hydrocarbon has the empirical formula CH2 and its molecule contains 4 carbon atoms. Deduce its molecular formula.
6 carbon atoms and 12 hydrogen atoms in one molecule give the molecular formula C6H12. The empirical formula is the simplest whole-number ratio of these atoms: dividing both numbers by their highest common factor, 6, gives a ratio of 1:2, so the empirical formula is CH2. For the second hydrocarbon, the empirical formula CH2 means the ratio of C:H is always 1:2. If the molecule contains 4 carbon atoms, multiply every number in the empirical formula by 4: C(1×4)H(2×4) = C4H8.
The table gives the formula and charge of four ions.
| Ion | Formula | Charge |
|---|---|---|
| Barium | Ba | 2+ |
| Sulfate | SO4 | 2− |
| Chloride | Cl | 1− |
| Sodium | Na | 1+ |
Deduce the formula of barium sulfate.
Deduce the formula of sodium chloride.
Aqueous barium chloride is mixed with aqueous sodium sulfate. A white precipitate of barium sulfate forms, leaving sodium chloride in solution. Write a balanced symbol equation for this reaction, including state symbols.
Ba2+ (charge 2+) and SO42− (charge 2−) combine 1:1, so barium sulfate is BaSO4. Na+ (charge 1+) and Cl− (charge 1−) combine 1:1, so sodium chloride is NaCl. For the reaction, barium ions from barium chloride swap partners with sulfate ions from sodium sulfate: BaCl2(aq) + Na2SO4(aq) → BaSO4(s) + 2NaCl(aq). Checking atoms: left-hand side has 1 Ba, 2 Cl, 2 Na, 1 S, 4 O; right-hand side has 1 Ba, 1 S, 4 O (in BaSO4) plus 2 Na, 2 Cl (in 2NaCl) — balanced. The precipitate BaSO4 is shown as a solid because it is insoluble, while the other species remain dissolved (aqueous).
Dilute hydrochloric acid reacts with aqueous sodium hydroxide in a neutralisation reaction, producing water and aqueous sodium chloride.
Write a full balanced symbol equation for this reaction, including state symbols.
Identify the two spectator ions in this reaction.
Write the ionic equation for this neutralisation reaction, including state symbols.
The full equation is HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l); it is already balanced, with 1 of each atom type on both sides. Sodium ions, Na+, and chloride ions, Cl−, appear unchanged on both sides of the equation (they start dissolved and end up dissolved, in the same form), so they are the spectator ions. Removing them from the full ionic version of the equation leaves only the ions that actually react: H+(aq) + OH−(aq) → H2O(l). This shows the essential chemistry of any strong acid–strong alkali neutralisation: hydrogen ions combine with hydroxide ions to form water.
Which statement correctly defines relative atomic mass, Ar?
Answer: B
Relative atomic mass compares the average mass of an element's isotopes to 1/12 of the mass of a carbon-12 atom (B). A confuses Ar with the actual mass of an atom in grams. C describes the mass number, not relative atomic mass. D describes the molar mass (mass of one mole), which has units of g/mol, whereas Ar has no units.
Using Ar: C = 12, O = 16, Ca = 40, what is the relative formula mass, Mr, of calcium carbonate, CaCO3?
Answer: D
Mr(CaCO3) = 40 + 12 + (16 × 3) = 40 + 12 + 48 = 100 (D). A (68) forgets to multiply oxygen by 3, using only one oxygen atom. B (84) uses only two oxygen atoms instead of three. C (116) uses four oxygen atoms instead of three.
Ar: H = 1, C = 12, N = 14, O = 16.
Calculate the relative molecular mass, Mr, of carbon dioxide, CO2.
Calculate the relative molecular mass, Mr, of glucose, C6H12O6.
Calculate the relative formula mass, Mr, of ammonium nitrate, NH4NO3.
For CO2: Mr = 12 + (16 × 2) = 12 + 32 = 44. For glucose, C6H12O6: Mr = (6 × 12) + (12 × 1) + (6 × 16) = 72 + 12 + 96 = 180. For ammonium nitrate, NH4NO3, count every atom in the formula: 2 nitrogen, 4 hydrogen, 3 oxygen. Mr = (2 × 14) + (4 × 1) + (3 × 16) = 28 + 4 + 48 = 80.
In the reaction 2Mg + O2 → 2MgO, 24 g of magnesium reacts completely with oxygen to form 40 g of magnesium oxide (relative formula mass of MgO = 40).
Calculate the mass of oxygen that reacts with 24 g of magnesium.
Calculate the mass of magnesium oxide formed when 9.6 g of magnesium is completely oxidised, using simple proportion.
Calculate the mass of oxygen needed to react completely with 9.6 g of magnesium.
By conservation of mass, the mass of oxygen that reacts equals the mass of magnesium oxide formed minus the mass of magnesium used: 40 − 24 = 16 g. To scale to 9.6 g of magnesium, find the scale factor: 9.6 ÷ 24 = 0.4. Multiply every mass in the original ratio by this factor: mass of MgO = 0.4 × 40 = 16.0 g, and mass of oxygen = 0.4 × 16 = 6.4 g. As a check, 9.6 + 6.4 = 16.0 g, which matches the mass of magnesium oxide calculated — mass is conserved.
Copper carbonate decomposes on heating: CuCO3 → CuO + CO2
The relative formula masses are: CuCO3 = 124, CuO = 80, CO2 = 44.
Show that these three relative formula masses are consistent with conservation of mass in this equation.
3.1 g of copper carbonate is heated until it decomposes completely. Calculate the mass of copper oxide formed.
Calculate the mass of carbon dioxide gas released in part (b).
The equation CuCO3 → CuO + CO2 must conserve mass, and indeed 80 + 44 = 124, matching Mr(CuCO3). The scale factor for 3.1 g of copper carbonate compared with the 124 g used to define the relative formula masses is 3.1 ÷ 124 = 0.025. Multiplying each relative formula mass by this factor gives the actual masses: mass of CuO = 0.025 × 80 = 2.0 g, and mass of CO2 = 0.025 × 44 = 1.1 g. Checking: 2.0 + 1.1 = 3.1 g, which equals the starting mass of copper carbonate, confirming the answer.
Which pair of units can both be used to express the concentration of a solution?
Answer: A
Concentration can be expressed either as a mass concentration, g/dm3, or as a molar concentration, mol/dm3 (A). B mixes a density-style unit (g/cm3) with an inverted molar-mass-style unit (mol/g), neither of which is a concentration unit used for solutions here. C inverts mol/dm3 to give dm3/mol, which is not a concentration unit. D (g/mol) is the unit of molar mass, not concentration.
What is the value of the Avogadro constant, and what does it represent?
Answer: C
The Avogadro constant is 6.02 × 10^23, and it gives the number of particles (atoms, ions or molecules) in one mole of a substance (C). A reverses the relationship between moles and particles. B has the correct idea but the wrong sign on the exponent, making the number extremely small rather than extremely large. D confuses the Avogadro constant with molar mass, which is measured in grams per mole, not as a pure number of particles.
Ar: Na = 23. What amount, in moles, is present in 11.5 g of sodium?
Answer: A
Amount (mol) = mass ÷ Ar = 11.5 ÷ 23 = 0.5 mol (A). B (0.05 mol) comes from dividing by 230 instead of 23, a decimal-place slip. C (2.0 mol) inverts the calculation (23 ÷ 11.5). D (5.0 mol) is a further decimal-place slip in the other direction.
The molar gas volume is 24 dm3 at room temperature and pressure (r.t.p.). What is the volume occupied by 0.25 mol of a gas at r.t.p.?
Answer: D
Volume = amount (mol) × molar gas volume = 0.25 × 24 = 6 dm3 (D). A simply restates the molar gas volume without multiplying by the amount in moles. B (96 dm3) comes from dividing 24 by 0.25 instead of multiplying. C (0.25 dm3) mistakenly restates the amount in moles as if it were already the volume.
0.25 mol of solute is dissolved in 250 cm3 of solution. What is the concentration of this solution, in mol/dm3?
Answer: B
Converting 250 cm3 to dm3 gives 0.250 dm3, so concentration = 0.25 mol ÷ 0.250 dm3 = 1.0 mol/dm3 (B). A (0.1 mol/dm3) comes from using an incorrect conversion factor of 100 instead of 1000 when converting cm3 to dm3 (250 cm3 wrongly treated as 2.5 dm3). C (4.0 mol/dm3) comes from inverting the calculation (dividing volume by amount instead of amount by volume). D (1000 mol/dm3) comes from not converting cm3 to dm3 at all and dividing the wrong way round.
A compound contains 82.4% nitrogen and 17.6% hydrogen by mass. Ar: N = 14, H = 1. What is the empirical formula of this compound?
Answer: C
Per 100 g: moles of N = 82.4 ÷ 14 = 5.89, moles of H = 17.6 ÷ 1 = 17.6. Dividing both by the smaller value (5.89) gives a ratio of N:H = 1:3, so the empirical formula is NH3 (C). A (NH) and B (NH2) do not use the correct mole ratio. D (N3H) has the ratio the wrong way round.
A student expected to obtain 12.0 g of a salt from a reaction but actually obtained 9.6 g. What is the percentage yield?
Answer: B
Percentage yield = (actual yield ÷ expected yield) × 100 = (9.6 ÷ 12.0) × 100 = 80% (B). A (20%) is the percentage "lost" (100 − 80), not the percentage yield itself. C (96%) does not match either value given. D (125%) comes from inverting the fraction (12.0 ÷ 9.6 × 100).
Ar: Na = 23, O = 16, H = 1.
A solution of sodium hydroxide has a concentration of 8.0 g/dm3.
Calculate the relative formula mass, Mr, of sodium hydroxide, NaOH.
Calculate the concentration of this solution in mol/dm3.
Calculate the amount, in moles, of sodium hydroxide present in 250 cm3 of this solution.
Mr(NaOH) = 23 + 16 + 1 = 40. To convert a mass concentration to a molar concentration, divide by the molar mass: concentration in mol/dm3 = concentration in g/dm3 ÷ Mr = 8.0 ÷ 40 = 0.20 mol/dm3. To find the amount of solute in 250 cm3, first convert the volume to dm3: 250 cm3 = 0.250 dm3. Then use amount (mol) = concentration (mol/dm3) × volume (dm3) = 0.20 × 0.250 = 0.05 mol.
Ar: Mg = 24.
Calculate the amount, in moles, of magnesium atoms in 3.0 g of magnesium.
Calculate the number of magnesium atoms in this 3.0 g sample.
Calculate the mass of 0.60 mol of magnesium atoms.
Amount (mol) = mass ÷ Ar = 3.0 ÷ 24 = 0.125 mol. To find the number of atoms, multiply the amount in moles by the Avogadro constant: number of atoms = 0.125 × 6.02 × 10^23 = 7.525 × 10^22, which rounds to 7.5 × 10^22 atoms. For part (c), rearrange the same relationship the other way: mass = amount × Ar = 0.60 × 24 = 14.4 g.
Ar: Mg = 24, H = 1, Cl = 35.5.
Magnesium reacts with hydrochloric acid: Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)
2.4 g of magnesium is added to 120 cm3 of hydrochloric acid of concentration 2.0 mol/dm3.
Calculate the amount, in moles, of magnesium used.
Calculate the amount, in moles, of hydrochloric acid used.
Determine which reactant is in excess. Show your working.
Calculate the maximum mass of magnesium chloride, MgCl2, that could form.
Amount of Mg = mass ÷ Ar = 2.4 ÷ 24 = 0.10 mol. Amount of HCl = concentration × volume (in dm3) = 2.0 × (120 ÷ 1000) = 2.0 × 0.120 = 0.24 mol. The equation shows that 1 mol Mg reacts with 2 mol HCl, so 0.10 mol Mg needs 0.10 × 2 = 0.20 mol HCl. Only 0.20 mol HCl is needed, but 0.24 mol is available, so hydrochloric acid is in excess and magnesium is the limiting reactant. Because magnesium is used up completely, it controls how much product forms: the equation shows a 1:1 ratio between Mg and MgCl2, so 0.10 mol MgCl2 forms. Mr(MgCl2) = 24 + (35.5 × 2) = 24 + 71 = 95, so mass of MgCl2 = 0.10 × 95 = 9.5 g.
Ar: Na = 23, C = 12, O = 16.
Excess dilute sulfuric acid reacts with 5.3 g of sodium carbonate:
Na2CO3(s) + H2SO4(aq) → Na2SO4(aq) + H2O(l) + CO2(g)
Calculate the amount, in moles, of sodium carbonate reacted.
Calculate the amount, in moles, of carbon dioxide gas produced.
Calculate the volume of carbon dioxide gas produced, measured at r.t.p.
Mr(Na2CO3) = (2 × 23) + 12 + (3 × 16) = 46 + 12 + 48 = 106. Amount of Na2CO3 = mass ÷ Mr = 5.3 ÷ 106 = 0.05 mol. The equation shows a 1:1 ratio between Na2CO3 and CO2, so the amount of CO2 produced is also 0.05 mol. Since sulfuric acid is in excess, all of the sodium carbonate reacts, so this figure is not limited by the acid. Using the molar gas volume, volume of CO2 at r.t.p. = amount × 24 = 0.05 × 24 = 1.2 dm3 (or 1200 cm3).
In a titration, 25.0 cm3 of a sodium hydroxide solution of unknown concentration was exactly neutralised by 22.5 cm3 of hydrochloric acid of concentration 0.100 mol/dm3.
NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)
Calculate the amount, in moles, of hydrochloric acid used.
Calculate the amount, in moles, of sodium hydroxide that reacted.
Calculate the concentration of the sodium hydroxide solution, in mol/dm3.
Ar: Na = 23, O = 16, H = 1. Calculate the concentration of the sodium hydroxide solution in g/dm3.
Amount of HCl = concentration × volume (in dm3) = 0.100 × (22.5 ÷ 1000) = 0.100 × 0.0225 = 0.00225 mol. The equation shows NaOH and HCl react in a 1:1 ratio, so the amount of NaOH that reacted is also 0.00225 mol. This amount of NaOH was in 25.0 cm3 = 0.0250 dm3 of solution, so concentration = amount ÷ volume = 0.00225 ÷ 0.0250 = 0.09 mol/dm3. To convert to g/dm3, multiply by the molar mass: Mr(NaOH) = 23 + 16 + 1 = 40, so concentration = 0.09 × 40 = 3.6 g/dm3.
A hydrocarbon contains 85.7% carbon and 14.3% hydrogen by mass. Its relative molecular mass, Mr, is 42.
Ar: C = 12, H = 1.
Calculate the empirical formula of this hydrocarbon. Show your working.
Calculate the molecular formula of this hydrocarbon.
To find the empirical formula, convert each percentage to an amount in moles using the Ar values, treating the percentages as masses in 100 g of compound: moles of C = 85.7 ÷ 12 = 7.14; moles of H = 14.3 ÷ 1 = 14.3. Dividing both by the smaller number, 7.14, gives a ratio of C : H = 1 : 2.00, so the empirical formula is CH2. The mass of this empirical formula unit is 12 + (2 × 1) = 14. Since the relative molecular mass of the actual compound is 42, the molecular formula must be a whole-number multiple of the empirical formula: 42 ÷ 14 = 3. Multiplying every number in CH2 by 3 gives the molecular formula, C3H6.