A Data Booklet may be used. Molar volume of a gas = 22.7 dm3 mol−1 at s.t.p.
The compound Bi2Sr2Ca2Cu3O10 is a superconductor. In it, the oxidation number of bismuth is +3, of strontium and calcium +2, and of oxygen −2.
What are the possible oxidation numbers of the three copper atoms?
Source: 2025 ACJC J2 Prelim P1 Q8
The table shows the absolute abundances of the isotopes in a sample of naturally occurring zinc.
| isotope | 64Zn | 65Zn | 67Zn | 68Zn |
|---|---|---|---|---|
| absolute abundance | 100 | 80 | 20 | 10 |
What is the relative atomic mass of this sample of zinc?
Source: NJC 2018 Promo, via EJC 2022 Mole Concept Topical Revision, Q2
CommonA disproportionation reaction occurs when an element is oxidised and reduced at the same time. Which equation describes a disproportionation reaction?
Source: IJC 2017 Promo, via EJC 2022 Mole Concept Topical Revision, Q18
FA 1 is a solution of hydrogen peroxide and FA 2 is a standard solution of potassium manganate(VII). A student found that 10 drops of FA 1, when acidified, needed 82 drops of FA 2 for complete reaction. She decided that FA 1 was too concentrated, and needed to be diluted before titration against FA 2.
Source: 2023 RI J1 Promo Section B Q3(a), (b), (d) and (e)
CommonA 3.20 g impure sample of calcium ethanedioate, CaC2O4, is shaken with 100.0 cm3 of 0.10 mol dm−3 acidified MnO4−. One of the products is CO2. The remaining MnO4− is reacted with excess iodide to liberate I2. A 25.0 cm3 aliquot of the resulting solution needs 24.50 cm3 of 0.20 mol dm−3 S2O32−.
2MnO4− + 10I− + 16H+ → 2Mn2+ + 5I2 + 8H2O 2S2O32− + I2 → S4O62− + 2I−
Source: 2025 ASRJC J2 Prelim P2 Q3(e)
The concentration of Al3+ in a water sample can be found by titration with EDTA4−: Al3+(aq) + EDTA4−(aq) → [Al(EDTA)]−(aq). A 10.0 cm3 water sample is made up to 250 cm3 with deionised water. A 25.0 cm3 aliquot needs 35.45 cm3 of 0.0200 mol dm−3 EDTA4−.
Source: 2024 DHS J2 Prelim P2 Q1(a)
CommonN-chlorosuccinimide, C4H4O2NCl, is used in water purification tablets. It dissolves in water to give chloric(I) acid, HClO. The amount in a tablet is found by adding excess acidified KI(aq): HClO is reduced to Cl− by I−, which is oxidised to I2. The I2 is titrated with sodium thiosulfate: I2 + 2S2O32− → 2I− + S4O62−.
Source: A Level 2011 P3 Q1(c), via EJC 2022 JC1 Mole Concept Tutorial, C8
CommonThe percentage of iron in an ore can be found by converting all the iron into Fe2+(aq) and carrying out a redox titration. A sample of iron ore weighing 11.15 g was converted to Fe2+(aq), mixed with dilute sulfuric acid and made up to 250 cm3. 25.0 cm3 portions were titrated with 0.100 mol dm−3 potassium dichromate(VI). The average titre was 21.45 cm3.
Source: A Level 2013 P2 Q2(c), modified, via EJC 2022 JC1 Mole Concept Tutorial, C9
When P2S5 is heated under a vacuum together with caesium sulfide, Cs2S, and sulfur, it produces an ionic compound R, which has the following composition by mass: Cs, 58.1%; P, 6.78%; S, 35.1%.
Source: 2024 CJC J2 Prelim P2 Q2(e)(i) and (ii)
Iron(III) oxide, Fe2O3, can be converted into iron metal by reacting it with carbon monoxide. Carbon dioxide is formed as the only by-product in this reaction.
0.15 kg of Fe2O3 and 60 dm3 of carbon monoxide at room temperature and pressure were used in the experiment. Calculate
Source: EJC 2022 JC1 Mole Concept Tutorial, A10
Source: RJC 2010/11 Moles and Stoichiometry Tutorial, Q15
B+1, +2, +3[1]
Let the sum of the oxidation numbers of the three Cu atoms be x.
2(+3) + 2(+2) + 2(+2) + x + 10(−2) = 0
x = +6. Only B adds up to +6 (A +4, C +4, D +5).
B64.9[1]
Relative atomic mass of Zn = 100 × 64 + 80 × 65 + 20 × 67 + 10 × 68210 = 64.9 (3 s.f.)
A3ClO− → ClO3− + 2Cl−[1]
Volume of FA 2 = 8210 × 25.0 = 205 cm3[1]
205.00 cm3 of FA 2 would be needed from the burette, so the titration would be very tedious, as the burette would have to be refilled carefully several times.[1]
Amount of H2O2 in 25.0 cm3 of FA 3 = 52 × y1000 × 0.0200 = 5.00 × 10−5y mol[1]
Amount of H2O2 in v cm3 of FA 1 = amount in w cm3 of FA 3 = w25.0 × 5.00 × 10−5y = 2.00 × 10−6wy mol
[H2O2] in FA 1 = 2.00 × 10−6wy ÷ v1000 = 0.00200 wyv mol dm−3[1]
Amount of O2 from 1 dm3 = 13.422.7 = 0.5903 mol
Amount of H2O2 = 2 × 0.5903 = 1.181 mol, so [H2O2] = 1.18 mol dm−3[1]
NoteThe volume is at s.t.p., so use 22.7 dm3 mol−1, not 24.0. The mole ratio comes from the decomposition 2H2O2 → 2H2O + O2, not from the equation in (a).
Reduction: MnO4− + 8H+ + 5e− → Mn2+ + 4H2O Oxidation: C2O42− → 2CO2 + 2e−
Overall: 5C2O42− + 2MnO4− + 16H+ → 10CO2 + 2Mn2+ + 8H2O[1]
Amount of S2O32− = 24.501000 × 0.20 = 4.90 × 10−3 mol
Amount of I2 = ½ × 4.90 × 10−3 = 2.45 × 10−3 mol
Amount of remaining MnO4− in 25.0 cm3 = 25 × 2.45 × 10−3 = 9.80 × 10−4 mol[1]
Remaining MnO4− in 100 cm3 = 9.80 × 10−4 × 10025.0 = 0.00392 mol
Initial MnO4− = 100.01000 × 0.10 = 0.0100 mol, so MnO4− reacted with C2O42− = 0.0100 − 0.00392 = 0.00608 mol[1]
Amount of CaC2O4 = 52 × 0.00608 = 0.0152 mol
Mass of pure CaC2O4 = 0.0152 × [40.1 + 2(12.0) + 4(16.0)] = 1.95 g
% purity = 1.953.20 × 100% = 60.9%[1]
No. of moles of Al3+ in 25.0 cm3 = 7.09 × 10−4 mol
No. of moles of Al3+ in 250 cm3 (the whole water sample) = 7.09 × 10−4 × 25025 = 7.09 × 10−3 mol[1]
Amount of S2O32− = 6.01000 × 0.0050 = 3.0 × 10−5 mol
Amount of I2 = amount of HClO = amount of N-chlorosuccinimide = 1.5 × 10−5 mol[1]
Mr(C4H4O2NCl) = 133.5, so mass = 1.5 × 10−5 × 133.5 = 2.0 × 10−3 g[1]
[R]: Cr2O72− + 14H+ + 6e− → 2Cr3+ + 7H2O (× 1) [O]: Fe2+ → Fe3+ + e− (× 6)
Cr2O72− + 6Fe2+ + 14H+ → 2Cr3+ + 6Fe3+ + 7H2O[1]
Amount of Cr2O72− reacted = 21.451000 × 0.100 = 2.145 × 10−3 mol
Since 6Fe2+ ≡ Cr2O72−, amount of Fe2+ in 25.0 cm3 of solution = 6 × 2.145 × 10−3 = 0.01287 mol[1]
Amount of Fe2+ in 250 cm3 of solution = 0.01287 × 25025.0 = 0.129 mol[1]
By the law of conservation of mass, amount of Fe in the iron ore = amount of Fe2+ in 250 cm3 of solution.
Mass of Fe in the iron ore sample = 0.1287 × 55.8 = 7.181 g[1]
% by mass of Fe in the sample = 7.18111.15 × 100% = 64.4%[1]
Let the mass of a sample of R be 100 g.
Amount of Cs = 58.1 ÷ 132.9 = 0.437 mol
Amount of P = 6.78 ÷ 31.0 = 0.219 mol
Amount of S = 35.1 ÷ 32.1 = 1.09 mol
Simplest ratio Cs : P : S = 2 : 1 : 5, so the empirical formula is Cs2PS5[1]
Let the chemical formula be (Cs2PS5)n.
(2(132.9) + 31.0 + 5(32.1))n = 914.6
457.3n = 914.6, so n = 2
The chemical formula of R is Cs4P2S10[1]
Amount of Fe2O3 used = 150 ÷ 159.6 = 0.940 mol
Amount of CO used = 60 ÷ 24.0 = 2.50 mol
Since 1 mol of Fe2O3 reacts with 3 mol of CO, 0.940 mol of Fe2O3 requires 2.82 mol of CO for complete reaction. Hence, CO is the limiting reagent.[1]
Since 3 mol of CO form 2 mol of Fe, the maximum amount of Fe formed = 2.50 × 23 = 1.667 mol
Maximum mass of Fe = 1.667 × 55.8 = 93.0 g[1]
Amount of CO2 formed = amount of CO used = 2.50 mol
Number of CO2 molecules = 2.50 × 6.02 × 1023 = 1.51 × 1024[1]
Only ethanedioic acid reacts with NaOH, and it is dibasic. nNaOH = 14.751000 × 0.100 = 1.475 × 10−3 mol
nH2C2O4 = ½ × 1.475 × 10−3 = 7.375 × 10−4 mol[1]
[H2C2O4] = 7.375 × 10−4 × 100025.0 = 0.0295 mol dm−3[1]
Both the acid and its salt are oxidised by MnO4−: 2MnO4− ≡ 5C2O42−. nMnO4− = 32.001000 × 0.0205 = 6.560 × 10−4 mol
Total ethanedioate = 52 × 6.560 × 10−4 = 1.640 × 10−3 mol[1]
nNa2C2O4 = 1.640 × 10−3 − 7.375 × 10−4 = 9.025 × 10−4 mol, so [Na2C2O4] = 0.0361 mol dm−3[1]