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DSE Chemistry Recap – Fundamental Chemistry | Tia (2026)

Fundamental Chemistry

DSE Chemistry

Tia (2026)

For use on Sunday, 6 September 2026

01

Calcium Carbonate Chemistry

Limestone, Chalk and Marble

Limestone, chalk and marble are naturally occurring materials containing calcium carbonate. A 25.0 g sample of impure limestone is heated strongly until its mass becomes constant. The solid residue has a mass of 15.2 g. Assume that calcium carbonate is the only component of the sample that decomposes on heating.

  1. Limestone, chalk and marble have different appearances and physical properties even though their main chemical component is the same. State the chemical formula of their main component and explain why the three materials may have different physical properties.
  2. Write a chemical equation, including state symbols, for the thermal decomposition of calcium carbonate.
  3. State one observable change when calcium carbonate is heated strongly.
  4. [Deleted] Calculate:
    1. the mass of carbon dioxide released;
    2. the number of moles of carbon dioxide released;
    3. the mass of calcium carbonate originally present;
    4. the percentage by mass of calcium carbonate in the limestone sample.
    Give the percentage to three significant figures.
    [Relative atomic masses: C = 12.0, O = 16.0, Ca = 40.0]
  5. Water is added carefully to the solid calcium oxide formed. State the observation and write the chemical equation for the reaction.
  6. Some of the resulting mixture is filtered. Carbon dioxide is passed into the clear filtrate for a short time.
    1. State the observation.
    2. Write an ionic equation for the reaction producing this observation.
  7. Carbon dioxide is then passed into the mixture for a prolonged period. The mixture eventually becomes clear. Explain this observation and write a chemical equation for the change.
  8. A separate sample of marble chips is added to dilute hydrochloric acid.
    1. State two observations.
    2. Write the chemical equation for the reaction.
    3. Describe a chemical test for the gas produced and state the positive result.
22 marks
02

Separation and Purification

Obtaining Sodium Chloride from Muddy Seawater

A sample of muddy seawater contains insoluble mud, sodium chloride and small amounts of other soluble salts. A student is required to obtain a sample of dry sodium chloride crystals of the highest practicable purity using common laboratory apparatus.

  1. Describe, in the correct sequence, a laboratory procedure for obtaining dry sodium chloride crystals from the muddy seawater. Your answer should include the purpose of each major step.
  2. Name the principal separation technique used to remove the mud. Identify the residue and the filtrate at this stage.
  3. Before crystallisation, the filtrate is heated gently until it becomes nearly saturated. Explain how the student can determine when the solution is sufficiently concentrated without evaporating it completely.
  4. Explain why the concentrated solution should be allowed to cool slowly and remain undisturbed.
  5. The crystals are separated from the remaining solution by filtration. State the name given to the solution that passes through the filter paper.
  6. The crystals are washed with a small amount of ice-cold distilled water rather than a large amount of warm water. Explain both parts of this instruction.
  7. Explain why heating the original seawater to complete dryness is not the best method of obtaining pure sodium chloride.
  8. A student claims:
    “Crystallisation must produce chemically pure sodium chloride because only sodium chloride crystals are visible.”
    Evaluate this claim and suggest one way to improve the purity of the product.
  9. State one suitable method for drying the crystals without causing significant loss or contamination of the product.
17 marks
03

Chemical Language

Common Cations, Anions and Their Colours

A strong command of the names, formulae, charges and colours of common ions is essential in DSE Chemistry. Unless otherwise stated, the colours in this question refer to ions in aqueous solution.

Ion Name of ion Ion Name of ion
H+ 1 OH− 9
NH4+ 2 NO3− 10
Mg2+ 3 CO32− 11
Al3+ 4 HCO3− 12
Cu2+ 5 SO42− 13
Fe2+ 6 SO32− 14
Fe3+ 7 MnO4− 15
Cr3+ 8 Cr2O72− 16
  1. Give the systematic name of each ion numbered 1–16 in the table.
  2. Write the formula, including the charge, of each of the following ions:
    1. sodium ion;
    2. potassium ion;
    3. calcium ion;
    4. barium ion;
    5. zinc ion;
    6. silver ion;
    7. lead(II) ion;
    8. chloride ion;
    9. bromide ion;
    10. iodide ion;
    11. chromate ion.
  3. State the colour of each of the following ions in aqueous solution:
    1. Cu2+;
    2. Fe2+;
    3. Fe3+;
    4. Cr3+;
    5. MnO4−;
    6. CrO42−;
    7. Cr2O72−.
  4. State the usual colour in aqueous solution of the other common ions listed in parts (a) and (b).
  5. A solution of copper(II) chloride is blue, while a solution of sodium chloride is colourless. Deduce which ion is responsible for the blue colour and explain your reasoning.
  6. Solution X is orange. It contains potassium ions and one type of anion only.
    1. Identify the coloured anion.
    2. Give the formula and name of compound X.
  7. Write the correct chemical formula for each compound:
    1. aluminium sulphate;
    2. ammonium carbonate;
    3. calcium hydrogencarbonate;
    4. iron(III) nitrate;
    5. copper(II) hydroxide;
    6. potassium permanganate;
    7. lead(II) iodide;
    8. chromium(III) sulphate.
  8. Give the correct name of each compound:
    1. FeCl2;
    2. FeCl3;
    3. Cu(NO3)2;
    4. (NH4)2SO3;
    5. K2CrO4;
    6. NaHCO3.
38 marks
04

Bonding and Structure

Formation of an Ionic Bond

Magnesium reacts with nitrogen to form magnesium nitride, an ionic compound. A magnesium atom has the electronic arrangement 2,8,2 and a nitrogen atom has the electronic arrangement 2,5.

  1. Explain why a magnesium atom tends to lose electrons, while a nitrogen atom tends to gain electrons, when magnesium nitride is formed.
  2. State the formula and electronic arrangement of:
    1. the magnesium ion formed;
    2. the nitride ion formed.
  3. Write separate equations to represent:
    1. the formation of a magnesium ion from a magnesium atom;
    2. the formation of a nitride ion from a nitrogen atom.
  4. Determine the number of magnesium atoms and nitrogen atoms needed so that the number of electrons lost equals the number of electrons gained. Hence, deduce the formula of magnesium nitride.
  5. Draw a complete electron diagram for the formation of magnesium nitride. Your diagram should:
    • show all outer-shell electrons;
    • distinguish electrons originally from magnesium from those originally from nitrogen;
    • show brackets and ionic charges;
    • show the correct ratio of ions.
  6. Explain precisely what is meant by an ionic bond in magnesium nitride.
  7. A student states:
    “One magnesium ion is joined to one nitride ion to form a magnesium nitride molecule.”
    Identify two errors in this statement and correct them.
  8. Explain, in terms of structure and bonding, why solid magnesium nitride:
    1. has a high melting point;
    2. does not conduct electricity;
    3. can conduct electricity when molten.
  9. Aluminium oxide is also an ionic compound. Aluminium forms Al3+ ions and oxygen forms O2− ions. Deduce the formula of aluminium oxide and explain how electrical neutrality is achieved.
24 marks

Question 1

  1. The main component is calcium carbonate, CaCO3. Limestone, chalk and marble may have different physical properties because they can have different crystal structures, particle sizes, textures and amounts or types of impurities. They are different natural forms of a material whose principal chemical component is calcium carbonate.
  2. CaCO3(s) &xrightarrow;Δ CaO(s) + CO2(g)
  3. Acceptable observation: a colourless gas is released and the mass of the solid decreases. Both calcium carbonate and calcium oxide are white solids, so no major colour change is expected.
    1. Mass of CO2 = 25.0 − 15.2
      = 9.80 g
    2. Molar mass of CO2 = 12.0 + 2(16.0) = 44.0 g mol−1.
      Number of moles of CO2 = 9.80 / 44.0
      = 0.2227 mol
    3. From the equation, the mole ratio CaCO3 : CO2 is 1 : 1.
      Number of moles of CaCO3 = 0.2227 mol

      Molar mass of CaCO3 = 40.0 + 12.0 + 3(16.0) = 100.0 g mol−1

      Mass of CaCO3 = 0.2227 × 100.0
      = 22.27 g
    4. Percentage by mass = (22.27 / 25.0) × 100%
      = 89.1%
  5. Water reacts vigorously with calcium oxide. The solid may crack or crumble, steam may be produced, and the mixture becomes hot because the reaction is exothermic.
    CaO(s) + H2O(l) → Ca(OH)2(aq)
    A suspension may form because calcium hydroxide is only slightly soluble in water.
    1. The clear filtrate is limewater. When carbon dioxide is passed through it for a short time, the limewater turns milky because a white precipitate forms.
    2. Ca2+(aq) + CO32−(aq) → CaCO3(s)
  7. In excess carbon dioxide, the insoluble calcium carbonate reacts with carbon dioxide and water to form soluble calcium hydrogencarbonate. The white precipitate therefore dissolves and the mixture becomes clear.
    CaCO3(s) + CO2(g) + H2O(l) → Ca(HCO3)2(aq)
    1. The marble chips become smaller or dissolve, and effervescence occurs.
    2. CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)
    3. Pass the gas through limewater. A positive result is that the limewater turns milky.
Calcium carbonate cycle: CaCO3 can be thermally decomposed into CaO. Calcium oxide reacts with water to produce Ca(OH)2. Carbon dioxide converts calcium hydroxide back into calcium carbonate.

Question 2

  1. A suitable procedure is:
    1. Allow large mud particles to settle if necessary.
    2. Filter the muddy seawater to remove insoluble mud.
    3. Collect the clear filtrate containing dissolved sodium chloride and other soluble salts.
    4. Heat the filtrate gently in an evaporating basin to evaporate some water and produce a hot, nearly saturated solution.
    5. Stop heating and allow the concentrated solution to cool slowly and remain undisturbed so that sodium chloride crystals form.
    6. Filter the mixture to separate the crystals from the mother liquor.
    7. Wash the crystals with a small amount of ice-cold distilled water.
    8. Dry the crystals between sheets of filter paper or in a warm, dry place.
    9. If higher purity is required, recrystallise the product.
  2. The technique is filtration.
    • Residue: insoluble mud.
    • Filtrate: seawater containing dissolved salts.
  3. Place a drop of the hot solution on a cool glass rod or watch glass. If crystals form as the drop cools, the solution is sufficiently concentrated. Alternatively, heating may be stopped when crystals begin to form at the edge or surface of the solution.
  4. Slow cooling allows ions to arrange themselves into a more regular crystal lattice and generally produces larger, better-formed crystals. Leaving the solution undisturbed reduces the formation of many small crystals and lowers the chance of trapping mother liquor and impurities inside the crystals.
  5. The solution passing through the filter paper after crystallisation is called the mother liquor.
  6. Only a small amount of water is used because sodium chloride is soluble in water. A large amount would dissolve and remove a significant amount of product. Ice-cold water is used to minimise the solubility of sodium chloride and hence reduce product loss. The water removes mother liquor and soluble impurities from the crystal surfaces.
  7. Heating to complete dryness would leave sodium chloride together with the other dissolved salts. These impurities would remain mixed with the product. It may also cause violent splashing near dryness, leading to loss of product. Crystallisation allows some dissolved impurities to remain in the mother liquor.
  8. The claim is not necessarily correct. Visible sodium chloride crystals may contain mother liquor on their surfaces or trapped within them. Other dissolved salts may also crystallise with the sodium chloride. Purity can be improved by:
    • washing the crystals with a small amount of ice-cold distilled water;
    • recrystallising the product;
    • rejecting crystals obtained during the very late stage of evaporation, when other salts may crystallise.
  9. Suitable methods include:
    • pressing the crystals gently between dry filter papers;
    • leaving them in a warm, dry place;
    • using a desiccator.
Important limitation: Real seawater contains several dissolved salts. Simple crystallisation can produce sodium chloride of high practicable purity, but it does not automatically guarantee an analytically pure product.

Question 3

  1. No. Ion Name No. Ion Name
    1 H+ Hydrogen ion 9 OH− Hydroxide ion
    2 NH4+ Ammonium ion 10 NO3− Nitrate ion
    3 Mg2+ Magnesium ion 11 CO32− Carbonate ion
    4 Al3+ Aluminium ion 12 HCO3− Hydrogencarbonate ion
    5 Cu2+ Copper(II) ion 13 SO42− Sulphate ion
    6 Fe2+ Iron(II) ion 14 SO32− Sulphite ion
    7 Fe3+ Iron(III) ion 15 MnO4− Permanganate ion
    8 Cr3+ Chromium(III) ion 16 Cr2O72− Dichromate ion
    1. Sodium ion: Na+
    2. Potassium ion: K+
    3. Calcium ion: Ca2+
    4. Barium ion: Ba2+
    5. Zinc ion: Zn2+
    6. Silver ion: Ag+
    7. Lead(II) ion: Pb2+
    8. Chloride ion: Cl−
    9. Bromide ion: Br−
    10. Iodide ion: I−
    11. Chromate ion: CrO42−
    1. Cu2+(aq): blue
    2. Fe2+(aq): pale green
    3. Fe3+(aq): yellow or yellow-brown
    4. Cr3+(aq): green
    5. MnO4−(aq): purple
    6. CrO42−(aq): yellow
    7. Cr2O72−(aq): orange
  4. The other common ions listed are normally treated as colourless in aqueous solution. A compound may nevertheless form a coloured solid or precipitate even if its separate aqueous ions are colourless.
  5. The ion responsible for the blue colour is Cu2+(aq). Sodium chloride solution contains Na+(aq) and Cl−(aq) and is colourless. Chloride ions are also present in copper(II) chloride solution. Since chloride ions do not make sodium chloride solution coloured, the blue colour is attributed to copper(II) ions.
    1. The orange anion is the dichromate ion, Cr2O72−.
    2. Two K+ ions are required to balance one Cr2O72− ion.
      K2Cr2O7
      Potassium dichromate
    1. Aluminium sulphate: Al2(SO4)3
    2. Ammonium carbonate: (NH4)2CO3
    3. Calcium hydrogencarbonate: Ca(HCO3)2
    4. Iron(III) nitrate: Fe(NO3)3
    5. Copper(II) hydroxide: Cu(OH)2
    6. Potassium permanganate: KMnO4
    7. Lead(II) iodide: PbI2
    8. Chromium(III) sulphate: Cr2(SO4)3
    1. FeCl2: iron(II) chloride
    2. FeCl3: iron(III) chloride
    3. Cu(NO3)2: copper(II) nitrate
    4. (NH4)2SO3: ammonium sulphite
    5. K2CrO4: potassium chromate
    6. NaHCO3: sodium hydrogencarbonate
Formula rule: The total positive charge and total negative charge in an ionic compound must be equal. Brackets are required when more than one polyatomic ion is present.

Question 4

  1. A magnesium atom has two outer-shell electrons. It can achieve a stable electronic arrangement by losing these two electrons. A nitrogen atom has five outer-shell electrons. It can achieve a stable electronic arrangement by gaining three electrons. Electron transfer therefore produces oppositely charged ions with stable outer electron shells.
    1. Magnesium ion:
      Mg2+, electronic arrangement 2,8
    2. Nitride ion:
      N3−, electronic arrangement 2,8
    1. Mg → Mg2+ + 2e−
    2. N + 3e− → N3−
  4. Each magnesium atom loses two electrons, while each nitrogen atom gains three electrons. The lowest common multiple of 2 and 3 is 6.
    • Three Mg atoms lose a total of six electrons.
    • Two N atoms gain a total of six electrons.
    3Mg → 3Mg2+ + 6e−
    2N + 6e− → 2N3−

    Formula of magnesium nitride = Mg3N2
  5. A correct electron diagram should show:
    3[Mg]2+     2[N]3−
    Each Mg2+ ion should be enclosed in brackets with no electrons shown in its original outer shell. Each N3− ion should be enclosed in brackets with eight outer-shell electrons. For each nitride ion, five outer-shell electrons should be shown as originating from nitrogen and three should be shown as transferred from magnesium. The complete diagram must contain three Mg2+ ions and two N3− ions.
  6. An ionic bond is the strong electrostatic attraction between oppositely charged ions. In magnesium nitride, it is the electrostatic attraction between Mg2+ ions and N3− ions throughout the giant ionic lattice.
  7. The statement contains two errors:
    • The ratio is not one magnesium ion to one nitride ion. The correct ratio is 3 Mg2+ : 2 N3−.
    • Magnesium nitride does not consist of separate molecules. It has a giant ionic lattice.
    A correct statement is:
    Magnesium nitride consists of Mg2+ and N3− ions arranged in a giant ionic lattice in the ratio 3 : 2.
    1. Magnesium nitride has a high melting point because there are strong electrostatic attractions between oppositely charged ions throughout the giant ionic lattice. A large amount of energy is required to overcome these attractions.
    2. In solid magnesium nitride, the ions are held in fixed positions in the lattice. There are no mobile charge carriers, so the solid does not conduct electricity.
    3. When magnesium nitride is molten, the ionic lattice is broken down and the ions are free to move. The mobile Mg2+ and N3− ions can carry electric charge, so the molten compound conducts electricity.
  9. The lowest total charge that can be produced by both 3+ and 2− ions is six.
    • Two Al3+ ions give a total charge of +6.
    • Three O2− ions give a total charge of −6.
    Total charge = 2(+3) + 3(−2) = 0

    Formula of aluminium oxide = Al2O3
Do not confuse the ideas: Electron transfer explains the formation of ions. The ionic bond itself is the electrostatic attraction between the oppositely charged ions after they have formed.
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