Think It Over, Activities & Pause-and-Ponder
Opening questions about water, oxygen and conductivity.
(a) Are all water samples chemically identical? Yes. After purification, water from any source always contains hydrogen and oxygen in the same fixed mass ratio of 1 : 8 (Law of Constant Proportions). So pure water is chemically identical everywhere; only the dissolved impurities differ between sources.
(b) Difference between O and O₂: O represents a single atom of oxygen, while O₂ represents a molecule of oxygen (two oxygen atoms joined by a double bond) — this is the form in which oxygen gas actually exists.
(c) Why does salt solution conduct but sugar doesn’t? Salt (NaCl) is ionic; in water it splits into free-moving ions (Na⁺ and Cl⁻) that carry electric current. Sugar is covalent; it dissolves as neutral molecules and gives no ions, so it cannot conduct electricity.
Key conclusions from the chapter’s activities.
- 9.1 (physical change): Mass of the salt solution = mass of water + mass of salt. So mass does not change in a physical change.
- 9.2 (chemical change): When the balloon is sealed to the flask (closed system), the mass before = mass after, because the CO₂ gas stays trapped. In the open set-up the gas escapes, so the reading appears lower — confirming mass is conserved only when gases are not allowed to escape.
- 9.3 (verifying the law): Mixing sodium sulfate and barium chloride gives a white precipitate of barium sulfate plus sodium chloride. The total mass stays the same before and after — verifying the Law of Conservation of Mass.
- 9.4 (properties): Ionic compounds (NaCl, CuSO₄) dissolve in water and conduct electricity in solution; covalent compounds (camphor, naphthalene) dissolve in kerosene/petrol and do not conduct; sugar dissolves in water but does not conduct (no ions).
Design an experiment to test that mass is conserved when zinc reacts with dilute HCl (forms zinc chloride + hydrogen gas).
- Take some dilute HCl in a conical flask and a few zinc granules inside a balloon.
- Fix the balloon tightly over the mouth of the flask (sealed) without letting the zinc fall in yet. Weigh the whole set-up and note the reading.
- Lift the balloon so the zinc falls into the acid. The reaction releases hydrogen gas, which inflates and is trapped in the balloon.
- Weigh the set-up again.
Expected result: the mass stays the same because the hydrogen gas is contained — verifying the Law of Conservation of Mass.
1. A student burns 10 g of ethanol in an open beaker. No residue is left. Is the Law of Conservation of Mass violated?
No, the law is not violated. Ethanol burns to form carbon dioxide and water vapour, which are gases that escape into the air from the open beaker. If these gases (plus the oxygen used) were collected, the total mass would equal the mass of ethanol + oxygen consumed. The mass only seems to vanish because the gaseous products leave the system.
2. When 20 g of hydrogen reacts completely with 160 g of oxygen, how much water is formed?
By the Law of Conservation of Mass, mass of products = mass of reactants.
(This also fits the 1 : 8 H : O ratio — 20 g hydrogen needs exactly 160 g oxygen.)
3. A compound is 40% sulfur and 60% oxygen by mass. In a sample containing 20 g of sulfur, what mass of oxygen must be present?
Ratio S : O $= 40 : 60 = 2 : 3$.
4. Carbon monoxide (CO) has carbon : oxygen $= 3 : 4$ by mass. How much oxygen combines with 9 g of carbon?
5. The Law of Definite Proportions holds for compounds but not mixtures. Give the reason.
In a compound, elements are chemically combined in a fixed ratio by mass, so its composition is always the same. In a mixture, the substances are only physically mixed and can be present in any proportion, so the ratio is not fixed. Hence the law applies only to compounds.
6. Student X uses copper : oxygen $= 4 : 1$ and student Y uses $8 : 2$. Do their results justify the Law of Constant Proportions?
Yes. The ratio $8 : 2$ simplifies to $4 : 1$, which is the same as student X’s ratio. So both prepared the same oxide of copper with the same fixed mass ratio of copper to oxygen, confirming the Law of Constant Proportions.
7. A: 2 g of hydrogen combines with 16 g of oxygen to form 18 g of water. R: By Dalton’s theory, atoms combine in a simple whole-number ratio to form compounds.
A is true (the masses fit the 1 : 8 ratio and total 18 g). R is also true. Because hydrogen and oxygen atoms combine in the simple whole-number ratio of 2 : 1 (as in H₂O), they always combine in the fixed mass ratio 2 : 16 (= 1 : 8). So R correctly explains A.
8. Nitrogen has 5 valence electrons. Draw the structure of the nitrogen molecule (N₂).
Each nitrogen atom needs 3 more electrons to complete its octet. So two nitrogen atoms share three pairs of electrons, forming a triple bond (N≡N). Each atom is then left with one lone pair.
9. Fluorine has atomic number 9. Explain the formation of the fluorine molecule (F₂).
Electronic configuration of fluorine $= 2, 7$, so it has 7 valence electrons and needs 1 more to complete its octet. Two fluorine atoms each share one electron, forming a single covalent bond (F–F). Both atoms now effectively have 8 electrons, making the molecule stable.
10. Show the formation of (i) CO₂, (ii) H₂S, (iii) NH₃.
- (i) CO₂: Carbon (4 valence electrons, needs 4) shares two electron pairs with each oxygen atom (each O needs 2). This forms two double bonds: O = C = O.
- (ii) H₂S: Sulfur (6 valence electrons, needs 2) shares one electron with each of two hydrogen atoms, forming two single bonds: H – S – H.
- (iii) NH₃: Nitrogen (5 valence electrons, needs 3) shares one electron with each of three hydrogen atoms, forming three single N–H bonds and leaving one lone pair on nitrogen.
11. Neon (atomic number 10) neither transfers nor shares its valence electrons. Explain.
Neon’s electronic configuration is $2, 8$ — its outermost shell already has a complete octet (8 electrons). Since it is already stable, it has no need to gain, lose, or share electrons. That is why neon (a noble gas) is chemically inert.
12. What kind of ion will oxygen (O) form?
Oxygen has 6 valence electrons and needs 2 more to complete its octet. By gaining 2 electrons, it forms a negative ion (anion): the oxide ion, O²⁻.
13. Fill in the blanks about magnesium and chlorine forming magnesium chloride.
Magnesium gives two electrons to become Mg²⁺. Chlorine takes only one electron to become Cl⁻. Now, one (1) ion of magnesium and two (2) ions of chlorine combine to give magnesium chloride (MgCl₂).
14. Show the formation of K⁺ and Ca²⁺ cations and their chlorides.
- Potassium (Z = 19, config 2,8,8,1): loses its 1 valence electron → K⁺ (2,8,8). One K⁺ combines with one Cl⁻ → KCl.
- Calcium (Z = 20, config 2,8,8,2): loses its 2 valence electrons → Ca²⁺ (2,8,8). One Ca²⁺ combines with two Cl⁻ → CaCl₂.
In each case the metal transfers electron(s) to chlorine atom(s), and the oppositely charged ions are held by an ionic bond.
15. Illustrate how sodium sulfide (Na₂S) is formed.
Sodium (2,8,1) loses 1 electron to become Na⁺. Sulfur (2,8,6) needs 2 electrons, so it takes one electron from each of two sodium atoms, becoming S²⁻. The two Na⁺ ions and one S²⁻ ion combine to form Na₂S (ionic bond).
16. Name the following: (i) CO₂ (ii) NO₂ (iii) SF₆ (iv) PCl₃.
- (i) CO₂ → Carbon dioxide
- (ii) NO₂ → Nitrogen dioxide
- (iii) SF₆ → Sulfur hexafluoride
- (iv) PCl₃ → Phosphorus trichloride
17. Write the formula: (i) Sodium hydrogencarbonate (ii) Sulfur dioxide (iii) Ferric chloride (iv) Cuprous oxide.
- (i) Sodium hydrogencarbonate → NaHCO₃
- (ii) Sulfur dioxide → SO₂
- (iii) Ferric chloride (Fe³⁺, Cl⁻) → FeCl₃
- (iv) Cuprous oxide (Cu⁺, O²⁻) → Cu₂O
18. Write the formulae from these ions: (i) Fe³⁺ and OH⁻ (ii) K⁺ and CO₃²⁻.
- (i) Fe³⁺ and OH⁻ → criss-cross charges → Fe(OH)₃
- (ii) K⁺ and CO₃²⁻ → criss-cross charges → K₂CO₃
19. What type of bond is in a solid that does not conduct in the solid state but conducts when dissolved in water?
An ionic bond (it is an ionic compound). In the solid state its ions are fixed in the lattice and cannot move, so it does not conduct. When dissolved in water, the ions become free to move and carry current, so the solution conducts electricity.
20. Metal M has 2 electrons in its valence (M) shell and reacts with oxygen to form a compound slightly soluble in water. Predict its formula, bond type and conductivity.
The valence (3rd / M) shell has 2 electrons, so M loses 2 electrons → M²⁺ (e.g., magnesium, config 2,8,2). Oxygen forms O²⁻.
- (i) Formula: M²⁺ + O²⁻ → MO (like MgO).
- (ii) Type of bond: ionic bond.
- (iii) Conductivity of aqueous solution: the small amount that dissolves provides free ions, so the solution conducts electricity.
21. Find the molecular mass of nitric acid (HNO₃). (H = 1, N = 14, O = 16 u)
22. Find the molecular mass of methane (CH₄). (C = 12, H = 1 u)
23. Find the formula unit mass of potassium chloride (KCl). (K = 39, Cl = 35.5 u)
24. Find the formula unit mass of magnesium hydroxide, Mg(OH)₂. (Mg = 24, O = 16, H = 1 u)
Revise, Reflect, Refine — Exercises
1. Element A has 1 electron in its third shell; element B has 6 electrons in its second shell.
A = config 2,8,1 → sodium (Na). B = config 2,6 → oxygen (O).
- (i) A tends to give 1 electron to become stable.
- (ii) A forms a cation (A⁺, i.e., Na⁺).
- (iii) B tends to take 2 electrons to complete its octet.
- (iv) B forms an anion (B²⁻, i.e., O²⁻).
- (v) A + B → an ionic bond (metal + non-metal, by electron transfer).
- (vi) Formula: criss-cross charges → A₂B (Na₂O).
2. Element X has 6 electrons in its outer shell and forms a diatomic molecule.
X = oxygen (O), config 2,6.
- (i) Why diatomic? Each X needs 2 more electrons for an octet, so two X atoms share two electron pairs with each other to become stable — forming X₂.
- (ii) Bond type: a double covalent bond (X = X).
- (iii) Structure of X₂: O = O (two shared pairs, two lone pairs on each atom) — shown below.
- (iv) X with Y (Y has 2 electrons in second shell, config 2,2, valency 2): Y shares its 2 electrons with X, which needs 2 — giving a double-bonded molecule Y = X.
3. For an ionic compound with total positive charge 6+ and total negative charge 6−, which combination is correct?
Only in option (iii) are the charges balanced: $2 \times (+3) = +6$ and $3 \times (-2) = -6$. The others leave a net charge, so they are not neutral compounds.
4. Choose the correct statement(s) and correct the false ones.
(i) “Elements are made of molecules and compounds are made of atoms.” → False. Correct: Both elements and compounds are made up of atoms; elements may exist as atoms or molecules, while compounds are made of atoms of different elements combined.
(ii) “A molecule of a compound is always made of two or more atoms of the same kind.” → False. Correct: A molecule of a compound is made of atoms of different kinds (different elements).
(iii) “One molecule of nitrogen gas contains three nitrogen atoms.” → False. Correct: Nitrogen gas (N₂) contains two nitrogen atoms.
(iv) “Water is made of two hydrogen atoms covalently bonded with one oxygen atom.” → True.
5. Write the chemical formulae for: (i) Aluminium nitrate (ii) Calcium oxide (iii) Ferric oxide.
- (i) Aluminium nitrate (Al³⁺, NO₃⁻) → Al(NO₃)₃
- (ii) Calcium oxide (Ca²⁺, O²⁻) → CaO
- (iii) Ferric oxide (Fe³⁺, O²⁻) → Fe₂O₃
6. Write the formulae of compounds from these ion pairs.
- (i) Ca²⁺ and Br⁻ → CaBr₂
- (ii) Al³⁺ and CO₃²⁻ → Al₂(CO₃)₃
- (iii) K⁺ and SO₄²⁻ → K₂SO₄
- (iv) NH₄⁺ and Cl⁻ → NH₄Cl
7. Which diagram correctly represents the Cl⁻ ion? (Atomic number of Cl = 17.)
A chlorine atom has 17 electrons (2,8,7). On gaining 1 electron, the Cl⁻ ion has 18 electrons, arranged as 2, 8, 8 (a complete octet in the outer shell). So the correct diagram is the one showing 2 electrons in the first shell, 8 in the second, and 8 in the third — shown below.
8. Determine the formula unit mass of (i) NH₄NO₃, (ii) H₃PO₄, (iii) NaHCO₃. (N=14, H=1, O=16, P=31, Na=23, C=12 u)
9. Write the formulae of compounds formed by: (i) Mg + N (ii) Li + N (iii) Na + S (iv) Al + O.
- (i) Mg²⁺ and N³⁻ → Mg₃N₂ (magnesium nitride)
- (ii) Li⁺ and N³⁻ → Li₃N (lithium nitride)
- (iii) Na⁺ and S²⁻ → Na₂S (sodium sulfide)
- (iv) Al³⁺ and O²⁻ → Al₂O₃ (aluminium oxide)
10. Complete the table of formulae (cations × anions).
| NO₃⁻ | SO₄²⁻ | PO₄³⁻ | |
|---|---|---|---|
| NH₄⁺ | NH₄NO₃ | (NH₄)₂SO₄ | (NH₄)₃PO₄ |
| Li⁺ | LiNO₃ | Li₂SO₄ | Li₃PO₄ |
| Al³⁺ | Al(NO₃)₃ | Al₂(SO₄)₃ | AlPO₄ |
| Cu²⁺ | Cu(NO₃)₂ | CuSO₄ | Cu₃(PO₄)₂ |
11. 5.3 g sodium carbonate + 6.0 g acetic acid → 2.2 g CO₂ + 0.9 g water + 8.2 g sodium acetate. Verify the law of conservation of mass.
Mass of reactants $=$ Mass of products, so the Law of Conservation of Mass is valid.
11.3 g = 11.3 g ✓12. A species has 11 protons, 12 neutrons and 10 electrons.
- (i) Atomic number $=$ protons $= 11$; Mass number $=$ protons + neutrons $= 11 + 12 = 23$.
- (ii) Protons (11) > electrons (10), so it has a net positive charge → it is a cation (charge +1).
- (iii) With 10 electrons, the electronic configuration is 2, 8.
- (iv) 11 protons → element is sodium, so the species is the sodium ion, Na⁺.
13. Two elements: A (2, 8, 5) and B (2, 8, 7).
A = phosphorus (needs 3 electrons), B = chlorine (needs only 1 electron).
- (i) B is more reactive — it needs only one electron to complete its octet, so it gains an electron very easily.
- (ii) Both A and B are non-metals, so they form a covalent bond by sharing electrons.
- (iii) A needs 3 electrons and B needs 1, so one A atom shares with three B atoms → AB₃ (like PCl₃).
14. A: Copper sulfate conducts in the molten state but not in the solid state. R: Cu and sulfate ions are fixed in the lattice in the molten state, while in the solid state they move freely.
A is true — ionic compounds conduct only when their ions can move. R is false because it is reversed: the ions are fixed in the solid state and become free to move in the molten state. So A is true but R is false.
15. The species ²⁷Al, ⁸⁰Br⁻ and ²⁰¹Hg²⁺ have 13, 35 and 80 protons. Find their electrons and neutrons.
Electrons: adjust protons for the charge. Neutrons = mass number − protons.
| Species | Protons | Electrons | Neutrons |
|---|---|---|---|
| ²⁷Al (neutral) | 13 | 13 | 27 − 13 = 14 |
| ⁸⁰Br⁻ (gained 1 e⁻) | 35 | 36 | 80 − 35 = 45 |
| ²⁰¹Hg²⁺ (lost 2 e⁻) | 80 | 78 | 201 − 80 = 121 |