Chemistry questions, chapter 3

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The Mole Concept, Chemical Formula and Equation

45 marks. Use Practice to check each answer as you go, or Exam to answer everything against the clock first. Use these values: H = 1, C = 12, N = 14, O = 16, Na = 23, Mg = 24, Al = 27, S = 32, Cl = 35.5, Ca = 40, Cu = 64, Zn = 65; N_A = 6.02 × 10²³ mol⁻¹; molar volume 22.4 dm³ mol⁻¹ at STP, 24 dm³ mol⁻¹ at room conditions.

Section A: objective questions

12 marks

Choose the best answer for each question.

Question 1

What is the relative atomic mass of an element?

The actual mass of one atom in grams
The average mass of one atom compared with 1/12 of the mass of one carbon-12 atom
The mass of one mole of atoms in grams
The number of atoms in 12 g of the element

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RAM compares the average mass of an atom with 1/12 of the mass of a carbon-12 atom (textbook p. 44). Option C is the molar mass.

For the teacher: Remembering, low difficulty, standard 3.1.1

Textbook p. 44

Question 2

What is the value of the Avogadro constant used at this level?

6.02 × 10²² mol⁻¹
6.20 × 10²³ mol⁻¹
6.02 × 10²³ mol⁻¹
6.02 × 10³² mol⁻¹

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N_A = 6.02 × 10²³ mol⁻¹, the number of particles in one mole.

For the teacher: Remembering, low difficulty, standard 3.2.2

Textbook p. 50

Question 3

The relative atomic mass of magnesium is 24. What does this mean?

One magnesium atom is 24 times as heavy as 1/12 of a carbon-12 atom
One magnesium atom has a mass of 24 g
One magnesium atom is 24 times as heavy as a carbon-12 atom
One magnesium atom has 24 protons

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A relative mass is a comparison with 1/12 of a carbon-12 atom and has no unit. Option C leaves out the 1/12.

For the teacher: Understanding, low difficulty, standard 3.1.1

Textbook p. 45

Question 4

What is the relative formula mass of magnesium nitrate, Mg(NO₃)₂?

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24 + 2[14 + 3(16)] = 24 + 2(62) = 148.

For the teacher: Applying, low difficulty, standard 3.1.2

Textbook p. 48

Question 5

How many atoms are there in 0.5 mol of sodium?

1.20 × 10²³
3.01 × 10²³
6.02 × 10²³
1.20 × 10²⁴

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0.5 × 6.02 × 10²³ = 3.01 × 10²³ atoms.

For the teacher: Applying, low difficulty, standard 3.2.2

Textbook p. 51

Question 6

What is the mass of 0.25 mol of carbon dioxide, CO₂?

11 g
22 g
44 g
176 g

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Molar mass of CO₂ = 12 + 2(16) = 44 g mol⁻¹. Mass = 0.25 × 44 = 11 g.

For the teacher: Applying, low difficulty, standard 3.2.4

Textbook p. 53

Question 7

What is the volume of 0.5 mol of oxygen gas at room conditions?

6 dm³
11.2 dm³
12 dm³
24 dm³

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0.5 × 24 dm³ mol⁻¹ = 12 dm³. 11.2 dm³ would be the volume at STP.

For the teacher: Applying, low difficulty, standard 3.2.6

Textbook p. 55

Question 8

Which sample contains the greatest total number of atoms?

1 mol of neon, Ne
1 mol of oxygen gas, O₂
1 mol of carbon dioxide, CO₂
1 mol of ammonia, NH₃

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Each sample has 6.02 × 10²³ particles. Atoms per particle: Ne 1, O₂ 2, CO₂ 3, NH₃ 4. So NH₃ has the most.

For the teacher: Analysing, medium difficulty, standard 3.2.2

Textbook p. 51

Question 9

A compound has the molecular formula C₄H₈O₂. What is its empirical formula?

C₂H₄O
C₄H₈O₂
CH₂O
C₂H₈O

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Divide the subscripts by their highest common factor, 2: C₂H₄O. CH₂O is wrong because it would be C₄H₈O₄, not C₄H₈O₂.

For the teacher: Understanding, low difficulty, standard 3.3.1

Textbook p. 60

Question 10

The empirical formula of a compound is CH₂ and its relative molecular mass is 70. What is its molecular formula?

C₃H₆
C₄H₈
C₅H₁₀
C₇H₁₄

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Mass of CH₂ = 14. n = 70 ÷ 14 = 5, so the formula is (CH₂)₅ = C₅H₁₀.

For the teacher: Applying, medium difficulty, standard 3.3.4

Textbook p. 65

Question 11

What is the chemical formula of calcium phosphate? [Ca²⁺, PO₄³⁻]

CaPO₄
Ca₂PO₄
Ca₂(PO₄)₃
Ca₃(PO₄)₂

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Cross-change the charges: 3 Ca²⁺ (+6) and 2 PO₄³⁻ (−6), so Ca₃(PO₄)₂.

For the teacher: Applying, low difficulty, standard 3.3.5

Textbook p. 67

Question 12

Propane burns as follows: C₃H₈ + O₂ → CO₂ + H₂O. What is the sum of the coefficients when the equation is balanced with the smallest whole numbers?

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C₃H₈ + 5O₂ → 3CO₂ + 4H₂O. Sum = 1 + 5 + 3 + 4 = 13.

For the teacher: Applying, medium difficulty, standard 3.4.1

Textbook p. 70

Section B: structured questions

33 marks

Answer all questions. Show your working.

Question 13

Calculate the relative molecular mass or relative formula mass of each substance. [H = 1, C = 12, O = 16, S = 32, Ca = 40, Cu = 64]

[2 marks] (a)Ethanol, C₂H₅OH

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2(12) + 6(1) + 16 (correct substitution)1
= 461

Tick each point you got.

[2 marks] (b)Hydrated copper(II) sulphate, CuSO₄·5H₂O

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64 + 32 + 4(16) + 5[2(1) + 16] (correct substitution)1
= 2501

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[1 mark] (c)Calcium hydroxide, Ca(OH)₂

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40 + 2(16 + 1) = 741

Tick each point you got.

For the teacher: Applying, low difficulty, standard 3.1.2

Textbook p. 47

Question 14

A gas jar contains 8.8 g of carbon dioxide gas, CO₂, at room conditions.

[1 mark] (a)What is the molar mass of carbon dioxide?

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44 g mol⁻¹1

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[1 mark] (b)Calculate the number of moles of carbon dioxide.

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8.8 ÷ 44 = 0.2 mol1

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[2 marks] (c)Calculate the number of molecules in the jar.

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0.2 × 6.02 × 10²³ (correct substitution)1
= 1.204 × 10²³ molecules1

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[1 mark] (d)Calculate the volume of the gas.

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0.2 × 24 = 4.8 dm³1

Tick each point you got.

For the teacher: Applying, low difficulty, standard 3.2.7

Textbook p. 57

Question 15

A hydrocarbon contains 0.72 g of carbon and 0.12 g of hydrogen. Its relative molecular mass is 84. [H = 1, C = 12]

[2 marks] (a)Calculate the number of moles of carbon atoms and hydrogen atoms.

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C: 0.72 ÷ 12 = 0.06 mol1
H: 0.12 ÷ 1 = 0.12 mol1

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[1 mark] (b)Determine the empirical formula.

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C : H = 0.06 : 0.12 = 1 : 2, so CH₂1

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[2 marks] (c)Determine the molecular formula.

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n = 84 ÷ 14 = 61
C₆H₁₂1

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For the teacher: Applying, medium difficulty, standard 3.3.4

Textbook p. 61

Question 16

Answer the questions on formulae and names of compounds.

[1 mark] (a)Write the formula of aluminium oxide. [Al³⁺, O²⁻]

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Al₂O₃1

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[1 mark] (b)Write the formula of sodium carbonate. [Na⁺, CO₃²⁻]

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Na₂CO₃1

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[1 mark] (c)Name the compound with the formula Fe₂O₃.

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Iron(III) oxide1

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[1 mark] (d)Name the compound with the formula SO₃.

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Sulphur trioxide1

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[1 mark] (e)Why is a Roman numeral needed in the name of Fe₂O₃?

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Iron forms more than one ion (Fe²⁺ and Fe³⁺), so the numeral shows which one1

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For the teacher: Understanding, low difficulty, standard 3.3.5

Textbook p. 68

Question 17

13 g of zinc reacts completely with excess hydrochloric acid. Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g). [Zn = 65, Cl = 35.5]

[1 mark] (a)Calculate the number of moles of zinc.

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13 ÷ 65 = 0.2 mol1

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[1 mark] (b)How many moles of hydrogen gas are produced?

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Mole ratio Zn : H₂ = 1 : 1, so 0.2 mol1

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[1 mark] (c)Calculate the volume of hydrogen at room conditions.

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0.2 × 24 = 4.8 dm³1

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[2 marks] (d)Calculate the mass of zinc chloride formed.

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Molar mass ZnCl₂ = 65 + 71 = 136 g mol⁻¹, and 0.2 mol formed1
0.2 × 136 = 27.2 g1

Tick each point you got.

For the teacher: Analysing, high difficulty, standard 3.4.3

Textbook p. 73

Question 18

Write balanced chemical equations with state symbols for these reactions.

[2 marks] (a)Aluminium powder burns in oxygen to form solid aluminium oxide.

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Correct formulae: Al, O₂, Al₂O₃1
4Al(s) + 3O₂(g) → 2Al₂O₃(s) (balanced, with states)1

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[2 marks] (b)Solid copper(II) carbonate decomposes on heating into solid copper(II) oxide and carbon dioxide gas.

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Correct formulae: CuCO₃, CuO, CO₂1
CuCO₃(s) → CuO(s) + CO₂(g) (balanced, with states)1

Tick each point you got.

For the teacher: Creating, medium difficulty, standard 3.4.1

Textbook p. 70

Question 19

A farmer wants the fertiliser with the higher percentage of nitrogen by mass: ammonium nitrate, NH₄NO₃, or urea, CO(NH₂)₂. [H = 1, C = 12, N = 14, O = 16]

[1 mark] (a)Calculate the relative formula mass of urea.

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12 + 16 + 2(14 + 2) = 601

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[1 mark] (b)Calculate the percentage of nitrogen in urea.

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28 ÷ 60 × 100 = 46.7%1

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[1 mark] (c)Calculate the percentage of nitrogen in ammonium nitrate.

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RFM = 80; 28 ÷ 80 × 100 = 35%1

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[1 mark] (d)Which fertiliser should the farmer choose? Give a reason.

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Urea, because it has the higher percentage of nitrogen (46.7% against 35%)1

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For the teacher: Evaluating, medium difficulty, standard 3.3.4

Textbook p. 66

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