Year 11 Chemistry Module 2 Checkpoint Quiz 1

Checkpoint Quiz 1

Inquiry Question 1 — How are measurements made in chemistry?

📚 Covers: L01–L05
❓ Questions: 15 (10 MC + 5 written)
🏆 Total marks: 30
⏱ Suggested time: 25 min

Your score

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Section A — Multiple Choice (10 questions × 1 mark)

Question 1
1 markL01MC

Which of the following best defines one mole of a substance?

A
The mass in grams equal to the relative atomic mass of the substance
B
6.022 × 10²³ particles of that substance
C
The volume occupied by 22.4 L of the substance at STP
D
The number of atoms in 12 g of carbon-12
Question 2
1 markL01MC

How many atoms are in 0.25 mol of helium gas?

A
6.022 × 10²³
B
2.409 × 10²³
C
1.505 × 10²³
D
1.204 × 10²³
Question 3
1 markL02MC

What is the molar mass of iron(III) sulfate, Fe₂(SO₄)₃? (Fe = 55.845, S = 32.06, O = 15.999)

A
199.91 g mol⁻¹
B
311.87 g mol⁻¹
C
399.91 g mol⁻¹
D
507.87 g mol⁻¹
Question 4
1 markL02MC

A sample of copper (Cu) has a mass of 31.77 g. How many moles of copper does it contain? (Cu = 63.546 g mol⁻¹)

A
0.250 mol
B
0.400 mol
C
0.500 mol
D
2.00 mol
Question 5
1 markL03MC

A compound contains 32.38% sodium, 22.57% sulfur, and 45.05% oxygen by mass. What is its empirical formula? (Na = 22.990, S = 32.06, O = 15.999)

A
NaSO₂
B
Na₂SO₃
C
Na₂SO₄
D
NaSO₄
Question 6
1 markL03MC

A compound has the empirical formula CH and a molar mass of 78.11 g mol⁻¹. What is its molecular formula? (C = 12.011, H = 1.008)

A
C₄H₄
B
C₅H₅
C
C₆H₆
D
C₇H₇
Question 7
1 markL04MC

What volume does 0.750 mol of argon gas (Ar) occupy at SATP?

A
16.8 L
B
18.6 L
C
22.4 L
D
24.8 L
Question 8
1 markL04MC

A 33.6 L sample of gas is collected at STP. Which of the following correctly calculates the number of moles?

A
n = 33.6 × 22.4 = 752.6 mol
B
n = 33.6 ÷ 24.8 = 1.35 mol
C
n = 33.6 ÷ 22.4 = 1.50 mol
D
n = 33.6 × 24.8 = 833.3 mol
Question 9
1 markL02 + L04MC

What mass of oxygen gas (O₂) occupies 12.4 L at SATP? (O = 15.999)

A
8.00 g
B
12.6 g
C
16.0 g
D
32.0 g
Question 10
1 markL01–L04MC

Equal volumes of two different gases are measured at the same temperature and pressure. Which of the following statements must be true?

A
They have the same mass
B
They have the same molar mass
C
They contain the same number of molecules
D
They contain the same total number of atoms

Section B — Short Answer (5 questions × 4 marks)

Question 11
4 marksL01 + L02Written

A student has a 9.03 g sample of water (H₂O). (H = 1.008, O = 15.999)

Calculate: (a) the molar mass of water, (b) the number of moles of water, (c) the number of water molecules, and (d) the total number of hydrogen atoms.

✅ Model Answer
a) MM(H₂O) = 2(1.008) + 15.999 = 18.015 g mol⁻¹ b) n = m ÷ MM = 9.03 ÷ 18.015 = 0.5013 mol ≈ 0.501 mol c) N = n × Nₐ = 0.5013 × 6.022 × 10²³ = 3.018 × 10²³ molecules d) Each H₂O has 2 H atoms → N(H) = 2 × 3.018 × 10²³ = 6.036 × 10²³ H atoms Award 1 mark per correct step with working shown.
Self-mark:
Question 12
4 marksL02Written

Calculate the mass of aluminium sulfate, Al₂(SO₄)₃, that contains 1.50 mol. Show all working including the molar mass calculation. (Al = 26.982, S = 32.06, O = 15.999)

✅ Model Answer
MM = 2(26.982) + 3(32.06) + 12(15.999) = 53.964 + 96.18 + 191.988 = 342.132 g mol⁻¹ m = n × MM = 1.50 × 342.132 = 513.20 g Award: 1 mark correct atom count (note: 3 × SO₄ = 3S + 12O); 1 mark correct MM; 1 mark correct formula rearrangement; 1 mark correct final answer with units.
Self-mark:
Question 13
4 marksL03Written

A compound contains 52.14% carbon, 13.13% hydrogen, and 34.73% oxygen by mass. Its molar mass is 46.07 g mol⁻¹. Determine the molecular formula of the compound. Show the full 4-step method. (C = 12.011, H = 1.008, O = 15.999)

✅ Model Answer
Step 1: C = 52.14 g, H = 13.13 g, O = 34.73 g Step 2: n(C) = 52.14 ÷ 12.011 = 4.341; n(H) = 13.13 ÷ 1.008 = 13.026; n(O) = 34.73 ÷ 15.999 = 2.171 Step 3: Divide by 2.171: C = 2.00, H = 6.00, O = 1.00 → Empirical formula: C₂H₆O Step 4: MM(C₂H₆O) = 2(12.011) + 6(1.008) + 15.999 = 46.069 g mol⁻¹ n = 46.07 ÷ 46.069 = 1.00 → Molecular formula: C₂H₆O (ethanol) Award 1 mark per step correctly completed.
Self-mark:
Question 14
4 marksL04 + L02Written

A gas jar contains 37.2 L of carbon dioxide (CO₂) at SATP.

Calculate: (a) the number of moles of CO₂, (b) the mass of CO₂ in the jar, and (c) the number of oxygen atoms present. (C = 12.011, O = 15.999)

✅ Model Answer
a) SATP → Vₘ = 24.8 L mol⁻¹; n = 37.2 ÷ 24.8 = 1.500 mol b) MM(CO₂) = 12.011 + 2(15.999) = 44.009 g mol⁻¹ m = n × MM = 1.500 × 44.009 = 66.01 g c) Each CO₂ has 2 O atoms; N(molecules) = 1.500 × 6.022 × 10²³ = 9.033 × 10²³ N(O atoms) = 2 × 9.033 × 10²³ = 1.807 × 10²⁴ O atoms
Self-mark:
Question 15
4 marksL01–L04Written

An unknown gas has a density of 1.964 g L⁻¹ at SATP. The gas contains only nitrogen and oxygen atoms in a 1:2 mole ratio.

(a) Use the density to calculate the molar mass of the gas. (b) Use the mole ratio and molar mass to determine the molecular formula. (c) Name this compound. (N = 14.007, O = 15.999)

✅ Model Answer
a) 1 L of gas at SATP has mass = 1.964 g; n = 1.000 ÷ 24.8 = 0.04032 mol MM = m ÷ n = 1.964 ÷ 0.04032 = 48.71 ≈ 48.0 g mol⁻¹ b) N:O mole ratio = 1:2 → empirical formula NO₂ MM(NO₂) = 14.007 + 2(15.999) = 46.005 g mol⁻¹ Multiplier n = 48.0 ÷ 46.005 ≈ 1.04 ≈ 1 → Molecular formula: NO₂ c) Nitrogen dioxide Award: 1 mark for MM calculation; 1 mark for correct empirical formula; 1 mark for multiplier and molecular formula; 1 mark for correct name.
Self-mark: