This checkpoint covers Lessons 1–5: the mole and Avogadro constant, molar mass, empirical and molecular formulas, and gases and molar volume. 20 multiple choice + 5 short answer questions.
Lesson Summaries, Quick Review
The mole is the SI unit for amount of substance; one mole contains exactly 6.022 × 10²³ specified particles (the Avogadro constant, NA). Number of particles N = n × NA, so n = N ÷ NA. Equal numbers of moles contain equal numbers of particles, regardless of the substance.
Molar mass (g mol⁻¹) is the mass of one mole of a substance, found by adding the relative atomic masses of every atom in the formula. Convert with n = m ÷ MM and m = n × MM. Relative atomic/molecular masses (Ar, Mr) are dimensionless; molar mass carries units of g mol⁻¹.
The empirical formula is the simplest whole-number ratio of atoms; the molecular formula is the actual number of atoms in a molecule. From % composition: divide each % by the atomic mass, then by the smallest result. The molecular formula needs a separately measured molar mass: multiplier = MM ÷ MM(empirical), which must be a whole number within experimental uncertainty.
By Avogadro's law, equal volumes of all ideal gases at the same temperature and pressure contain equal numbers of particles. One mole of any gas occupies the molar volume Vm: 22.71 L mol⁻¹ at STP (0 °C, 100 kPa, NESA standard) and 24.8 L mol⁻¹ at SATP/RTP (25 °C, 100 kPa). Convert with n = V ÷ Vm.
The mole is the central hub: mass ↔ n (÷ or × MM), particles ↔ n (÷ or × NA), gas volume ↔ n (÷ or × Vm), and empirical ↔ molecular formula (× integer multiplier). Every multi-step problem is solved by first converting the known quantity to moles, then converting moles to the target quantity.
Score updates as you answer each question. Short answer marks are self-assessed.
1. One mole of any substance contains how many specified particles? L01
2. How many molecules are present in 2.00 mol of carbon dioxide (CO₂)? L01
3. How many moles correspond to 3.011 × 10²³ atoms of neon? L01
4. A flask holds 2.0 mol of H₂O and another holds 2.0 mol of CO₂. Which statement is correct? L01
5. What is the molar mass of calcium carbonate, CaCO₃? (Ca = 40.08, C = 12.01, O = 16.00) L02
6. How many moles are in 36.0 g of water? (H₂O, MM = 18.0 g mol⁻¹) L02
7. What is the mass of 0.250 mol of sodium chloride? (NaCl, MM = 58.44 g mol⁻¹) L02
8. 0.100 mol of a pure metal M has a mass of 5.585 g. What is its molar mass (and likely identity)? L02
9. A compound is 40.0% C, 6.7% H and 53.3% O by mass. What is its empirical formula? (C = 12.01, H = 1.008, O = 16.00) L03
10. A sample contains 0.30 mol of carbon and 0.60 mol of hydrogen. What is its empirical formula? L03
11. A compound has empirical formula CH₂ and a measured molar mass of 56.0 g mol⁻¹. What is its molecular formula? (CH₂ = 14.03) L03
12. Which statement about empirical and molecular formulas is correct? L03
13. What volume does 0.500 mol of an ideal gas occupy at STP (0 °C, 100 kPa)? (Vm = 22.71 L mol⁻¹) L04
14. How many moles of gas are in 49.6 L at SATP (25 °C, 100 kPa)? (Vm = 24.8 L mol⁻¹) L04
15. Avogadro's law states that equal volumes of all ideal gases at the same temperature and pressure contain equal numbers of: L04
16. Which conditions correspond to a molar volume of 22.71 L mol⁻¹ (the NESA STP value)? L04
17. How many molecules are in 8.0 g of methane, CH₄? (MM = 16.0 g mol⁻¹, NA = 6.022 × 10²³) L05
18. What volume does 4.40 g of CO₂ occupy at STP? (MM = 44.0 g mol⁻¹, Vm = 22.71 L mol⁻¹) L05
19. What is the mass of 1.204 × 10²⁴ molecules of water? (MM = 18.0 g mol⁻¹) L05
20. 0.25 mol of an unknown gas has a mass of 11.0 g. What is its molar mass? L05
Attempt all five questions before checking answers.
21. Calculate the number of oxygen atoms in 9.00 g of water. (H₂O, MM = 18.0 g mol⁻¹; NA = 6.022 × 10²³) 3 MARKS
22. A sample of magnesium nitrate, Mg(NO₃)₂, has a mass of 14.82 g. Calculate the amount in moles. (Mg = 24.31, N = 14.01, O = 16.00) 3 MARKS
23. A hydrocarbon is 85.7% carbon and 14.3% hydrogen by mass, and its molar mass is 56.0 g mol⁻¹. Determine (a) the empirical formula and (b) the molecular formula. (C = 12.01, H = 1.008) 4 MARKS
24. Calculate the volume occupied by 7.00 g of nitrogen gas (N₂) at STP. (N = 14.01; STP Vm = 22.71 L mol⁻¹) 3 MARKS
25. A 2.20 g sample of dry ice (solid CO₂) sublimes completely. (a) Calculate the amount in moles. (b) Calculate the volume of CO₂ gas produced at SATP. (c) Calculate the number of CO₂ molecules. (CO₂ = 44.01; SATP Vm = 24.8 L mol⁻¹; NA = 6.022 × 10²³) 4 MARKS
1. A One mole = 6.022 × 10²³ specified particles (the Avogadro constant).
2. B N = n × NA = 2.00 × 6.022 × 10²³ = 1.204 × 10²⁴ molecules.
3. A n = N ÷ NA = 3.011 × 10²³ ÷ 6.022 × 10²³ = 0.500 mol.
4. C Equal moles contain equal numbers of particles (2.0 × NA), independent of molar mass.
5. A MM = 40.08 + 12.01 + 3(16.00) = 100.09 g mol⁻¹.
6. A n = m ÷ MM = 36.0 ÷ 18.0 = 2.00 mol.
7. A m = n × MM = 0.250 × 58.44 = 14.6 g.
8. A MM = m ÷ n = 5.585 ÷ 0.100 = 55.85 g mol⁻¹ → iron (Fe).
9. A C: 40.0/12.01 = 3.33; H: 6.7/1.008 = 6.65; O: 53.3/16.00 = 3.33 → 1 : 2 : 1 = CH₂O.
10. A Ratio C : H = 0.30 : 0.60 = 1 : 2 → CH₂.
11. A Multiplier = 56.0 ÷ 14.03 = 3.99 ≈ 4 → C₄H₈.
12. C The empirical formula is the simplest whole-number ratio; for some compounds (e.g. H₂O, CO₂) it equals the molecular formula.
13. A V = n × Vm = 0.500 × 22.71 = 11.4 L at STP.
14. A n = V ÷ Vm = 49.6 ÷ 24.8 = 2.00 mol.
15. B Avogadro's law: equal volumes contain equal numbers of particles (works for monatomic and molecular gases alike).
16. A The NESA STP value 22.71 L mol⁻¹ applies at 0 °C and 100 kPa. (22.4 L mol⁻¹ is the older 1 atm value.)
17. A n = 8.0 ÷ 16.0 = 0.50 mol; N = 0.50 × 6.022 × 10²³ = 3.011 × 10²³.
18. A n = 4.40 ÷ 44.0 = 0.100 mol; V = 0.100 × 22.71 = 2.27 L at STP.
19. A n = N ÷ NA = 1.204 × 10²⁴ ÷ 6.022 × 10²³ = 2.00 mol; m = 2.00 × 18.0 = 36.0 g.
20. A MM = m ÷ n = 11.0 ÷ 0.25 = 44 g mol⁻¹, consistent with CO₂.
Q21 (3 marks): n(H₂O) = 9.00 ÷ 18.0 = 0.500 mol [1]. molecules = 0.500 × 6.022 × 10²³ = 3.011 × 10²³ [1]. Each H₂O has 1 O atom, so O atoms = 3.011 × 10²³ [1].
Q22 (3 marks): MM = 24.31 + 2(14.01 + 3×16.00) = 24.31 + 2(62.01) = 148.33 g mol⁻¹ [2]. n = 14.82 ÷ 148.33 = 0.100 mol [1].
Q23 (4 marks): (a) n(C) = 85.7 ÷ 12.01 = 7.14; n(H) = 14.3 ÷ 1.008 = 14.19; ratio 1 : 1.99 → CH₂ [2]. (b) MM(CH₂) = 14.03; multiplier = 56.0 ÷ 14.03 = 3.99 ≈ 4 → C₄H₈ [2].
Q24 (3 marks): MM(N₂) = 28.02 g mol⁻¹ [1]; n = 7.00 ÷ 28.02 = 0.2498 mol [1]; V = 0.2498 × 22.71 = 5.67 L at STP [1].
Q25 (4 marks): (a) n = 2.20 ÷ 44.01 = 0.0500 mol [1]. (b) V = 0.0500 × 24.8 = 1.24 L at SATP [2]. (c) N = 0.0500 × 6.022 × 10²³ = 3.011 × 10²² molecules [1].
Tick when you have checked all your answers and noted areas to review.