How to Use This Worksheet
Work through the sections in order. Section A checks the ideas, Section B builds the calculation method, and Section C mixes everything the way an exam would.
Use a periodic table and a calculator. Show every step of your working, use dimensional analysis, and let the units cancel so you can see the method is right before you trust the number.
Do not look at the answer key until you have attempted every question. The key gives full working, so compare your method, not just your final number.
Section A: Recall and Concepts
A1. What is a mole? How many molecules are in 1 mol of CO₂, and how many atoms are there in total in 1 mol of CO₂?
A2. Sodium chloride, NaCl, is counted in formula units rather than molecules. Explain why.
A3. A student says: “One mole of H₂ and one mole of O₂ have the same mass, because both are one mole.” Correct the statement.
A4. Which sample contains more atoms: 1 mol of helium, He, or 1 mol of hydrogen gas, H₂? Explain your answer.
A5. A question gives a mass in grams and asks for a number of molecules. Explain why the calculation must pass through moles, and name the two conversion factors involved.
Section B: Core Calculations
B1. Calculate the amount, in moles, in a sample containing 3.011 × 10²³ atoms of iron.
B2. How many molecules are there in 2.50 mol of H₂O?
B3. Calculate the molar mass of calcium hydroxide, Ca(OH)₂.
B4. What is the mass of 0.250 mol of sodium carbonate, Na₂CO₃?
B5. How many moles are there in 50.0 g of calcium carbonate, CaCO₃?
B6. How many formula units are there in 5.85 g of NaCl?
Section C: Exam Style
C1. A sample of aluminium sulfate, Al₂(SO₄)₃, has a mass of 17.1 g. (a) Calculate the amount, in moles, of Al₂(SO₄)₃ in the sample. (b) Calculate the amount, in moles, of oxygen atoms in the sample. (c) Calculate the number of sulfate ions in the sample.
C2. Flask A contains 10.0 g of hydrogen gas, H₂. Flask B contains 10.0 g of oxygen gas, O₂. Which flask contains more molecules, and by roughly what factor? Show your working.
C3. A student is given 0.500 mol of the hydrate CuSO₄·5H₂O. (a) Calculate the mass of the sample. (b) Calculate the amount, in moles, of water molecules in the sample. (c) The student calculates the mass as 0.500 × 159.62 = 79.8 g. Explain the error.
Answer Key
Show the answer key and full working (only after you have attempted every question)
A1. A mole is the amount of substance that contains 6.022 × 10²³ representative particles (Avogadro’s number). One mole of CO₂ contains 6.022 × 10²³ molecules. Each molecule holds 3 atoms, so the sample contains 3 mol of atoms, which is 3 × 6.022 × 10²³ = 1.81 × 10²⁴ atoms in total.
A2. NaCl is an ionic compound. Its solid is a continuous lattice of Na⁺ and Cl⁻ ions in a 1:1 ratio, with no separate NaCl molecules anywhere in it. The particle that the formula describes is the formula unit, the simplest whole number ratio of the ions.
A3. One mole of each gas contains the same number of molecules, 6.022 × 10²³, but the masses differ because the molar masses differ. One mole of H₂ has a mass of 2.016 g, while one mole of O₂ has a mass of 32.00 g. The mole fixes the count, never the mass.
A4. The H₂ sample, with twice as many atoms. One mole of He is 1 mol of atoms, 6.022 × 10²³ atoms. One mole of H₂ is 1 mol of molecules, and each molecule holds 2 atoms, so it contains 2 mol of atoms, 1.20 × 10²⁴ atoms.
A5. There is no direct conversion between grams and particles. Mass converts to moles using the molar mass (n = m / M), and moles convert to particles using Avogadro’s number (N = n × Nₐ). The mole is the only quantity connected to both, so every route passes through it.
B1. n = N / Nₐ = (3.011 × 10²³) / (6.022 × 10²³ mol⁻¹) = 0.500 mol.
B2. N = n × Nₐ = 2.50 mol × 6.022 × 10²³ molecules/mol = 1.51 × 10²⁴ molecules.
B3. M(Ca(OH)₂) = 40.08 + 2(16.00) + 2(1.008) = 40.08 + 32.00 + 2.02 = 74.10 g/mol. The subscript 2 outside the bracket doubles both the oxygen and the hydrogen.
B4. M(Na₂CO₃) = 2(22.99) + 12.01 + 3(16.00) = 45.98 + 12.01 + 48.00 = 105.99 g/mol. m = n × M = 0.250 mol × 105.99 g/mol = 26.5 g.
B5. M(CaCO₃) = 40.08 + 12.01 + 3(16.00) = 100.09 g/mol. n = m / M = 50.0 g / 100.09 g/mol = 0.500 mol.
B6. M(NaCl) = 22.99 + 35.45 = 58.44 g/mol. n = 5.85 g / 58.44 g/mol = 0.100 mol. N = 0.100 mol × 6.022 × 10²³ formula units/mol = 6.03 × 10²² formula units.
C1. M(Al₂(SO₄)₃) = 2(26.98) + 3(32.06) + 12(16.00) = 53.96 + 96.18 + 192.00 = 342.14 g/mol. (a) n = 17.1 g / 342.14 g/mol = 0.0500 mol. (b) Each formula unit holds 12 oxygen atoms, so moles of O atoms = 0.0500 × 12 = 0.600 mol. (c) Each formula unit holds 3 sulfate ions, so moles of SO₄²⁻ = 0.0500 × 3 = 0.150 mol, and the number of ions = 0.150 × 6.022 × 10²³ = 9.03 × 10²² ions.
C2. Flask A: n(H₂) = 10.0 g / 2.016 g/mol = 4.96 mol. Flask B: n(O₂) = 10.0 g / 32.00 g/mol = 0.313 mol. Flask A contains more molecules, by a factor of 4.96 / 0.313 ≈ 16. Equal masses do not mean equal counts: the lighter gas packs far more molecules into the same mass.
C3. (a) M(CuSO₄·5H₂O) = 63.55 + 32.06 + 4(16.00) + 5(18.02) = 249.71 g/mol, so m = 0.500 mol × 249.71 g/mol = 124.9 g. (b) Each formula unit carries 5 water molecules, so moles of H₂O = 0.500 × 5 = 2.50 mol. (c) The student used the molar mass of anhydrous CuSO₄, 159.62 g/mol, and left out the five waters of hydration. The dot in CuSO₄·5H₂O is part of the formula, and its mass, 5 × 18.02 = 90.10 g/mol, must be included.