IB Chemistry Quiz: Apply The Mole Concept
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Apply The Mole ConceptQuestion 1 of 20

A 5.00 g sample of hydrated copper(II) sulfate, CuSO₄·xH₂O, is heated until all water of crystallization is driven off. The mass of the anhydrous copper(II) sulfate remaining is 3.20 g. What is the value of x? (Mᵣ(CuSO₄) = 159.6; Mᵣ(H₂O) = 18.0)

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IB Chemistry Quiz

IB Chemistry Quiz: Apply The Mole Concept

Practice Apply The Mole Concept in IB Chemistry with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Apply The Mole Concept, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Chemistry.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

A 5.00 g sample of hydrated copper(II) sulfate, CuSO₄·xH₂O, is heated until all water of crystallization is driven off. The mass of the anhydrous copper(II) sulfate remaining is 3.20 g. What is the value of x? (Mᵣ(CuSO₄) = 159.6; Mᵣ(H₂O) = 18.0)

  1. 2
  2. 3
  3. 4
  4. 5 (correct answer)
Explanation: First, determine the mass of water lost: mass(H₂O) = initial mass - final mass = 5.00 g - 3.20 g = 1.80 g. Next, convert the masses of anhydrous CuSO₄ and water to moles. n(CuSO₄) = 3.20 g / 159.6 g mol⁻¹ ≈ 0.02005 mol. n(H₂O) = 1.80 g / 18.0 g mol⁻¹ = 0.100 mol. The value of x is the mole ratio of water to copper(II) sulfate: x = n(H₂O) / n(CuSO₄) = 0.100 mol / 0.02005 mol ≈ 4.99, which rounds to 5.

Question 2

A student reacts 50.0 cm³ of liquid propan-1-ol (CH₃CH₂CH₂OH, density = 0.80 g cm⁻³) with excess oxidizing agent. The propan-1-ol is fully oxidized to propanoic acid (CH₃CH₂COOH). What is the theoretical mass of propanoic acid produced? (Mᵣ(propan-1-ol) = 60.1; Mᵣ(propanoic acid) = 74.1)

  1. 32.4 g
  2. 40.0 g
  3. 49.3 g (correct answer)
  4. 61.6 g
Explanation: This is a multi-step calculation. First, find the mass of propan-1-ol using its density and volume: mass = density × volume = 0.80 g cm⁻³ × 50.0 cm³ = 40.0 g. Second, calculate the moles of propan-1-ol: n = mass / Mᵣ = 40.0 g / 60.1 g mol⁻¹ = 0.6656 mol. The oxidation reaction has a 1:1 mole ratio, so 0.6656 mol of propanoic acid is produced. Finally, calculate the mass of propanoic acid: mass = n × Mᵣ = 0.6656 mol × 74.1 g mol⁻¹ = 49.3 g. Distractor D arises from incorrectly using 50.0 g as the mass of propan-1-ol. Distractor B is the initial mass of propan-1-ol. Distractor A results from an incorrect application of mass ratios.

Question 3

A solution containing 4.90 g of phosphoric acid, H₃PO₄, is added to a solution containing 4.00 g of sodium hydroxide, NaOH. The reaction is: H₃PO₄(aq) + 3NaOH(aq) → Na₃PO₄(aq) + 3H₂O(l). What is the maximum mass of sodium phosphate, Na₃PO₄, that can be formed? (Mᵣ(H₃PO₄) = 98.0; Mᵣ(NaOH) = 40.0; Mᵣ(Na₃PO₄) = 164.0)

  1. 16.4 g
  2. 8.20 g
  3. 5.47 g (correct answer)
  4. 49.2 g
Explanation: First, calculate the moles of each reactant. n(H₃PO₄) = 4.90 g / 98.0 g mol⁻¹ = 0.0500 mol. n(NaOH) = 4.00 g / 40.0 g mol⁻¹ = 0.100 mol. To determine the limiting reactant, compare the mole ratio. 0.0500 mol of H₃PO₄ requires 3 × 0.0500 = 0.150 mol of NaOH. Since only 0.100 mol of NaOH is available, NaOH is the limiting reactant. The moles of product are determined by the limiting reactant. From the stoichiometry, 3 moles of NaOH produce 1 mole of Na₃PO₄. So, n(Na₃PO₄) = 0.100 mol NaOH / 3 = 0.0333 mol. Finally, mass(Na₃PO₄) = 0.0333 mol × 164.0 g mol⁻¹ = 5.47 g. Distractor B incorrectly assumes H₃PO₄ is the limiting reactant. Distractor A ignores the 3:1 mole ratio for NaOH.

Question 4

100 cm³ of carbon monoxide gas and 100 cm³ of oxygen gas are mixed and ignited in a sealed container according to the equation: 2CO(g) + O₂(g) → 2CO₂(g). Assuming the reaction goes to completion and the final temperature and pressure are the same as the initial, what is the total volume of gas remaining in the container?

  1. 100 cm³
  2. 150 cm³ (correct answer)
  3. 200 cm³
  4. 50 cm³
Explanation: According to Avogadro's law, the ratio of volumes of reacting gases is equal to their mole ratio. The stoichiometric ratio is 2CO : 1O₂. To react with 100 cm³ of CO, 50 cm³ of O₂ is needed. Since there is 100 cm³ of O₂, O₂ is in excess and CO is the limiting reactant. After the reaction, no CO remains. The volume of O₂ remaining is 100 cm³ - 50 cm³ = 50 cm³. The volume of CO₂ produced is equal to the volume of CO reacted (due to the 2:2 ratio), which is 100 cm³. The total volume of gas remaining is the sum of the excess reactant and the product: 50 cm³ (O₂) + 100 cm³ (CO₂) = 150 cm³.

Question 5

An analysis of a carbohydrate shows it contains 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. Its molar mass is approximately 180 g mol⁻¹. What is its molecular formula?

  1. C₂H₄O₂
  2. C₆H₁₂O₆ (correct answer)
  3. C₅H₁₀O₅
  4. CH₂O
Explanation: First, determine the empirical formula from the percentage composition. Assuming a 100 g sample: n(C) = 40.0 g / 12.01 g mol⁻¹ ≈ 3.33 mol; n(H) = 6.7 g / 1.01 g mol⁻¹ ≈ 6.63 mol; n(O) = 53.3 g / 16.00 g mol⁻¹ ≈ 3.33 mol. Dividing by the smallest value (3.33) gives a ratio of C:H:O ≈ 1:2:1. The empirical formula is CH₂O. The molar mass of the empirical formula is 12.01 + 2(1.01) + 16.00 = 30.03 g mol⁻¹. To find the molecular formula, divide the molar mass of the compound by the empirical formula mass: n = 180 g mol⁻¹ / 30.03 g mol⁻¹ ≈ 6. Multiply the subscripts in the empirical formula by 6 to get the molecular formula: C₆H₁₂O₆.

Question 6

25.0 cm³ of a sodium chloride stock solution was diluted by adding water to a final volume of 500.0 cm³, creating a solution with a concentration of 0.0800 mol dm⁻³. What was the concentration of the original stock solution?

  1. 0.00400 mol dm⁻³
  2. 1.52 mol dm⁻³
  3. 1.60 mol dm⁻³ (correct answer)
  4. 4.00 mol dm⁻³
Explanation: This is a dilution calculation that can be solved using the formula M₁V₁ = M₂V₂, where M₁ is the unknown initial concentration. Rearranging for M₁ gives M₁ = (M₂V₂) / V₁. Plugging in the values: M₁ = (0.0800 mol dm⁻³ × 500.0 cm³) / 25.0 cm³ = 1.60 mol dm⁻³. Distractor A results from incorrectly arranging the formula as M₁ = (M₂V₁) / V₂. Distractor B arises from using the volume of water added (475 cm³) instead of the final volume in the calculation.

Question 7

The explosive compound RDX has a percentage composition by mass of 16.2% C, 2.7% H, 37.8% N, and 43.2% O. What is the empirical formula of RDX?

  1. CH₂N₂O₂ (correct answer)
  2. C₂H₃N₂O₂
  3. CHN₂O₂
  4. C₂H₄N₄O₄
Explanation: To find the empirical formula, assume a 100 g sample and convert the mass of each element to moles: n(C) = 16.2 g / 12.01 g mol⁻¹ ≈ 1.35 mol; n(H) = 2.7 g / 1.01 g mol⁻¹ ≈ 2.67 mol; n(N) = 37.8 g / 14.01 g mol⁻¹ ≈ 2.70 mol; n(O) = 43.2 g / 16.00 g mol⁻¹ ≈ 2.70 mol. Divide each mole value by the smallest value (1.35): C = 1.35/1.35 = 1; H = 2.67/1.35 ≈ 2; N = 2.70/1.35 = 2; O = 2.70/1.35 = 2. The simplest whole-number ratio is 1:2:2:2, so the empirical formula is CH₂N₂O₂. Distractor D is a multiple of the empirical formula, not the simplest ratio.

Question 8

A student dissolves 10.0 g of anhydrous sodium carbonate (Na₂CO₃) in water to make a 250.0 cm³ stock solution. They then take a 20.0 cm³ aliquot of this stock solution and dilute it to a final volume of 100.0 cm³. What is the concentration of the final diluted solution? (Mᵣ(Na₂CO₃) = 106.0)

  1. 0.0755 mol dm⁻³ (correct answer)
  2. 0.0377 mol dm⁻³
  3. 0.377 mol dm⁻³
  4. 1.89 mol dm⁻³
Explanation: Step 1: Calculate the concentration of the stock solution. Moles of Na₂CO₃ = 10.0 g / 106.0 g mol⁻¹ ≈ 0.09434 mol. Stock concentration M₁ = 0.09434 mol / 0.2500 dm³ ≈ 0.3774 mol dm⁻³. Step 2: Use the dilution formula M₁V₁ = M₂V₂ for the second step. (0.3774 mol dm⁻³) × (20.0 cm³) = M₂ × (100.0 cm³). M₂ = (0.3774 × 20.0) / 100.0 ≈ 0.0755 mol dm⁻³. Distractor C is the concentration of the initial stock solution. Distractor A arises from a calculation error. Distractor D arises from inverting the dilution factor.

Question 9

A reaction has a theoretical yield of 15.0 g. An experiment is performed which produces 12.3 g of the product. A second experiment with identical starting amounts produces 13.5 g of the product. What is the percentage yield of the second experiment?

  1. 82.0%
  2. 86.0%
  3. 90.0% (correct answer)
  4. 111%
Explanation: Percentage yield is calculated as (actual yield / theoretical yield) × 100%. The question asks for the percentage yield of the second experiment, so the data from the first experiment (12.3 g) is irrelevant information designed to be a distractor. For the second experiment: Percentage yield = (13.5 g / 15.0 g) × 100% = 90.0%. Distractor A is the percentage yield of the first experiment. Distractor B is the average of the two yields. Distractor D inverts the formula.

Question 10

Which of the following statements correctly compares the number of particles in a 10.0 g sample of calcium carbonate (CaCO₃) and a 10.0 g sample of calcium hydroxide (Ca(OH)₂)? (Mᵣ(CaCO₃)=100.1; Mᵣ(Ca(OH)₂)=74.1)

  1. The CaCO₃ sample contains more formula units and more oxygen atoms.
  2. The CaCO₃ sample contains fewer formula units but more oxygen atoms.
  3. The Ca(OH)₂ sample contains more formula units and more oxygen atoms.
  4. The Ca(OH)₂ sample contains more formula units but fewer oxygen atoms. (correct answer)
Explanation: First, compare the moles of formula units. Since Ca(OH)₂ has a lower molar mass, 10.0 g of it will contain more moles (and thus more formula units) than 10.0 g of CaCO₃. n(CaCO₃)=10.0/100.1≈0.10 mol. n(Ca(OH)₂)=10.0/74.1≈0.135 mol. So, the Ca(OH)₂ sample has more formula units. Next, compare the number of oxygen atoms. Moles of O in CaCO₃ = 0.10 mol × 3 = 0.30 mol. Moles of O in Ca(OH)₂ = 0.135 mol × 2 = 0.27 mol. Therefore, the CaCO₃ sample contains more oxygen atoms. Combining these findings, the Ca(OH)₂ sample contains more formula units but fewer oxygen atoms.

Question 11

0.840 g of a solid monoprotic acid, HA, is dissolved in water and titrated with 0.200 mol dm⁻³ NaOH solution. 20.00 cm³ of the NaOH solution is required to reach the equivalence point. What is the molar mass of the acid HA?

  1. 105 g mol⁻¹
  2. 84.0 g mol⁻¹
  3. 420 g mol⁻¹
  4. 210 g mol⁻¹ (correct answer)
Explanation: First, calculate the moles of NaOH used in the titration: n(NaOH) = concentration × volume = 0.200 mol dm⁻³ × 0.02000 dm³ = 0.00400 mol. Since the acid is monoprotic, the reaction stoichiometry is 1:1 (HA + NaOH → NaA + H₂O). Therefore, the moles of acid at the equivalence point are equal to the moles of base: n(HA) = 0.00400 mol. The molar mass of the acid is its mass divided by its moles: Mᵣ(HA) = mass / moles = 0.840 g / 0.00400 mol = 210 g mol⁻¹. Distractor A arises from incorrectly assuming a 1:2 acid:base stoichiometry. Distractor B results from a calculation error. Distractor C arises from incorrectly doubling the correct answer.

Question 12

An impure sample of magnesium carbonate, MgCO₃, has a mass of 10.0 g. When heated, it decomposes: MgCO₃(s) → MgO(s) + CO₂(g). After the reaction is complete, 4.00 g of magnesium oxide, MgO, is produced. What is the percentage purity of the MgCO₃ in the sample? (Mᵣ(MgCO₃) = 84.3; Mᵣ(MgO) = 40.3)

  1. 40.0%
  2. 47.8%
  3. 83.7% (correct answer)
  4. 16.3%
Explanation: First, work backwards from the product to find the mass of pure reactant. Calculate the moles of MgO produced: n(MgO) = 4.00 g / 40.3 g mol⁻¹ ≈ 0.09925 mol. The stoichiometry of the reaction is 1:1, so the moles of MgCO₃ that reacted is also 0.09925 mol. Calculate the mass of this pure MgCO₃: mass = n × Mᵣ = 0.09925 mol × 84.3 g mol⁻¹ ≈ 8.37 g. The percentage purity is the mass of the pure component divided by the total mass of the sample, multiplied by 100: % purity = (8.37 g / 10.0 g) × 100% = 83.7%. Distractor A incorrectly uses the ratio of product mass to sample mass. Distractor D calculates the percentage of impurity, not purity.

Question 13

50.0 cm³ of 0.100 mol dm⁻³ silver nitrate, AgNO₃, is mixed with 50.0 cm³ of 0.080 mol dm⁻³ magnesium chloride, MgCl₂. A precipitate of silver chloride, AgCl, is formed. What mass of precipitate is produced? (Mᵣ(AgCl) = 143.4)

  1. 0.359 g
  2. 0.717 g (correct answer)
  3. 1.15 g
  4. 1.43 g
Explanation: The balanced equation is 2AgNO₃(aq) + MgCl₂(aq) → 2AgCl(s) + Mg(NO₃)₂(aq). First, find the moles of each reactant: n(AgNO₃) = 0.100 M × 0.0500 dm³ = 0.00500 mol; n(MgCl₂) = 0.080 M × 0.0500 dm³ = 0.00400 mol. Determine the limiting reactant: 0.00500 mol AgNO₃ requires 0.00250 mol MgCl₂ (from 2:1 ratio). Since we have 0.00400 mol of MgCl₂, it is in excess, and AgNO₃ is limiting. The amount of product is determined by the limiting reactant. The mole ratio of AgNO₃ to AgCl is 2:2 or 1:1, so n(AgCl) = n(AgNO₃) = 0.00500 mol. Mass of AgCl = 0.00500 mol × 143.4 g mol⁻¹ = 0.717 g. Distractor C arises from incorrectly assuming MgCl₂ is the limiting reactant.

Question 14

A metal chloride has the formula MCl₂ and is found to contain 52.73% chlorine by mass. What is the identity of the metal M? (Aᵣ(Mg)=24.3; Aᵣ(Fe)=55.8; Aᵣ(Cu)=63.5; Aᵣ(Zn)=65.4)

  1. Magnesium (Mg)
  2. Iron (Fe)
  3. Copper (Cu) (correct answer)
  4. Zinc (Zn)
Explanation: Assume a 100 g sample of the compound. This sample contains 52.73 g of chlorine and 100 - 52.73 = 47.27 g of metal M. Calculate the moles of chlorine atoms: n(Cl) = 52.73 g / 35.45 g mol⁻¹ ≈ 1.487 mol. From the formula MCl₂, the mole ratio of M to Cl is 1:2. Therefore, n(M) = n(Cl) / 2 = 1.487 mol / 2 ≈ 0.7437 mol. The molar mass of M is calculated as M = mass / moles = 47.27 g / 0.7437 mol ≈ 63.5 g mol⁻¹. This relative atomic mass corresponds to Copper (Cu).

Question 15

A hydrated salt has the formula CuSO₄·xH₂O. When 12.49 g of this hydrated salt is heated to constant mass, 7.99 g of anhydrous CuSO₄ remains. Calculate the number of water molecules (x) in the original hydrated salt.

  1. x = 5 water molecules per formula unit (correct answer)
  2. x = 3 water molecules per formula unit
  3. x = 7 water molecules per formula unit
  4. x = 4 water molecules per formula unit
Explanation: Mass of water lost = 12.49 - 7.99 = 4.50 g H₂O. Moles of anhydrous CuSO₄ = 7.99 g ÷ 159.6 g/mol = 0.0501 mol. Moles of H₂O = 4.50 g ÷ 18.02 g/mol = 0.250 mol. The ratio of H₂O to CuSO₄ = 0.250/0.0501 = 4.99 ≈ 5. Therefore, x = 5.

Question 16

In the Haber process, nitrogen and hydrogen react to form ammonia: N₂ + 3H₂ → 2NH₃. If 28.0 g of N₂ reacts with excess H₂ and the reaction has an 85.0% yield, what mass of NH₃ is actually produced?

  1. 29.0 g of NH₃ is actually produced (correct answer)
  2. 34.1 g of NH₃ is actually produced
  3. 28.9 g of NH₃ is actually produced
  4. 24.6 g of NH₃ is actually produced
Explanation: Moles of N₂ = 28.0 g ÷ 28.0 g/mol = 1.00 mol. From stoichiometry, 1 mol N₂ produces 2 mol NH₃. Theoretical yield = 2.00 mol NH₃ × 17.0 g/mol = 34.0 g NH₃. Actual yield = 34.0 g × 0.850 = 28.9 g NH₃. The closest answer is 29.0 g.

Question 17

A gaseous compound contains 85.7% carbon and 14.3% hydrogen by mass. At STP, 2.80 g of this compound occupies 1.12 L. What is the molecular formula of the compound?

  1. C₄H₈ with molar mass 56.0 g/mol (correct answer)
  2. C₂H₄ with molar mass 28.0 g/mol
  3. C₆H₁₂ with molar mass 84.0 g/mol
  4. C₃H₆ with molar mass 42.0 g/mol
Explanation: First, find the empirical formula: C: (85.7 g/100 g) ÷ 12.01 g/mol = 7.14 mol; H: (14.3 g/100 g) ÷ 1.008 g/mol = 14.2 mol. Ratio C:H = 7.14:14.2 = 1:2, so empirical formula is CH₂ (14.0 g/mol). Next, find molar mass from gas data: at STP, 1.12 L contains 2.80 g, so 22.4 L contains (2.80 × 22.4/1.12) = 56.0 g. Therefore, molar mass = 56.0 g/mol. Since 56.0/14.0 = 4, the molecular formula is C₄H₈.

Question 18

What volume of water must be added to 20.0 cm³ of a 1.50 mol dm⁻³ NaOH solution to produce a solution with a concentration of 0.100 mol dm⁻³?

  1. 300 cm³
  2. 280 cm³ (correct answer)
  3. 28.0 cm³
  4. 30.0 cm³
Explanation: Use the dilution formula M₁V₁ = M₂V₂ to find the final volume (V₂). (1.50 mol dm⁻³) × (20.0 cm³) = (0.100 mol dm⁻³) × V₂. Solving for V₂ gives V₂ = (1.50 × 20.0) / 0.100 = 300 cm³. This is the final volume of the solution, not the volume of water added. The volume of water added is the final volume minus the initial volume: V(water) = V₂ - V₁ = 300 cm³ - 20.0 cm³ = 280 cm³. Distractor A is a common error where the final volume is given instead of the added volume. Distractors C and D result from decimal place errors during calculation.

Question 19

Which sample contains the greatest total number of atoms?

  1. 16.0 g of methane, CH₄ (correct answer)
  2. 18.0 g of water, H₂O
  3. 28.0 g of nitrogen, N₂
  4. 32.0 g of oxygen, O₂
Explanation: The mass given for each sample corresponds to one mole of the substance (Mᵣ(CH₄)≈16, Mᵣ(H₂O)≈18, Mᵣ(N₂)≈28, Mᵣ(O₂)≈32). Therefore, each sample contains 1.0 mol of molecules. To find the total number of atoms, multiply the moles of molecules by the number of atoms per molecule. A: 1.0 mol × (1+4) atoms/molecule = 5.0 mol atoms. B: 1.0 mol × (2+1) atoms/molecule = 3.0 mol atoms. C: 1.0 mol × 2 atoms/molecule = 2.0 mol atoms. D: 1.0 mol × 2 atoms/molecule = 2.0 mol atoms. Methane (CH₄) contains the greatest number of moles of atoms, and thus the greatest number of atoms.

Question 20

Four identical flasks at the same temperature and pressure are filled with four different gases. Which flask contains the greatest number of atoms?

  1. Flask A filled with neon (Ne)
  2. Flask B filled with ethane (C₂H₆) (correct answer)
  3. Flask C filled with carbon dioxide (CO₂)
  4. Flask D filled with dinitrogen monoxide (N₂O)
Explanation: According to Avogadro's law, equal volumes of gases at the same temperature and pressure contain an equal number of molecules (moles). To find the flask with the most atoms, we need to identify the gas with the most atoms per molecule. Neon (Ne) is monatomic (1 atom). Ethane (C₂H₆) has 2 + 6 = 8 atoms. Carbon dioxide (CO₂) has 1 + 2 = 3 atoms. Dinitrogen monoxide (N₂O) has 2 + 1 = 3 atoms. Since ethane has the most atoms per molecule (8), the flask containing ethane will have the greatest total number of atoms.