Amount of Substance

How many molecules are present in 0.25 mol of carbon dioxide gas? (Avogadro constant = 6.02 × 10²³ mol⁻¹)

• A1.51 × 10²³✓
• B6.02 × 10²³
• C2.41 × 10²⁴
• D1.51 × 10²⁴

Answer: Number of molecules = moles × Avogadro constant = 0.25 × 6.02 × 10²³ = 1.505 × 10²³ ≈ 1.51 × 10²³ molecules.

What is the amount, in moles, of 13.7 g of calcium hydroxide, Ca(OH)₂? (Ar: Ca = 40.1, O = 16.0, H = 1.0)

• A0.135 mol
• B0.500 mol
• C0.185 mol✓
• D0.740 mol

Answer: Mr of Ca(OH)₂ = 40.1 + 2(16.0 + 1.0) = 40.1 + 34.0 = 74.1. Moles = mass/Mr = 13.7/74.1 = 0.185 mol.

A 0.500 mol sample of an ideal gas occupies a volume of 1.20 × 10⁻² m³ at a pressure of 1.00 × 10⁵ Pa. What is the temperature of the gas? (R = 8.31 J K⁻¹ mol⁻¹)

• A144 K
• B289 K✓
• C346 K
• D578 K

Answer: Using pV = nRT, T = pV/(nR) = (1.00 × 10⁵ × 1.20 × 10⁻²)/(0.500 × 8.31) = 1200/4.155 = 288.8 ≈ 289 K.

25.0 cm³ of a sodium hydroxide solution required 20.0 cm³ of 0.100 mol dm⁻³ hydrochloric acid for complete neutralisation. What is the concentration of the sodium hydroxide solution? (NaOH + HCl → NaCl + H₂O)

• A0.100 mol dm⁻³
• B0.0800 mol dm⁻³✓
• C0.125 mol dm⁻³
• D0.250 mol dm⁻³

Answer: Moles HCl = 0.0200 dm³ × 0.100 = 2.00 × 10⁻³ mol. The ratio is 1:1, so moles NaOH = 2.00 × 10⁻³ mol. Concentration = moles/volume = 2.00 × 10⁻³ / 0.0250 = 0.0800 mol dm⁻³.

A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. What is its empirical formula? (Ar: C = 12.0, H = 1.0, O = 16.0)

• ACH₂O✓
• BC₂H₄O₂
• CCHO
• DC₂H₆O

Answer: Divide % by Ar: C = 40.0/12.0 = 3.33; H = 6.7/1.0 = 6.7; O = 53.3/16.0 = 3.33. Divide by smallest (3.33): C = 1, H = 2.0, O = 1. Empirical formula = CH₂O.

The empirical formula of a compound is CH₂ and its relative molecular mass is 56.0. What is its molecular formula? (Ar: C = 12.0, H = 1.0)

• AC₂H₄
• BC₃H₆
• CC₄H₈✓
• DC₅H₁₀

Answer: Empirical formula mass of CH₂ = 12.0 + 2(1.0) = 14.0. Number of units = Mr/empirical mass = 56.0/14.0 = 4. Molecular formula = (CH₂)₄ = C₄H₈.

In an industrial process, 64.0 g of methanol (CH₃OH, Mr = 32.0) reacted to give 75.0 g of methyl ethanoate (CH₃COOCH₃, Mr = 74.0). Assuming methanol is the limiting reagent and reacts in a 1:1 mole ratio with the product, what is the percentage yield?

• A86.5%
• B92.5%
• C117%
• D50.7%✓

Answer: Moles methanol = 64.0/32.0 = 2.00 mol, so the theoretical yield of product = 2.00 mol × 74.0 = 148 g. Percentage yield = (75.0/148) × 100 = 50.7%.

Ethanol can be made by hydration of ethene: C₂H₄ + H₂O → C₂H₅OH. Calculate the atom economy of this reaction. (Mr: C₂H₄ = 28.0, H₂O = 18.0, C₂H₅OH = 46.0)

• A39.1%
• B60.9%
• C100%✓
• D76.1%

Answer: In an addition reaction there is only one product, so all reactant atoms end up in the desired product. Atom economy = (Mr of desired product / sum of Mr of all products) × 100 = (46.0/46.0) × 100 = 100%.

What mass of oxygen is required to completely burn 1.00 mol of propane? (C₃H₈ + 5O₂ → 3CO₂ + 4H₂O; Ar: O = 16.0)

• A32.0 g
• B80.0 g
• C160 g✓
• D320 g

Answer: From the equation, 1 mol C₃H₈ reacts with 5 mol O₂. Mass O₂ = moles × Mr = 5 × 32.0 = 160 g.

A 2.40 dm³ sample of an ideal gas at 100 kPa and 300 K contains how many moles? (R = 8.31 J K⁻¹ mol⁻¹)

• A0.0962 mol✓
• B0.0481 mol
• C0.192 mol
• D0.962 mol

Answer: Convert: V = 2.40 × 10⁻³ m³, p = 1.00 × 10⁵ Pa, T = 300 K. n = pV/(RT) = (1.00 × 10⁵ × 2.40 × 10⁻³)/(8.31 × 300) = 240/2493 = 0.0963 ≈ 0.0962 mol.