AS Chemistry Unit 2 Calculations: June 2022 Exam Questions Explained | AS化学第二单元计算题型：2022年6月真题解析

📚 AS Chemistry Unit 2 Calculations: June 2022 Exam Questions Explained | AS化学第二单元计算题型：2022年6月真题解析

Calculation questions are a major component of AS Chemistry Unit 2, and the June 2022 paper was no exception. This walkthrough revisits the key calculation types that appeared, breaking down each step with clear explanations so you can master moles, equations, energetics, and yields. Whether you are preparing for a resit or reinforcing your skills, these worked examples will boost your confidence.

计算题是AS化学第二单元的重要组成部分，2022年6月的试卷也不例外。本文重新演练了试卷中出现的主要计算类型，用清晰的解释逐步分解，助你掌握摩尔、方程式、能量学以及产率计算。无论你是在准备重考还是巩固技能，这些详尽的例题都将增强你的自信。

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1. Moles and Avogadro’s Constant | 摩尔与阿伏伽德罗常数

The mole is the unit for amount of substance, containing 6.02 × 10²³ entities. In the June 2022 paper, a typical starter question asked students to convert mass to moles.

摩尔是物质的量的单位，包含 6.02 × 10²³ 个实体。在2022年6月的试卷中，一道典型的开篇题要求学生将质量转换为摩尔数。

Example: “Calculate the number of moles in 2.46 g of magnesium. (Mg = 24.3)”

例题：”计算 2.46 g 镁中的摩尔数。（Mg = 24.3）”

Step 1: Write down the mass and the molar mass. Molar mass of Mg is 24.3 g mol⁻¹.

步骤1：写下质量和摩尔质量。镁的摩尔质量是 24.3 g mol⁻¹。

Step 2: Use the formula n = m / M, where n is amount in mol, m is mass in g, and M is molar mass.

步骤2：使用公式 n = m / M，其中 n 为摩尔数，m 为质量（g），M 为摩尔质量。

n = m / M

Step 3: Substitute the values: n = 2.46 g ÷ 24.3 g mol⁻¹ = 0.10123… mol, which rounds to 0.101 mol (3 s.f.).

步骤3：代入数值：n = 2.46 g ÷ 24.3 g mol⁻¹ = 0.10123… mol，四舍五入为 0.101 mol（三位有效数字）。

Always check that your answer is given to the correct number of significant figures, as marks are often allocated for this.

务必检查答案的有效数字位数是否正确，因为阅卷常常为此分配分值。

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2. Empirical and Molecular Formula | 经验式与分子式

Empirical formula shows the simplest whole-number ratio of atoms in a compound. The June 2022 paper included a combustion analysis question requiring the empirical and molecular formula.

经验式表示化合物中原子的最简整数比。2022年6月的试卷中有一道燃烧分析题，要求得出经验式和分子式。

Example: “A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Its relative molecular mass is 180. Determine the empirical and molecular formula.”

例题：”某化合物含碳40.0%、氢6.7%、氧53.3%（质量分数），其相对分子质量为180。求经验式和分子式。”

Step 1: Assume 100 g of the compound, so masses are: C = 40.0 g, H = 6.7 g, O = 53.3 g.

步骤1：假设化合物为 100 g，则各元素质量为：C = 40.0 g，H = 6.7 g，O = 53.3 g。

Step 2: Convert masses to moles by dividing by atomic masses (C=12.0, H=1.0, O=16.0): C: 40.0/12.0 = 3.33 mol; H: 6.7/1.0 = 6.7 mol; O: 53.3/16.0 = 3.33 mol.

步骤2：除以原子量得到摩尔数（C=12.0, H=1.0, O=16.0）：C: 40.0/12.0 = 3.33 mol；H: 6.7/1.0 = 6.7 mol；O: 53.3/16.0 = 3.33 mol。

Step 3: Divide by the smallest number of moles (3.33) to get the ratio: C=1, H=2, O=1. Empirical formula = CH₂O.

步骤3：除以最小摩尔数（3.33）得到比例：C=1，H=2，O=1。经验式为 CH₂O。

Step 4: Find the empirical formula mass: 12 + (2×1) + 16 = 30. The multiplier is 180/30 = 6, so molecular formula = C₆H₁₂O₆.

步骤4：求经验式质量：12 + (2×1) + 16 = 30。倍数为 180/30 = 6，因此分子式为 C₆H₁₂O₆。

This approach works for any percentage composition data and is highly examinable.

此方法适用于任何百分组成数据，是高频考点。

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3. Reacting Masses | 反应质量计算

Reacting mass calculations link the masses of reactants and products using the balanced equation and mole ratios. A question from the June 2022 paper is reviewed here.

反应质量计算通过配平方程式和摩尔比，将反应物与产物的质量联系起来。以下回顾2022年6月的一道题。

Example: “2Mg(s) + O₂(g) → 2MgO(s). What mass of magnesium oxide is produced when 1.20 g of magnesium is completely burnt in oxygen?”

例题：”2Mg(s) + O₂(g) → 2MgO(s)。1.20 g 镁在氧气中完全燃烧，生成的氧化镁质量是多少？”

Step 1: Calculate moles of Mg: n(Mg) = 1.20 g / 24.3 g mol⁻¹ = 0.04938 mol.

步骤1：计算 Mg 的摩尔数：n(Mg) = 1.20 g / 24.3 g mol⁻¹ = 0.04938 mol。

Step 2: From the equation, 2 mol Mg produce 2 mol MgO, so mole ratio is 1:1. Thus n(MgO) = 0.04938 mol.

步骤2：由方程式，2 mol Mg 生成 2 mol MgO，摩尔比为1:1，故 n(MgO) = 0.04938 mol。

Step 3: Convert moles of MgO to mass: M(MgO) = 24.3 + 16.0 = 40.3 g mol⁻¹; mass = 0.04938 × 40.3 = 1.99 g (3 s.f.).

步骤3：将 MgO 的摩尔数转化为质量：M(MgO) = 24.3 + 16.0 = 40.3 g mol⁻¹；质量 = 0.04938 × 40.3 = 1.99 g（三位有效数字）。

Remember to use the balanced equation to deduce mole ratios; without it, the calculation is impossible.

切记利用配平方程式推出摩尔比；否则无法进行计算。

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4. Gas Volumes and Molar Volume | 气体体积与摩尔体积

At room temperature and pressure (RTP), 1 mole of any gas occupies 24.0 dm³. The June 2022 paper included a gas volume calculation from a reaction.

在室温和常压（RTP）下，1摩尔任何气体占据24.0 dm³。2022年6月试卷涵盖了一道由反应求气体体积的计算题。

Example: “Calcium carbonate decomposes: CaCO₃(s) → CaO(s) + CO₂(g). Calculate the volume of CO₂ produced at RTP when 0.250 mol CaCO₃ decomposes completely.”

例题：”碳酸钙分解：CaCO₃(s) → CaO(s) + CO₂(g)。0.250 mol CaCO₃ 完全分解，计算在RTP下生成的 CO₂ 体积。”

Step 1: Mole ratio from equation: 1 mol CaCO₃ gives 1 mol CO₂, so n(CO₂) = 0.250 mol.

步骤1：由方程式，1 mol CaCO₃ 产生 1 mol CO₂，因此 n(CO₂) = 0.250 mol。

Step 2: Volume of gas = n × 24.0 dm³ mol⁻¹ at RTP.

步骤2：气体体积 = n × 24.0 dm³ mol⁻¹（RTP条件下）。

V = n × 24.0

Step 3: V(CO₂) = 0.250 × 24.0 = 6.00 dm³.

步骤3：V(CO₂) = 0.250 × 24.0 = 6.00 dm³。

If the question specified a different temperature and pressure, the ideal gas equation pV = nRT would be needed, but the June 2022 question used RTP.

若题目指定不同温度和压力，需使用理想气体状态方程 pV = nRT，但2022年6月的题目采用的是RTP。

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5. Titration Calculations | 滴定计算

Titration calculations determine an unknown concentration using a neutralisation reaction. The June 2022 paper contained a standard acid-base titration problem.

滴定计算利用中和反应求出未知浓度。2022年6月试卷包含一道典型的酸碱滴定题。

Example: “25.0 cm³ of NaOH solution required 22.5 cm³ of 0.100 mol dm⁻³ HCl for complete neutralisation. Calculate the concentration of the NaOH solution.”

例题：”25.0 cm³ NaOH 溶液需 22.5 cm³ 0.100 mol dm⁻³ HCl 完全中和。计算 NaOH 溶液的浓度。”

Step 1: Write the balanced equation: HCl + NaOH → NaCl + H₂O. Mole ratio is 1:1.

步骤1：写出配平方程式：HCl + NaOH → NaCl + H₂O。摩尔比为1:1。

Step 2: Calculate moles of HCl used: n(HCl) = c × V = 0.100 mol dm⁻³ × (22.5 / 1000) dm³ = 0.00225 mol.

步骤2：计算所用 HCl 的摩尔数：n(HCl) = c × V = 0.100 mol dm⁻³ × (22.5 / 1000) dm³ = 0.00225 mol。

Step 3: From ratio, n(NaOH) = n(HCl) = 0.00225 mol. Volume of NaOH = 25.0 cm³ = 0.0250 dm³.

步骤3：由比例，n(NaOH) = n(HCl) = 0.00225 mol。NaOH 体积 = 25.0 cm³ = 0.0250 dm³。

Step 4: Concentration of NaOH = n/V = 0.00225 mol / 0.0250 dm³ = 0.0900 mol dm⁻³.

步骤4：NaOH 浓度 = n/V = 0.00225 mol / 0.0250 dm³ = 0.0900 mol dm⁻³。

Always convert volumes to dm³ when using mol dm⁻³ concentration units.

使用 mol dm⁻³ 浓度单位时，务必将体积转换为 dm³。

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6. Enthalpy Change: Calorimetry | 焓变计算：量热法

Calorimetry experiments measure temperature changes to calculate enthalpy changes. A typical June 2022 question involved a metal-acid reaction.

量热实验通过测量温度变化来计算焓变。2022年6月的一道典型题目涉及金属与酸的反应。

Example: “0.50 g of magnesium was added to an excess of HCl in a polystyrene cup. The temperature of the solution rose by 15.2°C. The mass of the solution was 100 g and its specific heat capacity is 4.18 J g⁻¹°C⁻¹. Calculate the enthalpy change, ΔH, in kJ mol⁻¹.”

例题：”将0.50 g镁加入过量HCl中（聚苯乙烯杯）。溶液温度上升15.2°C。溶液质量为100 g，比热容为4.18 J g⁻¹°C⁻¹。计算焓变 ΔH，单位为 kJ mol⁻¹。”

Step 1: Calculate heat absorbed by the solution, q = mcΔT = 100 g × 4.18 J g⁻¹°C⁻¹ × 15.2°C = 6353.6 J ≈ 6.35 kJ.

步骤1：计算溶液吸收的热量，q = mcΔT = 100 g × 4.18 J g⁻¹°C⁻¹ × 15.2°C = 6353.6 J ≈ 6.35 kJ。

q = mcΔT

Step 2: Calculate moles of Mg: n(Mg) = 0.50 g / 24.3 g mol⁻¹ = 0.02058 mol.

步骤2：计算 Mg 的摩尔数：n(Mg) = 0.50 g / 24.3 g mol⁻¹ = 0.02058 mol。

Step 3: The reaction is exothermic, so ΔH is negative. ΔH = –q / n = –6.35 kJ / 0.02058 mol = –309 kJ mol⁻¹ (3 s.f.).

步骤3：反应放热，故 ΔH 为负值。ΔH = –q / n = –6.35 kJ / 0.02058 mol = –309 kJ mol⁻¹（三位有效数字）。

Ensure the sign is correct and that heat lost by the reaction equals heat gained by the solution.

确保符号正确，且反应放出的热量等于溶液吸收的热量。

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7. Hess’s Law | 赫斯定律

Hess’s Law allows the calculation of enthalpy changes for reactions that are difficult to measure directly, by combining known enthalpy changes of formation or combustion.

赫斯定律通过组合已知的生成焓或燃烧焓，计算难以直接测量的反应焓变。

Example from June 2022: “Use the following standard enthalpy changes of formation to calculate the standard enthalpy of combustion of methane, CH₄. ΔHf°(CO₂) = –394 kJ mol⁻¹, ΔHf°(H₂O) = –286 kJ mol⁻¹, ΔHf°(CH₄) = –75 kJ mol⁻¹.”

2022年6月例题：”利用下列标准生成焓变，计算甲烷 CH₄ 的标准燃烧焓。ΔHf°(CO₂) = –394 kJ mol⁻¹，ΔHf°(H₂O) = –286 kJ mol⁻¹，ΔHf°(CH₄) = –75 kJ mol⁻¹。”

Step 1: Write the combustion equation: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l).

步骤1：写出燃烧方程式：CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l

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