IB & AQA Chemistry: Calculation Practice | IB & AQA 化学：计算题专项训练

📚 IB & AQA Chemistry: Calculation Practice | IB & AQA 化学：计算题专项训练

Success in IB Chemistry and AQA A-level Chemistry hinges on confident, accurate numerical problem-solving. From mole conversions and titrations to enthalpy cycles and equilibrium constants, calculation questions carry significant marks on both syllabi. This article consolidates the most frequently tested calculation types, presenting core formulas, worked examples, and common pitfalls to help you train methodically.

在IB化学和AQA A-Level化学中，精准的数值计算能力是取得高分的关键。无论是摩尔换算、滴定分析还是焓变循环与平衡常数，计算题都占据重要分值。本文梳理了两套课程中最常考的计算类型，给出核心公式、典型例题和常见错误，帮助你有条不紊地开展专项训练。

1. Moles and Molar Mass | 摩尔与摩尔质量

The mole is the central unit in quantitative chemistry. The amount of substance, n (mol), relates mass m (g) and molar mass M (g mol⁻¹) through n = m / M. Always show unit cancellation and round to the appropriate number of significant figures. In IB, you are expected to use values from the data booklet; AQA gives relative atomic masses on the periodic table provided.

摩尔是化学计量的核心单位。物质的量 n（mol）通过 n = m / M 将质量 m（g）与摩尔质量 M（g mol⁻¹）联系起来。务必展示单位消去并保留合适的有效数字。在IB考试中使用数据手册中的相对原子质量，AQA则提供元素周期表上的数值。

n = m / M M = m / n m = n × M

Worked example: Calculate the amount of sodium carbonate in 5.30 g of anhydrous Na₂CO₃. (Aᵣ: Na = 23.0, C = 12.0, O = 16.0)

典型例题：计算 5.30 g 无水碳酸钠 Na₂CO₃ 的物质的量。（相对原子质量：Na 23.0, C 12.0, O 16.0）

M of Na₂CO₃ = (2×23.0) + 12.0 + (3×16.0) = 106.0 g mol⁻¹. Then n = 5.30 g / 106.0 g mol⁻¹ = 0.0500 mol. Notice the answer is given to three significant figures because the mass was given to three significant figures.

Na₂CO₃ 的摩尔质量 M = (2×23.0) + 12.0 + (3×16.0) = 106.0 g mol⁻¹。则 n = 5.30 g / 106.0 g mol⁻¹ = 0.0500 mol。注意结果保留三位有效数字，因为质量数据为三位有效数字。

2. Empirical and Molecular Formulae | 实验式与分子式

The empirical formula gives the simplest whole‑number ratio of atoms in a compound. From percentage composition or combustion data, convert masses to moles, divide by the smallest number of moles, and obtain integer ratios. The molecular formula is a multiple of the empirical formula: molecular formula = (empirical formula)ₙ, where n = Mᵣ (molecular) / Mᵣ (empirical). Both IB and AQA ask students to deduce these from experimental data.

实验式表示化合物中各原子最简整数比。根据元素质量分数或燃烧数据，将质量换算为物质的量，除以最小物质的量得到整数比。分子式是实验式的整数倍：分子式 = (实验式)ₙ，其中 n = 相对分子质量 / 实验式相对质量。IB和AQA均会要求考生根据实验数据推导。

Worked example: A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Its relative molecular mass is 180. Determine its molecular formula.

典型例题：某化合物含碳 40.0%、氢 6.7% 和氧 53.3%，相对分子质量为 180。确定其分子式。

Element	% mass	moles (÷ Aᵣ)	ratio (÷ smallest)
C	40.0	40.0/12.0 = 3.33	1
H	6.7	6.7/1.0 = 6.7	2
O	53.3	53.3/16.0 = 3.33	1

Empirical formula is CH₂O; empirical mass = 12.0 + (2×1.0) + 16.0 = 30.0. n = 180 / 30.0 = 6, so molecular formula = C₆H₁₂O₆.

实验式为 CH₂O；实验式质量= 30.0。n = 180 / 30.0 = 6，分子式为 C₆H₁₂O₆。

3. Reacting Masses and Limiting Reactants | 反应质量与限量试剂

Stoichiometric calculations link masses of reactants and products via the mole ratio in the balanced equation. Always identify the limiting reactant first: convert all given masses to moles, then compare the actual mole ratio with the theoretical ratio. The reactant that produces the least amount of product is limiting. AQA often embeds this in structured questions; IB may include it in data‑based problems.

化学计量计算通过配平方程式中的摩尔比关联反应物与产物的质量。务必先找出限量试剂：将所有给定量换算为物质的量，将实际摩尔比与理论摩尔比进行比较，生成产物物质的量最少的反应物即为限量试剂。AQA常在结构化试题中考查，IB可能以数据题形式出现。

Mass A → mol A → mol B → mass B

Worked example: 2.00 g of magnesium reacts with 5.00 g of oxygen to form magnesium oxide. Which reactant is in excess? (Mg = 24.3, O = 16.0).

典型例题：2.00 g镁与 5.00 g氧气反应生成氧化镁。哪种反应物过量？

n(Mg) = 2.00/24.3 = 0.0823 mol; n(O₂) = 5.00/32.0 = 0.156 mol. Equation 2Mg + O₂ → 2MgO requires Mg:O₂ = 2:1. For 0.0823 mol Mg, O₂ needed = 0.0412 mol. Available O₂ is 0.156 mol, so O₂ is in excess, Mg is limiting. Maximum mass of MgO = 0.0823 mol × (24.3+16.0) g mol⁻¹ = 3.32 g.

n(Mg) = 2.00/24.3 = 0.0823 mol；n(O₂) = 0.156 mol。反应 2Mg + O₂ → 2MgO，Mg 与 O₂ 摩尔比 2:1。0.0823 mol Mg 需 O₂ 0.0412 mol，现有 O₂ 过量，Mg 为限量试剂。MgO 最大质量 = 0.0823 × 40.3 = 3.32 g。

4. Gas Volume and the Ideal Gas Law | 气体体积与理想气体定律

For gases at room temperature and pressure (RTP), molar volume is 24 dm³ mol⁻¹ (AQA) or 22.7 dm³ mol⁻¹ at STP (IB uses 273 K, 100 kPa). When conditions differ, apply the ideal gas equation pV = nRT. Use R = 8.31 J K⁻¹ mol⁻¹ for pressure in Pa and volume in m³, or R = 0.0821 L atm K⁻¹ mol⁻¹. Convert units carefully—a common pitfall in both IB and AQA papers.

在常温常压下，气体摩尔体积为 24 dm³ mol⁻¹（AQA），IB 在标准状况（0 °C, 100 kPa）下使用 22.7 dm³ mol⁻¹。若条件不同，则使用理想气体状态方程 pV = nRT。压强单位为 Pa、体积为 m³ 时 R = 8.31 J K⁻¹ mol⁻¹；若压强为 atm、体积为 L，则 R = 0.0821 L atm K⁻¹ mol⁻¹。务必小心换算单位——这是两套试卷中的常见失分点。

pV = nRT

Worked example: Calculate the volume occupied by 0.500 mol of N₂ at 35 °C and 105 kPa. (R = 8.31 J K⁻¹ mol⁻¹)

典型例题：计算 0.500 mol N₂ 在 35 °C、105 kPa 下所占体积。

T = 35 + 273 = 308 K; p = 105 000 Pa. V = nRT / p = (0.500 × 8.31 × 308) / 105000 = 0.0122 m³ = 12.2 dm³. Always express volume in the requested unit.

T = 35 + 273 = 308 K；p = 105000 Pa。V = nRT / p = (0.500 × 8.31 × 308) / 105000 = 0.0122 m³ = 12.2 dm³。最后按题目要求单位作答。

5. Solution Concentration and Dilution | 溶液浓度与稀释

Concentration c (mol dm⁻³) links amount of solute and volume: c = n / V. For dilution, the number of moles stays the same: c₁V₁ = c₂V₂. IB often asks students to prepare a standard solution and to carry out dilution steps; AQA may embed dilution in titration contexts. Always convert volumes to dm³ (1 dm³ = 1000 cm³).

浓度 c（mol dm⁻³）联系溶质的物质的量与溶液体积：c = n / V。稀释时溶质的物质的量不变：c₁V₁ = c₂V₂。IB 常要求配制标准溶液并进行稀释计算；AQA 则在滴定场景中考查稀释。注意将体积单位统一为 dm³（1 dm³ = 1000 cm³）。

c = n / V c₁V₁ = c₂V₂

Worked example: What volume of 2.00 mol dm⁻³ HCl is needed to prepare 250 cm³ of 0.100 mol dm⁻³ HCl?

典型例题：配制 250 cm³ 0.100 mol dm⁻³ HCl 需要多少体积的 2.00 mol dm⁻³ HCl？

V₁ = (c₂V₂) / c₁ = (0.100 × 0.250) / 2.00 = 0.0125 dm³ = 12.5 cm³. Pipette 12.5 cm³ of stock acid into a 250 cm³ volumetric flask and make up to the mark with distilled water.

V₁ = (c₂V₂) / c₁ = (0.100 × 0.250) / 2.00 = 0.0125 dm³ = 12.5 cm³。用移液管量取 12.5 cm³ 浓酸至 250 cm³ 容量瓶中，加蒸馏水至刻度线即可。

6. Titration Calculations | 滴定计算

Titrations test your ability to use concordant results to determine an unknown concentration. The general path: calculate average titre (discarding anomalous results), use the known concentration to find the moles of the standard solution, apply the reacting mole ratio, and finally calculate the concentration of the analyte. AQA practical assessments heavily feature acid–base titrations; IB also requires a titration as part of the internal assessment portfolio.

滴定考查利用合理重复数据计算未知浓度的能力。一般步骤为：计算平均滴定体积（剔除异常值），利用标准液浓度求其物质的量，根据反应摩尔比确定分析物的物质的量，最后求出其浓度。AQA 的实验测评常考酸碱滴定；IB 同样要求将滴定作为内部评估的一部分。

Worked example: 25.0 cm³ of NaOH solution required a mean titre of 22.40 cm³ of 0.100 mol dm⁻³ HCl for neutralisation. Find the concentration of NaOH.

典型例题：25.0 cm³ NaOH 溶液用 0.100 mol dm⁻³ HCl 滴定，平均体积为 22.40 cm³。求 NaOH 的浓度。

n(HCl) = 0.100 × 0.02240 = 2.24×10⁻³ mol. Reaction HCl + NaOH → NaCl + H₂O; mole ratio 1:1. n(NaOH) = 2.24×10⁻³ mol. c(NaOH) = 2.24×10⁻³ / 0.0250 = 0.0896 mol dm⁻³. Show consistent significant figures based on the titre.

n(HCl) = 0.100 × 0.02240 = 2.24×10⁻³ mol。反应计量比 1:1，故 n(NaOH) = 2.24×10⁻³ mol。c(NaOH) = 2.24×10⁻³ / 0.0250 = 0.0896 mol dm⁻³。有效数字应与滴定体积一致。

7. Enthalpy Change and Calorimetry | 焓变与量热法

Calorimetry calculations use q = mcΔT, where q is the heat energy (J), m is the mass of the solution (g; approximate density as 1 g cm⁻³), c is the specific heat capacity (4.18 J g⁻¹ K⁻¹), and ΔT is the temperature change. The enthalpy change per mole, ΔH, is found by dividing q by the number of moles of the limiting reactant and adding the correct sign (exothermic negative, endothermic positive). IB expects uncertainty propagation with the ΔT measurement; AQA focuses on accurate processing of the data.

量热法计算基于 q = mcΔT，其中 q 为热能（J），m 为溶液质量（g，密度近似 1 g cm⁻³），c 为比热容（4.18 J g⁻¹ K⁻¹），ΔT 为温度变化。摩尔焓变 ΔH 由 q 除以限量试剂物质的量得到，并赋予正确符号（放热为负，吸热为正）。IB 要求给出温度变化的不确定度传递，AQA 侧重数据处理的精确性。

q = mcΔT ΔH = –q / n (exothermic)

Worked example: 0.80 g of magnesium was added to 100 cm³ of excess HCl in a polystyrene cup. The temperature rose by 15.5 °C. Calculate ΔH per mole of Mg. (c = 4.18 J g⁻¹ K⁻¹, Mg = 24.3)

典型例题：将 0.80 g 镁加入盛有 100 cm³ 过量盐酸的聚苯乙烯杯中，温度升高 15.5 °C。计算每摩尔 Mg 的 ΔH。

m(solution) ≈ 100 g; q = 100 × 4.18 × 15.5 = 6479 J = 6.48 kJ. n(Mg) = 0.80 / 24.3 = 0.0329 mol. ΔH = –6.48 kJ / 0.0329 mol = –197 kJ mol⁻¹ (to 3 s.f.). The negative sign indicates an exothermic reaction.

溶液质量≈100 g；q = 100 × 4.18 × 15.5 = 6479 J = 6.48 kJ。n(Mg) = 0.80 / 24.3 = 0.0329 mol。ΔH = –6.48 kJ / 0.0329 mol = –197 kJ mol⁻¹ (三位有效数字)。负号表示放热反应。

8. Hess’s Law and Enthalpy Cycles | 赫斯定律与焓变循环

Hess’s Law states that the enthalpy change of a reaction is independent of the route taken. Construct a cycle linking ΔHf° or ΔHc° values to the target reaction. IB may require Born‑Haber cycles for lattice enthalpy; AQA also tests Hess cycles using enthalpies of formation, combustion, or mean bond enthalpies. The key equation: ΔHᵣₑₐ꜀ₜᵢₒₙ = Σ ΔHf°(products) – Σ ΔHf°(reactants).

赫斯定律指出反应的焓变与途径无关。可通过标准生成焓或燃烧焓构建能量循环，求解目标反应。IB 可能要求通过玻恩-哈伯循环计算晶格焓；AQA 则常见生成焓、燃烧焓或平均键焓的赫斯循环。关键公式：ΔHᵣₑₐ꜀ₜᵢₒₙ = Σ ΔHf°(生成物) – Σ ΔHf°(反应物)。

ΔHᵣₑₐ꜀ₜᵢₒₙ = Σ ΔHf°(products) – Σ ΔHf°(reactants)

Worked example: Use the following data to find ΔH for the reaction 2SO₂(g) + O₂(g) → 2SO₃(g). ΔHf°(SO₂) = –297 kJ mol⁻¹, ΔHf°(SO₃) = –396 kJ mol⁻¹.

典型例题：利用以下数据计算反应 2SO₂(g) + O₂(g) → 2SO₃(g) 的 ΔH。ΔHf°(SO₂) = –297 kJ mol⁻¹, ΔHf°(SO₃) = –396 kJ mol⁻¹。

ΔHᵣₑₐ꜀ₜᵢₒₙ = [2×(–396)] – [2×(–297) + 0] = –792 – (–594) = –198 kJ mol⁻¹. The value is per mole of reaction as written (2 mol SO₃). Always specify the reaction equation.

ΔHᵣₑₐ꜀ₜᵢₒₙ = [2×(–396)] – [2×(–297) + 0] = –792 – (–594) = –198 kJ mol⁻¹。该值基于所写方程式（生成 2 mol SO₃）。务必注明对应的反应方程式。

9. Kc and Kp Equilibrium Calculations | 平衡常数 Kc 与 Kp 计算

Equilibrium constants quantify the position of equilibrium. For homogeneous systems, Kc uses concentrations (mol dm⁻³); Kp uses partial pressures (Pa, atm, or bar). The expression follows the stoichiometric coefficients as exponents. Use an ICE (Initial, Change, Equilibrium) table to organise data. IB covers both Kc and Kp; AQA includes Kc in AS and Kp in A2 topics. Remember: solids and pure liquids are omitted from the expression.

平衡常数用于量化平衡位置。均相体系中，Kc 基于浓度（mol dm⁻³），Kp 基于分压（Pa、atm 或 bar）。平衡表达式中各物质的指数等于其化学计量数。使用 ICE 表（初始、变化、平衡）整理数据。IB 同时涉及 Kc 和 Kp；AQA 在 AS 阶段考查 Kc，A2 阶段加入 Kp。切记固体和纯液体不写入表达式。

aA + bB ⇌ cC + dD Kc = [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ

Worked example: For the equilibrium 2HI(g) ⇌ H₂(g) + I₂(g) at 700 K, 0.200 mol HI was placed

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IB & AQA Chemistry: Calculation Practice | IB & AQA 化学：计算题专项训练