A-Level Edexcel Chemistry: Calculation Practice Drill | A-Level Edexcel 化学：计算题专项训练

📚 A-Level Edexcel Chemistry: Calculation Practice Drill | A-Level Edexcel 化学：计算题专项训练

Mastering calculations is essential for achieving top grades in A-Level Edexcel Chemistry. This drill focuses on the quantitative skills required by the specification, from simple mole conversions to complex equilibrium and pH calculations. Each section provides key formulas, worked examples, and common mistakes to avoid.

掌握计算对于在 A-Level Edexcel 化学中取得高分至关重要。本专项训练聚焦教学大纲所要求的定量技能，从简单的摩尔换算到复杂的平衡与 pH 计算。每一部分都提供关键公式、例题以及需避免的常见错误。

1. Mole Concepts and Basic Calculations | 摩尔概念与基本计算

The mole is the SI unit for amount of substance. One mole contains 6.02 × 10²³ particles (Avogadro constant). The fundamental equation linking mass and moles is n = m / M, where n is amount (mol), m is mass (g), and M is molar mass (g mol⁻¹).

摩尔是物质的量的国际单位。1 摩尔包含 6.02 × 10²³ 个粒子 (阿伏伽德罗常数)。联系质量与摩尔的基本方程为 n = m / M，其中 n 为物质的量 (mol)，m 为质量 (g)，M 为摩尔质量 (g mol⁻¹)。

Concentration c (mol dm⁻³) is linked by n = cV, where V must be in dm³. Remember that 1 dm³ = 1000 cm³.

溶液浓度 c (mol dm⁻³) 通过 n = cV 关联，其中 V 必须使用 dm³ 作单位。注意 1 dm³ = 1000 cm³。

Worked example: Calculate the number of moles in 8.5 g of NaNO₃ (M = 85 g mol⁻¹).

例题：计算 8.5 g NaNO₃ 中的物质的量 (M = 85 g mol⁻¹)。

Step 1: Use n = m / M. n = 8.5 g / 85 g mol⁻¹ = 0.10 mol.

步骤 1：使用 n = m / M。n = 8.5 g / 85 g mol⁻¹ = 0.10 mol。

Step 2: Check units and significant figures. The answer 0.10 mol is given to two significant figures.

步骤 2：检查单位和有效数字。答案 0.10 mol 取两位有效数字。

For solution calculations, convert cm³ to dm³ by dividing by 1000. 250 cm³ = 0.250 dm³. If 0.050 mol is dissolved in 0.250 dm³, c = 0.050 / 0.250 = 0.20 mol dm⁻³.

在溶液计算中，将 cm³ 除以 1000 转化为 dm³。250 cm³ = 0.250 dm³。若 0.050 mol 溶于 0.250 dm³，则 c = 0.050 / 0.250 = 0.20 mol dm⁻³。

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

The empirical formula is the simplest whole-number ratio of atoms in a compound. It can be found from percentage composition by mass or combustion data.

实验式是化合物中各原子最简单整数比，可通过质量百分组成或燃烧分析数据求得。

Worked example: A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Determine its empirical formula. (Aᵣ: H = 1, C = 12, O = 16)

例题：某化合物含碳 40.0%、氢 6.7%、氧 53.3% (质量分数)。求其实验式。(Aᵣ: H = 1, C = 12, O = 16)

Step 1: Assume 100 g sample, so masses: C = 40.0 g, H = 6.7 g, O = 53.3 g.

步骤 1：假设样品为 100 g，则各元素质量：C = 40.0 g，H = 6.7 g，O = 53.3 g。

Step 2: Convert masses to moles: C: 40.0/12 = 3.33 mol; H: 6.7/1 = 6.7 mol; O: 53.3/16 = 3.33 mol.

步骤 2：将质量转化为物质的量：C: 40.0/12 = 3.33 mol；H: 6.7/1 = 6.7 mol；O: 53.3/16 = 3.33 mol。

Step 3: Divide by the smallest (3.33): C: 1, H: 2, O: 1. Empirical formula is CH₂O.

步骤 3：除以最小值 (3.33) 得：C: 1, H: 2, O: 1。实验式为 CH₂O。

The molecular formula is a multiple of the empirical formula (n = Mᵣ / empirical formula mass). If the relative molecular mass is 180, the empirical formula mass of CH₂O is 30, so n = 6 and molecular formula is C₆H₁₂O₆.

分子式是实验式的整数倍 (n = Mᵣ / 实验式量)。若相对分子质量为 180，CH₂O 式量为 30，则 n = 6，分子式为 C₆H₁₂O₆。

3. Gas Volume Calculations | 气体体积计算

At room temperature and pressure (RTP, 20 °C and 1 atm), the molar volume of an ideal gas is 24 dm³ mol⁻¹. Under standard conditions (0 °C, 1 atm) it is 22.4 dm³ mol⁻¹. Always check the condition specified in the question.

在常温常压下 (RTP, 20 °C, 1 atm)，理想气体摩尔体积为 24 dm³ mol⁻¹；在标准状况下 (0 °C, 1 atm) 则为 22.4 dm³ mol⁻¹。务必确认题目给定的条件。

The ideal gas equation is

PV = nRT

, where P is pressure in Pa, V is volume in m³, n is moles, T is temperature in K, and R = 8.31 J K⁻¹ mol⁻¹.

理想气体状态方程为 PV = nRT，其中 P 为压力 (Pa)，V 为体积 (m³)，n 为物质的量，T 为温度 (K)，R = 8.31 J K⁻¹ mol⁻¹。

To convert °C to K, add 273.

将 °C 转化为 K 需加上 273。

Worked example: Calculate the volume of 0.200 mol of gas at 100 kPa and 25 °C. (R = 8.31)

例题：计算 0.200 mol 气体在 100 kPa 和 25 °C 下的体积。(R = 8.31)

Step 1: Convert units: P = 100 kPa = 100,000 Pa; T = 25 + 273 = 298 K.

步骤 1：单位换算：P = 100 kPa = 100 000 Pa；T = 25 + 273 = 298 K。

Step 2: Rearrange PV = nRT to V = nRT/P.

步骤 2：将 PV = nRT 整理为 V = nRT/P。

Step 3: V = (0.200 × 8.31 × 298) / 100,000 = 0.00495 m³. To convert to dm³, multiply by 1000: 4.95 dm³.

步骤 3：V = (0.200 × 8.31 × 298) / 100 000 = 0.00495 m³。转化为 dm³ (乘以 1000) 得 4.95 dm³。

Equal volumes of gases at the same temperature and pressure contain equal numbers of moles. This allows volume ratios to be used directly in stoichiometry.

同温同压下，相同体积的气体含有相同的物质的量，因此在化学计量中可直接使用体积比。

4. Solution Concentration and Titrations | 溶液浓度与滴定

Concentration can be expressed in mol dm⁻³ or g dm⁻³. The conversion is: concentration (g dm⁻³) = concentration (mol dm⁻³) × M.

浓度可用 mol dm⁻³ 或 g dm⁻³ 表示，换算关系为：浓度 (g dm⁻³) = 浓度 (mol dm⁻³) × M。

In a titration, the equivalence point is reached when the moles of acid and base have reacted completely according to the stoichiometric ratio.

在滴定中，当酸与碱按化学计量比完全反应时即达到等当点。

Worked example: 25.0 cm³ of 0.100 mol dm⁻³ HCl neutralises 20.0 cm³ of NaOH solution. Calculate the concentration of NaOH.

例题：25.0 cm³ 的 0.100 mol dm⁻³ HCl 中和了 20.0 cm³ NaOH 溶液。计算 NaOH 的浓度。

Step 1: Reaction: HCl + NaOH → NaCl + H₂O. Mole ratio 1:1.

步骤 1：反应：HCl + NaOH → NaCl + H₂O，物质的量比为 1:1。

Step 2: Moles of HCl = 0.100 × (25.0/1000) = 0.00250 mol.

步骤 2：HCl 的物质的量 = 0.100 × (25.0/1000) = 0.00250 mol。

Step 3: Moles of NaOH = 0.00250 mol (1:1). Volume of NaOH = 20.0/1000 = 0.0200 dm³. Concentration = 0.00250 / 0.0200 = 0.125 mol dm⁻³.

步骤 3：NaOH 的物质的量 = 0.00250 mol (1:1)。NaOH 体积 = 20.0/1000 = 0.0200 dm³。浓度 = 0.00250 / 0.0200 = 0.125 mol dm⁻³。

When the ratio is not 1:1, multiply by the appropriate factor. For example, 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O, so n(H₂SO₄) = ½ n(NaOH).

当计量比不是 1:1 时，须乘以相应的系数。例如 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O，因此 n(H₂SO₄) = ½ n(NaOH)。

5. Enthalpy Changes | 焓变计算

The heat change in a reaction is often measured using a calorimeter. The equation Q = mcΔT gives the heat transferred, where m is the mass of solution (g), c is specific heat capacity (commonly 4.18 J g⁻¹ °C⁻¹ for water), and ΔT is the temperature change.

反应的热量变化常用量热计测量。公式 Q = mcΔT 计算传递的热量，其中 m 为溶液质量 (g)，c 为比热容 (水通常取 4.18 J g⁻¹ °C⁻¹)，ΔT 为温度变化。

To find ΔH (kJ mol⁻¹), divide Q by moles of limiting reactant and adjust sign (negative for exothermic, positive for endothermic).

为得到 ΔH (kJ mol⁻¹)，需将 Q 除以极限反应物的物质的量，并根据放热 (负值) 或吸热 (正值) 加上符号。

Worked example: When 0.0500 mol of acid is added to 50.0 g of water, the temperature rises by 6.0 °C. Calculate ΔH.

例题：当 0.0500 mol 酸加入到 50.0 g 水中，温度上升 6.0 °C。计算 ΔH。

Step 1: Q = 50.0 × 4.18 × 6.0 = 1254 J = 1.254 kJ. Step 2: ΔH = -1.254 kJ / 0.0500 mol = -25.1 kJ mol⁻¹ (exothermic).

步骤 1：Q = 50.0 × 4.18 × 6.0 = 1254 J = 1.254 kJ。步骤 2：ΔH = -1.254 kJ / 0.0500 mol = -25.1 kJ mol⁻¹ (放热)。

Hess’s Law states that the total enthalpy change for a reaction is independent of the route taken. Use enthalpy of formation or combustion data to construct cycles.

赫斯定律指出，反应的总焓变与途径无关。可利用生成焓或燃烧焓数据构造循环计算。

Bond enthalpy calculations: ΔH = Σ (bond enthalpies of bonds broken) – Σ (bond enthalpies of bonds formed). Be careful with values for average bond enthalpies in gaseous state.

键焓计算：ΔH = Σ (断裂键的键焓) − Σ (生成键的键焓)。注意使用气态下的平均键焓数据。

6. Equilibrium Constants (Kc and Kp) | 平衡常数 (Kc 与 Kp)

For a reaction aA + bB ⇌ cC + dD, the equilibrium constant in terms of concentration is Kc = [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ, using equilibrium concentrations in mol dm⁻³.

对于反应 aA + bB ⇌ cC + dD，以浓度表示的平衡常数 Kc = [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ，式中均使用平衡浓度 (mol dm⁻³)。

Worked example: For H₂ + I₂ ⇌ 2HI at equilibrium: [H₂] = 0.10, [I₂] = 0.20, [HI] = 0.40 mol dm⁻³. Calculate Kc.

例题：对于 H₂ + I₂ ⇌ 2HI，平衡浓度：[H₂] = 0.10，[I₂] = 0.20，[HI] = 0.40 mol dm⁻³。计算 Kc。

Kc = [HI]² / ([H₂][I₂]) = (0.40)² / (0.10 × 0.20) = 0.16 / 0.020 = 8.0. Units: (mol dm⁻³)² / (mol dm⁻³)², so Kc has no units here.

Kc = [HI]² / ([H₂][I₂]) = (0.40)² / (0.10 × 0.20) = 0.16 / 0.020 = 8.0。量纲：(mol dm⁻³)² / (mol dm⁻³)²，此处 Kc 无量纲。

For gas-phase equilibria, Kp uses partial pressures. Mole fraction of gas A = n(A) / total moles. Partial pressure p(A) = mole fraction × total pressure. Then Kp = p(C)ᶜp(D)ᵈ / p(A)ᵃp(B)ᵇ.

对于气相平衡，Kp 使用分压表示。A 的摩尔分数 = n(A) / 总物质的量；分压 p(A) = 摩尔分数 × 总压。然后 Kp = p(C)ᶜp(D)ᵈ / p(A)ᵃp(B)ᵇ。

Always express Kp with appropriate units, e.g., atm⁻¹ if the sum of stoichiometric coefficients on product side minus reactant side is negative.

始终注意 Kp 的相应单位，例如当生成物与反应物计量系数之和差值为负时，单位可能为 atm⁻¹。

7. Acid-Base pH Calculations | 酸碱 pH 计算

For strong acids, [H⁺] = acid concentration (for monoprotic). pH = -log₁₀[H⁺]. For strong bases, find [OH⁻] then use Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 298 K to get [H⁺].

对于强酸 (一元酸)，[H⁺] 等于酸浓度。pH = -log₁₀[H⁺]。对于强碱，先求 [OH⁻]，再利用 298 K 时 Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ 计算 [H⁺]。

Worked example: Calculate the pH of 0.0500 mol dm⁻³ HCl. [H⁺] = 0.0500, pH = -log₁₀(0.0500) = 1.30.

例题：计算 0.0500 mol dm⁻³ HCl 的 pH。[H⁺] = 0.0500，pH = -log₁₀(0.0500) = 1.30。

For a weak acid HA ⇌ H⁺ + A⁻, Kₐ = [H⁺][A⁻]/[HA]. Assuming [H⁺] = [A⁻] and that dissociation is small, [H⁺] ≈ √(Kₐ × c). Always check the approximation: if c/Kₐ > 500, it is valid.

对于弱酸 HA ⇌ H⁺ + A⁻，Kₐ = [H⁺][A⁻]/[HA]。假设 [H⁺] = [A⁻] 且解离度较小，则 [H⁺] ≈ √(Kₐ × c)。务必检验近似条件：c/Kₐ > 500 时近似成立。

Worked example: Calculate the pH of 0.100 mol dm⁻³ CH₃COOH (Kₐ = 1.8 × 10⁻⁵). [H⁺] = √(1.8×10⁻⁵ × 0.100) = 1.34×10⁻³, pH = 2.87.

例题：计算 0.100 mol dm⁻³ CH₃COOH 的 pH (

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A-Level Edexcel Chemistry: Calculation Practice Drill | A-Level Edexcel 化学：计算题专项训练