📚 IB & Edexcel Chemistry: Calculation Drill | IB与Edexcel化学计算题专项训练
Calculation questions form the backbone of chemistry assessments in both IB and Edexcel specifications. From mole conversions to pH of buffer solutions, mastering quantitative problem‑solving is essential for reaching top grades. This intensive drill covers the core calculation topics that appear repeatedly in Paper 2 (IB), Paper 1 & 2 (Edexcel), and the individual investigation. Each section breaks down the key formula, walks through a typical worked example, and highlights common pitfalls. The aim is to build the speed and accuracy you need for the exam, whether you are sitting IB SL/HL or Edexcel AS/A Level.
计算题是 IB 和 Edexcel 化学考试的核心组成部分。从摩尔换算到缓冲溶液的 pH，掌握定量问题解决能力是取得高分的关键。本专项训练涵盖了 IB 试卷二、Edexcel 试卷一与二中反复出现的核心计算题型，以及个人研究项目中涉及的运算。每一节都会分解核心公式，逐步展示典型例题，并指出常见陷阱。无论你参加的是 IB SL/HL 还是 Edexcel AS/A Level 考试，目标都是帮你提高解题速度和准确性。
1. Mole Concepts and Avogadro’s Number | 摩尔概念与阿伏伽德罗常数
The mole is the central unit for chemical quantity. One mole of any substance contains 6.02 × 10²³ elementary entities. In calculations, always write down which definition of the mole you are using: from mass, gas volume, or number of particles.
摩尔是化学量的核心单位。1 摩尔任何物质含有 6.02 × 10²³ 个基本单元。计算时一定要先明确你使用的是哪种摩尔定义：由质量定义、由气体体积定义，还是由粒子数定义。
The most frequently used relationship is n = m / M, where m is the mass in grams and M is the molar mass in g mol⁻¹. If a sample has m = 5.85 g of NaCl and M = 58.5 g mol⁻¹, then n = 5.85 / 58.5 = 0.100 mol. For gases at a given temperature and pressure, n = V / Vₘ, where Vₘ is the molar volume. IB students commonly use Vₘ = 22.7 dm³ mol⁻¹ at STP (0 °C, 1 atm); Edexcel uses 24.0 dm³ mol⁻¹ at RTP (25 °C, 1 atm). Always check the conditions given in the question.
最常用的关系是 n = m / M，其中 m 为质量（g），M 为摩尔质量（g mol⁻¹）。例如，若 m = 5.85 g NaCl，M = 58.5 g mol⁻¹，则 n = 5.85 / 58.5 = 0.100 mol。对于给定温度和压力下的气体，n = V / Vₘ，其中 Vₘ 为摩尔体积。IB 考生通常在 STP（0 °C、1 atm）下使用 Vₘ = 22.7 dm³ mol⁻¹；Edexcel 在 RTP（25 °C、1 atm）下使用 24.0 dm³ mol⁻¹。务必确认题目给出的条件。
Common mistake: using a molar mass with incorrect units or confusing the molar volume of a gas with that of a liquid. Always convert mass to grams and volume to dm³ before substituting into the formula.
常见错误：摩尔质量单位错误，或把气体的摩尔体积与液体体积混淆。代入公式前，务必将质量换算为克、体积换算为 dm³。
2. Empirical and Molecular Formulae | 实验式与分子式
An empirical formula shows the simplest whole‑number ratio of atoms in a compound, while the molecular formula shows the actual number of atoms of each element in a molecule. The empirical formula is determined from the percentage composition by mass or from combustion data.
实验式表示化合物中各原子的最简整数比，分子式则表示一个分子中各原子的实际数目。实验式由质量百分组成或燃烧数据确定。
The stepwise approach is: (i) divide the mass or percentage of each element by its relative atomic mass to obtain the mole ratio; (ii) divide each mole value by the smallest to get the simplest ratio; (iii) if the ratio is not whole, multiply by a suitable factor. The molecular formula is then found by dividing the relative molecular mass by the mass of the empirical formula unit, and multiplying the subscripts by this integer.
步骤为：(i) 将每种元素的质量或百分比除以其相对原子质量，得到摩尔比；(ii) 将各摩尔值除以其中的最小值，得到最简比；(iii) 若比值不是整数，则乘以适当因子。再以相对分子质量除以实验式单元质量，将下标乘以该整数，即得分子式。
For example, a compound contains 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. Assuming 100 g, moles: C = 40.0 / 12.0 = 3.33, H = 6.7 / 1.0 = 6.7, O = 53.3 / 16.0 = 3.33. Dividing by 3.33 gives ratio C:H:O = 1:2:1, so empirical formula is CH₂O. If the molecular mass is 180 g mol⁻¹, CH₂O mass = 30, multiple = 6, molecular formula = C₆H₁₂O₆.
例如，某化合物含碳 40.0%、氢 6.7%、氧 53.3%。假设质量为 100 g，物质的量：C = 40.0/12.0 = 3.33，H = 6.7/1.0 = 6.7，O = 53.3/16.0 = 3.33。除以 3.33 得比例 C:H:O = 1:2:1，实验式为 CH₂O。若其分子量为 180 g mol⁻¹，CH₂O 质量为 30，倍数 = 6，分子式即为 C₆H₁₂O₆。
3. Mass–Mole–Volume Relationships in Reactions | 反应中的质量-摩尔-体积关系
Stoichiometric coefficients in a balanced equation give the mole ratio between reactants and products. To solve mass‑volume problems, change the given quantity into moles, use the mole ratio, and convert the unknown into the desired unit.
配平方程中的化学计量系数给出了反应物与产物之间的摩尔比。解决质量‑体积问题时，先将已知量化为物质的量，利用摩尔比，再将未知量换算为目标单位。
A typical IB / Edexcel question: what volume of CO₂ (at RTP) is produced when 10.0 g of CaCO₃ decomposes? CaCO₃ → CaO + CO₂. n(CaCO₃) = 10.0 / 100.1 = 0.0999 mol. 1:1 ratio gives n(CO₂) = 0.0999 mol. At RTP, V = n × 24.0 = 2.40 dm³. At STP the volume would be 0.0999 × 22.7 = 2.27 dm³. Always state the temperature and pressure because the molar volume differs.
典型的 IB / Edexcel 问题：10.0 g CaCO₃ 分解可产生多少体积的 CO₂（RTP 条件下）? 反应 CaCO₃ → CaO + CO₂。n(CaCO₃) = 10.0 / 100.1 = 0.0999 mol。1:1 得 n(CO₂) = 0.0999 mol。在 RTP 下，V = n × 24.0 = 2.40 dm³。若在 STP 下则为 0.0999 × 22.7 = 2.27 dm³。必须注明温度和压力，因为摩尔体积不同。
For solutions, we use concentration and volume: n = c × V (in dm³). Combining this with the mole ratio allows calculations involving titrations and precipitations.
对于溶液，使用浓度与体积：n = c × V（单位 dm³）。将此与摩尔比结合，即可进行涉及滴定和沉淀的反应计算。
4. Limiting Reactants and Percentage Yield | 限量反应物与产率
The limiting reactant is the substance that is completely consumed first, determining the maximum amount of product. The other reactants are in excess. To identify the limiting reactant, calculate the moles of each reactant and divide by its stoichiometric coefficient; the smallest value indicates the limiting species.
限量反应物是最先完全消耗的物质，决定了产物的最大量。其他反应物则为过量。确定限量反应物的方法是：计算每种反应物的物质的量，除以其计量系数，比值最小者即为限量物。
For instance, 2Al + 3Cl₂ → 2AlCl₃. If 0.50 mol Al is mixed with 0.60 mol Cl₂: Al ratio = 0.50/2 = 0.25; Cl₂ ratio = 0.60/3 = 0.20. Cl₂ gives the smaller value, so Cl₂ is limiting. Theoretical yield of AlCl₃ = 0.60 × (2 mol AlCl₃ / 3 mol Cl₂) = 0.40 mol. If the actual yield is 0.35 mol, percentage yield = (0.35/0.40) × 100% = 87.5%.
例如，2Al + 3Cl₂ → 2AlCl₃。若将 0.50 mol Al 与 0.60 mol Cl₂ 混合：Al 的比值 = 0.50/2 = 0.25；Cl₂ 的比值 = 0.60/3 = 0.20。Cl₂ 较小，故为限量。AlCl₃ 的理论产量 = 0.60 × (2/3) = 0.40 mol。若实际产量为 0.35 mol，则产率 = (0.35/0.40) × 100% = 87.5%。
Exam tip: never assume the reactant with smaller mass is limiting – always compare moles. Also, percentage yield is reduced by side reactions, incomplete reaction, and losses during purification.
考试技巧：切勿假设质量较小的反应物即为限量，必须比较物质的量。同时，副反应、反应不完全以及纯化过程中的损失都会降低产率。
5. Gas Law Calculations | 气体定律计算
The ideal gas equation pV = nRT links pressure, volume, temperature, and moles. R is the gas constant, 8.31 J mol⁻¹ K⁻¹. All quantities must be in SI units: pressure in pascals (Pa), volume in m³, temperature in kelvin, and n in mol.
理想气体状态方程 pV = nRT 将压强、体积、温度与物质的量关联起来。R 为气体常数，8.31 J mol⁻¹ K⁻¹。所有物理量必须使用 SI 单位：压强用帕斯卡（Pa），体积用立方米（m³），温度用开尔文，n 用 mol。
Many candidates lose marks by forgetting to convert cm³ to m³ (divide by 10⁶) or kPa to Pa (multiply by 10³). For example, find the volume of 0.0500 mol of an ideal gas at 100 kPa and 298 K. p = 100 × 10³ Pa, T = 298 K. V = nRT/p = (0.0500 × 8.31 × 298) / (100 × 10³) = 1.24 × 10⁻³ m³ = 1.24 dm³. Alternatively, using the molar volume at RTP (24 dm³ mol⁻¹) gives 0.0500 × 24 = 1.20 dm³; the slight difference arises because RTP is defined as 20 °C in some contexts, so always use the given data if p and T are specified.
许多考生因忘记将 cm³ 转换为 m³（除以 10⁶）或将 kPa 转换为 Pa（乘以 10³）而丢分。例如，求 0.0500 mol 理想气体在 100 kPa、298 K 时的体积：p = 100 × 10³ Pa，T = 298 K。V = nRT/p = (0.0500 × 8.31 × 298) / (100 × 10³) = 1.24 × 10⁻³ m³ = 1.24 dm³。若用 RTP 下的摩尔体积（24 dm³ mol⁻¹）则得 0.0500 × 24 = 1.20 dm³，微小差异源于 RTP 有时定义为 20 °C，因此若题目给出了 p 和 T，务必采用给定数值。
When a question asks for the relative molecular mass of a volatile liquid using the Dumas method, rearrange pV = nRT and n = m/M to give M = mRT / pV. Always measure the mass of the gas, not the liquid, and ensure the flask is completely filled with vapour.
当题目涉及使用杜马法测定挥发性液体的相对分子质量时，可联立 pV = nRT 与 n = m/M，得到 M = mRT / pV。一定要测量气体的质量，而非液体的质量，并确保烧瓶完全充满蒸气。
6. Solution Stoichiometry and Titration | 溶液计量学与滴定
Titration calculations rely on the ability to find the exact number of moles of the standard solution and then apply the balanced equation. The basic formula is n = c × V, where V is in dm³. For example, 25.0 cm³ of 0.100 mol dm⁻³ HCl is titrated against NaOH. At equivalence, n(HCl) = 0.0250 dm³ × 0.100 = 0.00250 mol. If the reaction is 1:1, n(NaOH) = 0.00250 mol.
滴定计算需要准确求出标准溶液的物质的量，再结合配平方程式。基本公式为 n = c × V，其中 V 单位为 dm³。例如，用 25.0 cm³ 0.100 mol dm⁻³ HCl 滴定 NaOH。在等当点时，n(HCl) = 0.0250 dm³ × 0.100 = 0.00250 mol。若反应为 1:1，则 n(NaOH) = 0.00250 mol。
Back titration is used when the analyte is volatile, insoluble, or reacts slowly. A known excess of reagent A is added, and the unreacted A is titrated with reagent B. The moles of analyte = (initial moles of A) − (moles of B used to react with excess A). Always draw a diagram of the two steps to avoid sign errors.
返滴定适用于分析物挥发性强、难溶或反应缓慢的情况。先加入已知过量的试剂 A，再用试剂 B 滴定未反应的 A。分析物的物质的量 =（A 的初始物质的量）−（与过量 A 反应的 B 的物质的量）。建议画出两步示意图，以免符号出错。
Common pitfalls: misreading the meniscus, using the titre volume without subtracting the initial reading, and forgetting to divide cm³ by 1000 to get dm³. Practice obtaining concordant titres within 0.10 cm³.
常见陷阱：读数错误、未用滴定管读数差、忘记将 cm³ 除以 1000 化为 dm³。平时应练习获得 0.10 cm³ 以内的吻合滴定体积。
7. Enthalpy Change and Calorimetry | 焓变与量热法
The heat absorbed or released in a reaction is calculated using q = mcΔT, where m is the mass of the solution (usually water), c is specific heat capacity (4.18 J g⁻¹ K⁻¹ for water), and ΔT is the temperature change. The enthalpy change per mole is then ΔH = −q / n, where n is the number of moles of the limiting reactant. The negative sign indicates that in an exothermic reaction, heat is released to the surroundings.
反应吸收或放出的热量通过 q = mcΔT 计算，其中 m 为溶液质量（通常为水），c 为比热容（水 4.18 J g⁻¹ K⁻¹），ΔT 为温度变化。摩尔焓变则为 ΔH = −q / n，n 为限量反应物的物质的量。负号表示放热反应中将热量释放给环境。
When 50.0 cm³ of 1.00 mol dm⁻³ HCl is added to 50.0 cm³ of 1.00 mol dm⁻³ NaOH, the temperature rises from 21.0 °C to 27.5 °C. Assuming the density of the solution is 1.00 g cm⁻³, total mass = 100 g. q = 100 × 4.18 × (27.5 − 21.0) = 2717 J. Moles of water formed = 0.0500 mol. ΔH = −2717 / 0.0500 = −54340 J mol⁻¹ ≈ −54.3 kJ mol⁻¹. Both IB and Edexcel allow this simplified treatment, as long as assumptions are stated.
将 50.0 cm³ 1.00 mol dm⁻³ HCl 与 50.0 cm³ 1.00 mol dm⁻³ NaOH 混合，温度从 21.0 °C 升至 27.5 °C。假设溶液密度为 1.00 g cm⁻³，总质量 = 100 g。q = 100 × 4.18 × (27.5 − 21.0) = 2717 J。生成水的物质的量 = 0.0500 mol。ΔH = −2717 / 0.0500 = −54340 J mol⁻¹ ≈ −54.3 kJ mol⁻¹。只要注明假设，IB 和 Edexcel 都接受这种简化的处理方法。
For combustion reactions, the heat released is proportional to the mass of fuel burned. Use a spirit burner and measure the temperature rise of a known mass of water. A common improvement is to use a draft shield to minimise heat loss to the surroundings.
对于燃烧反应，放出的热量与燃烧的燃料质量成正比。可使用酒精灯加热已知质量的水，并测量水温上升值。常见的改进方法是使用防风罩以减少热量流失。
8. Reaction Rate and Rate Equations | 反应速率与速率方程
The rate of a chemical reaction can be measured by monitoring the change in concentration of a reactant or product over time. The rate equation has the form: Rate = k [A]ᵐ [B]ⁿ, where m and n are the orders of reaction determined experimentally, not from the stoichiometric coefficients.
化学反应速率可通过监测反应物或产物浓度随时间的变化来测量。速率方程形如 Rate = k [A]ᵐ [B]ⁿ，其中 m 和 n 为由实验确定的反应级数，并非来自化学计量系数。
To determine the order with respect to a reactant, use the initial rates method: for two experiments where the concentration of one reactant changes while others are constant, the change in initial rate gives the order. If doubling [A] causes the initial rate to double, the order is 1; if the rate quadruples, order is 2; if the rate remains unchanged, order is 0. Once orders are known, k can be calculated using any set of data.
确定某一反应物的反应级数可使用初始速率法：在两次实验中，保持其他反应物浓度不变，只改变该反应物的浓度，观察初始速率的变化。若 [A] 加倍导致初速率加倍，则为一级；若速率变为四倍，则为二级；若速率不变，则为零级。得知级数后，代入任一组数据即可求出 k。
The units of k depend on the overall order. For a zero‑order reaction, k has units mol dm⁻³ s⁻¹; first‑order, s⁻¹; second‑order, dm³ mol⁻¹ s⁻¹. Edexcel and IB both require candidates to work out the units of k from the rate equation.
k 的单位取决于总反应级数。零级反应，k 的单位为 mol dm⁻³ s⁻¹；一级反应为 s⁻¹；二级反应为 dm³ mol⁻¹ s⁻¹。Edexcel 和 IB 都要求考生能根据速率方程推导 k 的单位。
9. Equilibrium Constant Kc and Kp | 平衡常数 Kc 与 Kp
For a homogeneous reaction at equilibrium, the equilibrium constant in terms of concentration, Kc, is given by the product of the concentrations of the products (each raised to its coefficient) divided by the product of the concentrations of the reactants. Solids and pure liquids are omitted.
对于均相反应，基于浓度的平衡常数 Kc 等于生成物浓度以计量系数为指数的乘积除以反应物浓度以计量系数为指数的乘积。固体和纯液体不写入表达式。
For gaseous equilibria, Kp uses partial pressures instead of concentrations. For example, in the Haber process N₂(g) + 3H₂(g) ⇌ 2NH₃(g), Kp = (p(NH₃))² / (p(N₂) × (p(H₂))³). Partial pressure = mole fraction × total pressure. The value of Kc and Kp depends only on temperature. A large K indicates the equilibrium lies far to the right; a small K indicates the reverse.
对于气体平衡，Kp 使用分压代替浓度。例如哈伯法 N₂(g) + 3H₂(g) ⇌ 2NH₃(g)，Kp = (p(NH₃))² / (p(N₂) × (p(H₂))³)。分压 = 摩尔分数 × 总压。Kc 与 Kp 的值只受温度影响。K 值很大表示平衡强烈偏向生成物一方；K 值很小则相反。
Calculating equilibrium compositions often involves setting up an ICE table (Initial, Change, Equilibrium). Use the equilibrium mole values to calculate concentrations or partial pressures, then substitute into the K expression. Students frequently forget to convert moles to concentrations before inserting into Kc or to divide by the total moles when working out mole fractions. Practice is key.
计算平衡组成时常使用 ICE 表格（初始、变化、平衡）。用平衡时的物质的量计算浓度或分压，再代入 K 的表达式。学生常犯的错误是：在代入 Kc 表达式前忘记将物质的量换算为浓度；或计算摩尔分数时忘记除以总物质的量。练习至关重要。
10. Acid–Base pH and Buffer Calculations | 酸碱pH与缓冲溶液计算
For strong monoprotic acids, pH = −log[H⁺], where [H⁺] equals the acid concentration. For strong bases, calculate [OH⁻] from concentration and stoichiometry, then use pOH = −log[OH⁻] and pH = 14 − pOH at 25 °C. Weak acids require the acid dissociation constant Ka: for HA ⇌ H⁺ + A⁻, Ka = [H⁺][A⁻] / [HA]. Assuming [H⁺] = [A⁻] and the change in [HA] is negligible, [H⁺] = √(Ka × [HA]₀).
对于强一元酸，pH = −log[H⁺]，其中 [H⁺] 等于酸的浓度。对于强碱，根据浓度和化学计量比求出 [OH⁻]，再通过 pOH = −log[OH⁻] 及 pH = 14 − pOH（25 °C）计算。弱酸需要使用酸解离常数 Ka：HA ⇌ H⁺ + A⁻，Ka = [H⁺][A⁻] / [HA]。假设 [H⁺] = [A⁻] 且 [HA] 的变化可忽略，则 [H⁺] = √(Ka × [HA]₀)。
Buffer solutions resist changes in pH. Their pH can be found using the Henderson–Hasselbalch equation: pH = pKa + log([A⁻]/[HA]), where [A⁻] is the concentration of the conjugate base and [HA] is the concentration of the weak acid. For a buffer made by partially neutralising a weak acid, calculate the moles of salt formed and the remaining weak acid after reaction with the strong base.
缓冲溶液能抵抗 pH 变化。其 pH 可通过亨德森‑哈塞尔巴尔赫方程计算：pH = pKa + log([A⁻]/[HA])，其中 [A⁻] 为共轭碱浓度，[HA] 为弱酸浓度。对于部分中和弱酸制备的缓冲溶液，需先行计算生成的盐的物质的量以及剩余弱酸的物质的量。
An example: a buffer is made from 0.100 mol ethanoic acid and 0.0500 mol sodium ethanoate in 1.00 dm³. Ka = 1.75 × 10⁻⁵, so pKa = 4.76. pH = 4.76 + log(0.0500/0.100) = 4.76 − 0.30 = 4.46. Both IB and Edexcel expect you to be able to derive the buffer equation from the Ka expression, not just quote it.
示例：用 0.100 mol 乙酸与 0.0500 mol 乙酸钠配制 1.00 dm³ 缓冲液。Ka = 1.75 × 10⁻⁵，故 pKa = 4.76。pH = 4.76 + log(0.0500/0.100) = 4.76 − 0.30 = 4.46。IB 和 Edexcel 都要求考生能从 Ka 表达式推导出缓冲方程，而非简单套用。
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