📚 IB & CIE Chemistry: Calculation Training | IB & CIE 化学：计算题专项训练
Calculation forms the backbone of IB and CIE Chemistry examinations. Whether you are tackling stoichiometric ratios, energy cycles, or equilibrium constants, numerical fluency and a logical approach are essential. This article consolidates the most common calculation types, spanning from moles to electrochemistry, with step-by-step bilingual explanations and targeted practice strategies. Master these topics to boost both speed and accuracy on exam day.
计算是 IB 和 CIE 化学考试的支柱。无论是处理化学计量比、能量循环还是平衡常数，数字流畅度和逻辑方法都至关重要。本文汇总了最常见的计算题型，涵盖从摩尔到电化学的各个领域，提供逐步的双语解释和针对性练习策略。掌握这些专题，助你在考试当天提升速度与准确度。
1. Moles & Stoichiometry | 摩尔与化学计量
The mole (mol) is the SI unit for amount of substance. One mole contains exactly 6.02 × 10²³ elementary entities (Avogadro constant, L). The central equation linking mass, molar mass and moles is n = m / M, where m is mass in grams and M is molar mass in g mol⁻¹.
摩尔（mol）是物质的量的国际单位。1 摩尔恰好包含 6.02 × 10²³ 个基本单元（阿伏伽德罗常数，L）。连接质量、摩尔质量和摩尔数的核心公式是 n = m / M，其中 m 为质量（克），M 为摩尔质量（克/摩尔）。
n = m / M
For gases at room temperature and pressure (RTP), molar volume Vₘ ≈ 24 dm³ mol⁻¹. Thus, n = V(gas) / 24 dm³. At standard temperature and pressure (STP), Vₘ = 22.7 dm³ mol⁻¹. Always check the conditions specified in the question.
对于室温常压（RTP）下的气体，摩尔体积 Vₘ ≈ 24 dm³ mol⁻¹。因此 n = V（气体）/ 24 dm³。在标准状况（STP）下，Vₘ = 22.7 dm³ mol⁻¹。务必根据题目给出的条件选择。
Balanced equations provide the mole ratios of reactants and products. For the reaction 2H₂ + O₂ → 2H₂O, 2 mol H₂ react with 1 mol O₂ to yield 2 mol H₂O. Use these ratios to convert between masses, gas volumes and solution concentrations.
配平的方程式给出反应物和生成物的摩尔比。对于反应 2H₂ + O₂ → 2H₂O，2 mol H₂ 与 1 mol O₂ 反应生成 2 mol H₂O。利用这些比值在质量、气体体积和溶液浓度之间进行换算。
2. Empirical & Molecular Formula | 经验式与分子式
The empirical formula gives the simplest whole-number ratio of atoms in a compound. To find it, convert percentage or mass data into moles by dividing by atomic masses, then divide all mole values by the smallest number of moles to obtain integer ratios.
经验式表示化合物中各原子的最简整数比。求算时，先将质量百分数或直接质量除以相对原子质量得到摩尔数，然后将所有摩尔数除以最小值，化为最简整数比。
The molecular formula is a whole-number multiple of the empirical formula. Determine the multiplier by dividing the relative molecular mass (Mᵣ) by the empirical formula mass. For example, if empirical formula is CH₂ and Mᵣ = 56, then n = 56 / 14 = 4, giving molecular formula C₄H₈.
分子式是经验式的整数倍。用相对分子质量（Mᵣ）除以经验式质量求得倍数。例如，若经验式为 CH₂，Mᵣ = 56，则倍数 n = 56 / 14 = 4，分子式为 C₄H₈。
Combustion analysis data can also be used. The masses of CO₂ and H₂O produced allow the determination of moles of C and H, from which the empirical formula is derived.
燃烧分析数据同样适用。通过生成的 CO₂ 和 H₂O 质量可推算出 C 和 H 的摩尔数，进而推导经验式。
3. Gas Calculations | 气体计算
The ideal gas equation, pV = nRT, relates pressure (Pa), volume (m³), moles, and temperature (K). R = 8.31 J K⁻¹ mol⁻¹. Always convert units: 1 atm = 1.01 × 10⁵ Pa, 1 dm³ = 10⁻³ m³, and °C to K by adding 273.
理想气体状态方程为 pV = nRT，涉及压力（Pa）、体积（m³）、摩尔数和温度（K）。R = 8.31 J K⁻¹ mol⁻¹。务必换算单位：1 atm = 1.01 × 10⁵ Pa，1 dm³ = 10⁻³ m³，摄氏温度加 273 转为开尔文。
pV = nRT
Example: Calculate the volume of 0.500 mol of gas at 298 K and 100 kPa. V = nRT/p = (0.500 × 8.31 × 298) / (100 × 10³) = 0.0124 m³ = 12.4 dm³.
例题：计算 0.500 mol 气体在 298 K、100 kPa 下的体积。V = nRT/p = (0.500 × 8.31 × 298) / (100 × 10³) = 0.0124 m³ = 12.4 dm³。
Where RTP is assumed, you may use the simpler molar volume (24 dm³) to save time. Combine gas volumes with stoichiometric coefficients in reactions.
若题目假定 RTP，可直接使用摩尔体积（24 dm³）以节省时间。气体体积计算常与反应的化学计量系数结合。
4. Solution Concentration & Dilution | 溶液浓度与稀释
Concentration (c) is amount of solute per unit volume, normally mol dm⁻³. The key formula is c = n / V. Be careful with volume units: 1 dm³ = 1000 cm³. To prepare a solution, dissolve a known mass in solvent and transfer to a volumetric flask.
浓度（c）是单位体积溶液中溶质的量，通常单位为 mol dm⁻³。核心公式为 c = n / V。注意体积单位：1 dm³ = 1000 cm³。配制溶液时，将已知质量溶解后转移至容量瓶定容。
For dilutions, the number of moles stays constant: c₁V₁ = c₂V₂. This is used when making a weaker solution from a stock solution. Always use the same volume unit on both sides.
稀释时溶质的摩尔数保持不变：c₁V₁ = c₂V₂。从浓溶液配制稀溶液时常用此公式。等式两边体积单位须一致。
c₁V₁ = c₂V₂
Mass concentration (g dm⁻³) can be linked to molar concentration by multiplying by molar mass: mass conc. = c × M.
质量浓度（g dm⁻³）与摩尔浓度的关系：质量浓度 = c × M（摩尔质量）。
5. Titration Calculations | 滴定计算
In a titration, the unknown concentration is found from the known concentration, the reacting volumes, and the mole ratio from the balanced equation. For acid–base titrations, the simple formula (cₐVₐ) / (c₆V₆) = nₐ/n₆ can be applied, where a = acid, b = base.
通过滴定，根据已知浓度、反应体积和配平方程式的摩尔比，可求出未知浓度。酸碱滴定中，常用公式 (cₐVₐ) / (c₆V₆) = nₐ/n₆，其中 a 为酸，b 为碱。
Always convert cm³ to dm³ before calculation. For redox titrations, such as MnO₄⁻ / Fe²⁺, write the half-equations to determine the electron mole ratio. Example: 5Fe²⁺ + MnO₄⁻ + 8H⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O. Here 5 mol Fe²⁺ react with 1 mol MnO₄⁻.
计算前务必将 cm³ 转换为 dm³。对于氧化还原滴定，如 KMnO₄ 滴定 Fe²⁺，需写出半反应以确定电子转移摩尔比。例如：5Fe²⁺ + MnO₄⁻ + 8H⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O，5 mol Fe²⁺ 与 1 mol MnO₄⁻ 反应。
Back titrations are used when the substance is insoluble or volatile. Excess reagent is added, then the leftover is titrated. Moles reacted = initial moles − moles leftover.
返滴定适用于难溶或挥发性物质。先加入过量试剂，再滴定剩余的量。反应的摩尔数 = 初始摩尔数 − 剩余摩尔数。
6. Yield & Atom Economy | 产率与原子经济
Percentage yield = (actual yield / theoretical yield) × 100. Theoretical yield is the maximum mass of product predicted from the limiting reactant. Actual yield is the mass obtained from the experiment. Low yields may be due to incomplete reaction, side reactions, or purification losses.
产率百分数 = (实际产量 / 理论产量) × 100。理论产量是根据限制反应物预测的最大产物质量。实际产量是实验得到的产品质量。产率低可能由于反应不完全、副反应或纯化损失。
Percentage yield = (actual yield / theoretical yield) × 100
Atom economy measures the efficiency of a reaction in incorporating atoms into the desired product. Atom economy = (molar mass of desired product / total molar mass of all reactants) × 100. High atom economy is desirable for green chemistry.
原子经济性衡量反应中原子有效融入目标产物的效率。原子经济性 = (目标产物摩尔质量 / 所有反应物摩尔质量总和) × 100。高原子经济性符合绿色化学的要求。
7. Energetics Calculations | 能量学计算
Calorimetry experiments give temperature changes (ΔT) from which heat transferred is calculated: q = m × c × ΔT, where c is specific heat capacity (usually 4.18 J g⁻¹ K⁻¹ for water). The enthalpy change is then ΔH = −q / n, where n is moles of limiting reactant.
量热实验得到温度变化（ΔT），由此计算热传递：q = m × c × ΔT，其中 c 是比热容（水通常为 4.18 J g⁻¹ K⁻¹）。焓变 ΔH = −q / n，n 为限制反应物的摩尔数。
q = m c ΔT
Hess’s law states that the total enthalpy change for a reaction is independent of the path. Construct cycles using enthalpies of formation or combustion to find unknown ΔH values.
盖斯定律指出，反应的总焓变与途径无关。可利用生成焓或燃烧焓构建循环，计算未知的 ΔH。
Calculate ΔH from bond enthalpies: ΔH = Σ (bond energies broken) − Σ (bond energies formed). Remember that bond enthalpies are average values and may differ from experimental results.
通过键能计算 ΔH：ΔH = Σ（断裂键键能总和）− Σ（形成键键能总和）。注意键能是平均值，可能与实验结果有差异。
8. Chemical Equilibrium | 化学平衡
The equilibrium constant Kc is expressed in terms of concentrations: for aA + bB ⇌ cC + dD, Kc = [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ. Only gases and aqueous species appear; solids and pure liquids are omitted. Units of Kc depend on the stoichiometry.
平衡常数 Kc 以浓度表示：对于 aA + bB ⇌ cC + dD，Kc = [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ。只包含气体和水溶液物种，固体和纯液体不列入。Kc 的单位取决于化学计量数。
When given initial amounts and equilibrium concentration of one species, use an ICE table (Initial, Change, Equilibrium) to find equilibrium concentrations, then substitute into Kc expression.
若已知初始量和一种物质的平衡浓度，使用 ICE 表格（起始、变化、平衡）求算所有平衡浓度，再代入 Kc 表达式。
The reaction quotient Qc has the same form as Kc but uses initial concentrations. Compare Qc with Kc to predict the direction of shift: if Qc < Kc, reaction proceeds forward; if Qc > Kc, reverse.
反应商 Qc 与 Kc 形式相同，但使用起始浓度。比较 Qc 与 Kc 可判断平衡移动方向：Qc < Kc，正向移动；Qc > Kc，逆向移动。
9. Acid–Base Calculations | 酸碱计算
The pH of a strong acid is given by pH = −log[H⁺]. For strong monoprotic acids, [H⁺] equals the acid concentration. The hydrogen ion concentration is [H⁺] = 10⁻ᵖᴴ. For strong bases, find [OH⁻] from concentration, then use Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 298 K to get pH.
强酸溶液的 pH = −log[H⁺]。对于一元强酸，[H⁺] 等于酸的浓度。氢离子浓度 [H⁺] = 10⁻ᵖᴴ。对于强碱，先由浓度求 [OH⁻]，再通过 Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴（298 K）计算 pH。
Weak acids partially dissociate; use the acid dissociation constant Ka. Ka = [H⁺][A⁻] / [HA]. Often, [H⁺] ≈ √(Ka × Cₐ). Similar treatment applies to weak bases with Kb.
弱酸部分电离，使用酸解离常数 Ka。Ka = [H⁺][A⁻] / [HA]。常见近似公式为 [H⁺] ≈ √(Ka × Cₐ)。弱碱用 Kb 作类似处理。
Buffer pH can be calculated with the Henderson–Hasselbalch equation: pH = pKa + log([A⁻]/[HA]). This is useful for designing buffers.
缓冲溶液的 pH 可用亨德森-哈塞尔巴尔赫方程计算：pH = pKa + log([A⁻]/[HA])，常用于设计缓冲体系。
10. Electrochemistry Calculations | 电化学计算
Faraday’s laws link electric charge and chemical amount. Charge Q = I × t, where I is current in amperes and t is time in seconds. The moles of electrons n(e⁻) = Q / F, where Faraday constant F = 96 500 C mol⁻¹. Use the half-equation to relate moles of electrons to moles of substance.
法拉第定律联系电荷量与化学量。电荷量 Q = I × t，I 为电流（安培），t 为时间（秒）。电子的摩尔数 n(e⁻) = Q / F，法拉第常数 F = 96 500 C mol⁻¹。通过半反应式建立电子与物质的摩尔关系。
Example: Calculate the mass of copper deposited when 2.00 A flows for 30 minutes through Cu²⁺ solution. n(e⁻) = (2.00 × 30 × 60) / 96 500 = 0.0373 mol. Cu²⁺ + 2e⁻ → Cu, so n(Cu) = n(e⁻)/2 = 0.0186 mol, mass = 0.0186 × 63.5 = 1.18 g.
例题：2.00 A 的电流通过 Cu²⁺ 溶液 30 分钟，求析出铜的质量。n(e⁻) = (2.00 × 30 × 60) / 96 500 = 0.0373 mol。Cu²⁺ + 2e⁻ → Cu，所以 n(Cu) = n(e⁻)/2 = 0.0186 mol，质量 = 0.0186 × 63.5 = 1.18 g。
Cell potential E⁰_cell = E⁰_cathode − E⁰_anode. A positive E⁰ indicates a spontaneous reaction. The relationship ΔG⁰ = −nFE⁰ links thermodynamics and electrochemistry.
电池电动势 E⁰_cell = E⁰_阴极 − E⁰_阳极。正值表明反应自发。关系式 ΔG⁰ = −nFE⁰ 将热力学与电化学联系起来。
11. Rate of Reaction & Kinetics | 反应速率与动力学
The rate of reaction can be expressed as change in concentration per unit time, e.g., rate = −Δ[A]/Δt. The rate equation, rate = k[A]ᵐ[B]ⁿ, must be determined experimentally, not from the stoichiometric coefficients. Overall order = m + n.
反应速率可用浓度随时间的变化表示，如 rate = −Δ[A]/Δt。速率方程 rate = k[A]ᵐ[B]ⁿ 必须通过实验确定，不能直接从化学计量系数推出。总级数 = m + n。
The units of rate constant k depend on overall order. For zero order: mol dm⁻³ s⁻¹; first order: s⁻¹; second order: dm³ mol⁻¹ s⁻¹. Calculate k by substituting rate and concentrations into the rate equation.
速率常数 k 的单位取决于总级数。零级：mol dm⁻³ s⁻¹；一级：s⁻¹；二级：dm³ mol⁻¹ s⁻¹。将速率和浓度代入速率方程即可求出 k。
For the initial rates method, compare experiments where only one concentration changes. The ratio of rates gives the order with respect to that reactant. Graphical methods such as concentration–time and rate–concentration graphs also help.
初始速率法中，比较只有一个浓度改变的实验，速率比可得到该反应物的级数。也可用浓度–时间图和速率–浓度图等图像法确定。
12. Mixed Calculation Drills & Exam Tips | 混合计算训练与考试技巧
In exams, a single question often weaves together several calculation types. For instance: a hydrated salt is heated, water lost measured, then anhydrous salt used in a titration, requiring moles, empirical formula, concentration and stoichiometry steps.
考试中一道题常会组合多种计算类型。例如：水合盐加热失水，测量失水质量，再用无水盐进行滴定，这需要摩尔、经验式、浓度和化学计量等步骤。
Always show working clearly: write the balanced equation, state the formula, convert units, and label each step. Even if the final answer is wrong, marks are awarded for correct reasoning and substitution.
务必清晰展示推导过程：写出配平方程式、列出公式、换算单位、标注每一步。即便最终答案有误，正确的推理和代入也能得到步骤分。
Practice with past IB and CIE papers, focusing on time management. For multi-step calculations, use the “plan–execute–check” approach: plan the route, perform the calculation, then check if the numerical answer makes sense chemically.
练习 IB 和 CIE 历年真题，重视时间管理。对于多步计算，采用“规划–执行–检查”策略：规划路径，执行计算，然后检查答案数值是否在化学上合理。
Common pitfalls include forgetting to convert cm³ to dm³, using mass instead of moles in ratios, omitting × 100 for percentage, and incorrect R units. Systematic revision builds confidence and exam readiness.
常见错误包括忘记将 cm³ 换算成 dm³、在比率中使用质量而非摩尔数、遗漏百分比中的乘以 100、R 值单位错误等。系统复习能建立信心，从容应考。
Published by TutorHao | Chemistry Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导