📚 Mastering Calculation Questions in IGCSE CIE Chemistry | IGCSE CIE 化学计算题型精讲
Calculation questions form the backbone of IGCSE CIE Chemistry papers, consistently appearing in both Core and Extended theory exams. Success demands not only memorising formulas but also developing a structured, step‑by‑step approach to data, ratios and units. This guide covers every major calculation type, from basic mole concepts to multi‑step titration and energy problems, equipping you with the tools to score full marks.
计算题是IGCSE CIE化学考试的核心题型，无论在核心还是拓展理论卷中都稳定出现。想在计算题上拿满分，仅仅记住公式远远不够——你需要对数据、比率和单位建立起清晰有序的解题思路。本文覆盖从基础摩尔概念到多步滴定和能量计算的所有重要类型，帮你构建系统化的应考能力。
1. The Mole Concept and Molar Mass | 摩尔概念与摩尔质量
umber). For elements and compounds, the molar mass (M) is the mass of one mole in grams per mole (g mol⁻¹), numerically equal to the relative atomic mass (Aᵣ) or relative formula mass (Mᵣ).
摩尔是化学家的计数单位：1摩尔任何物质含有6.02 × 10²³个粒子（阿伏伽德罗常数）。对单质和化合物而言，摩尔质量（M）是1摩尔物质的质量，单位为克每摩尔（g mol⁻¹），数值上等于相对原子质量（Aᵣ）或相对式量（Mᵣ）。
The key relationship linking mass, moles and molar mass is the foundation of nearly every calculation in the syllabus. Examiners often embed it inside larger problems, so you must be able to rearrange the formula instantly.
质量、摩尔数、摩尔质量之间的核心关系式是几乎所有计算的基石。考官经常把它嵌套在综合性大题中，因此你必须做到能瞬间变形使用。
n = m / M
n = m ÷ M
Always convert mass to grams and molar mass to g mol⁻¹ before substituting. When the question gives mass in kilograms, multiply by 1000.
代入前一定要将质量换算为克，摩尔质量使用g mol⁻¹。若题目给出的质量单位是千克，必须先乘以1000。
2. Calculating Moles from Mass | 由质量计算摩尔数
A classic opening step is determining the amount of substance present. Worked example: "Calculate the number of moles in 8.0 g of sodium hydroxide, NaOH." ᵣ: Na=23, O=16, H=1). Mᵣ of NaOH = 23 + 16 + 1 = 40, so n = 8.0 g ÷ 40 g mol⁻¹ = 0.20 mol.
典型的解题第一关是计算物质所含的摩尔数。例题：“计算8.0 g氢氧化钠（NaOH）的摩尔数。”（Aᵣ: Na=23, O=16, H=1）。NaOH的Mᵣ = 23 + 16 + 1 = 40，因此n = 8.0 g ÷ 40 g mol⁻¹ = 0.20 mol。
÷ 71 = 0.20 mol. Never forget that gaseous elements are diatomic in their standard state.
当处理O₂、N₂、Cl₂等双原子分子气体时，这一技能同样重要。对14.2 g氯气（Cl₂），Mᵣ = 2 × 35.5 = 71，n = 14.2 ÷ 71 = 0.20 mol。务必记住气态非金属单质在标准状态下以双原子分子存在。
3. Empirical and Molecular Formulae | 经验式与分子式
The empirical formula gives the simplest whole‑ asses) to moles, then divide by the smallest mole value. Example: a compound contains 40.0% carbon, 6.7% hydrogen and 53.3% 7; O = 53.3/16 = 3.33. Divide by 3.33 → ratio C:H:O = 1:2:1, so empirical formula is CH₂O.
经验式给出化合物中各原子最简整数比。求法是将质量百分比（或已知质量）换算成摩尔数，再除以最小的摩尔值。例题：某化合物含碳40.0%、氢6.7%、氧53.3%。摩尔数：C = 40.0/12 = 3.33；H = 6.7/1 = 6.7；O = 53.3/16 = 3.33。各除以3.33 → 原子比C:H:O = 1:2:1，经验式为CH₂O。
above, if the relative molecular mass is 180, then Mᵣ(CH₂O) = 30, multiplier = 180 ÷ 30 = 6, giving the molecular formula C₆H₁₂O₆.
分子式是经验式的整数倍。倍数n = 分子Mᵣ ÷ 经验式Mᵣ。上例中如果分子相对质量为180，经验式CH₂O的Mᵣ=30，倍数=180÷30=6，分子式即为C₆H₁₂O₆。
4. Reacting Masses and Stoichiometry | 反应质量与化学计量
Using a balanced equation, we predict the masses of reactants consumed and products formed. The three‑step method: (1) convert given mass to moles; (2) use the mole ratio from the equation to find moles of the target substance; (3) convert back to mass.
利用配平的化学方程式，我们可以预测消耗的反应物质量和生成的产物质量。三步法：(1) 将已知质量换算为摩尔数；(2) 利用方程式中的摩尔比求目标物质的摩尔数；(3) 换回质量。
Consider: "What mass of water is produced when 4.0 g of hydrogen burns completely?" ₂ : H₂O = 2 : 2 = 1 : 1, so n(H₂O) = 2.0 mol. M(H₂O) = 18 g mol⁻¹, mass = 2.0 × 18 = 36 g.
例题：“4.0 g氢气完全燃烧生成多少克水？”2H₂ + O₂ → 2H₂O。H₂摩尔数 = 4.0 g ÷ 2.0 g mol⁻¹ = 2.0 mol。摩尔比H₂:H₂O = 2:2 = 1:1，因此n(H₂O)=2.0 mol。M(H₂O)=18 g mol⁻¹，质量 = 2.0 × 18 = 36 g。
Watch out for limiting reactants. If masses of two reactants are given, calculate moles of each and compare using the ratio. The reactant that produces fewer moles of product is the limiting one.
要特别注意限量反应物。如果给出两种反应物的质量，应分别计算摩尔数，并按方程式摩尔比比较——产生较少产物摩尔数的反应物即为限量反应物。
5. Gas Volumes and Molar Volume | 气体体积与摩尔体积
At room temperature and pressure (r.t.p., 20 °C, 1 atm), one mole of any gas occupies 24 dm³ (24 000 cm³). This value is given on the CIE data sheet but you must use it fluently. For volumes measured in other conditions, the question will supply the new molar volume.
在常温常压下（r.t.p., 20 °C, 1 atm），1摩尔任何气体的体积为24 dm³（24 000 cm³）。CIE数据手册会提供这个数值，但你必须熟练运用。如果题目设定为其它条件，会给出相应的摩尔体积。
To find the volume of a gas produced: first calculate moles of the gas, then multiply by 24 dm³ mol⁻¹ (or 24 000 cm³ mol⁻¹). Unit consistency is vital: if the answer is required in cm³, use 24 000; if in dm³, use 24.
计算生成气体的体积时，先求出气体的摩尔数，再乘以24 dm³ mol⁻¹（或24 000 cm³ mol⁻¹）。单位统一至关重要：若答案要求用cm³表达，用24 000；若用dm³，用24。
Worked example: "What volume of CO₂ (r.t.p.) is released when 10.0 g of CaCO₃ is thermally decomposed?" ₃ = 10.0 / 100 = 0.100 mol. 1:1 ratio → n(CO₂) = 0.100 mol. Volume = 0.100 × 24 = 2.4 dm³.
例题：“10.0 g CaCO₃热分解时，释放出多少体积的CO₂（r.t.p.）？”CaCO₃ → CaO + CO₂。CaCO₃摩尔数 = 10.0/100 = 0.100 mol。1:1 摩尔比 → n(CO₂)=0.100 mol。体积 = 0.100 × 24 = 2.4 dm³。
6. Concentration and Solution Calculations | 浓度与溶液计算
Concentration is usually expressed in mol dm⁻³ or g dm⁻³. The central formula links moles, concentration and volume of solution. In the CIE exam, volume must almost always be in dm³; convert cm³ to dm³ by dividing by 1000.
浓度通常用mol dm⁻³或g dm⁻³表示。核心公式将摩尔数、浓度和溶液体积联系起来。CIE 考试中体积几乎都要用dm³为单位；若以cm³给出，需除以1000换算。
n = c × V
摩尔数 = 浓度 × 体积（dm³）
Example: "How many moles of HCl are in 25.0 cm³ of 0.200 mol dm⁻³ hydrochloric acid?" V = 25.0 / 1000 = 0.0250 dm³; n = 0.200 × 0.0250 = 0.00500 mol.
例题：“25.0 cm³ 浓度为0.200 mol dm⁻³的盐酸中含有多少摩尔HCl？”V = 25.0/1000 = 0.0250 dm³；n = 0.200 × 0.0250 = 0.00500 mol。
se the conservation of moles: c₁V₁ = c₂V₂, but ensure both V₁ and V₂ are in the same unit.
用固体配制溶液时，可逆向计算：先由n = c × V (dm³) 求摩尔数，再由质量 = n × M 算出所需固体质量。稀释问题利用摩尔守恒：c₁V₁ = c₂V₂，但要保证V₁和V₂单位一致。
7. Titration Calculations | 滴定计算
Titration is a direct application of stoichiometry to solutions. The goal is to find the unknown concentration of a solution using a standard solution of known concentration. The typical pathway: (1) calculate moles of the known reactant (use its concentration and titre volume); (2) apply the mole ratio from the balanced equation; (3) find concentration of the unknown.
滴定是化学计量在溶液中的直接应用，旨在用已知浓度的标准溶液求未知溶液的浓度。典型路径：(1) 计算已知反应物的摩尔数（用其浓度和滴定体积）；(2) 运用配平方程式中的摩尔比；(3) 求出未知物的浓度。
20.0/1000) = 0.00200 mol. Ratio 1:1, so n(NaOH) = 0.00200 mol. c(NaOH) = 0.00200 / (25.0/1000) = 0.0800 mol dm⁻³.
例题：25.0 cm³ NaOH溶液恰好被20.0 cm³ 0.100 mol dm⁻³ HCl中和。方程式：HCl + NaOH → NaCl + H₂O。HCl摩尔数 = 0.100 × (20.0/1000) = 0.00200 mol。摩尔比1:1，因此n(NaOH)=0.00200 mol。c(NaOH) = 0.00200/(25.0/1000) = 0.0800 mol dm⁻³。
→ Na₂SO₄ + 2H₂O. Here, n(NaOH) = 2 × n(H₂SO₄). Always check the number of acidic protons.
如果酸是二元酸（如H₂SO₄），摩尔比会改变：2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O。此时n(NaOH) = 2 × n(H₂SO₄)。务必查验酸分子可提供的质子数。
8. Percentage Yield and Purity | 产率与纯度
The theoretical yield is the maximum calculated product mass assuming complete conversion and no losses. Percentage yield compares the actual mass obtained to the theoretical mass. In industry and the lab, yields below 100% are common due to reversible reactions, side products or mechanical losses.
理论产量是假设完全转化、无损失的前提下计算出的最大产物质量。产率是将实际获得的质量与理论质量进行比较。无论是工业还是实验室，因为可逆反应、副反应或转移损耗，产率低于100%十分常见。
% yield = (actual yield ÷ theoretical yield) × 100
产率 = (实际产量 ÷ 理论产量) × 100%
ms involve calculating the mass of a pure substance within an impure sample. For instance, "25.0 g of impure limestone yielded 8.8 g of CO₂ on heating. What is the % purity of CaCO₃?" O₃ = 0.20 mol; mass of pure CaCO₃ = 0.20 × 100 = 20.0 g; purity = (20.0/25.0) × 100 = 80.0%.
纯度问题需要计算不纯样品中纯物质的质量。例如：“25.0 g 不纯石灰石加热后产生8.8 g CO₂，求CaCO₃的纯度。”先求CO₂摩尔数 = 8.8/44 = 0.20 mol；CaCO₃摩尔数 = 0.20 mol；纯CaCO₃质量 = 0.20 × 100 = 20.0 g；纯度 = (20.0/25.0) × 100 = 80.0%。
9. Energy Changes and Calorimetry | 能量变化与量热计算
lculated using q = m × c × ΔT, where m is the mass of water/solution (in g, assume 1 cm³ ≈ 1 g), c is the specific heat capacity (4.2 J g⁻¹ °C⁻¹ for water) and ΔT is the temperature change.
焓变通常通过量热器测量温度变化来确定。溶液或水吸收或放出的热量用 q = m × c × ΔT 计算，其中m是水/溶液的质量（g，假设1 cm³ ≈ 1 g），c是比热容（水为4.2 J g⁻¹ °C⁻¹），ΔT是温度变化。
To find ΔH in kJ mol⁻¹: calculate q in joules, convert to kJ (÷ 1000), then divide by the moles of limiting reactant. The sign of ΔH is negative for exothermic reactions (temperature rises) and positive for endothermic reactions (temperature falls).
求以kJ mol⁻¹为单位的ΔH：先计算q（焦耳），转换为kJ（÷ 1000），再除以限量反应物的摩尔数。放热反应ΔH为负（温度上升），吸热反应ΔH为正（温度下降）。
ΔH = – (m × c × ΔT) / n (exothermic)
ΔH = – (m × c × ΔT) ÷ n（放热）
. q = 100 × 4.2 × 6.5 = 2730 J = 2.73 kJ. Moles of HCl = 0.050 dm³ × 1.0 = 0.050 mol. ΔH = –2.73 kJ ÷ 0.050 mol = –54.6 kJ mol⁻¹.
例题：50 cm³ 1.0 mol dm⁻³ HCl与50 cm³ 1.0 mol dm⁻³ NaOH混合，温度上升6.5 °C。总体积 = 100 cm³，质量 ≈ 100 g。q = 100 × 4.2 × 6.5 = 2730 J = 2.73 kJ。HCl摩尔数 = 0.050 dm³ × 1.0 = 0.050 mol。ΔH = –2.73 kJ ÷ 0.050 mol = –54.6 kJ mol⁻¹。
10. All-in-One Problem Solving | 综合计算题策略
High‑mark questions often combine several concepts: you might need to use gas volumes, reacting masses and concentration in sequence. The secret is to break the journey into familiar blocks — find moles first, then use ratios, then convert to the required unit at the end. Keeping a clear, labelled working is the best defence against slip‑ups.
高分大题常常将多个概念串联：你可能需要依次运用气体体积、反应质量和浓度计算。秘诀是把复杂路径拆解为熟悉的模块——先求出摩尔数，再运用摩尔比，最后转换成题目要求的单位。书写清晰、注明已知量和未知量的解题过程是防止失误的最好方法。
Common pitfalls include forgetting to balance equations before using mole ratios, mixing cm³ and dm³, and failing to convert masses to grams. A quick sense‑check: "Does my answer make chemical sense? Is the mass of product smaller than total reactant mass?" goes a long way.
常见失分点包括使用摩尔比之前未配平方程式、混淆cm³和dm³、质量未转换为克。一个快速的合理性检验——“我的答案在化学上合理吗？产物质量小于反应物总质量吗？”——能有效避免低级错误。
- Memorise the core formulas actively: write them down at the start of the exam. 主动默写核心公式：考试一开始就把它们写在草稿纸上。
- Show every step: CIE awards marks for correct working even if the final answer is wrong. 写出每一步过程：CIE按步骤给分，即使最终答案有误也能得分。
- Practise past papers under timed conditions: speed and accuracy come from familiarity. 计时刷真题：速度和准确度来自熟练度。
Published by TutorHao | Chemistry Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图