📚 IGCSE Chemistry 0620: Calculation Question Types | IGCSE 化学 0620 计算题型
Calculation questions represent a core component of the IGCSE Chemistry 0620 examination, especially in the 2026-2028 syllabus. These quantitative problems test your ability to apply the mole concept, manipulate formulae, and interpret experimental data. Whether appearing in multiple-choice or structured papers, a systematic approach to numerical questions can reliably earn you high marks. This guide compiles the most important types of calculations you will encounter and demonstrates how to solve them step by step, with examples and clear reasoning.
计算题是IGCSE化学0620考试（2026-2028大纲）的核心组成部分。这些定量问题考查你是否能运用摩尔概念、处理公式以及解读实验数据。不论是选择题还是结构化试卷，只要掌握系统化的解题方法，就能稳稳拿下高分。本指南汇集了你将遇到的最重要的计算题型，并通过示例和清晰思路逐步展示如何解答。
1. Moles and Molar Mass | 摩尔与摩尔质量
The mole is the fundamental unit for amount of substance. One mole of any substance contains 6.02 × 10²³ particles (Avogadro’s number). The molar mass (M) is the mass of one mole, expressed in g mol⁻¹, and is numerically equal to the relative atomic mass (Aᵣ) or relative formula mass (Mᵣ) taken from the Periodic Table.
摩尔是物质的量的基本单位。1摩尔任何物质都含有6.02 × 10²³个微粒（阿伏伽德罗常数）。摩尔质量（M）是1摩尔物质的质量，单位为g mol⁻¹，数值上等于从周期表中获取的相对原子质量（Aᵣ）或相对式量（Mᵣ）。
The central formula connecting mass, moles and molar mass is:
连接质量、摩尔和摩尔质量的核心公式是：
n = m / M or number of moles = mass (g) ÷ molar mass (g mol⁻¹)
For example, to find the number of moles in 8.0 g of oxygen gas (O₂, M = 32 g mol⁻¹): n = 8.0 / 32 = 0.25 mol. Always check your units and make sure you use the formula mass of the correct species (e.g., O₂ rather than O).
例如，计算8.0 g氧气（O₂，M = 32 g mol⁻¹）的物质的量：n = 8.0 / 32 = 0.25 mol。解题时务必核对单位，并使用正确物种的式量（比如用O₂而不是O）。
2. Empirical and Molecular Formulae | 经验式与分子式
The empirical formula gives the simplest whole-number ratio of atoms in a compound. The molecular formula shows the actual number of atoms of each element in a molecule. Determining these formulae from experimental data is a classic IGCSE calculation.
经验式表示化合物中原子最简整数比。分子式则表示一个分子中各原子的实际数目。根据实验数据确定这些化学式是IGCSE的经典计算题型。
To find the empirical formula: (1) Convert the mass or percentage of each element to moles by dividing by its atomic mass. (2) Divide each mole value by the smallest number of moles obtained. (3) If necessary, multiply to get whole numbers. For example, a compound contains 2.4 g of carbon and 0.8 g of hydrogen. Moles of C = 2.4 / 12 = 0.20 mol; moles of H = 0.8 / 1 = 0.80 mol. Ratio = 0.20 : 0.80 = 1 : 4, so the empirical formula is CH₄.
求经验式的步骤：（1）将每种元素的质量或百分含量除以各自的原子质量，转化为物质的量。（2）将各物质的量值除以其中的最小值。（3）必要时乘以倍数得到整数。例如，某化合物含碳2.4 g和氢0.8 g。C的物质的量 = 2.4 / 12 = 0.20 mol；H的物质的量 = 0.8 / 1 = 0.80 mol。比例 = 0.20 : 0.80 = 1 : 4，因此经验式为CH₄。
The molecular formula is found by comparing the empirical formula mass with the given molar mass. If the empirical formula mass of CH₄ is 16, and the actual molar mass is 16 g mol⁻¹, the molecular formula remains CH₄. If the molar mass were 32 g mol⁻¹, the multiplier would be 32 / 16 = 2, giving C₂H₈.
通过比较经验式质量与给出的摩尔质量可求得分子式。若CH₄的经验式质量为16，而实际摩尔质量为16 g mol⁻¹，则分子式依然是CH₄。若摩尔质量为32 g mol⁻¹，倍数就是32 / 16 = 2，得到C₂H₈。
3. Reacting Masses | 反应质量
Reacting mass calculations use the balanced chemical equation to find the mass of a reactant or product. The key is to work in moles: convert the known mass to moles, use the mole ratio from the equation, and then convert moles back to mass.
反应质量计算利用已配平的化学方程式求出反应物或生成物的质量。关键是以摩尔为单位进行换算：将已知质量转化为物质的量，利用方程式中的摩尔比，再将物质的量换算回质量。
For example, calcium carbonate decomposes: CaCO₃ → CaO + CO₂. What mass of CaCO₃ is needed to produce 11.0 g of CO₂? (Mᵣ: CaCO₃ = 100, CO₂ = 44). First, moles of CO₂ = 11.0 / 44 = 0.25 mol. The equation shows a 1:1 molar ratio between CaCO₃ and CO₂, so 0.25 mol of CaCO₃ is required. Mass of CaCO₃ = 0.25 × 100 = 25.0 g.
例如，碳酸钙分解：CaCO₃ → CaO + CO₂。要产生11.0 g CO₂需要多少质量的CaCO₃？（Mᵣ：CaCO₃ = 100，CO₂ = 44）。首先，CO₂的物质的量 = 11.0 / 44 = 0.25 mol。方程式显示CaCO₃与CO₂的摩尔比为1:1，因此需要0.25 mol CaCO₃。CaCO₃的质量 = 0.25 × 100 = 25.0 g。
If the equation has a ratio other than 1:1, always apply the balancing coefficients. For 2Mg + O₂ → 2MgO, 2 moles of Mg produce 2 moles of MgO, so the ratio is 1:1, but between Mg and O₂ it is 2:1. Careful conversion avoids common errors.
如果方程式中的比例不是1:1，始终要使用配平系数。在2Mg + O₂ → 2MgO中，2 mol Mg生成2 mol MgO，因此比例为1:1，但Mg与O₂的比例为2:1。仔细换算能避免常见错误。
4. Gas Volumes | 气体体积
At room temperature and pressure (r.t.p., 20°C and 1 atmosphere), one mole of any gas occupies a volume of 24 dm³. This molar gas volume allows you to relate moles of a gas to its volume directly, provided the conditions are r.t.p.
在常温常压下（r.t.p.，20°C，1个大气压），1摩尔任何气体所占体积为24 dm³。利用这个气体摩尔体积，只要条件为r.t.p.，就可以直接将气体的物质的量与体积关联起来。
volume of gas (dm³) = number of moles × 24 or n = volume / 24
For instance, what volume does 0.50 mol of hydrogen gas occupy at r.t.p.? Volume = 0.50 × 24 = 12 dm³. Conversely, 48 dm³ of carbon dioxide corresponds to 48 / 24 = 2.0 mol. You may also be asked to calculate the volume of gas produced in a reaction, by first finding the moles of the gaseous product and then multiplying by 24.
例如，0.50 mol氢气在r.t.p.下占据多大体积？体积 = 0.50 × 24 = 12 dm³。反过来，48 dm³ 二氧化碳对应48 / 24 = 2.0 mol。考试中也可能要求计算反应生成的气体体积，只需先求出气态产物的物质的量，再乘以24即可。
Remember to convert volume units if necessary: 1 dm³ = 1000 cm³. If a question gives a volume in cm³, convert to dm³ by dividing by 1000 before applying the molar volume.
必要时记得转换体积单位：1 dm³ = 1000 cm³。如果题目给出的体积单位是cm³，在应用摩尔体积前先除以1000换算为dm³。
5. Concentration of Solutions | 溶液浓度
Concentration can be expressed in mol dm⁻³ (molar concentration) or in g dm⁻³ (mass concentration). The two are linked by the molar mass of the solute. The most common formula is:
浓度可用mol dm⁻³（摩尔浓度）或g dm⁻³（质量浓度）表示。两者通过溶质的摩尔质量关联起来。最常用的公式是：
concentration (mol dm⁻³) = number of moles / volume (dm³) c = n / V
To prepare a solution, you may need to dissolve a certain mass of solid in a solvent. For example, to make 250 cm³ (0.250 dm³) of 0.100 mol dm⁻³ sodium hydroxide (NaOH, M = 40 g mol⁻¹), first find moles needed: n = c × V = 0.100 × 0.250 = 0.0250 mol. Then mass = n × M = 0.0250 × 40 = 1.00 g. Dissolve 1.00 g of NaOH and make up to 250 cm³ with water.
配制溶液时，可能需要将一定质量的固体溶解在溶剂中。例如，要配制250 cm³（0.250 dm³）的0.100 mol dm⁻³氢氧化钠溶液（NaOH，M = 40 g mol⁻¹），先求所需物质的量：n = c × V = 0.100 × 0.250 = 0.0250 mol。然后质量 = n × M = 0.0250 × 40 = 1.00 g。称取1.00 g NaOH加水定容至250 cm³。
Converting between mass concentration and molar concentration is straightforward: molar concentration (mol dm⁻³) = mass concentration (g dm⁻³) / molar mass (g mol⁻¹). If a solution contains 4.0 g dm⁻³ of NaOH, its molar concentration = 4.0 / 40 = 0.10 mol dm⁻³.
质量浓度与摩尔浓度的换算很直接：摩尔浓度（mol dm⁻³）= 质量浓度（g dm⁻³）/ 摩尔质量（g mol⁻¹）。若某溶液含4.0 g dm⁻³ NaOH，其摩尔浓度 = 4.0 / 40 = 0.10 mol dm⁻³。
6. Titration Calculations | 滴定计算
Titration experiments are used to determine an unknown concentration. When a neutralisation reaction (e.g., acid + base) has a 1:1 molar ratio, the relationship c₁V₁ = c₂V₂ can be applied, where c is concentration in mol dm⁻³ and V is volume. For reactions with different stoichiometry, adjust accordingly.
滴定实验用于测定未知浓度。当中和反应（如酸+碱）的摩尔比为1:1时，可使用关系式c₁V₁ = c₂V₂，其中c为摩尔浓度（mol dm⁻³），V为体积。对于计量比不同的反应，需要相应调整。
For the titration of 25.0 cm³ of sodium hydroxide with 0.100 mol dm⁻³ hydrochloric acid, the average titre is 20.0 cm³. The equation is NaOH + HCl → NaCl + H₂O (1:1 ratio). Moles of HCl used = 0.100 × (20.0/1000) = 0.00200 mol. Because the ratio is 1:1, moles of NaOH in 25.0 cm³ also = 0.00200 mol. Therefore, concentration of NaOH = 0.00200 / (25.0/1000) = 0.0800 mol dm⁻³.
用0.100 mol dm⁻³盐酸滴定25.0 cm³氢氧化钠溶液，平均滴定体积为20.0 cm³。反应方程式为NaOH + HCl → NaCl + H₂O（1:1比例）。所用HCl的物质的量 = 0.100 × (20.0/1000) = 0.00200 mol。因比例为1:1，25.0 cm³中的NaOH物质的量也为0.00200 mol。因此NaOH浓度 = 0.00200 / (25.0/1000) = 0.0800 mol dm⁻³。
If the acid is diprotic, such as H₂SO₄, the ratio changes: H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Here, 1 mol of acid reacts with 2 mol of base. In this case, moles of NaOH = 2 × moles of H₂SO₄ at the endpoint. Always write the balanced equation and use the mole ratio.
若酸为二元酸，例如H₂SO₄，比例会变化：H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O。这里1 mol酸与2 mol碱反应。此时，终点时NaOH的物质的量 = 2 × H₂SO₄的物质的量。务必先写出配平方程式，再使用摩尔比。
7. Percentage Yield and Purity | 产率与纯度
The percentage yield compares the actual amount of product obtained to the theoretical amount expected from stoichiometric calculations. It is a measure of reaction efficiency.
产率（百分产率）将实际得到的产品量与根据化学计量计算的理论产量进行比较。它是反应效率的量度。
% yield = (actual mass / theoretical mass) × 100%
For example, if a reaction is calculated to produce 50.0 g of a salt but only 40.0 g is collected, the percentage yield = (40.0 / 50.0) × 100% = 80.0%. Yields below 100% may arise from incomplete reaction, side reactions, or loss during purification.
例如，某反应计算应产生50.0 g盐，但仅收集到40.0 g，则产率 = (40.0 / 50.0) × 100% = 80.0%。产率低于100%可能源于反应不完全、副反应或提纯过程中的损失。
Percentage purity is used when a sample is impure. It tells you what fraction of the sample is the desired substance.
当样品不纯时使用纯度百分比。它告诉你样品中目标物质的比例。
% purity = (mass of pure substance / mass of impure sample) × 100%
If 10.0 g of an impure limestone sample contains 8.5 g of CaCO₃, the percentage purity of CaCO₃ = (8.5 / 10.0) × 100% = 85.0%. These concepts often appear combined with reacting mass questions where you must first find the mass of pure reactant that actually takes part in the reaction.
若10.0 g不纯石灰石样品中含有8.5 g CaCO₃，则CaCO₃的纯度 = (8.5 / 10.0) × 100% = 85.0%。这类概念常与反应质量题结合，你需要先求出实际参与反应的纯反应物质量。
8. Energy Changes (Calorimetry) | 能量变化（量热法）
Calorimetry experiments allow you to calculate the enthalpy change (ΔH) for a reaction. The heat energy transferred is calculated from the temperature change in the water or solution using q = mcΔT.
量热实验可用于计算反应的焓变（ΔH）。利用q = mcΔT，可通过水或溶液的温度变化计算传递的热能。
q = m × c × ΔT
where m is the mass of water or solution (usually in g, and assuming the density is 1.00 g cm⁻³ for dilute solutions), c is the specific heat capacity (for water, c = 4.18 J g⁻¹ °C⁻¹), and ΔT is the temperature change (°C). 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 the direction of heat flow (exothermic reactions have negative ΔH).
其中m是水或溶液的质量（通常以g为单位，对于稀溶液假设密度为1.00 g cm⁻³），c是比热容（对于水，c = 4.18 J g⁻¹ °C⁻¹），ΔT是温度变化（°C）。每摩尔的焓变则为ΔH = -q / n，其中n是限制反应物的物质的量。负号表示热流方向（放热反应的ΔH为负）。
In a typical question, 50.0 cm³ of 1.0 mol dm⁻³ HCl is mixed with 50.0 cm³ of 1.0 mol dm⁻³ NaOH. The temperature rises by 6.5°C. The total mass of the mixture is approximately 100 g. So q = 100 × 4.18 × 6.5 = 2717 J. Moles of HCl = 1.0 × (50.0/1000) = 0.050 mol. Since HCl and NaOH react in a 1:1 ratio, the limiting reactant is 0.050 mol. ΔH = -2717 J / 0.050 mol = -54340 J mol⁻¹ ≈ -54.3 kJ mol⁻¹. The answer is normally expressed in kJ mol⁻¹.
典型的题目中，将50.0 cm³ 1.0 mol dm⁻³ HCl与50.0 cm³ 1.0 mol dm⁻³ NaOH混合，温度上升6.5°C。混合物总质量约为100 g。因此q = 100 × 4.18 × 6.5 = 2717 J。HCl物质的量 = 1.0 × (50.0/1000) = 0.050 mol。由于HCl与NaOH按1:1比例反应，限制反应物为0.050 mol。ΔH = -2717 J / 0.050 mol = -54340 J mol⁻¹ ≈ -54.3 kJ mol⁻¹。答案通常以kJ mol⁻¹表示。
9. Limiting Reactants | 限制反应物
The limiting reactant is the substance that is completely consumed in a reaction and thus determines the amount of product formed. The other reactants are present in excess. To identify the limiting reactant, calculate the moles of each reactant and compare them using the stoichiometric ratio from the balanced equation.
限制反应物是在反应中完全消耗的物质，因此决定了产物的生成量。其他反应物则过量存在。要确定限制反应物，需计算各反应物的物质的量，并根据配平方程式中的计量比进行比较。
Consider the reaction: 2H₂ + O₂ → 2H₂O. If you have 4.0 mol of H₂ and 1.0 mol of O₂, the equation says 2 mol H₂ react with 1 mol O₂. So 4.0 mol H₂ would require 2.0 mol O₂. Since only 1.0 mol O₂ is available, O₂ is the limiting reactant. The amount of H₂O produced = 2 × moles of O₂ = 2 × 1.0 = 2.0 mol. Excess H₂ remains unreacted.
考虑反应：2H₂ + O₂ → 2H₂O。若有4.0 mol H₂和1.0 mol O₂，方程式显示2 mol H₂与1 mol O₂反应。因此4.0 mol H₂需要2.0 mol O₂。由于只有1.0 mol O₂可用，O₂是限制反应物。生成H₂O的量 = 2 × O₂的物质的量 = 2 × 1.0 = 2.0 mol。过量的H₂未反应。
Often you are given masses rather than moles. Convert masses to moles first, then find the limiting reagent. The subsequent calculation of product mass or volume must be based on the moles of the limiting reactant, not the ones in excess.
通常题目给出的是质量而非物质的量。先换算成物质的量，再找出限制试剂。后续计算产物质量或体积时必须基于限制反应物的物质的量，而不是过量反应物。
10. Calculations from Equations (Using Molar Ratios) | 根据方程式计算（摩尔比）
Most quantitative problems ultimately require you to use the molar ratios given by the coefficients in a balanced equation. Once you have identified the number of moles of a known substance, you can determine the moles of any other substance in the reaction by multiplying by the appropriate ratio.
大多数定量问题最终都要求运用配平方程式中各物质的系数给出的摩尔比。一旦确定了已知物质的物质的量，你就可以乘以适当的比例求得反应中任何其他物质的物质的量。
The general method for any stoichiometry problem is: (i) write the balanced equation, (ii) list the known data and molar masses, (iii) convert the given quantity to moles, (iv) use the mole ratio to find moles of the target substance, (v) convert these moles to the required quantity (mass, volume, or concentration). Practise this sequence so that it becomes automatic in the exam.
任何化学计量问题的通用步骤为：(i) 写出配平方程式，(ii) 列出已知数据和摩尔质量，(iii) 将已知量转化为物质的量，(iv) 利用摩尔比求出目标物质的物质的量，(v) 将这些物质的量换算为所需量（质量、体积或浓度）。反复练习这套流程，以便在考试中自然运用。
For example, when 13.0 g of zinc reacts with excess hydrochloric acid (Zn + 2HCl → ZnCl₂ + H₂), calculate the volume of hydrogen produced at r.t.p. Moles of Zn = 13.0 / 65.0 = 0.200 mol. From the equation, 1 mol Zn produces 1 mol H₂, so moles of H₂ = 0.200 mol. Volume = 0.200 × 24 = 4.80 dm³. You can see how all the individual skills join together in a single, multi-step problem.
例如，当13.0 g锌与过量盐酸反应（Zn + 2HCl → ZnCl₂ + H₂），计算在r.t.p.下产生的氢气体积。Zn的物质的量 = 13.0 / 65.0 = 0.200 mol。由方程式，1 mol Zn生成1 mol H₂，因此H₂的物质的量为0.200 mol。体积 = 0.200 × 24 = 4.80 dm³。可以看到，各项独立技能是如何汇聚到一道多步问题中的。
Mastering these calculation types requires consistent practice. The 2026-2028 syllabus continues to emphasise the application of the mole concept across all areas of chemistry, from titrations to energetics. Keep a formula sheet handy and always double-check your unit conversions.
掌握这些计算题型需要持续的练习。2026-2028大纲继续强调摩尔概念在化学各领域的应用，从滴定到能量学。手边常备公式表，并始终仔细核对单位换算。
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