📚 IGCSE CIE Chemistry ATP Calculation Questions | IGCSE CIE 化学 实验替代 计算题型
In the CIE IGCSE Chemistry 0620 syllabus, Paper 6 (Alternative to Practical) tests your ability to handle experimental data without actually performing the experiments. A significant portion of this paper involves numerical calculations based on given measurements, observations, and experimental setups. Mastering these calculation questions is essential for achieving a high grade. This article breaks down the most common calculation types you will encounter, providing step-by-step explanations and practical examples.
在CIE IGCSE化学0620课程中，试卷6（实验替代）考查你无需实际操作实验而处理实验数据的能力。这张试卷的很大一部分涉及基于给定测量、观察和实验装置的数字计算。掌握这些计算题型对于取得高分至关重要。本文将分解你最可能遇到的最常见计算类型，提供逐步解释和实际示例。
1. Understanding the ATP Exam | 理解ATP考试
The Alternative to Practical paper lasts 1 hour and carries 40 marks. Questions are based on common practical activities such as titrations, temperature changes, gas collection, and salt preparation. You will be asked to read instruments, plot graphs, draw conclusions, and perform calculations. The calculation questions often require you to use formulas and apply the mole concept to find unknown quantities like concentration, molar mass, or percentage purity.
实验替代试卷时长为1小时，总分40分。题目基于常见的实践活动，如滴定、温度变化、气体收集和盐制备。你需要读取仪器、绘制图表、得出结论并进行计算。计算题通常要求你运用公式并应用摩尔概念来求出未知量，如浓度、摩尔质量或百分纯度。
2. Key Formulas and Units | 关键公式和单位
Before tackling calculations, you must memorise and understand these fundamental relationships. Always pay attention to units and convert where necessary.
在着手计算之前，你必须记住并理解这些基本关系。始终注意单位，并在必要时进行换算。
number of moles = mass (g) / molar mass (g/mol)
摩尔数 = 质量 (克) / 摩尔质量 (克/摩尔)
number of moles = concentration (mol/dm³) × volume (dm³)
摩尔数 = 浓度 (摩尔/立方分米) × 体积 (立方分米)
number of moles = volume of gas (dm³) / 24 dm³ (at r.t.p.)
摩尔数 = 气体体积 (立方分米) / 24 立方分米 (在常温常压下)
percentage yield = (actual yield / theoretical yield) × 100%
产率百分比 = (实际产量 / 理论产量) × 100%
percentage purity = (mass of pure substance / total mass) × 100%
百分纯度 = (纯物质质量 / 总质量) × 100%
energy change = mass × specific heat capacity × temperature change
能量变化 = 质量 × 比热容 × 温度变化
In the ATP paper, volumes may be given in cm³; you must convert to dm³ by dividing by 1000. Time measurements are commonly in seconds when calculating rates.
在ATP试卷中，体积可能以cm³给出；你必须除以1000转换为dm³。计算速率时，时间测量通常以秒为单位。
3. Mole Calculations from Experimental Data | 从实验数据计算摩尔
Many ATP problems start by asking you to calculate the number of moles of a reactant or product from a measured mass, solution volume, or gas volume. For example, if 2.8 g of iron is used, moles of Fe = 2.8 / 56 = 0.05 mol. If 50 cm³ of 0.1 mol/dm³ HCl is used, moles = 0.1 × (50/1000) = 0.005 mol. If 120 cm³ of hydrogen gas is collected, moles = 0.12 / 24 = 0.005 mol. Always show your working clearly; the number of marks often depends on the steps.
许多ATP问题一开始就要求你根据测定的质量、溶液体积或气体体积计算反应物或生成物的摩尔数。例如，若使用2.8克铁，Fe的摩尔数 = 2.8 / 56 = 0.05 mol。若使用50 cm³ 0.1 mol/dm³ HCl，摩尔数 = 0.1 × (50/1000) = 0.005 mol。若收集到120 cm³氢气，摩尔数 = 0.12 / 24 = 0.005 mol。始终清晰地展示你的计算步骤；分数往往取决于步骤。
Remember the link to the balanced equation: the reacting mole ratio allows you to find unknown moles. For instance, Mg + 2HCl → MgCl₂ + H₂. If 0.005 mol of H₂ is collected, the moles of Mg used is also 0.005 (1:1 ratio), and moles of HCl is 0.010 (1:2).
记住与配平方程式的联系：反应的摩尔比可让你求出未知的摩尔数。例如，Mg + 2HCl → MgCl₂ + H₂。若收集到0.005 mol H₂，则使用的Mg摩尔数也是0.005（1:1），HCl摩尔数为0.010（1:2）。
4. Titration Calculations | 滴定计算
Titration is a core practical skill assessed in ATP. You will be provided with burette readings (initial and final) and the volume of one solution used to neutralise another. The typical task is to find the concentration of an unknown solution. First, calculate the average titre volume (usually in cm³) and convert to dm³. Then find the moles of the known solution using n = c × V. Using the balanced equation, determine the moles of the unknown solution. Finally, divide moles by its volume (in dm³) to find its concentration.
滴定是ATP中考查的核心实验技能。你会得到滴定管读数（初始和最终）和用于中和另一种溶液的某溶液的体积。典型任务是求出未知溶液的浓度。首先，计算平均滴定体积（通常以cm³计）并转换为dm³。然后利用n = c × V求出已知溶液的摩尔数。根据配平方程式，求出未知溶液的摩尔数。最后，将摩尔数除以它的体积（dm³）以求得其浓度。
Example: 25.0 cm³ of NaOH solution is titrated with 0.100 mol/dm³ HCl. The average titre is 20.0 cm³. Find the concentration of NaOH. HCl moles = 0.100 × 0.020 = 0.0020 mol. Equation: NaOH + HCl → NaCl + H₂O, mole ratio 1:1, so NaOH moles = 0.0020. Concentration of NaOH = 0.0020 / 0.025 = 0.080 mol/dm³.
示例：用0.100 mol/dm³ HCl滴定25.0 cm³ NaOH溶液。平均滴定体积为20.0 cm³。求NaOH的浓度。HCl摩尔数 = 0.100 × 0.020 = 0.0020 mol。方程式：NaOH + HCl → NaCl + H₂O，摩尔比1:1，所以NaOH摩尔数 = 0.0020。NaOH浓度 = 0.0020 / 0.025 = 0.080 mol/dm³。
You may also be asked to calculate the concentration in g/dm³. Multiply the molar concentration by the molar mass of the solute.
你可能还被要求计算以g/dm³为单位的浓度。将摩尔浓度乘以溶质的摩尔质量即可。
5. Gas Volume Calculations | 气体体积计算
Collecting a gas over water or by syringe is common in ATP scenarios. You will need to use the molar gas volume, 24 dm³/mol at room temperature and pressure (r.t.p.). If the gas is collected at conditions other than r.t.p., the question will provide the necessary conversion, but usually you can assume r.t.p. Do not forget to convert volumes in cm³ to dm³ by dividing by 1000. Also, you may need to account for water vapour pressure if the gas is collected over water, but this is usually indicated in the question if required.
用排水法或注射器收集气体在ATP场景中很常见。你需要使用摩尔气体体积，常温常压（r.t.p.）下为24 dm³/mol。如果气体在非r.t.p.条件下收集，题目会提供必要的转换，但通常你可以假设r.t.p.。不要忘记将cm³体积转换为dm³，即除以1000。另外，如果气体是用排水法收集的，可能需要考虑水蒸气压力，但如果需要，题目通常会指明。
Typical calculation: In an experiment, excess magnesium is added to 25 cm³ of 1.0 mol/dm³ sulfuric acid. Calculate the volume of hydrogen produced at r.t.p. Moles of H₂SO₄ = 1.0 × 0.025 = 0.025 mol. Equation: Mg + H₂SO₄ → MgSO₄ + H₂. Moles of H₂ = 0.025 mol. Volume of H₂ = 0.025 × 24 = 0.60 dm³ or 600 cm³.
典型计算：在实验中，将过量的镁加入25 cm³ 1.0 mol/dm³硫酸中。计算在r.t.p.下产生的氢气体积。H₂SO₄的摩尔数 = 1.0 × 0.025 = 0.025 mol。方程式：Mg + H₂SO₄ → MgSO₄ + H₂。H₂的摩尔数 = 0.025 mol。H₂体积 = 0.025 × 24 = 0.60 dm³ 或 600 cm³。
6. Enthalpy Change Calculations | 焓变计算
Simple calorimetry experiments appear frequently. You may be given temperature changes when a solid dissolves or a neutralisation reaction occurs. Use the formula Q = mcΔT, where m is the mass of the solution (usually water, density 1 g/cm³, so volume in cm³ equals mass in g), c is the specific heat capacity of water (4.2 J/g/°C), and ΔT is the temperature change. Q is the heat energy in joules. Then calculate the enthalpy change per mole (ΔH) by dividing Q by the number of moles of the limiting reactant, and adjust the sign (+ for endothermic, – for exothermic). Report your answer in kJ/mol (divide by 1000).
简单的量热实验经常出现。当固体溶解或发生中和反应时，你可能需要用到温度变化数据。使用公式Q = mcΔT，其中m是溶液质量（通常是水，密度1 g/cm³，因此cm³体积等于克质量），c是水的比热容（4.2 J/g/°C），ΔT是温度变化。Q是以焦耳为单位的热能。然后，将Q除以限制反应物的摩尔数，计算每摩尔的焓变（ΔH），并调整符号（+表示吸热，-表示放热）。答案以kJ/mol报告（除以1000）。
Example: 50 cm³ of 1.0 mol/dm³ HCl is mixed with 50 cm³ of 1.0 mol/dm³ NaOH. Temperature rises from 22.0 °C to 28.5 °C. Total volume = 100 cm³, mass = 100 g. ΔT = 6.5 °C. Q = 100 × 4.2 × 6.5 = 2730 J. Moles of HCl = 1.0 × 0.050 = 0.05 mol (NaOH also 0.05 mol, both limiting). ΔH = -2730 / 0.05 = -54600 J/mol = -54.6 kJ/mol (negative because temperature increased).
示例：将50 cm³ 1.0 mol/dm³ HCl与50 cm³ 1.0 mol/dm³ NaOH混合。温度由22.0 °C升至28.5 °C。总体积 = 100 cm³，质量 = 100 g。ΔT = 6.5 °C。Q = 100 × 4.2 × 6.5 = 2730 J。HCl摩尔数 = 1.0 × 0.050 = 0.05 mol（NaOH也是0.05 mol，两者都是限制反应物）。ΔH = -2730 / 0.05 = -54600 J/mol = -54.6 kJ/mol（负号因为温度升高）。
7. Percentage Yield and Purity | 产率和纯度计算
In synthesis experiments described in ATP, you might be asked to calculate the percentage yield. First, use the mass of the limiting reactant to calculate the theoretical yield of the product (in grams) via stoichiometry. Then divide the actual mass of product obtained by the theoretical mass and multiply by 100%. The actual mass will be given in the question. Percentage purity is similar: you may be given the mass of an impure sample and the mass of pure substance obtained after purification. Divide the pure mass by the impure mass and multiply by 100%.
在ATP描述的合成实验中，你可能需要计算产率百分比。首先，利用限制反应物的质量经化学计量计算出产品的理论产量（以克为单位）。然后将实际获得的产品质量除以理论质量并乘以100%。实际质量将在题目中给出。百分纯度类似：你可能会得到不纯样品的质量和经纯化后获得的纯物质质量。将纯物质质量除以不纯样品质量再乘以100%。
For example: In an experiment, 5.6 g of iron is reacted with excess copper sulfate solution. 6.0 g of copper metal is obtained. Calculate the percentage yield. Theoretical moles of Fe = 5.6 / 56 = 0.10 mol. Fe + CuSO₄ → FeSO₄ + Cu, mole ratio 1:1, so theoretical moles of Cu = 0.10 mol. Theoretical mass of Cu = 0.10 × 63.5 = 6.35 g. Percentage yield = (6.0 / 6.35) × 100% = 94.5%.
例如：在实验中，将5.6 g铁与过量的硫酸铜溶液反应。获得6.0 g铜金属。计算产率百分比。Fe的理论摩尔数 = 5.6 / 56 = 0.10 mol。Fe + CuSO₄ → FeSO₄ + Cu，摩尔比1:1，所以Cu的理论摩尔数 = 0.10 mol。Cu的理论质量 = 0.10 × 63.5 = 6.35 g。产率百分比 = (6.0 / 6.35) × 100% = 94.5%。
8. Relative Atomic Mass from Experimental Data | 从实验数据求相对原子质量
A classic ATP question involves the reaction of a metal with an acid to produce hydrogen gas. From the measured volume of hydrogen and the mass of the metal, you can determine the relative atomic mass (Aᵣ) of the metal. Steps: calculate moles of H₂ using volume/24, then use the balanced equation (metal + acid → salt + H₂) to find the mole ratio. Typically, a Group 2 metal: M + 2HCl → MCl₂ + H₂, so moles of metal = moles of H₂. Then Aᵣ = mass of metal / moles of metal.
经典的ATP题目涉及金属与酸反应产生氢气。通过测定的氢气体积和金属质量，你可以确定该金属的相对原子质量（Aᵣ）。步骤：用体积/24计算H₂的摩尔数，然后利用配平方程式（金属 + 酸 → 盐 + H₂）找出摩尔比。通常，对于II族金属：M + 2HCl → MCl₂ + H₂，所以金属的摩尔数 = H₂的摩尔数。然后Aᵣ = 金属质量 / 金属摩尔数。
Example: 0.24 g of an unknown Group 2 metal reacts with excess acid to produce 100 cm³ of hydrogen at r.t.p. Find Aᵣ. Moles of H₂ = 0.100 / 24 = 0.004167 mol. Moles of metal = 0.004167 mol (1:1). Aᵣ = 0.24 / 0.004167 ≈ 57.6. This is close to the Aᵣ of iron (56) but iron is not Group 2; however, some metals like manganese (55) might appear, or it highlights the need to identify the metal from the calculated Aᵣ.
示例：0.24 g某种未知的II族金属与过量酸反应，在r.t.p.下产生100 cm³氢气。求Aᵣ。H₂的摩尔数 = 0.100 / 24 = 0.004167 mol。金属的摩尔数 = 0.004167 mol（1:1）。Aᵣ = 0.24 / 0.004167 ≈ 57.6。这接近铁（56）的Aᵣ，但铁不是II族；然而，有些金属如锰（55）可能出现，或者这凸显了需要根据计算出的Aᵣ来确定金属。
9. Concentration and Dilution Calculations | 浓度和稀释计算
In ATP, you might need to prepare a solution by dilution or calculate the new concentration after mixing. The key idea is that the number of moles of solute does not change during dilution: c₁V₁ = c₂V₂ (where volumes must be in the same unit, usually dm³ or cm³). For mixing two solutions, you can find the total moles of solute and divide by total volume.
在ATP中，你可能需要通过稀释制备溶液或计算混合后的新浓度。关键思路是：溶质的摩尔数在稀释过程中保持不变：c₁V₁ = c₂V₂（其中体积必须使用相同单位，通常是dm³或cm³）。对于混合两种溶液，你可以求出总溶质摩尔数再除以总体积。
Example: 10.0 cm³ of 2.0 mol/dm³ HCl is diluted to 100.0 cm³. What is the new concentration? Moles of HCl = 2.0 × 0.010 = 0.020 mol. New volume = 0.100 dm³. New concentration = 0.020 / 0.100 = 0.20 mol/dm³. Alternative using c₁V₁ = c₂V₂: 2.0 × 10 = c₂ × 100 → c₂ = 0.20 mol/dm³.
示例：将10.0 cm³ 2.0 mol/dm³ HCl稀释至100.0 cm³。新浓度是多少？HCl摩尔数 = 2.0 × 0.010 = 0.020 mol。新体积 = 0.100 dm³。新浓度 = 0.020 / 0.100 = 0.20 mol/dm³。另一种方法使用c₁V₁ = c₂V₂：2.0 × 10 = c₂ × 100 → c₂ = 0.20 mol/dm³。
10. Common Pitfalls and Tips | 常见错误和技巧
One common mistake is failing to convert cm³ to dm³ for volume-based calculations. Always check the units required in the answer. Another pitfall is using the wrong mole ratio; you must write a correct balanced equation first. In thermochemistry, forgetting to divide by moles to get kJ/mol is frequent. Also, when calculating average titre, exclude the rough trial or any anomalous readings. Show all steps clearly, as even with an incorrect final answer you can gain marks for correct working. Finally, time management is crucial: allocate about 15–20 minutes for the calculation-heavy parts of the ATP paper.
一个常见错误是未能将cm³转换为dm³进行基于体积的计算。务必检查答案所需的单位。另一个陷阱是使用错误的摩尔比；你必须先写出正确的配平方程式。在热化学中，经常忘记除以摩尔数以获得kJ/mol。此外，在计算平均滴定时，应排除粗略试验或任何异常读数。清晰地展示所有步骤，因为即使最终答案错误，你也能因正确的运算步骤得分。最后，时间管理至关重要：为ATP试卷中计算密集的部分分配大约15–20分钟。
Practice numerical questions repeatedly from past papers to build confidence. Keep a formula sheet and practise until you can recall them instantly. Calculation questions are highly predictable, so thorough preparation will pay off.
反复练习往年真题中的数字题以建立信心。保留一份公式表并练习直到你能马上回忆起来。计算题的可预测性很强，因此充分的准备会带来回报。
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