📚 Mastering Chemistry Calculation Questions from Activate Student Book | 掌握 Activate 学生用书中的化学计算题型
The Activate Chemistry Student Book for Key Stage 3 lays the groundwork for all future chemistry learning by blending core principles with essential calculation skills. Questions often cover relative atomic mass, formula mass, conservation of mass, percentage composition, and mass relationships in equations. This article offers a bilingual, step‑by‑step guide to the major calculation types, with worked examples and clear strategies to help students gain confidence and accuracy.
《Activate 化学学生用书》为关键阶段3（KS3）奠定了化学学习的基础，将核心原理与基本的计算技能融合在一起。书中的题目经常涉及相对原子质量、式量、质量守恒、百分组成以及方程中的质量关系。本文提供双语的分步指南，涵盖主要计算题型，配有示例和清晰的策略，帮助学生建立信心并提高准确性。
1. Understanding Relative Atomic Mass (Ar) | 理解相对原子质量（Ar）
Relative atomic mass (Ar) compares the average mass of an atom of an element to 1/12 of the mass of a carbon‑12 atom. In the Activate course, Ar values are usually whole numbers taken from the periodic table: hydrogen is 1, carbon is 12, oxygen is 16, and iron is 56. These numbers are used to work out the mass of molecules and compounds.
相对原子质量（Ar）将元素的一个原子的平均质量与碳‑12原子质量的1/12进行比较。在Activate课程中，Ar值通常取周期表上的整数：氢为1，碳为12，氧为16，铁为56。这些数值用于计算分子和化合物的质量。
For example, the Ar of chlorine is 35.5 because chlorine has two common isotopes. Students do not need to recall isotopic masses but must be able to read Ar values from the periodic table and use them in calculations.
例如，氯的Ar为35.5，因为氯有两种常见的同位素。学生不需要记住同位素质量，但必须能够从周期表中读取Ar值并将其用于计算。
2. Calculating Relative Formula Mass (Mr) | 计算相对式量（Mr）
Relative formula mass (Mr) is the sum of the relative atomic masses of all the atoms in a formula unit. For a covalent molecule like water, H₂O, the Mr = (2 × 1) + (1 × 16) = 18. For an ionic compound such as calcium carbonate, CaCO₃, the Mr = 40 + 12 + (3 × 16) = 100.
相对式量（Mr）是一个式单元中所有原子的相对原子质量之和。对于共价分子水 H₂O，Mr = (2 × 1) + (1 × 16) = 18。对于离子化合物碳酸钙 CaCO₃，Mr = 40 + 12 + (3 × 16) = 100。
Always look for brackets in formulae like Mg(OH)₂: first calculate the mass inside the bracket (O + H = 16 + 1 = 17), then multiply by the subscript outside: 2 × 17 = 34. Add the Mg (24) to get Mr = 58.
遇到带有括号的化学式如 Mg(OH)₂ 时，先计算括号内的质量（O + H = 16 + 1 = 17），再乘以外面的下标：2 × 17 = 34。加上 Mg (24) 得到 Mr = 58。
A common exercise in Activate is to fill in a table of Mr values for a list of compounds. Students should practise with water, carbon dioxide, sodium chloride, sulfuric acid and copper sulfate crystals.
Activate 中常见的练习是填写化合物式量数值的表格。学生应练习计算水、二氧化碳、氯化钠、硫酸和硫酸铜晶体的式量。
3. Conservation of Mass in Reactions | 化学反应中的质量守恒
The law of conservation of mass states that the total mass of reactants equals the total mass of products in a chemical reaction. No atoms are lost or created; they are only rearranged. This principle is fundamental when checking balanced equations and solving calculation problems.
质量守恒定律指出，在化学反应中，反应物的总质量等于生成物的总质量。没有原子消失或产生，它们只是重新排列。这个原理是检验配平方程式和解答计算题的基础。
If 5.6 g of iron reacts with excess sulfur to form iron(II) sulfide, the product mass will be greater than 5.6 g because sulfur atoms have added to the iron. The mass increase matches the mass of sulfur that bonded. In a closed system, the total mass stays constant.
如果5.6 g铁与过量的硫反应生成硫化亚铁，产物的质量将大于5.6 g，因为硫原子与铁结合。增加的质量等于结合的硫的质量。在封闭系统中，总质量保持不变。
When a gas is produced and escapes, the measured mass may appear to decrease. Activate experiments often explore this by reacting acid with limestone in an open flask and then repeating with a balloon to trap the gas, showing that mass is conserved.
当产生的气体逸出时，测得的表观质量可能会下降。Activate 常通过开放烧瓶中酸与石灰石的反应来探索这一现象，随后用气球收集气体重复实验，以此证明质量守恒。
4. Using Mass Conservation to Find Unknown Masses | 利用质量守恒求未知质量
A typical question provides the masses of all reactants and all products except one, and asks students to calculate the missing value. The sum of reactant masses must equal the sum of product masses.
一道典型的题目会给出除一个物质之外的所有反应物和生成物的质量，要求学生计算缺失的数值。反应物质量之和必须等于生成物质量之和。
Example: 12 g of magnesium reacts with oxygen to produce 20 g of magnesium oxide. How much oxygen reacted? Reactants total = mass of Mg + mass of O₂ = products mass. 12 g + mass O₂ = 20 g, so mass of O₂ = 8 g.
例题：12 g镁与氧气反应生成20 g氧化镁。反应了多少氧气？反应物总质量 = Mg 质量 + O₂ 质量 = 生成物质量。12 g + O₂ 质量 = 20 g，因此 O₂ 质量 = 8 g。
When two solutions react to form a precipitate and a gas, students should list all substances, add known masses, and then subtract from the total to find the unknown. Always check that the final answer makes sense and units are consistent.
当两种溶液反应生成沉淀和气体时，学生应列出所有物质，将已知质量相加，再从总质量中减去，以求得未知质量。务必检查最终结果是否合理且单位一致。
5. Calculating Percentage by Mass | 计算质量百分比
Percentage by mass tells you how much of a compound’s mass comes from a particular element. The formula is: % mass = (total mass of the element in the formula ÷ Mr of the compound) × 100%.
质量百分比表示化合物中有多少质量来自某一特定元素。计算公式为：质量% = (化学式中该元素的总质量 ÷ 化合物的 Mr) × 100%。
For iron(III) oxide, Fe₂O₃, Mr = (2 × 56) + (3 × 16) = 112 + 48 = 160. The mass of iron in the formula is 112. Percentage of iron = (112 ÷ 160) × 100% = 70%. This means every 100 g of iron ore of pure Fe₂O₃ contains 70 g of iron.
对于氧化铁 Fe₂O₃，Mr = (2 × 56) + (3 × 16) = 112 + 48 = 160。化学式中铁的质量为112。铁的质量百分比 = (112 ÷ 160) × 100% = 70%。这意味着每100 g纯净的 Fe₂O₃ 铁矿石中含70 g铁。
Activate often includes questions on fertilisers such as ammonium nitrate, NH₄NO₃, asking for the percentage of nitrogen. Mr = 80, mass of nitrogen = 2 × 14 = 28, so % N = (28 ÷ 80) × 100% = 35%.
Activate 中经常出现有关化肥如硝酸铵 NH₄NO₃ 的题目，要求计算氮的百分含量。Mr = 80，氮的质量 = 2 × 14 = 28，因此 N% = (28 ÷ 80) × 100% = 35%。
6. Interpreting Chemical Equations | 解读化学方程式
A balanced chemical equation shows the ratio of reacting particles and the ratio of masses. For 2Mg + O₂ → 2MgO, the equation tells us that two magnesium atoms react with one oxygen molecule to form two formula units of magnesium oxide.
配平的化学方程式显示了反应微粒的个数比和质量比。对于 2Mg + O₂ → 2MgO，该方程式告诉我们两个镁原子与一个氧分子反应，生成两个式单位的氧化镁。
The mass relationship can be found using Ar values: 2Mg has mass 2 × 24 = 48, O₂ is 2 × 16 = 32, so 48 g of magnesium reacts with 32 g of oxygen to give 80 g of magnesium oxide. The mass ratio is 48:32:80, which simplifies to 3:2:5.
质量关系可以用 Ar 值求得：2Mg 的质量为 2 × 24 = 48，O₂ 为 2 × 16 = 32，因此48 g镁与32 g氧气反应生成80 g氧化镁。质量比为 48:32:80，化简为 3:2:5。
Students must be able to recognise that coefficients in an equation refer to numbers of atoms, molecules or formula units, not directly to grams. The step from ‘chemical amounts’ to mass requires multiplying by the Mr of each substance.
学生需要认识到，方程式中的系数表示原子、分子或式单元的个数，而非直接代表克数。从“化学计量数”到质量的换算需要乘以每种物质的 Mr。
7. Balancing Equations by Counting Atoms | 通过原子计数配平方程式
Before any calculation can be done, the equation must be balanced. Count the number of each type of atom on the left and right. Add coefficients only in front of the chemical formulas—never change the subscript numbers inside a formula.
在进行任何计算之前，必须配平方程式。数一下左边和右边每种原子的个数。只能在化学式前面添加系数——切勿改动化学式内部的下标数字。
Example: H₂ + Cl₂ → HCl. Left: 2H, 2Cl; right: 1H, 1Cl. Place a coefficient 2 before HCl: H₂ + Cl₂ → 2HCl. Now both sides have 2H and 2Cl. The equation is balanced.
示例：H₂ + Cl₂ → HCl。左边：2H, 2Cl；右边：1H, 1Cl。在 HCl 前加上系数2：H₂ + Cl₂ → 2HCl。现在两边均有 2H 和 2Cl。方程式已配平。
For more complex equations like C₂H₆ + O₂ → CO₂ + H₂O, balance C first, then H, and finally O. C₂H₆ + O₂ → 2CO₂ + 3H₂O gives 2C, 6H, and right now 7O. To balance O, place 3½ before O₂, then multiply all by 2: 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O.
对于较复杂的方程式，如 C₂H₆ + O₂ → CO₂ + H₂O，先配平 C，再配平 H，最后配平 O。C₂H₆ + O₂ → 2CO₂ + 3H₂O 得到 2C, 6H，右边 O 为 7。为配平 O，在 O₂ 前加 3½，然后将所有系数乘以2：2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O。
8. Mass Relationships from Balanced Equations | 从配平方程式看质量关系
Once the equation is balanced, calculate the total mass of reactants and products using Mr values. This allows us to predict how much product forms from a given mass of reactant, or how much reactant is needed to make a desired mass of product.
方程式配平后，利用 Mr 值计算反应物和生成物的总质量。这使我们能够预测从给定的反应物质量能得到多少产物，或者需要多少反应物才能制得期望的产物质量。
Consider the thermal decomposition of calcium carbonate: CaCO₃ → CaO + CO₂. Mr: CaCO₃ = 100, CaO = 56, CO₂ = 44. Mass is conserved: 100 g CaCO₃ yields 56 g CaO and 44 g CO₂. So if a student starts with 25 g of CaCO₃, the mass of CaO produced is (56/100) × 25 = 14 g.
以碳酸钙的热分解为例：CaCO₃ → CaO + CO₂。Mr: CaCO₃ = 100，CaO = 56，CO₂ = 44。质量守恒：100 g CaCO₃ 产生 56 g CaO 和 44 g CO₂。因此如果学生从 25 g CaCO₃ 开始，产生的 CaO 质量为 (56/100) × 25 = 14 g。
Activate problems often ask: ‘What mass of carbon dioxide is produced when 10 g of carbon is burnt in excess oxygen?’ C + O₂ → CO₂. Ar(C)=12, Mr(CO₂)=44. Ratio: 12 g C gives 44 g CO₂. For 10 g C, mass of CO₂ = (44/12) × 10 = 36.7 g.
Activate 中的问题经常问：“10 g 碳在过量氧气中燃烧会产生多少质量的二氧化碳？” C + O₂ → CO₂。Ar(C)=12，Mr(CO₂)=44。比例关系：12 g 碳生成 44 g CO₂。10 g C 产生的 CO₂ 质量 = (44/12) × 10 = 36.7 g。
9. Reacting Masses and Limiting Reactants | 反应质量与限量反应物
When amounts of both reactants are given, one will be used up completely – the limiting reactant – while the other is in excess. The limiting reactant determines the maximum amount of product that can form.
当给出两种反应物的质量时，其中一种会完全消耗——即限量反应物，而另一种则是过量的。限量反应物决定了能生成的产物的最大量。
Example: 6 g of magnesium and 4 g of oxygen are heated together. 2Mg + O₂ → 2MgO. From the balanced equation, 48 g Mg reacts with 32 g O₂. So 1 g Mg needs 32/48 = 0.667 g O₂. 6 g Mg would need 6 × 0.667 = 4 g O₂ exactly. The given oxygen is 4 g, so both react completely with no excess. If only 3 g O₂ were provided, oxygen would be limiting and magnesium would be in excess.
例题：将6 g镁与4 g氧气一起加热。2Mg + O₂ → 2MgO。根据配平方程式，48 g Mg 与 32 g O₂ 完全反应。因此1 g Mg 需要 32/48 = 0.667 g O₂。6 g Mg 需要 6 × 0.667 = 4 g O₂，恰好等于提供的氧气量。因此两者完全反应，没有过量。如果只提供3 g O₂，则氧气是限量反应物，镁会过量。
Students should practise identifying limiting reactants by calculating how much of one reactant is needed to react with the given mass of the other, then compare with what is available. The smaller calculated ‘need’ indicates the limiting substance.
学生应练习通过计算一种反应物与另一种给定质量完全反应所需的质量，再与实际可用量进行对比，以确定限量反应物。计算结果中较小的“需求量”表明该物质是限量反应物。
10. Practical Calculation Questions: Example Walkthrough | 实际计算题：示例讲解
A popular Activate investigation measures the mass of magnesium oxide formed by burning magnesium ribbon in a crucible. Students start with a known mass of magnesium, heat it strongly with the lid slightly open, and reweigh until constant mass. The difference gives the mass of oxygen that combined.
Activate 中一个很受欢迎的实验探究是测量镁条在坩埚中燃烧生成的氧化镁的质量。学生从已知质量的镁开始，强烈加热并微开盖子，反复称量至恒重。质量差即化合的氧气的质量。
Worked example: 0.48 g of magnesium ribbon is heated. The final mass of white magnesium oxide is 0.80 g. Calculate the empirical formula and verify mass conservation. Mass of oxygen = 0.80 − 0.48 = 0.32 g. Moles of Mg = 0.48/24 = 0.02, moles of O = 0.32/16 = 0.02. Ratio Mg:O = 1:1, so formula is MgO. Total reactant mass = 0.48 + 0.32 = 0.80 g, consistent with product mass.
示例解析：0.48 g镁条被加热，最终白色氧化镁的质量为0.80 g。计算实验式并验证质量守恒。氧气的质量 = 0.80 − 0.48 = 0.32 g。Mg的物质的量 = 0.48/24 = 0.02，O的物质的量 = 0.32/16 = 0.02。Mg:O的比例为1:1，因此化学式为 MgO。反应物总质量 = 0.48 + 0.32 = 0.80 g，与产物质量一致。
Another typical question: ‘A student heated 3.25 g of zinc in a stream of chlorine gas, obtaining 6.80 g of zinc chloride. Find the mass of chlorine that reacted and the empirical formula.’ Chlorine mass = 6.80 − 3.25 = 3.55 g. Moles Zn = 3.25/65 = 0.05; moles Cl = 3.55/35.5 = 0.10. Ratio 0.05:0.10 = 1:2, giving ZnCl₂.
另一道常见题：“一名学生在氯气流中加热3.25 g锌，得到6.80 g氯化锌。求参加反应的氯气的质量和实验式。”氯的质量 = 6.80 − 3.25 = 3.55 g。Zn的物质的量 = 3.25/65 = 0.05；Cl的物质的量 = 3.55/35.5 = 0.10。比例 0.05:0.10 = 1:2，得出 ZnCl₂。
These calculations blend conservation of mass with formula determination, giving students a real sense of how quantitative chemistry works in the laboratory.
这些计算将质量守恒与化学式的确定结合在一起，让学生真实感受到定量化学在实验室中是如何运作的。
11. Top Tips for Success in Activate Calculation Questions | Activate 计算题的高分技巧
Always begin by writing down what you are given and what you must find. Underline the key numbers and units. Next, write a balanced equation if the question involves a reaction. Show all your working clearly so that marks can be awarded for method even if a final answer slips.
始终先写下已知条件和需要求解的内容。在关键数字和单位下划线。接着，如果题目涉及反应，写出配平的方程式。清晰地展示所有解题步骤，这样即使最终答案有误，方法步骤仍能得分。
Memorise common Ar values (H=1, C=12, N=14, O=16, Na=23, Mg=24, S=32, Cl=35.5, Ca=40, Fe=56, Cu=63.5) to speed up work. Practise using a standard calculator and double‑check Mr calculations – a small arithmetic error can lead to an entirely wrong answer.
记住常见的 Ar 值（H=1, C=12, N=14, O=16, Na=23, Mg=24, S=32, Cl=35.5, Ca=40, Fe=56, Cu=63.5）以提高解题速度。练习使用标准计算器并反复检查 Mr 的计算——小小的算术错误可能导致完全错误的答案。
Finally, use proportionality reasoning: once you know the mass ratio from the balanced equation, you can scale it up or down using simple division and multiplication. Keep the logic visible, and you will build a solid quantitative foundation.
最后，使用比例推理：一旦从配平方程式得到质量比，便可通过简单的除法和乘法进行缩放。让解题逻辑清晰可见，你就能打下坚实的定量基础。
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