Calculation Questions in Cambridge Lower Secondary Complete Chemistry SB 2nd | 剑桥初中完全化学第二版计算题型解析

📚 Calculation Questions in Cambridge Lower Secondary Complete Chemistry SB 2nd | 剑桥初中完全化学第二版计算题型解析

In the Cambridge Lower Secondary Complete Chemistry Student Book (2nd Edition), calculation questions are designed to build foundational skills in quantitative chemistry. These problems often involve relative atomic masses, formula masses, percentage composition, reacting masses, and solution concentrations. Mastering these calculations early not only helps students excel in lower secondary exams but also prepares them for IGCSE and beyond. This article provides a structured overview of the key calculation types, step-by-step methods, and common pitfalls, with both English and Chinese explanations.

在《剑桥初中完全化学学生用书》（第二版）中，计算题旨在建立定量化学的基础技能。这些题目通常涉及相对原子质量、式量、百分比组成、反应质量和溶液浓度。尽早掌握这些计算不仅有助于学生赢得初中考试，也为IGCSE及更高阶段打下基础。本文以英中双语系统梳理关键题型、分步方法和常见错误。

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1. Types of Calculation Questions in the Textbook | 教材中的计算题型概览

The calculation questions in this textbook can be grouped into several core topics: finding relative formula mass, calculating percentage by mass of an element, mass ratios, using chemical equations to relate masses of reactants and products, determining concentration in g/dm³, and finding the simplest formula from experimental data. Each type requires careful use of Ar values from the periodic table and an understanding of chemical formulae.

该教材中的计算题可归纳为几大核心主题：求相对分子质量、计算元素质量分数、质量比、利用化学方程式联系反应物与生成物的质量、计算g/dm³浓度，以及根据实验数据求最简式。每类题型都需要仔细运用元素周期表中的相对原子质量Ar，并理解化学式的含义。

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2. Relative Atomic Mass (Ar) and Relative Formula Mass (Mr) | 相对原子质量与相对分子质量

The relative atomic mass (Ar) is the average mass of an atom of an element compared to 1/12 of the mass of a carbon-12 atom. It has no units. Students find Ar values on the periodic table. For example, Ar of hydrogen is 1, carbon is 12, oxygen is 16. The relative formula mass (Mr) of a compound is the sum of the Ar of all atoms in its formula.

相对原子质量(Ar)是一个元素的原子平均质量与碳‑12原子质量的1/12相比较的比值，没有单位。学生在周期表中找到Ar值。例如，氢的Ar=1，碳=12，氧=16。化合物的相对分子质量(Mr)是其化学式中所有原子Ar的总和。

Mr = sum of (Ar × number of atoms of that element)

Mr = 各元素（Ar × 原子个数）之和

For water, H₂O: Mr = (2 × 1) + 16 = 18. For calcium carbonate, CaCO₃: Mr = 40 + 12 + (3 × 16) = 100. For magnesium nitrate, Mg(NO₃)₂: Mr = 24 + 2 × (14 + 3×16) = 24 + 2×(14+48) = 24 + 2×62 = 24 + 124 = 148. When a formula contains brackets, multiply the inside total by the subscript outside. Always double-check atomic counts.

水H₂O：Mr = (2×1) + 16 = 18。碳酸钙CaCO₃：Mr = 40 + 12 + (3×16) = 100。硝酸镁Mg(NO₃)₂：Mr = 24 + 2×(14 + 3×16) = 24 + 2×62 = 148。当化学式含括号时，将括号内总和乘以外侧下标。务必核实原子个数。

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3. Percentage by Mass of an Element in a Compound | 化合物中元素的质量分数

To calculate the percentage by mass of an element, use the formula:

计算元素的质量分数，公式为：

% element = (total Ar of the element in the formula ÷ Mr of compound) × 100

元素% = (该元素在化学式中的总Ar ÷ 化合物的Mr) × 100

Example: Find the percentage of oxygen in sulfuric acid, H₂SO₄. Mr = (2×1) + 32 + (4×16) = 2 + 32 + 64 = 98. Total mass of oxygen = 4×16 = 64. %O = (64 / 98) × 100 = 65.3%. Another common question: calculate the percentage of nitrogen in ammonium nitrate, NH₄NO₃. Mr = 14+(4×1)+14+(3×16) = 14+4+14+48 = 80. Total N = 2×14 = 28. %N = (28/80)×100 = 35%.

例：求硫酸H₂SO₄中氧的质量分数。Mr = 98，氧的总Ar = 64。%O = (64/98)×100 = 65.3%。另一个常见题：求硝酸铵NH₄NO₃中氮的质量分数。Mr = 80，氮的总Ar = 28，%N = (28/80)×100 = 35%。

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4. Mass Ratios in Compounds | 化合物中的质量比

The mass ratio of elements in a compound is simply the ratio of their total relative atomic masses. For CO₂, the mass ratio of carbon to oxygen is 12 : (2×16) = 12 : 32 = 3 : 8 when simplified. For Al₂O₃, mass ratio Al : O = (2×27) : (3×16) = 54 : 48 = 9 : 8. Mass ratios are useful for determining how much of each element is needed to make a certain amount of compound, and they often appear in questions about ‘calculating the mass of element required’.

化合物中元素的质量比即各元素总Ar之比。对CO₂，碳与氧的质量比 = 12:32 = 3:8。对Al₂O₃，铝与氧的质量比 = (2×27):(3×16) = 54:48 = 9:8。质量比有助于判断制备一定量化合物所需各元素的质量，也常在“计算所需元素质量”类问题中出现。

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5. Calculating Mass of an Element in a Given Mass of Compound | 计算给定质量化合物中元素的质量

Once the percentage of an element is known, you can find its mass in any sample. For instance, what mass of iron is present in 200 g of iron(III) oxide, Fe₂O₃? Mr of Fe₂O₃ = (2×56) + (3×16) = 112 + 48 = 160. %Fe = (112/160)×100 = 70%. Mass of Fe = 70% of 200 g = 0.70 × 200 = 140 g. Alternatively, use the ratio method directly: mass of Fe = (total Ar of Fe / Mr) × sample mass = (112/160) × 200 = 140 g. This method saves time.

知道元素的质量分数后，可求出任意样品中该元素的质量。例如，200 g氧化铁Fe₂O₃中含有多少克铁？Fe₂O₃的Mr = 160，%Fe = 70%，故铁的质量 = 200 g × 0.70 = 140 g。也可直接用比值法：铁质量 = (112/160)×200 = 140 g，这种方法更快捷。

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6. Conservation of Mass in Reactions | 反应中的质量守恒

In a chemical reaction, atoms are rearranged, but no atoms are created or destroyed. Therefore the total mass of reactants always equals the total mass of products. For example, when 12 g of carbon reacts with 32 g of oxygen to form carbon dioxide, the mass of CO₂ produced is 12 + 32 = 44 g. If a reaction is carried out in an open container and a gas escapes, the measured mass of the remaining solid or liquid may decrease. Students must be able to explain such changes using the idea of uncollected gas, not a violation of mass conservation.

在化学反应中，原子重新排列，但原子不会凭空产生或消失。因此反应物的总质量始终等于生成物的总质量。例如，12 g碳与32 g氧气反应生成二氧化碳，所得CO₂的质量为44 g。若在敞开容器中进行反应且有气体逸出，剩余固体或液体的实测质量会减少。学生必须能够利用未收集气体来解释此变化，而非错误地认为质量守恒定律被打破。

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7. Reacting Mass Calculations using Relative Formula Mass | 利用相对分子质量进行反应质量计算

Even without the mole concept, students can use relative formula masses to calculate reacting masses. The key is to write a balanced equation and then find the mass ratio from the coefficients and Mr values. For the reaction: 2Mg + O₂ → 2MgO. The total mass of reactants: 2×24 + 32 = 48 + 32 = 80. The total mass of products: 2×(24+16) = 80. The mass ratio of Mg to MgO is 48:80 = 3:5. Thus 24 g of Mg produces (80/48)×24 = 40 g of MgO.

即使不引入摩尔概念，学生也可用相对分子质量计算反应质量。关键是要写出配平的化学方程式，再根据系数和Mr求质量比。以反应2Mg + O₂ → 2MgO为例，Mg与MgO的质量比为48:80=3:5，因此24 g Mg生成40 g MgO。

Another example: decomposition of calcium carbonate, CaCO₃ → CaO + CO₂. Mr of CaCO₃=100, Mr of CaO=56. If 50 g of CaCO₃ is heated, the mass of CaO formed = (56/100)×50 = 28 g. For the reaction 2H₂ + O₂ → 2H₂O, the mass ratio H₂ : O₂ : H₂O = (2×2) : 32 : (2×18) = 4:32:36 = 1:8:9. So 6 g of hydrogen will react with 48 g of oxygen to produce 54 g of water.

再如碳酸钙分解：CaCO₃ → CaO + CO₂，Mr：CaCO₃=100，CaO=56。加热50

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