📚 IGCSE WJEC Chemistry: Mastering Calculation Questions | IGCSE WJEC 化学计算题专项训练
Calculation questions form a significant part of the IGCSE WJEC Chemistry examination, testing your ability to apply quantitative reasoning to chemical reactions, formulas, and energetics. This revision guide provides targeted practice and step-by-step methods for every calculation type you are likely to encounter, from basic mole concepts to atom economy and titration analyses. By mastering these techniques, you will build confidence in handling numbers precisely and efficiently under exam conditions. Each section below pairs the working method with a fully commented example, so you can revise the logic as well as the arithmetic.
计算题是 IGCSE WJEC 化学考试的重要组成部分，考查你将定量推理应用于化学反应、化学式和能量变化的能力。本复习指南针对你可能遇到的每一类计算题提供专项训练和分步方法，从基础的摩尔概念到原子经济性和滴定分析。通过掌握这些技巧，你将建立信心，在考试条件下精确高效地处理数字。以下每一节都把解题方法与一个完整范例配对讲解，让你既复习思路也复习运算。
1. The Mole Concept and Molar Mass | 摩尔概念与摩尔质量
The mole is the chemist' rams, is its molar mass, which is numerically equal to its relative formula mass (Mᵣ). To find the amount in moles (n), use n = mass (g) ÷ molar mass (g/mol).
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）。1 摩尔物质的质量以克为单位即为其摩尔质量，其数值等于该物质的相对式量（Mᵣ）。求物质的量 n，使用 n = 质量（g）÷ 摩尔质量（g/mol）。
n = m ÷ Mᵣ · N = n × 6.02 × 10²³
= 0.100 mol, so the number of molecules is 0.100 × 6.02 × 10²³ = 6.02 × 10²². Notice that "mass" here must be pure substance mass, not the mass of a solution.
范例：4.40 g 二氧化碳（Mᵣ = 44.0）。物质的量 = 4.40 ÷ 44.0 = 0.100 mol，故分子数为 0.100 × 6.02 × 10²³ = 6.02 × 10²²。注意这里的"质量"必须是纯物质的质量，而不是溶液的质量。
÷ n when a question gives you a mass and a mole value and asks you to identify a formula.
考试中经常使用的两个变形是：已知物质的量时 m = n × Mᵣ；题目给出质量和物质的量、要求推断化学式时 Mᵣ = m ÷ n。
2. Calculating Reacting Masses from Equations | 由化学方程式计算反应质量
Reacting-mass questions always follow the same five-step route: write and balance the equation; calculate the moles of the substance whose mass is given; use the mole ratio from the balanced equation; convert to moles of the required substance; and finally convert moles back to mass.
反应质量类题目始终遵循同样的五步法：写出并配平方程式；计算已知质量物质的物质的量；利用配平方程式的摩尔比；换算成所求物质的物质的量；最后把物质的量换算回质量。
- Step 1: Write the balanced symbol equation, including state symbols if asked.
第 1 步：写出配平的符号方程式，如要求则注明状态符号。 - Step 2: n = m ÷ Mᵣ for the known substance.
第 2 步：对已知物质使用 n = m ÷ Mᵣ。 - Step 3: Apply the mole ratio from the coefficients.
第 3 步：根据系数确定摩尔比。 - Step 4: m = n × Mᵣ for the required substance.
第 4 步：对所求物质使用 m = n × Mᵣ。
CaCO₃ → CaO + CO₂ (Mᵣ: 100, 56, 44)
d, giving 0.100 × 56 = 5.6 g of calcium oxide, and 0.100 × 44 = 4.4 g of carbon dioxide. The masses add up to 10.0 g, which confirms your answer by conservation of mass.
范例：加热分解 10.0 g 碳酸钙。n(CaCO₃) = 10.0 ÷ 100 = 0.100 mol。摩尔比为 1 : 1 : 1，故生成 0.100 mol CaO，即 0.100 × 56 = 5.6 g 氧化钙，以及 0.100 × 44 = 4.4 g 二氧化碳。两者质量之和为 10.0 g，可用质量守恒检验答案。
3. Concentration, Moles and Solution Volume | 浓度、物质的量与溶液体积
ecause 1 dm³ = 1000 cm³, always divide a volume given in cm³ by 1000 before substituting it.
浓度以 mol/dm³ 为单位，把溶质的物质的量与溶液体积联系起来。核心关系式为 c = n ÷ V，其中 V 必须以 dm³ 为单位。由于 1 dm³ = 1000 cm³，凡以 cm³ 给出的体积，代入前务必先除以 1000。
÷ V · n = c × V · V (dm³) = V (cm³) ÷ 1000
100 ÷ 0.250 = 0.400 mol/dm³. If a question instead asks for mass concentration in g/dm³, multiply the molar concentration by Mᵣ: 0.400 × 40.0 = 16.0 g/dm³.
范例：将 4.00 g 氢氧化钠（Mᵣ = 40.0）溶于水配成 250 cm³ 溶液。n = 4.00 ÷ 40.0 = 0.100 mol；V = 250 ÷ 1000 = 0.250 dm³；c = 0.100 ÷ 0.250 = 0.400 mol/dm³。若题目要求质量浓度（g/dm³），则把物质的量浓度乘以 Mᵣ：0.400 × 40.0 = 16.0 g/dm³。
0 cm³ of 0.400 mol/dm³ diluted to 100 cm³ gives c₂ = (0.400 × 25.0) ÷ 100 = 0.100 mol/dm³.
常见的延伸考法是稀释。加水时溶质的物质的量不变，故 c₁V₁ = c₂V₂。例如把 25.0 cm³ 的 0.400 mol/dm³ 溶液稀释至 100 cm³，则 c₂ = (0.400 × 25.0) ÷ 100 = 0.100 mol/dm³。
4. Gas Volumes and the Molar Gas Volume | 气体体积与摩尔气体体积
At room temperature and pressure (r.t.p., taken as 25 °C and 1 atm in WJEC questions), one mole of any gas occupies 24 dm³, which is 24 000 cm³. This lets you convert directly between moles and gas volume without knowing which gas it is.
在室温常压（r.t.p.，WJEC 题目中取 25 °C 和 1 atm）下，1 摩尔任何气体都占据 24 dm³，即 24 000 cm³。因此你无需知道是哪种气体，就能在物质的量与气体体积之间直接换算。
n = V (dm³) ÷ 24 · V (dm³) = n × 24
Worked example: what volume does 8.80 g of carbon dioxide occupy at r.t.p.? ³. Remember that the molar gas volume applies only to gases, so a solid reactant must always be converted through moles first.
范例：8.80 g 二氧化碳在 r.t.p. 下占多大体积？先求 n = 8.80 ÷ 44.0 = 0.200 mol，再求 V = 0.200 × 24 = 4.80 dm³，即 4800 cm³。请记住摩尔气体体积只适用于气体，因此固体反应物必须先换算成物质的量。
In gas-ratio questions you can often work with volumes directly, because equal volumes of gases at the same temperature and pressure contain equal numbers of moles. For instance, 20 cm³ of hydrogen reacts with 10 cm³ of oxygen to give 20 cm³ of water vapour, matching the ratio 2 : 1 : 2 in the equation.
在气体比例题中，你常常可以直接用体积计算，因为在相同温度和压强下，相同体积的气体含有相同的物质的量。例如 20 cm³ 氢气与 10 cm³ 氧气反应生成 20 cm³ 水蒸气，与方程式中的 2 : 1 : 2 一致。
5. Percentage Yield and Atom Economy | 产率与原子经济性
Percentage yield compares the mass you actually obtain with the maximum mass predicted by the equation. Atom economy compares the molar mass of the desired product with the total molar mass of all reactants, and it measures how efficiently atoms are used rather than how much product was lost.
产率把实际得到的质量与方程式预测的最大质量进行比较。原子经济性则把目标产物的摩尔质量与全部反应物的摩尔质量之和进行比较，衡量的是原子利用的效率，而不是产物的损失量。
% yield = (actual mass ÷ theoretical mass) × 100
atom economy = (Mᵣ of desired product ÷ Σ Mᵣ of all reactants) × 100
Worked example (yield): theory predicts 5.6 g of CaO, but only 4.2 g is collected, so % yield = (4.2 ÷ 5.6) × 100 = 75.0% (atom economy): for CaCO₃ → CaO + CO₂, the atom economy for CaO is (56 ÷ 100) × 100 = 56.0%.
范例（产率）：理论上应得到 5.6 g CaO，实际只收集到 4.2 g，则产率 = (4.2 ÷ 5.6) × 100 = 75.0%。范例（原子经济性）：对 CaCO₃ → CaO + CO₂，生成 CaO 的原子经济性 = (56 ÷ 100) × 100 = 56.0%。
A yield above 100% means the sample is impure or still wet, and a low atom economy means most atoms end up in unwanted by-products. Reaction conditions and the choice of route affect both values, and exam questions often ask you to explain the difference between them.
产率大于 100% 说明样品不纯或仍含水分；原子经济性低说明大多数原子进入了不需要的副产物。反应条件和路线选择都会影响这两个数值，考试中常要求你解释二者的区别。
6. Empirical and Molecular Formulae | 实验式与分子式
Percentage-composition data are converted into a formula by dividing each percentage by the relative atomic mass of that element, then dividing all the results by the smallest value to obtain the simplest whole-number ratio. That ratio gives the empirical formula.
把百分组成数据换算成化学式的做法是：将每个百分数除以该元素的相对原子质量，再把所有结果除以其中最小的数值，得到最简整数比。这个比值就是实验式。
ratio = (mass or % ÷ Aᵣ) for each element
Worked example: a compound contains 40.0% carbon, 6.7% hydrogen and 53.3% 3.33. Dividing by 3.33 gives 1 : 2 : 1, so the empirical formula is CH₂O, with Mᵣ = 30.0.
范例：某化合物含 40.0% 碳、6.7% 氢和 53.3% 氧。除以 Aᵣ 得 C：40.0 ÷ 12 = 3.33，H：6.7 ÷ 1 = 6.7，O：53.3 ÷ 16 = 3.33。再除以 3.33 得 1 : 2 : 1，故实验式为 CH₂O，Mᵣ = 30.0。
cular formula is C₆H₁₂O₆. Always state the multiplier explicitly, because method marks are awarded for showing the division.
若题目同时给出 Mᵣ = 180，则倍数为 180 ÷ 30 = 6，分子式即为 C₆H₁₂O₆。务必明确写出倍数，因为除以最小值的步骤本身也有过程分。
7. Water of Crystallisation and Formula from Experiment | 结晶水与由实验确定化学式
Heating a hydrated salt drives off water of crystallisation, and the mass loss equals the mass of water removed. Dividing the mass of water by 18 and the mass of the anhydrous salt by its Mᵣ, then forming a ratio, gives the value of x in a formula such as CuSO₄·xH₂O.
加热水合盐会赶出结晶水，失去的质量即被除去的水的质量。把水的质量除以 18、把无水盐的质量除以其 Mᵣ，再求比值，就能得到 CuSO₄·xH₂O 这类化学式中的 x 值。
(H₂O) = mass of water ÷ 18 · n(salt) = mass of anhydrous salt ÷ Mᵣ
0 ÷ 18 = 0.0500 mol. For the anhydrous salt, n(CuSO₄) = 1.60 ÷ 160 = 0.0100 mol. The ratio is 0.0500 : 0.0100 = 5 : 1, so x = 5.
范例：加热 2.50 g 水合硫酸铜至恒重，剩余 1.60 g。水的质量 = 2.50 − 1.60 = 0.90 g，故 n(H₂O) = 0.90 ÷ 18 = 0.0500 mol。无水盐方面，n(CuSO₄) = 1.60 ÷ 160 = 0.0100 mol。比值为 0.0500 : 0.0100 = 5 : 1，故 x = 5。
"Heating to constant mass" is an essential phrase: it means no further mass loss occurs, guaranteeing that all the water has been removed.
"加热至恒重"是必须出现的关键表述：它意味着质量不再继续减少，从而保证水已全部除去。
8. Titration Calculations | 滴定计算
Titration calculations combine the concentration equation with the mole ratio from the balanced equation. Convert the volume of the known solution to dm³, find its moles, apply the ratio, then divide the moles of the unknown by its own volume to obtain its concentration.
滴定计算把浓度公式与配平方程式的摩尔比结合起来。先把已知溶液的体积换算成 dm³，求出其物质的量，再应用摩尔比，最后用未知物的物质的量除以其自身体积，得到其浓度。
c₁V₁ ÷ n₁ = c₂V₂ ÷ n₂
.00250 mol. Since NaOH + HCl → NaCl + H₂O is 1 : 1, n(HCl) = 0.00250 mol, so c(HCl) = 0.00250 ÷ 0.02350 = 0.106 mol/dm³.
范例（1 : 1 比）：25.0 cm³ 的 0.100 mol/dm³ 氢氧化钠被 23.50 cm³
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