IGCSE OCR Chemistry: Calculations Practice | IGCSE OCR 化学：计算题专项训练

📚 IGCSE OCR Chemistry: Calculations Practice | IGCSE OCR 化学：计算题专项训练

Success in IGCSE OCR Chemistry requires confidence with quantitative problems. This revision guide breaks down the essential calculation skills tested in the exams, from moles and reacting masses to titrations and energy changes. Work through each topic systematically, practise the worked examples, and you will be ready to tackle any numbers-based question.

想在 IGCSE OCR 化学中取得好成绩，必须熟练掌握计算题。本复习指南将逐一拆解考试中必考的核心计算技能，包括摩尔、反应质量、滴定及能量变化。只要按专题系统练习、吃透例题，你就能自信应对任何数字类题目。

1. Relative Atomic Mass and Relative Formula Mass | 相对原子质量与相对式量

Relative atomic mass (Aᵣ) is the average mass of an atom of an element compared to 1/12th the mass of a carbon‑12 atom. For chlorine, Aᵣ = 35.5 because the natural mixture of ³⁵Cl and ³⁷Cl gives this weighted average.

相对原子质量 (Aᵣ) 是一个原子的平均质量与一个碳‑12 原子质量的 1/12 的比值。氯的 Aᵣ = 35.5，因为自然界中 ³⁵Cl 和 ³⁷Cl 的混合给出了这个加权平均值。

Relative formula mass (Mᵣ) applies to ionic compounds and is the sum of the Aᵣ values of all atoms in the formula. For magnesium hydroxide, Mg(OH)₂, Mᵣ = 24.3 + 2×(16.0 + 1.0) = 58.3.

相对式量 (Mᵣ) 适用于离子化合物，是化学式中所有原子 Aᵣ 的总和。例如氢氧化镁 Mg(OH)₂，Mᵣ = 24.3 + 2×(16.0 + 1.0) = 58.3。

2. The Mole and Molar Mass | 摩尔与摩尔质量

One mole of any substance contains 6.02 × 10²³ particles (Avogadro’s number). The mass of one mole of a substance is its molar mass (M) in g mol⁻¹, which is numerically equal to its Aᵣ or Mᵣ. The amount of substance n (mol) is found using: n = m / M, where m is mass in grams.

任何物质的 1 mol 都含有 6.02 × 10²³ 个微粒（阿伏伽德罗常数）。1 mol 物质的质量就是其摩尔质量 (M)，单位 g mol⁻¹，数值上等于其 Aᵣ 或 Mᵣ。物质的量 n (mol) 用公式 n = m / M 计算，其中 m 是质量（克）。

Example: How many moles are in 8.0 g of sulfur (S₈)? Mᵣ of S₈ = 8 × 32.1 = 256.8, so M ≈ 257 g mol⁻¹. n = 8.0 / 257 ≈ 0.0311 mol.

例题：8.0 g 硫（S₈）是多少摩尔？S₈ 的 Mᵣ = 8 × 32.1 = 256.8，因此 M ≈ 257 g mol⁻¹。n = 8.0 / 257 ≈ 0.0311 mol。

3. Empirical Formula and Molecular Formula | 经验式与分子式

An empirical formula gives the simplest whole‑number ratio of atoms in a compound. It is found by converting the masses (or percentages) of each element to moles, then dividing by the smallest mole value to get the ratio.

经验式表示化合物中各原子的最简整数比。计算方法是把每元素的质量（或百分含量）换算成物质的量，再除以其中最小的摩尔值得到比例。

Example: A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Moles: C = 40.0/12.0 = 3.33, H = 6.7/1.0 = 6.7, O = 53.3/16.0 = 3.33. Divide by 3.33 → ratio 1 : 2 : 1, so empirical formula is CH₂O.

例题：某化合物含碳 40.0%、氢 6.7%、氧 53.3%。物质的量：C = 40.0/12.0 = 3.33，H = 6.7/1.0 = 6.7，O = 53.3/16.0 = 3.33。除以 3.33 → 比为 1 : 2 : 1，经验式为 CH₂O。

The molecular formula is a whole‑number multiple of the empirical formula. It is determined from the Mᵣ of the compound. If the Mᵣ of the above compound is 60.0, then the multiplier is 60.0 / 30.0 = 2, giving C₂H₄O₂.

分子式是经验式的整数倍，由化合物的 Mᵣ 确定。若上述化合物的 Mᵣ = 60.0，则倍数 = 60.0 / 30.0 = 2，分子式为 C₂H₄O₂。

4. Reacting Mass Calculations | 反应质量计算

To find the mass of a product formed from a known mass of reactant, first write the balanced equation. Convert the given mass to moles, use the mole ratio to find moles of the unknown substance, then convert back to mass using molar mass.

要从已知反应物质量求生成物质量，首先写出配平的化学方程式。将已知质量换算成物质的量，利用摩尔比求出未知物的物质的量，再用摩尔质量换算回质量。

Example: What mass of carbon dioxide is produced when 1.00 g of calcium carbonate decomposes? CaCO₃ → CaO + CO₂. Mᵣ of CaCO₃ = 100.1, so n(CaCO₃) = 1.00 / 100.1 = 0.00999 mol. Mole ratio CaCO₃ : CO₂ = 1 : 1, so n(CO₂) = 0.00999 mol. Mᵣ of CO₂ = 44.0, mass = 0.00999 × 44.0 = 0.440 g.

例题：1.00 g 碳酸钙分解生成多少克二氧化碳？CaCO₃ → CaO + CO₂。CaCO₃ 的 Mᵣ = 100.1， n(CaCO₃) = 1.00 / 100.1 = 0.00999 mol。摩尔比 1:1，所以 n(CO₂) = 0.00999 mol。CO₂ 的 Mᵣ = 44.0，质量 = 0.00999 × 44.0 = 0.440 g。

5. Gas Volumes at RTP | 常温常压下气体体积

At room temperature and pressure (RTP: 20 °C, 1 atm), one mole of any gas occupies 24.0 dm³ (or 24 000 cm³). The volume V (dm³) can be calculated from moles using V = n × 24.0, or moles from volume using n = V / 24.0.

在常温常压下（RTP：20 °C，1 atm），1 mol 任何气体的体积为 24.0 dm³（或 24 000 cm³）。可用 V = n × 24.0 由物质的量求体积，或用 n = V / 24.0 由体积求物质的量。

Example: What volume of hydrogen is produced at RTP when 0.500 g of magnesium reacts with excess acid? Mg + 2HCl → MgCl₂ + H₂. Mᵣ of Mg = 24.3, n(Mg) = 0.500 / 24.3 = 0.0206 mol. Mole ratio 1:1, so n(H₂) = 0.0206 mol. V = 0.0206 × 24.0 = 0.494 dm³ (494 cm³).

例题：0.500 g 镁与过量酸反应，在 RTP 下可产生多少体积的氢气？Mg + 2HCl → MgCl₂ + H₂。Mg 的 Mᵣ = 24.3，n(Mg) = 0.500 / 24.3 = 0.0206 mol。摩尔比 1:1， n(H₂) = 0.0206 mol。V = 0.0206 × 24.0 = 0.494 dm³ (494 cm³)。

6. Concentration of Solutions | 溶液浓度

Concentration can be expressed in mol dm⁻³ (molarity) or in g dm⁻³. Molarity c (mol dm⁻³) = n / V, where V is the volume of solution in dm³. To convert between the two, use mass concentration (g dm⁻³) = c × M.

浓度可用 mol dm⁻³（摩尔浓度）或 g dm⁻³ 表示。摩尔浓度 c (mol dm⁻³) = n / V，其中 V 是溶液体积（dm³）。两种浓度可以通过质量浓度 (g dm⁻³) = c × M 互相转换。

Example: 5.85 g of NaCl is dissolved in 250 cm³ of water. Calculate the concentration in mol dm⁻³. Mᵣ of NaCl = 58.5, n = 5.85 / 58.5 = 0.100 mol. V = 250 / 1000 = 0.250 dm³. c = 0.100 / 0.250 = 0.400 mol dm⁻³.

例题：将 5.85 g NaCl 溶于 250 cm³ 水。计算摩尔浓度。NaCl 的 Mᵣ = 58.5，n = 5.85 / 58.5 = 0.100 mol。V = 250 / 1000 = 0.250 dm³。c = 0.100 / 0.250 = 0.400 mol dm⁻³。

7. Titration Calculations | 滴定计算

Titration results are used to find an unknown concentration. At the endpoint, the moles of acid and base are related by the stoichiometric ratio. The key formula is nₐcₐVₐ / n_bc_bV_b = 1 after adjusting for the ratio nₐ : n_b from the balanced equation.

滴定结果用于计算未知浓度。在终点时，酸和碱的物质的量符合化学计量比。根据配平方程式中 nₐ : n_b 的比例调整后，关键公式可写为 nₐcₐVₐ / n_bc_bV_b = 1。

Example: 25.0 cm³ of 0.100 mol dm⁻³ NaOH is neutralised by 22.5 cm³ of HCl. Find the concentration of HCl. NaOH + HCl → NaCl + H₂O, ratio 1:1. Moles of NaOH = 0.100 × (25.0/1000) = 0.00250 mol = moles of HCl. c(HCl) = 0.00250 / (22.5/1000) = 0.111 mol dm⁻³.

例题：25.0 cm³ 的 0.100 mol dm⁻³ NaOH 被 22.5 cm³ 的 HCl 中和。求 HCl 的浓度。NaOH + HCl → NaCl + H₂O，比例 1:1。NaOH 物质的量 = 0.100 × (25.0/1000) = 0.00250 mol = HCl 物质的量。c(HCl) = 0.00250 / (22.5/1000) = 0.111 mol dm⁻³。

Always use a concordant average titre, discarding any rough or non‑concordant results. Volumes should be recorded to the nearest 0.05 cm³ and read at eye level.

始终使用吻合的平均体积，弃去粗测或不相符的结果。体积应记录至 0.05 cm³，并在平视高度读数。

8. Percentage Yield | 产率

Percentage yield compares the actual amount of product obtained to the theoretical amount predicted from stoichiometry. It is calculated using: % yield = (actual mass / theoretical mass) × 100.

产率用于比较实际获得的产品量与按化学计量预测的理论量。计算公式为：产率 (%) = (实际质量 / 理论质量) × 100。

Reasons for a yield less than 100% include incomplete reactions, side reactions, losses during purification and handling. In industrial processes, yield directly affects cost and sustainability.

产率低于 100% 的原因包括反应不完全、副反应发生、纯化及转移过程中的损失。在工业生产中，产率直接影响成本和可持续性。

Example: From 10.0 g of reactant the theoretical mass of product is 14.5 g, but only 11.6 g is collected. % yield = (11.6 / 14.5) × 100 = 80.0%.

例题：用 10.0 g 反应物，理论可制得产品 14.5 g，但实际只收集到 11.6 g。产率 = (11.6 / 14.5) × 100 = 80.0%。

9. Atom Economy | 原子经济

Atom economy measures the efficiency of a reaction by showing what percentage of the atoms from all reactants is incorporated into the desired product. % atom economy = (Mᵣ of desired product / sum of Mᵣ of all reactants) × 100.

原子经济衡量反应的原子利用率，表示所有反应物中有多大比例的原子进入了目标产物。原子经济 (%) = (目标产物的 Mᵣ / 所有反应物的 Mᵣ 总和) × 100。

A high atom economy means less waste and greater sustainability. For example, the production of ethene from ethanol via dehydration (C₂H₅OH → C₂H₄ + H₂O) has atom economy = Mᵣ(C₂H₄) / Mᵣ(C₂H₅OH) × 100 = 28.0 / 46.0 × 100 = 60.9%.

原子经济高意味着废弃物少、可持续性更好。例如乙醇脱水制乙烯 (C₂H₅OH → C₂H₄ + H₂O) 的原子经济 = Mᵣ(C₂H₄) / Mᵣ(C₂H₅OH) × 100 = 28.0 / 46.0 × 100 = 60.9%。

Only the balanced equation matters; catalysts, solvents or excess reagents are not included in the calculation. Addition reactions often have 100% atom economy, whereas substitution reactions can produce by‑products and therefore lower atom economy.

计算只取决于配平的方程式，催化剂、溶剂和过量试剂不列入计算。加成反应通常原子经济可达 100%，而取代反应会生成副产物，因此原子经济较低。

10. Bond Energy Calculations | 键能计算

The overall energy change for a reaction can be estimated using mean bond energies. Energy is absorbed to break bonds (endothermic) and released when new bonds form (exothermic). The overall enthalpy change ΔH is:

ΔH = Σ(bond energies broken) − Σ(bond energies formed)

反应的总体能量变化可以通过平均键能估算。断键吸收能量（吸热），成键释放能量（放热）。总焓变 ΔH 为：ΔH = Σ(断裂键的总键能) − Σ(形成键的总键能)。

A positive ΔH indicates an endothermic reaction; a negative ΔH indicates an exothermic reaction. All bond energies are given in kJ mol⁻¹, and structural diagrams are essential to identify the bonds present.

ΔH 为正表示吸热反应；ΔH 为负表示放热反应。所有键能以 kJ mol⁻¹ 为单位，务必使用结构图识别存在的化学键。

Example: For the reaction H₂ + Cl₂ → 2HCl, bonds broken: 1 mol H−H (436 kJ) and 1 mol Cl−Cl (242 kJ) → total 678 kJ. Bonds formed: 2 mol H−Cl (2 × 431 = 862 kJ). ΔH = 678 − 862 = −184 kJ, so the reaction is exothermic.

例题：反应 H₂ + Cl₂ → 2HCl，断键：1 mol H−H (436 kJ) 和 1 mol Cl−Cl (242 kJ) → 总计 678 kJ。成键：2 mol H−Cl (2 × 431 = 862 kJ)。ΔH = 678 − 862 = −184 kJ，反应放热。

Published by TutorHao | Chemistry Revision Series | aleveler.com

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IGCSE OCR Chemistry: Calculations Practice | IGCSE OCR 化学：计算题专项训练