📚 569 Chemical Calculations: Moles, Mass, and Gas Volumes | 569 化学计算：摩尔、质量与气体体积
Mastering chemical calculations is essential for success in IGCSE Edexcel Science. This article takes you through the core concepts of relative masses, the mole, reacting masses, gas volumes, solution concentrations, and formula determination – all reinforced with step‑by‑step examples and practical tips.
掌握化学计算对于在 IGCSE Edexcel 科学中取得成功至关重要。本文带你梳理相对质量、摩尔、反应质量、气体体积、溶液浓度和化学式确定等核心概念，并通过分步示例与实用技巧加以强化。
1. Relative Atomic and Formula Masses | 相对原子质量与式量
Relative atomic mass (Aᵣ) is the weighted mean mass of an atom compared to 1/12 of the mass of a carbon‑12 atom. It has no units. For a compound, the relative formula mass (Mᵣ) is the sum of the Aᵣ values of all the atoms in the formula.
相对原子质量 (Aᵣ) 是原子的加权平均质量与一个碳‑12 原子质量的 1/12 之比，单位为一。对于化合物，相对式量 (Mᵣ) 是化学式中所有原子 Aᵣ 值的总和。
- Example: H₂O → (2 × 1) + 16 = 18
- Example: CaCO₃ → 40 + 12 + (3 × 16) = 100
| Element | Symbol | Aᵣ |
| Hydrogen | H | 1 |
| Carbon | C | 12 |
| Oxygen | O | 16 |
| Calcium | Ca | 40 |
2. The Mole and Avogadro's Constant | 摩尔与阿伏伽德罗常数
tains 6.02 × 10²³ particles (atoms, molecules, ions or electrons). This number is Avogadro's constant. The mass of one mole of a substance is its Mᵣ expressed in grams.
一摩尔任何物质都含有 6.02 × 10²³ 个粒子（原子、分子、离子或电子）。这个数即是阿伏伽德罗常数。一摩尔物质的质量即为其以克为单位的 Mᵣ 值。
number of moles = mass (g) ÷ Mᵣ or n = m / M
摩尔数 = 质量 (g) ÷ Mᵣ 即 n = m / M
- If you have 18 g of water (Mᵣ = 18), you have 1 mole of water molecules.
- 若拥有 18 g 水 (Mᵣ = 18)，则含有 1 摩尔水分子。
3. Calculating Moles from Mass | 从质量计算摩尔
Always start by identifying the substance' nown. If you are given 8.8 g of CO₂ (Mᵣ = 44), the number of moles is 8.8 ÷ 44 = 0.20 mol.
始终先确定物质的化学式及其 Mᵣ，然后通过公式 n = m / M 求取未知量。若给出 8.8 g CO₂ (Mᵣ = 44)，则摩尔数为 8.8 ÷ 44 = 0.20 mol。
This relationship is the cornerstone of all quantitative chemistry. Whether you are calculating reacting masses, gas volumes, or solution concentrations, the mole acts as the bridge between the macroscopic and particulate worlds.
这一关系是所有定量化学的基石。无论计算反应质量、气体体积还是溶液浓度，摩尔都在宏观与粒子世界之间架起桥梁。
4. Moles and Gas Volumes | 摩尔与气体体积
At room temperature and pressure (r.t.p., 25 °C and 1 atm), one mole of any gas occupies 24 dm³ (24 000 cm³). This is the molar gas volume. The formula linking moles, volume and molar volume is:
在室温和常压 (r.t.p., 25 °C 与 1 atm) 下，任何气体一摩尔的体积为 24 dm³ (24 000 cm³)，即摩尔气体体积。相关公式为：
moles of gas = volume (dm³) ÷ 24 dm³ mol⁻¹
气体摩尔数 = 体积 (dm³) ÷ 24 dm³ mol⁻¹
- If 48 dm³ of oxygen is collected, the amount present is 48 ÷ 24 = 2.0 mol.
- 若收集到 48 dm³ 氧气，则其物质的量为 48 ÷ 24 = 2.0 mol。
0 cm³, so 240 cm³ = 0.240 dm³. For volumes measured in cm³, use 24 000 cm³ mol⁻¹ directly.
注意单位换算：1 dm³ = 1000 cm³，因此 240 cm³ = 0.240 dm³。对于以 cm³ 计量的体积，可直接使用 24 000 cm³ mol⁻¹。
5. Moles and Concentration of Solutions | 摩尔与溶液浓度
The concentration of a solution, usually expressed in mol dm⁻³ (molarity, M), tells us how many moles of solute are dissolved in 1 dm³ of solution. The key equation is:
溶液浓度通常以 mol dm⁻³（摩尔浓度）表示，指明 1 dm³ 溶液中溶解了多少摩尔溶质。关键公式为：
moles = concentration (mol dm⁻³) × volume (dm³)
摩尔数 = 浓度 (mol dm⁻³) × 体积 (dm³)
- A hydrochloric acid solution labelled '0.500 mol dm⁻³' contains 0.500 mol of HCl in every 1 dm³ of solution.
- 标记为“0.500 mol dm⁻³”的盐酸溶液表示每 1 dm³ 溶液中含 0.500 mol HCl。
Often volumes are given in cm³. Remember to convert cm³ to dm³ by dividing by 1000. For a titration, if 25.0 cm³ of acid is used, the volume in dm³ is 0.0250 dm³.
体积常以 cm³ 给出。记得通过除以 1000 转换为 dm³。在滴定中，若用去 25.0 cm³ 酸，其体积单位为 dm³ 即 0.0250 dm³。
6. Empirical and Molecular Formulae | 实验式与分子式
The empirical formula shows the simplest whole‑number ratio of atoms in a compound. The molecular formula gives the actual number of atoms. To find the empirical formula from mass data, convert masses to moles and then find the simplest ratio.
实验式表示化合物中各原子最简整数比。分子式则给出实际原子数目。若由质量数据求实验式，需先将质量转为摩尔，再确定最简比。
= 0.8 ÷ 1 = 0.80 mol. Divide by the smallest (0.20): C = 1, H = 4 → empirical formula CH₄.
示例：某化合物含碳 2.4 g、氢 0.8 g。摩尔：C = 2.4 ÷ 12 = 0.20 mol；H = 0.8 ÷ 1 = 0.80 mol。除以最小值 (0.20)：C = 1, H = 4 → 实验式 CH₄。
u need the Mᵣ. If the Mᵣ is 16, (CH₄)ₙ = 16 → n = 1, so the molecular formula is also CH₄.
确定分子式需要 Mᵣ。若 Mᵣ = 16，(CH₄)ₙ = 16 → n = 1，故分子式亦为 CH₄。
7. Reacting Masses and Limiting Reactants | 反应质量与限量反应物
Stoichiometry uses balanced equations to calculate the masses of reactants and products. The step‑by‑step method: (1) write the balanced equation, (2) calculate moles of the known substance, (3) use the mole ratio to find moles of the unknown, (4) convert to mass.
化学计量法利用配平方程式计算反应物与产物的质量。分步方法如下：(1) 写出配平方程式，(2) 计算已知物质的摩尔数，(3) 用摩尔比求未知物摩尔数，(4) 转化为质量。
mass of magnesium oxide will be produced when 6.0 g of magnesium is burnt in excess oxygen? MgO = 1 : 1, so moles of MgO = 0.25. Mass of MgO = 0.25 × (24 + 16) = 0.25 × 40 = 10.0 g.
示例：6.0 g 镁在过量氧气中燃烧能生成多少氧化镁？2Mg + O₂ → 2MgO。Mg 摩尔 = 6.0 ÷ 24 = 0.25 mol。Mg : MgO 摩尔比 = 1 : 1，故 MgO 摩尔 = 0.25。MgO 质量 = 0.25 × (24 + 16) = 0.25 × 40 = 10.0 g。
The limiting reactant is the one that is completely used up first, limiting the amount of product formed. The other reactant is said to be in excess.
限量反应物是最先耗尽、限制了产物生成量的物质。另一种反应物则称为过量。
8. Percentage Yield and Atom Economy | 百分产率与原子经济性
Percentage yield compares the actual yield of a product to the theoretical yield. In industrial processes, a high yield reduces waste and saves money.
百分产率比较实际产量与理论产量。在工业过程中，高产率可减少浪费并节约成本。
percentage yield = (actual yield ÷ theoretical yield) × 100%
百分产率 = (实际产量 ÷ 理论产量) × 100%
Atom economy measures the efficiency of a reaction in terms of how much of the reactants ends up in the desired product. A higher atom economy means a greener process.
原子经济性衡量有多少反应物转为所需产物。原子经济性越高，过程越绿色。
atom economy = (Mᵣ of desired product ÷ sum of Mᵣ of all reactants) × 100%
原子经济性 = (期望产物 Mᵣ ÷ 所有反应物 Mᵣ 总和) × 100%
- Synthesising a useful medicine with a high atom economy minimises by‑products and waste disposal.
- 以高原子经济性合成有用药物，可最大限度减少副产品与废物处理。
9. Titration Calculations | 滴定计算
Titration is a practical method used to find the concentration of an acid (or alkali) by neutralisation. In Edexcel IGCSE, you are expected to handle straightforward titration results. The key steps: convert volumes to dm³, calculate moles of the known solution, use the neutralisation mole ratio, find concentration of the unknown.
滴定是一种通过中和反应求酸（或碱）浓度的实验方法。在 Edexcel IGCSE 中，需能处理简单的滴定结果。关键步骤：将体积转为 dm³，计算已知溶液的摩尔数，利用中和摩尔比，求出未知液浓度。
oles of NaOH = 0.00200. Concentration of NaOH = 0.00200 ÷ (25.0 ÷ 1000) = 0.0800 mol dm⁻³.
示例：25.0 cm³ NaOH（浓度未知）被 20.0 cm³ 0.100 mol dm⁻³ HCl 中和。方程式：NaOH + HCl → NaCl + H₂O。HCl 摩尔 = 0.100 × (20.0 ÷ 1000) = 0.00200 mol。1:1 比，故 NaOH 摩尔 = 0.00200。NaOH 浓度 = 0.00200 ÷ 0.0250 = 0.0800 mol dm⁻³。
10. Handling Uncertainties and Practical Skills | 处理不确定度与实验技能
te reading the uncertainty is typically ±0.05 cm³. Repeated titrations improve reliability – lts (within 0.10 cm³). Always record readings to the appropriate number of decimal places.
任何测量均带有不确定度。滴定管读数的不确定度通常为 ±0.05 cm³。重复滴定可提高可靠性——必须获得吻合结果（差值在 0.10 cm³ 内）。始终以适当的小数位数记录读数。
Other key competency: using a pipette filler safely, swirling the conical flask, and recognising the end‑point sharply with an indicator such as phenolphthalein (pink to colourless) or methyl orange. Good technique is vital for accurate results in your practical assessment.
其他关键技能：安全使用移液管吸耳球、旋摇锥形瓶、通过指示剂（如酚酞由粉红变无色，或甲基橙）敏锐判断终点。良好的操作对实验评估中获得准确结果至关重要。
11. Common Pitfalls and How to Avoid Them | 常见错误及避免方法
Many students lose marks by forgetting unit conversions, especially between cm³ and dm³. Always write units in your working. Another common error is misreading the mole ratio from the balanced equation – it is not always 1:1!
许多学生因忘记单位换算、尤其是 cm³ 与 dm³ 之间的转换而失分。演算中务必写出单位。另一常见错误是误解配平方程中的摩尔比——并不总是 1:1！
When calculating empirical formulae, always divide by the smallest number of moles. If you get a ratio like 1 : 1.33, multiply everything by 3 to get whole numbers. Rounding too early can lead to wrong formulae.
计算实验式时，总是除以最小的摩尔数。若得到如 1 : 1.33 的比例，则所有数字乘以 3 以得到整数。过早四舍五入可能导致错误的化学式。
12. Revision Checklist for IGCSE Chemical Calculations | IGCSE 化学计算复习清单
- Can you calculate Aᵣ and Mᵣ from a given formula? 能否根据化学式计算 Aᵣ 和 Mᵣ？
- Do you know the value of Avogadro's constant and the molar gas volume at r.t.p.? 是否知道阿伏伽德罗常数与 r.t.p. 下的摩尔气体体积？
- Are you confident using n = m/M, n = V/(24 dm³), and n = c × V? 能否熟练运用 n = m/M、n = V/(24 dm³) 及 n = c × V？
- Can you determine empirical and molecular formulae from experimental data? 能否从实验数据确定实验式和分子式？
- Can you calculate reacting masses and identify the limiting reactant? 能否计算反应质量并判断限量反应物？
- Do you understand percentage yield and atom economy? 是否理解百分产率与原子经济性？
- Can you interpret titration data to find an unknown concentration? 能否解析滴定数据求未知浓度？
Review these checklist items alongside past paper questions to build speed and accuracy. Mastering these calculations will not only boost your exam performance but also deepen your appreciation of how chemists quantify the world around us.
对照这份清单，结合历年真题进行复习，以提高速度与准确度。掌握这些计算不仅会提升你的考试成绩，也能加深你对化学家如何量化我们周围世界的理解。
Published by TutorHao | IGCSE Edexcel Science Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图