Mole Calculations in IGCSE AQA Chemistry: Key Points | IGCSE AQA 化学：摩尔计算考点精讲

📚 Mole Calculations in IGCSE AQA Chemistry: Key Points | IGCSE AQA 化学：摩尔计算考点精讲

Mole calculations sit at the heart of quantitative chemistry. In IGCSE AQA Chemistry, you are expected to move confidently between mass, moles, concentration and gas volumes. This article walks you through every essential formula, worked example and common pitfall, so you can tackle any mole question with clarity and precision.

摩尔计算是定量化学的核心。在 IGCSE AQA 化学中，你需要熟练地在质量、摩尔、浓度和气体体积之间进行转换。本文将逐一梳理所有关键公式、典型例题和常见错误，帮助你清晰、精准地解决任何摩尔计算问题。

1. The Mole Concept | 摩尔的概念

A mole is the amount of substance that contains exactly 6.02 × 10²³ particles. These particles can be atoms, molecules, ions or electrons. The number 6.02 × 10²³ is called the Avogadro constant, symbol Nₐ. One mole of any substance always contains this fixed number of formula units.

摩尔是含有恰好 6.02 × 10²³ 个微粒的物质的量。这些微粒可以是原子、分子、离子或电子。6.02 × 10²³ 这个数被称为阿伏伽德罗常数，符号为 Nₐ。任何物质的 1 摩尔都含有这个固定数量的基本单元。

The mass of one mole of a substance in grams is numerically equal to its relative atomic mass (Aᵣ) or relative formula mass (Mᵣ). This means you can link laboratory masses directly to numbers of particles through the mole. The unit of molar mass is g/mol.

以克为单位的 1 摩尔物质的质量在数值上等于其相对原子质量 (Aᵣ) 或相对式量 (Mᵣ)。这意味着你可以通过摩尔将实验室称量的质量与微粒数目直接联系起来。摩尔质量的单位是 g/mol。

The mole equation that unites mass and molar mass is:

联系质量和摩尔质量的基本公式为：

n = m ÷ M

where n is the number of moles (mol), m is the mass (g) and M is the molar mass (g/mol). Memorise this triangle relationship – it will underpin almost every calculation in this topic.

其中 n 是摩尔数 (mol)，m 是质量 (g)，M 是摩尔质量 (g/mol)。请记住这个三角关系——它几乎支撑了本主题的每一道计算。

2. Molar Mass (Mᵣ) | 摩尔质量

Molar mass is the mass of one mole of a substance. To calculate it, add up the relative atomic masses of all the atoms in the formula. For elements, the molar mass is simply the Aᵣ taken from the Periodic Table. For compounds, it is the sum of Aᵣ values of each element multiplied by its subscript in the formula.

摩尔质量是 1 摩尔物质的质量。计算时，将化学式中所有原子的相对原子质量相加即可。对于单质，摩尔质量就是周期表上的 Aᵣ；对于化合物，则是每种元素的 Aᵣ 乘以其化学式中的下标数后的总和。

Example: Calculate the molar mass of magnesium nitrate, Mg(NO₃)₂. Aᵣ values: Mg = 24, N = 14, O = 16. Mᵣ = 24 + (14 × 2) + (16 × 6) = 24 + 28 + 96 = 148 g/mol. Always remember to multiply the nitrogen and oxygen atoms by the brackets appropriately.

例题：计算硝酸镁 Mg(NO₃)₂ 的摩尔质量。Aᵣ 值：Mg = 24，N = 14，O = 16。Mᵣ = 24 + (14 × 2) + (16 × 6) = 24 + 28 + 96 = 148 g/mol。务必注意括号内的氮原子和氧原子要根据下标正确相乘。

Common mistake: forgetting to multiply atoms inside brackets, or using the wrong atomic mass. Always double‑check the Periodic Table and the formula layout before adding.

常见错误：忘记乘括号内原子的个数，或者用错了原子量。在相加之前，一定要仔细核对周期表和化学式的结构。

3. Calculating Moles from Mass | 由质量计算摩尔数

To find the number of moles from a given mass, divide the mass (in grams) by the molar mass. This is the most frequent step in stoichiometry. Write: n = m / M, substitute values and calculate. Always include units in your working to catch errors.

要从已知质量求出摩尔数，用质量（克）除以摩尔质量。这是化学计量中最常见的一步。写出 n = m / M，代入数值并计算。解题过程中始终带单位，有助于发现错误。

Worked example: How many moles are there in 5.00 g of calcium carbonate, CaCO₃? Aᵣ: Ca = 40, C = 12, O = 16, so Mᵣ = 40 + 12 + (3 × 16) = 100 g/mol. n = 5.00 / 100 = 0.0500 mol. Notice the answer is given to 3 significant figures, matching the mass data.

例题：5.00 g 碳酸钙 (CaCO₃) 中含有多少摩尔？Aᵣ：Ca = 40，C = 12，O = 16，故 Mᵣ = 40 + 12 + (3 × 16) = 100 g/mol。n = 5.00 / 100 = 0.0500 mol。注意答案保留 3 位有效数字，与质量数据一致。

If the mass is given in kilograms, convert to grams first (×1000). If the mass is in milligrams, convert to grams (÷1000). Never plug non‑gram units directly into n = m / M unless you have adjusted the molar mass accordingly.

如果质量以千克给出，先转换为克 (×1000)。如果质量以毫克给出，转换为克 (÷1000)。切勿将非克单位的数值直接代入 n = m / M，除非你相应调整了摩尔质量的单位。

4. Calculating Mass from Moles | 由摩尔数计算质量

To find the mass of a given number of moles, multiply the amount in moles by the molar mass: m = n × M. This is the reverse of the previous calculation. Watch your significant figures and use the molar mass to at least one decimal place more than the given data if needed.

要从已知摩尔数求质量，用摩尔数乘以摩尔质量：m = n × M。这是上一节计算的逆运算。注意有效数字，必要时摩尔质量比已知数据多保留一位小数。

Worked example: What mass of sodium hydroxide (NaOH) is needed to obtain 0.200 mol? Aᵣ: Na = 23, O = 16, H = 1, so Mᵣ = 23 + 16 + 1 = 40 g/mol. m = 0.200 × 40 = 8.00 g. Again, three significant figures because 0.200 has three.

例题：要得到 0.200 mol 氢氧化钠 (NaOH)，需要多少克？Aᵣ：Na = 23，O = 16，H = 1，故 Mᵣ = 23 + 16 + 1 = 40 g/mol。m = 0.200 × 40 = 8.00 g。这里也取 3 位有效数字，因为 0.200 有三位。

In multi‑step synthesis problems, you often convert a known mass of reactant to moles, use the mole ratio from the balanced equation, then convert the target moles back to mass. Practice this flow until it becomes automatic.

在多步合成问题中，你会经常先将已知的反应物质量换算为摩尔数，利用配平方程式中的摩尔比，再将目标产物的摩尔数转换回质量。反复练习这一流程，直到成为条件反射。

5. Moles and Gas Volumes | 摩尔与气体体积

At room temperature and pressure (RTP, about 20°C and 1 atm), one mole of any gas occupies a volume of 24 dm³. This is called the molar volume, Vₘ. The relationship is:

在常温常压 (RTP，约 20°C 和 1 atm) 下，任何气体的 1 摩尔都占据 24 dm³ 的体积，称为摩尔体积 Vₘ。关系式为：

Volume (dm³) = n × 24

Or in cm³, since 1 dm³ = 1000 cm³:

或者使用 cm³，因为 1 dm³ = 1000 cm³：

Volume (cm³) = n × 24000

Always check whether the question expects the volume in dm³ or cm³. IGCSE AQA papers frequently switch between these units.

始终检查题目要求体积的单位是 dm³ 还是 cm³。IGCSE AQA 试卷经常在这两个单位之间切换。

Worked example: Calculate the volume occupied by 0.500 mol of nitrogen gas at RTP in dm³. Volume = 0.500 × 24 = 12.0 dm³. For cm³, multiply by 24000: 0.500 × 24000 = 12000 cm³.

例题：计算 0.500 mol 氮气在 RTP 下的体积（单位 dm³）。体积 = 0.500 × 24 = 12.0 dm³。若求 cm³，则乘以 24000：0.500 × 24000 = 12000 cm³。

This law applies only to gases, not to liquids or solids. Also, the value 24 dm³/mol is only valid at RTP. If the question specifies different conditions, you may be told to use a different molar volume – read carefully.

这一定律仅适用于气体，不适用于液体或固体。此外，24 dm³/mol 的取值仅在 RTP 下有效。如果题目指定了不同条件，你可能会得到不同的摩尔体积数据——请仔细审题。

6. Connecting Mass, Moles and Gas Volume | 质量、摩尔与气体体积的综合计算

Many problems require you to chain conversions. A typical route is: mass of solid → moles of solid → mole ratio → moles of gas → volume of gas. Alternatively, you might start with a gas volume and work backwards to a solid mass. The principle is the same: always go through moles.

许多问题要求你进行连锁换算。典型的路径为：固体质量 → 固体摩尔数 → 摩尔比 → 气体摩尔数 → 气体体积。有时你也可以从气体体积出发，逆向计算固体质量。核心原则不变：始终以摩尔为中转站。

Worked example: When 2.00 g of calcium carbonate is heated, it decomposes: CaCO₃ → CaO + CO₂. What volume of CO₂ is produced at RTP? Mᵣ CaCO₃ = 100 g/mol, so moles of CaCO₃ = 2.00 / 100 = 0.0200 mol. The balanced equation shows a 1:1 mole ratio, thus moles of CO₂ = 0.0200 mol. Volume of CO₂ = 0.0200 × 24 = 0.480 dm³ (or 480 cm³).

例题：加热 2.00 g 碳酸钙，发生分解：CaCO₃ → CaO + CO₂。在 RTP 下生成的 CO₂ 体积是多少？Mᵣ CaCO₃ = 100 g/mol，故 CaCO₃ 的摩尔数 = 2.00 / 100 = 0.0200 mol。配平方程式显示摩尔比为 1:1，所以 CO₂ 的摩尔数也是 0.0200 mol。CO₂ 体积 = 0.0200 × 24 = 0.480 dm³ (或 480 cm³)。

It is vital to have the correctly balanced equation before using the mole ratio. A wrong coefficient will throw off the entire calculation. Practice balancing and mole‑ratio extraction until it becomes second nature.

在使用摩尔比之前，务必确保方程式已正确配平。一个错误的系数会毁掉整道计算。请反复练习配平与摩尔比的提取，直到形成条件反射。

7. Concentration and Moles | 浓度与摩尔

The concentration of a solution tells you how many moles of solute are dissolved in 1 dm³ of solution. The formula linking moles, concentration and volume is:

溶液的浓度表示在 1 dm³ 溶液中溶解了多少摩尔溶质。联系摩尔数、浓度和体积的公式为：

n = c × V

where n is the number of moles (mol), c is the concentration (mol/dm³), and V is the volume in dm³. If the volume is given in cm³, convert to dm³ by dividing by 1000. For example, 25.0 cm³ = 0.0250 dm³.

其中 n 是摩尔数 (mol)，c 是浓度 (mol/dm³)，V 是体积 (dm³)。如果体积以 cm³ 给出，先除以 1000 换算成 dm³。例如 25.0 cm³ = 0.0250 dm³。

The same relationship can be rearranged to find concentration: c = n / V, or to find the volume needed to obtain a certain number of moles: V = n / c. Familiarity with all three forms is essential.

该关系式可变形为求浓度：c = n / V，或求取特定摩尔数的所需体积：V = n / c。熟练掌握这三种形式至关重要。

Worked example: What is the concentration of a solution containing 0.100 mol of sodium chloride in 500 cm³ of water? Convert 500 cm³ to 0.500 dm³. c = n / V = 0.100 / 0.500 = 0.200 mol/dm³.

例题：将 0.100 mol 氯化钠溶于 500 cm³ 水，所得溶液的浓度是多少？将 500 cm³ 转换为 0.500 dm³。c = n / V = 0.100 / 0.500 = 0.200 mol/dm³。

Concentration can also be given in g/dm³. In that case, convert to mol/dm³ by dividing the mass concentration by the molar mass: c (mol/dm³) = mass concentration (g/dm³) / Mᵣ. Being able to switch between the two concentration units is a key skill.

浓度有时也以 g/dm³ 表示。此时，用质量浓度除以摩尔质量即可转换为 mol/dm³：c (mol/dm³) = 质量浓度 (g/dm³) / Mᵣ。能够在这两种浓度单位间自由切换是一项关键技能。

8. Titration Calculations | 滴定计算

Titration calculations combine all the previous skills. You are typically given two volumes and one concentration, and asked to find the unknown concentration. The key is to use the balanced equation to establish the mole ratio between the two reactants.

滴定计算综合了前述所有技能。通常题目会给出两种液体的体积以及一种溶液的浓度，要求计算未知浓度。关键在于利用配平方程式确定两种反应物之间的摩尔比。

Step‑by‑step approach:

1. Calculate the moles of the known substance using n = c × V (volume in dm³).

2. Use the balanced equation to find the mole ratio, and determine moles of the unknown substance.

3. Convert the volume of the unknown solution to dm³.

4. Calculate the unknown concentration using c = n / V.

分步解题法：

1. 利用 n = c × V （体积单位为 dm³）计算已知物质的摩尔数。

2. 根据配平方程式确定摩尔比，求出未知物质的摩尔数。

3. 将未知溶液的体积换算成 dm³。

4. 利用 c = n / V 计算未知浓度。

Worked example: 25.0 cm³ of sodium hydroxide solution is neutralised by 20.0 cm³ of 0.100 mol/dm³ hydrochloric acid. Find the concentration of NaOH. Equation: NaOH + HCl → NaCl + H₂O. Moles of HCl = 0.100 × (20.0/1000) = 0.00200 mol. Mole ratio NaOH:HCl = 1:1, so moles of NaOH = 0.00200 mol. Volume of NaOH = 0.0250 dm³. Concentration of NaOH = 0.00200 / 0.0250 = 0.0800 mol/dm³.

例题：25.0 cm³ 氢氧化钠溶液恰好被 20.0 cm³ 0.100 mol/dm³ 盐酸中和。求 NaOH 溶液的浓度。方程式：NaOH + HCl → NaCl + H₂O。HCl 的摩尔数 = 0.100 × (20.0/1000) = 0.00200 mol。摩尔比 NaOH:HCl = 1:1，所以 NaOH 的摩尔数为 0.00200 mol。NaOH 体积 = 0.0250 dm³。NaOH 浓度 = 0.00200 / 0.0250 = 0.0800 mol/dm³。

A common pitfall is forgetting to convert cm³ to dm³. You must divide by 1000. Also, always check if the acid is diprotic (like H₂SO₄), because then the mole ratio changes: 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O gives a 2:1 ratio.

常见陷阱是忘记将 cm³ 转换为 dm³，必须除以 1000。此外，务必检查酸是否为二元酸（如 H₂SO₄），因为那样摩尔比会改变：2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O 的比例是 2:1。

9. Reacting Mass Calculations | 反应质量计算

Reacting mass problems ask you to calculate the mass of a product formed from a given mass of reactant, or the mass of reactant needed to produce a certain mass of product. The bridge is always the mole ratio from the balanced equation.

反应质量问题要求你计算由一定质量反应物生成产物的质量，或者产生某一定质量产物所需反应物的质量。桥梁始终是配平方程式中的摩尔比。

Steps:

1. Write the balanced equation.

2. Calculate moles of the known substance (n = m / M).

3. Use the mole ratio to find moles of the target substance.

4. Convert target moles to mass (m = n × M).

步骤：

1. 写出配平的化学方程式。

2. 计算已知物质的摩尔数 (n = m / M)。

3. 利用摩尔比求出目标物质的摩尔数。

4. 将目标摩尔数转换为质量 (m = n × M)。

Worked example: What mass of magnesium oxide forms when 3.00 g of magnesium burns completely in oxygen? 2Mg + O₂ → 2MgO. Moles of Mg = 3.00 / 24.3 = 0.123 mol. Mole ratio Mg:MgO = 2:2 = 1:1, so moles of MgO = 0.123 mol. Mᵣ of MgO = 24.3 + 16.0 = 40.3 g/mol. Mass of MgO = 0.123 × 40.3 = 4.96 g (3 s.f.).

例题：3.00 g 镁在氧气中完全燃烧，生成多少克氧化镁？2Mg + O₂ → 2MgO。Mg 的摩尔数 = 3.00 / 24.3 = 0.123 mol。摩尔比 Mg:MgO = 2:2 = 1:1，所以 MgO 的摩尔数 = 0.123 mol。MgO 的 Mᵣ = 24.3 + 16.0 = 40.3 g/mol。MgO 质量 = 0.123 × 40.3 = 4.96 g (3 s.f.)。

When the mole ratio is not 1:1, a simple proportion is used. For instance, if the ratio is 2:3, you multiply the known moles by 3/2 to get the target moles. Always derive the ratio from the balanced equation – never guess.

当摩尔比不是 1:1 时，使用简单比例。例如，若比例为 2:3，将已知摩尔数乘以 3/2 即得目标摩尔数。始终从配平方程式中推导比例，切勿猜测。

10. Limiting Reactants | 限量反应物

In many reactions, one reactant is used up before the others. This is the limiting reactant – it determines the maximum amount of product that can form. The other reactants are said to be in excess.

在许多反应中，某一种反应物会先被耗尽。这种反应物就是限量反应物——它决定了能生成的产物的最大量。其他反应物则被称为过量。

To identify the limiting reactant:

1. Calculate the moles of each reactant separately.

2. Divide each by its stoichiometric coefficient in the balanced equation.

3. The reactant giving the smallest resulting number is the limiting reactant.

确定限量反应物的方法：

1. 分别计算每种反应物的摩尔数。

2. 将每种摩尔数除以其在配平方程式中的化学计量数。

3. 所得数值最小的那个反应物就是限量反应物。

Worked example: 2.0 g of hydrogen gas (H₂) reacts with 16.0 g of oxygen gas (O₂). Equation: 2H₂ + O₂ → 2H₂O. Moles of H₂ = 2.0 / 2.0 = 1.0 mol; moles of O₂ = 16.0 / 32.0 = 0.50 mol. Divide by coefficients: H₂: 1.0 / 2 = 0.50; O₂: 0.50 / 1 = 0.50. Both give the same value, so neither is in excess; they are in exact stoichiometric proportion. If the values differ, the smallest one indicates the limiting reactant.

例题：2.0 g 氢气 (H₂) 与 16.0 g 氧气 (O₂) 反应。方程式：2H₂ + O₂ → 2H₂O。H₂ 的摩尔数 = 2.0 / 2.0 = 1.0 mol；O₂ 的摩尔数 = 16.0 / 32.0 = 0.50 mol。除以计量数：H₂：1.0 / 2 = 0.50；O₂：0.50 / 1 = 0.50。两者数值相等，因此均未过量；它们恰好为化学计量比。若数值不同，最小者指向限量反应物。

All further product calculations must be based on the moles of the limiting reactant. Using the excess reactant will give an impossibly large product mass – a common exam trap.

此后所有产物计算都必须基于限量反应物的摩尔数。若使用过量反应物计算，会得出一个不可能的大产物质量——这是考试中常见的陷阱。

11. Percentage Yield and Atom Economy | 百分产率与原子经济性

Percentage yield compares the actual mass of product obtained to the theoretical maximum mass. Use the formula:

百分产率将实际获得的产品质量与理论最大质量进行比较。使用公式：

Percentage yield = (actual yield / theoretical yield) × 100%

The theoretical yield is calculated from the limiting reactant and balanced equation. Actual yield is always less than theoretical yield due to incomplete reactions, side reactions or product lost during purification.

理论产量通过限量反应物和配平方程式计算得出。由于反应不完全、发生副反应或产物在提纯过程中损失，实际产量总是低于理论产量。

Atom economy measures the efficiency of a reaction in terms of atoms used to make the desired product. The formula is:

原子经济性衡量原子被用于生成目标产物的反应效率。公式为：

Atom economy = (Mᵣ of desired product / sum of Mᵣ of all reactants) × 100%

A higher atom economy means less waste and a more sustainable process. IGCSE AQA questions may ask you to calculate atom economy or to compare two synthetic routes based on it.

原子经济性越高，意味着废弃物越少，过程越可持续。IGCSE AQA 题目可能会要求你计算原子经济性，或根据原子经济性比较两条合成路线。

12. Exam Tips and Common Mistakes | 考试技巧与常见错误

1. Always write the balanced equation first – without it, mole ratios are guesses.

1. 始终先写出配平的方程式——没有它，摩尔比就是瞎猜。

2. Convert all masses to grams and volumes to dm³ before plugging into formulas. Losing a factor of 1000 is the most frequent arithmetic error.

2. 代入公式前，将所有质量换算为克、体积换算为 dm³。丢失因子 1000 是最常见的算术错误。

3. Show your working step by step. Even if the final answer is wrong, correct intermediate steps can earn method marks.

3. 逐步展示计算过程。即使最终答案错误，正确的中间步骤也能获得方法分。

4. Use the periodic table provided in the exam for Aᵣ values; do not rely on memory. Round to a sensible number of significant figures based on the given data.

4. 使用考试提供的周期表查找 Aᵣ 值，不要凭记忆。根据所给数据保留合理的有效数字位数。

5. Check that the molar mass unit (g/mol) cancels correctly in your calculation. If your answer for moles has units of g²/mol, you have inverted the formula.

5. 检查摩尔质量的单位 (g/mol) 在计算中是否正确约分。如果你求出的摩尔数单位是 g²/mol，说明公式用反了。

6. Practice with a variety of contexts – combustion, neutralisation, metal extraction – to become comfortable picking out the relevant numbers from a wordy question.

6. 在各种情境中练习——燃烧、中和、金属提取——直到你能够从冗长的题干中熟练提取相关数字。

7. In titration calculations, if the acid is diprotic, remember the 2:1 ratio. Underline or circle the acid formula to remind yourself.

7. 在滴定计算中，若酸为二元酸，请记住 2:1 的比例。在酸的化学式下划线或画圈以提醒自己。

Published by TutorHao | Chemistry Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

Mole Calculations in IGCSE AQA Chemistry: Key Points | IGCSE AQA 化学：摩尔计算 考点精讲