Mole Calculations | 摩尔计算考点精讲

📚 Mole Calculations | 摩尔计算考点精讲

The mole is the chemist’s counting unit. In WJEC IGCSE Chemistry, mastering mole calculations is essential for quantitative analysis, from reacting masses to solution concentrations and gas volumes. This revision guide breaks down every core concept, providing clear step-by-step methods and worked examples to help you tackle any mole-related problem with confidence.

摩尔是化学家的计数单位。在 WJEC IGCSE 化学中，掌握摩尔计算对于定量分析至关重要，涉及反应质量、溶液浓度和气体体积等。本复习指南分解每个核心概念，提供清晰的逐步方法和典型例题，帮助你自信地解决任何摩尔相关问题。

1. The Mole Concept | 摩尔概念

The mole (symbol: mol) is the SI unit for amount of substance. One mole of any substance contains exactly 6.02 × 10²³ particles. This number, 6.02 × 10²³ mol⁻¹, is called the Avogadro constant (Nₐ). These particles can be atoms, molecules, ions, or formula units depending on the substance.

摩尔（符号：mol）是物质的量的国际单位。一摩尔任何物质恰好包含 6.02 × 10²³ 个粒子。这个数字 6.02 × 10²³ mol⁻¹ 被称为阿伏伽德罗常数 (Nₐ)。这些粒子可以是原子、分子、离子或式单位，取决于具体物质。

For example, 1 mol of carbon atoms contains 6.02 × 10²³ C atoms; 1 mol of water (H₂O) contains 6.02 × 10²³ H₂O molecules; 1 mol of sodium chloride (NaCl) contains 6.02 × 10²³ NaCl formula units.

例如，1 mol 碳原子含有 6.02 × 10²³ 个碳原子；1 mol 水 (H₂O) 含有 6.02 × 10²³ 个水分子；1 mol 氯化钠 (NaCl) 含有 6.02 × 10²³ 个 NaCl 式单位。

2. Molar Mass and Mᵣ | 摩尔质量与式量

Molar mass (M) is the mass of one mole of a substance, measured in grams per mole (g mol⁻¹). Numerically, it equals the relative atomic mass (Aᵣ) or relative formula mass (Mᵣ). For atoms, use Aᵣ from the Periodic Table; for compounds, add up the Aᵣ values of all atoms in the formula.

摩尔质量 (M) 是一摩尔物质的质量，单位为克每摩尔 (g mol⁻¹)。数值上，它等于相对原子质量 (Aᵣ) 或相对式量 (Mᵣ)。对于原子，使用周期表中的 Aᵣ；对于化合物，将化学式中所有原子的 Aᵣ 值相加。

The core formula linking mass, molar mass and moles is:

连接质量、摩尔质量和摩尔的核心公式为：

moles (n) = mass (m) ÷ molar mass (M)

n = m / M

Worked example: Calculate the mass of 0.500 mol of sodium hydroxide (NaOH). Mᵣ(NaOH) = 23.0 + 16.0 + 1.0 = 40.0 g mol⁻¹. m = n × M = 0.500 × 40.0 = 20.0 g.

典型例题：计算 0.500 mol 氢氧化钠 (NaOH) 的质量。Mᵣ(NaOH) = 23.0 + 16.0 + 1.0 = 40.0 g mol⁻¹。m = n × M = 0.500 × 40.0 = 20.0 g。

3. Moles and Number of Particles | 摩尔与粒子数

To convert between moles and the number of atoms, molecules or ions, use the relationship:

在摩尔与原子、分子或离子数之间转换，使用以下关系：

number of particles = moles × 6.02 × 10²³

This applies equally in reverse: moles = number of particles ÷ 6.02 × 10²³. Pay attention to what the particle is (e.g. O atoms vs O₂ molecules).

反向同样适用：摩尔 = 粒子数 ÷ 6.02 × 10²³。注意粒子的种类（例如是 O 原子还是 O₂ 分子）。

Example: How many hydrogen atoms are present in 0.250 mol of methane (CH₄)? Each CH₄ molecule contains 4 H atoms. Number of CH₄ molecules = 0.250 × 6.02 × 10²³ = 1.505 × 10²³. Number of H atoms = 4 × 1.505 × 10²³ = 6.02 × 10²³.

例题：0.250 mol 甲烷 (CH₄) 中含有多少个氢原子？每个 CH₄ 分子含有 4 个 H 原子。CH₄ 分子数 = 0.250 × 6.02 × 10²³ = 1.505 × 10²³。H 原子数 = 4 × 1.505 × 10²³ = 6.02 × 10²³。

4. Moles and Gas Volumes | 摩尔与气体体积

At room temperature and pressure (RTP), which is about 20 °C and 1 atmosphere, one mole of any gas occupies a volume of 24 dm³. The equivalent value in cm³ is 24 000 cm³ mol⁻¹. The molar volume at RTP is 24 dm³ mol⁻¹.

在常温常压 (RTP) 下，大约 20 °C 和 1 个大气压，一摩尔任何气体的体积为 24 dm³。用 cm³ 表示的等效值为 24 000 cm³ mol⁻¹。RTP 下的气体摩尔体积为 24 dm³ mol⁻¹。

volume (dm³) = moles × 24 dm³ mol⁻¹

If a volume is given in cm³, first convert to dm³ by dividing by 1000. WJEC questions usually specify which temperature and pressure conditions are used – stick to RTP unless stated otherwise.

如果体积以 cm³ 给出，先除以 1000 转换为 dm³。WJEC 的题目通常会指明使用的温度和压强条件——若无特殊说明，一律使用 RTP。

Example: What is the volume of 0.800 mol of nitrogen gas (N₂) at RTP? Volume = 0.800 × 24 = 19.2 dm³. In cm³, that is 19.2 × 1000 = 19 200 cm³.

例题：0.800 mol 氮气 (N₂) 在 RTP 下的体积是多少？体积 = 0.800 × 24 = 19.2 dm³。换算为 cm³，即 19.2 × 1000 = 19 200 cm³。

5. Molar Concentration of Solutions | 溶液的摩尔浓度

Concentration (c) tells us how much solute is dissolved in a given volume of solution. The most common unit is mol dm⁻³, often shortened to M. The fundamental formula is:

浓度 (c) 告诉我们单位体积溶液中溶解了多少溶质。最常见的单位是 mol dm⁻³，常简写为 M。基本公式为：

concentration (mol dm⁻³) = moles ÷ volume (dm³)

c = n / V

Remember to convert volumes: 1 dm³ = 1000 cm³. To find the mass of solute in a given volume, first calculate moles using n = c × V, then convert to mass using m = n × M.

注意体积转换：1 dm³ = 1000 cm³。计算给定溶液中溶质的质量时，先用 n = c × V 求摩尔，再用 m = n × M 换算为质量。

Worked example: 4.00 g of sodium hydroxide (NaOH) is dissolved in water to make 250 cm³ of solution. Calculate the concentration in mol dm⁻³. Mᵣ(NaOH) = 40.0. Moles NaOH = 4.00 ÷ 40.0 = 0.100 mol. Volume = 250 ÷ 1000 = 0.250 dm³. Concentration = 0.100 ÷ 0.250 = 0.400 mol dm⁻³.

典型例题：将 4.00 g 氢氧化钠 (NaOH) 溶于水，配成 250 cm³ 溶液。计算浓度（mol dm⁻³）。Mᵣ(NaOH) = 40.0。NaOH 的摩尔数 = 4.00 ÷ 40.0 = 0.100 mol。体积 = 250 ÷ 1000 = 0.250 dm³。浓度 = 0.100 ÷ 0.250 = 0.400 mol dm⁻³。

6. Chemical Equations and Mole Ratios | 化学方程式与摩尔比

A balanced chemical equation tells you the ratio in which reactants react and products form – this ratio is always in moles. The large numbers in front of each formula (coefficients) represent the mole amounts.

一个配平的化学方程式说明了反应物消耗和产物生成的比例——该比例总是以摩尔为单位。每个化学式前的数字（系数）代表摩尔数。

For instance, the equation 2H₂ + O₂ → 2H₂O shows that 2 moles of hydrogen react with 1 mole of oxygen to produce 2 moles of water. The mole ratio is 2 : 1 : 2.

例如，方程式 2H₂ + O₂ → 2H₂O 表示 2 摩尔氢气与 1 摩尔氧气反应生成 2 摩尔水。摩尔比为 2 : 1 : 2。

When doing calculations, identify the known and unknown substances, write down their mole ratio from the balanced equation, and use the ratio to scale your known number of moles to the unknown.

计算时，先确定已知物和未知物，从配平方程式中找出它们的摩尔比，再利用比例将已知摩尔数换算为未知物的摩尔数。

7. Reacting Mass Calculations | 反应质量计算

The classic three-step method for reacting mass problems: (1) convert the given mass to moles using n = m / M; (2) use the mole ratio from the balanced equation to find moles of the target substance; (3) convert moles back to mass using m = n × M.

反应质量计算的经典三步法：(1) 利用 n = m / M 将已知质量转换为摩尔数；(2) 使用配平方程式中的摩尔比，求目标物质的摩尔数；(3) 用 m = n × M 将摩尔数转换回质量。

Example: What mass of magnesium oxide (MgO) is formed when 6.00 g of magnesium burns completely in air? Equation: 2Mg + O₂ → 2MgO. Mᵣ(Mg)=24.3. Moles Mg = 6.00 ÷ 24.3 = 0.247 mol. Mole ratio Mg : MgO = 2 : 2, so moles MgO = 0.247 mol. Mᵣ(MgO) = 24.3 + 16.0 = 40.3. Mass MgO = 0.247 × 40.3 = 9.95 g (≈10.0 g).

例题：6.00 g 镁在空气中完全燃烧，生成多少克氧化镁 (MgO)？方程式：2Mg + O₂ → 2MgO。Mᵣ(Mg)=24.3。Mg 的摩尔数 = 6.00 ÷ 24.3 = 0.247 mol。摩尔比 Mg : MgO = 2 : 2，因此 MgO 摩尔数 = 0.247 mol。Mᵣ(MgO) = 24.3 + 16.0 = 40.3。MgO 质量 = 0.247 × 40.3 = 9.95 g（约 10.0 g）。

Always check your mole ratio carefully – a common mistake is using a molar mass where a mole ratio is needed, or vice versa.

务必仔细核对摩尔比——常见错误是将需用摩尔比的地方误用摩尔质量，反之亦然。

8. Titration Calculations | 滴定计算

Titration experiments are used to find an unknown concentration by reacting it with a standard solution of known concentration. For a 1:1 reaction (like HCl + NaOH → NaCl + H₂O), the simple formula applies:

滴定实验通过使未知浓度溶液与已知浓度的标准溶液反应，从而求出未知浓度。对于 1:1 的反应（如 HCl + NaOH → NaCl + H₂O），可直接使用以下公式：

cₐ × Vₐ = c_b × V_b

where cₐ and Vₐ are the concentration and volume of the acid, and c_b and V_b for the base. Both volumes must be in the same unit (usually dm³, but cm³ works when both sides are in cm³ and the ratio is 1:1).

其中 cₐ、Vₐ 为酸的浓度和体积，c_b、V_b 为碱的浓度和体积。两个体积单位必须一致（通常用 dm³，但当摩尔比为 1:1 且两侧均用 cm³ 时也成立）。

If the mole ratio is not 1:1, you must account for the ratio. For example, in the neutralisation of sulfuric acid: H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Here the acid:base ratio is 1:2. The correct relationship is: cₐ × Vₐ = (c_b × V_b) / 2, or more generally:

如果摩尔比不是 1:1，则必须计入比例。例如硫酸的中和反应：H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O。此反应中酸与碱的摩尔比为 1:2。正确的关系为：cₐ × Vₐ = (c_b × V_b) / 2，写成更通用的形式：

(c₁ × V₁) / n₁ = (c₂ × V₂) / n₂

where n₁ and n₂ are the mole ratios from the equation. This method is safer for all titration calculations.

其中 n₁ 和 n₂ 为方程式中的摩尔比。这种方法对所有滴定计算都更为安全。

Worked example: 25.0 cm³ of NaOH solution is titrated against 0.100 mol dm⁻³ HCl. The average titre of HCl is 22.40 cm³. Calculate the concentration of NaOH. Reaction is 1:1, so c(NaOH) × 25.0 = 0.100 × 22.40, c(NaOH) = (0.100 × 22.40) ÷ 25.0 = 0.0896 mol dm⁻³.

典型例题：25.0 cm³ NaOH 溶液用 0.100 mol dm⁻³ HCl 滴定，平均读数为 22.40 cm³。计算 NaOH 的浓度。反应为 1:1，因此 c(NaOH) × 25.0 = 0.100 × 22.40，c(NaOH) = (0.100 × 22.40) ÷ 25.0 = 0.0896 mol dm⁻³。

9. Percentage Yield and Atom Economy | 产率与原子经济

Percentage yield compares the actual mass of product obtained in an experiment with the theoretical mass predicted by mole calculations.

产率将实验中实际获得的产品质量与通过摩尔计算预测的理论质量进行比较。

percentage yield = (actual yield ÷ theoretical yield) × 100%

The theoretical yield is always calculated from the limiting reactant using stoichiometry. Percentage yield can be less than 100% due to incomplete reactions, side reactions, or losses during purification.

理论产量总是根据限量反应物通过化学计量计算得出。由于反应不完全、副反应或提纯过程中的损失，产率可能低于 100%。

Atom economy measures how efficiently atoms from the reactants are incorporated into the desired product. It is calculated using formula masses:

原子经济衡量反应物中的原子有多少有效进入了目标产物。它利用式量进行计算：

atom economy = (total Mᵣ of desired product ÷ total Mᵣ of all reactants) × 100%

Reactions with high atom economy produce less waste and are more sustainable. Both percentage yield and atom economy are key indicators in green chemistry and are examined regularly in WJEC papers.

原子经济性高的反应产生的废弃物较少，更加可持续。产率和原子经济都是绿色化学的关键指标，在 WJEC 试卷中经常考查。

10. Empirical and Molecular Formulae | 实验式与分子式

The empirical formula is the simplest whole-number ratio of atoms of each element in a compound. The molecular formula shows the actual number of atoms of each element in a molecule.

实验式是化合物中各元素原子的最简整数比。分子式则显示一个分子中各元素的实际原子个数。

To find the empirical formula: (1) convert the mass (or percentage) of each element to moles; (2) divide each by the smallest number of moles to get the simplest ratio; (3) if necessary, multiply to obtain whole numbers. The molecular formula is then (empirical formula)ₙ, where n = relative molecular mass ÷ relative mass of empirical formula.

求实验式的步骤：(1) 将各元素的质量（或百分含量）转换为摩尔数；(2) 分别除以最小的摩尔数，得到最简比；(3) 如有必要，乘以整数使其成为整数比。分子式即为 (实验式)ₙ，其中 n = 相对分子质量 ÷ 实验式的式量。

Example: A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Its relative molecular mass is 180. Find the molecular formula. Assume 100 g sample: C: 40.0 ÷ 12 = 3.33 mol; H: 6.7 ÷ 1 = 6.7 mol; O: 53.3 ÷ 16 = 3.33 mol. Divide by 3.33: ratio 1 : 2 : 1, so empirical formula is CH₂O. Empirical formula mass = 12 + 2 + 16 = 30. n = 180 ÷ 30 = 6. Molecular formula = C₆H₁₂O₆.

例题：某化合物含碳 40.0%、氢 6.7%、氧 53.3%。其相对分子质量为 180。求分子式。假设 100 g 样品：C: 40.0 ÷ 12 = 3.33 mol；H: 6.7 ÷ 1 = 6.7 mol；O: 53.3 ÷ 16 = 3.33 mol。各除以

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Mole Calculations | 摩尔计算考点精讲