IGCSE CCEA Chemistry: Stoichiometry Key Points | IGCSE CCEA 化学：化学计量考点精讲

📚 IGCSE CCEA Chemistry: Stoichiometry Key Points | IGCSE CCEA 化学：化学计量考点精讲

Stoichiometry is the quantitative study of reactants and products in chemical reactions. In the CCEA IGCSE Chemistry course, you must be confident using the mole concept to calculate reacting masses, gas volumes, solution concentrations and percentage yields. This revision guide breaks down every essential skill into clear, step-by-step points so you can tackle any stoichiometry problem in your exam.

化学计量学是研究化学反应中反应物与产物之间定量关系的学科。在 CCEA IGCSE 化学课程中，你必须熟练运用摩尔概念计算反应质量、气体体积、溶液浓度和产率百分比。这份复习指南将每一个核心技能分解为清晰的步骤化要点，帮助你在考试中解决任何化学计量问题。

1. Relative Atomic Mass & Relative Molecular Mass | 相对原子质量与相对分子质量

Relative atomic mass (Aᵣ) is the average mass of an atom of an element compared to 1/12 the mass of a carbon‑12 atom. It has no units.

相对原子质量 (Aᵣ) 是元素的一个原子的平均质量与一个碳‑12 原子质量的 1/12 的比值，没有单位。

Relative molecular mass (Mᵣ) is the sum of the relative atomic masses of all atoms present in a molecular formula. For ionic compounds we often use the term relative formula mass instead.

相对分子质量 (Mᵣ) 是分子式中所有原子的相对原子质量之和。对于离子化合物，我们常用相对式量这个术语。

You must be able to calculate Mᵣ quickly. For example, the Mᵣ of H₂SO₄ is (2×1) + 32 + (4×16) = 98.

你必须能够快速计算 Mᵣ。例如，H₂SO₄ 的 Mᵣ 为 (2×1) + 32 + (4×16) = 98。

2. The Mole Concept & Avogadro’s Constant | 摩尔概念与阿伏伽德罗常数

One mole of any substance contains exactly 6.02 × 10²³ particles (atoms, molecules, ions or electrons). This number is called Avogadro’s constant.

任何物质的一摩尔恰好含有 6.02 × 10²³ 个微粒（原子、分子、离子或电子）。这个数字称为阿伏伽德罗常数。

The mole allows chemists to count particles by weighing. The mass of one mole of a substance is its molar mass, which has units of g mol⁻¹ and is numerically equal to its Aᵣ or Mᵣ.

摩尔让化学家可以通过称重来计量微粒。一摩尔物质的质量就是其摩尔质量，单位为 g mol⁻¹，数值上等于其 Aᵣ 或 Mᵣ。

Use the formula number of moles = number of particles ÷ (6.02 × 10²³) to convert between number of entities and amount in moles.

使用公式 摩尔数 = 微粒数量 ÷ (6.02 × 10²³) 在微粒数目与摩尔量之间进行转换。

3. Molar Mass & Mass-Mole Conversions | 摩尔质量与质量‑摩尔转换

The central relationship in all stoichiometry calculations is:

所有化学计量计算的核心关系是：

amount of substance (mol) = mass (g) ÷ molar mass (g mol⁻¹)

物质的数量 (mol) = 质量 (g) ÷ 摩尔质量 (g mol⁻¹)

Example: How many moles are present in 20 g of NaOH? Mᵣ of NaOH = 23 + 16 + 1 = 40, so n = 20 ÷ 40 = 0.50 mol.

例题：20 g NaOH 中含有多少摩尔？NaOH 的 Mᵣ = 23 + 16 + 1 = 40，因此 n = 20 ÷ 40 = 0.50 mol。

You can also calculate mass from moles: mass = moles × molar mass. Always show your working clearly and include units.

你也可以由摩尔数求质量：质量 = 摩尔数 × 摩尔质量。务必清晰地写出计算过程并标明单位。

4. Calculations Using Chemical Equations | 使用化学方程式的计算

A balanced equation gives the mole ratio of reactants and products. The coefficients tell you how many moles of each substance are involved.

配平的化学方程式给出了反应物与产物的摩尔比。系数告诉你每种物质参与反应的摩尔数。

Standard method: 1) Write the balanced equation. 2) Work out moles of the known substance (using mass or concentration). 3) Use the mole ratio to find moles of the unknown. 4) Convert moles of unknown into the required quantity (mass, volume, concentration).

标准方法：1) 写出配平的方程式。2) 计算出已知物质的摩尔数（利用质量或浓度）。3) 利用摩尔比求出未知物的摩尔数。4) 将未知物的摩尔数转换为所需的物理量（质量、体积或浓度）。

Example: 2Mg + O₂ → 2MgO. What mass of MgO is formed when 48 g of Mg is burnt? Mᵣ Mg = 24, so n(Mg) = 48/24 = 2.0 mol. Mole ratio Mg : MgO = 1 : 1, so n(MgO) = 2.0 mol. Mᵣ MgO = 40, so mass MgO = 2.0 × 40 = 80 g.

例题：2Mg + O₂ → 2MgO。燃烧 48 g Mg 可生成多少质量的 MgO？Mᵣ Mg = 24，n(Mg) = 48/24 = 2.0 mol。摩尔比 Mg : MgO = 1 : 1，所以 n(MgO) = 2.0 mol。Mᵣ MgO = 40，因此 MgO 质量 = 2.0 × 40 = 80 g。

5. Limiting Reactants | 限量反应物

The limiting reactant is the substance that is completely used up in a reaction; it determines the maximum amount of product that can form. The other reactant is in excess.

限量反应物是在反应中完全消耗掉的物质，它决定了能够生成的最大产物量。另一种反应物则为过量。

To identify the limiting reactant, calculate the mole of each reactant and then divide by its coefficient in the balanced equation. The substance giving the smaller value is the limiting reactant.

要确定限量反应物，先计算每种反应物的摩尔数，再除以其在配平方程式中的系数。所得数值较小的物质即为限量反应物。

Example: 2H₂ + O₂ → 2H₂O. If 10 mol H₂ is mixed with 4 mol O₂, H₂ gives 10/2 = 5, O₂ gives 4/1 = 4; therefore O₂ is limiting. Only 8 mol H₂ will react and 2 mol H₂ remains in excess.

例题：2H₂ + O₂ → 2H₂O。若将 10 mol H₂ 与 4 mol O₂ 混合，H₂ 的比值是 10/2 = 5，O₂ 的比值是 4/1 = 4；因此 O₂ 是限量反应物。只有 8 mol H₂ 会参与反应，剩余 2 mol H₂ 过量。

6. Reacting Masses & Yield | 反应质量与产率

The theoretical yield is the maximum mass of product calculated from the limiting reactant using stoichiometry. In practice, the actual yield is often less.

理论产率是根据限量反应物用化学计量计算出的最大产物质量。在实际中，实际产率往往更低。

Percentage yield = (actual yield ÷ theoretical yield) × 100%. This is a key concept in industrial chemistry and appears regularly in CCEA papers.

产率百分比 = (实际产率 ÷ 理论产率) × 100%。这是工业化学的一个重要概念，在 CCEA 考试中经常出现。

Low yields can be caused by incomplete reactions, side reactions, or loss of product during purification. You must be able to suggest and explain such reasons.

产率偏低可能由反应不完全、副反应或纯化过程中产物的损失造成。你必须能够提出并解释这些原因。

7. Empirical & Molecular Formulae | 实验式与分子式

The empirical formula is the simplest whole‑number ratio of atoms 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原子的最简整数比。分子式则表示一个分子中每种原子的真实个数。

To find the empirical formula: convert given masses or percentages to moles, divide by the smallest number of moles, and adjust to whole numbers if needed.

求实验式的方法：将给出的质量或百分比转换为摩尔数，除以最小的摩尔数，并根据需要调整为整数。

Example: A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Assume 100 g: 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 gives 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%（质量分数）。假设样品 100 g：C = 40.0/12 = 3.33 mol；H = 6.7/1 = 6.7 mol；O = 53.3/16 = 3.33 mol。除以 3.33 得到比例 1 : 2 : 1，故实验式为 CH₂O。

The molecular formula is found by dividing the relative molecular mass by the empirical formula mass and multiplying the empirical subscripts by that factor.

分子式由相对分子质量除以实验式质量，再用该倍数乘实验式中的下标得到。

8. Water of Crystallisation | 结晶水

Many salts contain water molecules as part of their crystal structure, e.g. CuSO₄·5H₂O. Heating drives off the water, leaving the anhydrous salt.

许多盐的晶体结构中含有水分子，例如 CuSO₄·5H₂O。加热可除去水分，剩下无水盐。

CCEA questions often give mass data: mass of hydrated salt, mass after heating. Calculate the mass of water lost, then find the mole ratio of water to anhydrous salt to determine x.

CCEA 考题常给出质量数据：水合盐的质量和加热后的质量。先计算失去的水的质量，再求出水与无水盐的摩尔比，从而确定 x。

Example: 4.99 g of hydrated copper sulfate was heated, leaving 3.19 g of anhydrous CuSO₄. Mass of water = 4.99 – 3.19 = 1.80 g. n(CuSO₄) = 3.19/159.5 = 0.0200 mol; n(H₂O) = 1.80/18 = 0.100 mol. Ratio = 0.100/0.0200 = 5, so formula is CuSO₄·5H₂O.

例题：4.99 g 水合硫酸铜加热后剩下 3.19 g 无水 CuSO₄。水的质量 = 4.99 – 3.19 = 1.80 g。n(CuSO₄) = 3.19/159.5 = 0.0200 mol；n(H₂O) = 1.80/18 = 0.100 mol。比值 = 0.100/0.0200 = 5，因此化学式为 CuSO₄·5H₂O。

9. Gas Volumes & Molar Volume | 气体体积与摩尔体积

At room temperature and pressure (RTP, 20 °C and 1 atmosphere), one mole of any gas occupies a volume of 24 dm³ (24 000 cm³). This is the molar gas volume.

在室温和常压（RTP，20 °C，1 个大气压）下，任何气体的一摩尔所占体积为 24 dm³（24 000 cm³）。这就是气体摩尔体积。

Use the formula volume of gas (dm³) = amount (mol) × 24 dm³ mol⁻¹ for calculations. Remember to convert cm³ to dm³ by dividing by 1000 if necessary.

计算时使用公式 气体体积 (dm³) = 物质的数量 (mol) × 24 dm³ mol⁻¹。如需要，记得将 cm³ 换算为 dm³（除以 1000）。

Example: What volume of CO₂ is produced when 0.50 mol of CaCO₃ decomposes? CaCO₃ → CaO + CO₂, mole ratio 1:1, so n(CO₂) = 0.50 mol. Volume = 0.50 × 24 = 12 dm³.

例题：0.50 mol CaCO₃ 分解时生成的 CO₂ 体积是多少？CaCO₃ → CaO + CO₂，摩尔比 1:1，因此 n(CO₂) = 0.50 mol。体积 = 0.50 × 24 = 12 dm³。

10. Concentration of Solutions | 溶液浓度

Concentration is the amount of solute dissolved in a given volume of solution. It is usually expressed in mol dm⁻³ or g dm⁻³.

浓度是指溶解在一定体积溶液中的溶质数量，通常用 mol dm⁻³ 或 g dm⁻³ 表示。

The key formula is concentration (mol dm⁻³) = amount of solute (mol) ÷ volume of solution (dm³). You can rearrange this to find moles or volume.

关键公式为 浓度 (mol dm⁻³) = 溶质的量 (mol) ÷ 溶液的体积 (dm³)。你可以对该式变形来求摩尔数或体积。

Example: 0.200 mol of NaCl is dissolved to make 500 cm³ of solution. Volume in dm³ = 500/1000 = 0.500 dm³. Concentration = 0.200/0.500 = 0.400 mol dm⁻³.

例题：将 0.200 mol NaCl 溶解并配制成 500 cm³ 溶液。体积（dm³）= 500/1000 = 0.500 dm³。浓度 = 0.200/0.500 = 0.400 mol dm⁻³。

11. Titration Calculations | 滴定计算

Titration is used to find the unknown concentration of a solution by reacting it with a standard solution of known concentration. The volumes of both solutions and the balanced equation are required.

滴定法通过将待测液与已知浓度的标准溶液反应，来求得待测液的浓度。需要两种溶液的体积以及配平的化学方程式。

At the equivalence point, the mole ratio in the equation links the amounts of the two reactants. Use n = c × V for each solution and compare them using the mole ratio.

在等当点，方程式中的摩尔比将两种反应物的量联系起来。对每种溶液使用 n = c × V，并通过摩尔比进行比较。

Example: 25.0 cm³ of H₂SO₄ requires 30.0 cm³ of 0.100 mol dm⁻³ NaOH for neutralisation. 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O. n(NaOH) = 0.100 × 0.0300 = 0.00300 mol. Mole ratio NaOH : H₂SO₄ = 2:1, so n(H₂SO₄) = 0.00300/2 = 0.00150 mol. c(H₂SO₄) = 0.00150/0.0250 = 0.0600 mol dm⁻³.

例题：25.0 cm³ H₂SO₄ 需要 30.0 cm³ 0.100 mol dm⁻³ NaOH 进行中和。2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O。n(NaOH) = 0.100 × 0.0300 = 0.00300 mol。摩尔比 NaOH : H₂SO₄ = 2:1，因此 n(H₂SO₄) = 0.00300/2 = 0.00150 mol。c(H₂SO₄) = 0.00150/0.0250 = 0.0600 mol dm⁻³。

Always check that units are consistent (cm³ to dm³). In back‑titration questions, one reactant is added in excess and the excess is then titrated; the same mole‑ratio logic applies.

始终检查单位是否一致（cm³ 转 dm³）。在返滴定问题中，先加入过量的一种反应物，再滴定过量的部分；同样运用摩尔比的逻辑。

12. Percentage Composition & Purity | 百分比组成与纯度

Percentage composition by mass of an element in a compound = (number of atoms of the element × Aᵣ ÷ Mᵣ of compound) × 100%. This is frequently examined in the context of fertilisers and ores.

某元素在化合物中的质量百分比 = (该元素的原子个数 × Aᵣ ÷ 化合物的 Mᵣ) × 100%。这在肥料和矿石的背景下经常被考查。

Example: Calculate the percentage of iron in Fe₂O₃. Aᵣ Fe = 56, Mᵣ Fe₂O₃ = 160. %Fe = (2×56/160)×100 = 70.0%.

例题：计算 Fe₂O₃ 中铁的百分比。Aᵣ Fe = 56，Mᵣ Fe₂O₃ = 160。铁的质量分数 = (2×56/160)×100 = 70.0%。

Purity of a sample is often expressed as a percentage. % purity = (mass of pure substance ÷ total mass of sample) × 100%. This is important when a reactant is not 100% pure and you need to calculate the mass needed or the expected yield.

样品的纯度通常用百分比表示。纯度% = (纯物质的质量 ÷ 样品总质量) × 100%。当反应物不是 100% 纯净且你需要计算所需质量或预期产率时，这非常重要。

Using percentage composition to identify a compound or determine the formula of a mineral is a common CCEA question style.

利用百分比组成来鉴别化合物或确定矿物化学式是 CCEA 常见的题型。

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IGCSE CCEA Chemistry: Stoichiometry Key Points | IGCSE CCEA 化学：化学计量 考点精讲