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

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

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in chemical reactions. It links the microscopic world of atoms and molecules to the macroscopic world of grams and litres. Mastering stoichiometry is essential for tackling mole calculations, predicting yields, and solving volumetric problems in IGCSE Chemistry.

化学计量学是化学中处理化学反应中反应物与生成物之间定量关系的分支。它将原子和分子的微观世界与克和升的宏观世界联系起来。掌握化学计量学对于解决IGCSE化学中的摩尔计算、产率预测和容量问题至关重要。

1. The Mole Concept | 摩尔概念

The mole is the SI unit for ‘amount of substance’. One mole of any substance contains exactly 6.02 × 10²³ elementary entities (Avogadro constant). These entities can be atoms, molecules, ions, or formula units.

摩尔是“物质的量”的国际单位。一摩尔的任何物质都精确地含有6.02 × 10²³ 个基本单元（阿伏伽德罗常数）。这些基本单元可以是原子、分子、离子或化学式单元。

The mass of one mole of a substance in grams is numerically equal to its relative atomic mass (Aᵣ) or relative formula mass (Mᵣ).

一摩尔物质的质量以克为单位时，其数值等于其相对原子质量（Aᵣ）或相对化学式质量（Mᵣ）。

Amount in moles (n) = mass (m) ÷ molar mass (M). Rearranging this formula allows you to find mass or molar mass as needed.

摩尔数（n）= 质量（m）÷ 摩尔质量（M）。重新排列此公式，您可以按需求出质量或摩尔质量。

2. Molar Mass and Calculations | 摩尔质量与计算

Molar mass (M) is the mass of one mole of a substance, expressed in g/mol. For elements, it is the relative atomic mass in grams; for compounds, it is the sum of the relative atomic masses of all atoms in the formula.

摩尔质量（M）是一摩尔物质的质量，以克/摩尔为单位。对于元素，它是用克表示的相对原子质量；对于化合物，它是化学式中所有原子相对原子质量的总和。

Example: Calculate the molar mass of Mg(NO₃)₂. Mg = 24, N = 14 × 2 = 28, O = 16 × 6 = 96; total = 148 g/mol.

示例：计算Mg(NO₃)₂的摩尔质量。Mg = 24，N = 14 × 2 = 28，O = 16 × 6 = 96；总计 = 148 g/mol。

Key conversions: n = m / M, m = n × M, M = m / n. Always ensure units are in grams and g/mol.

关键转换：n = m / M，m = n × M，M = m / n。始终确保单位是克和克/摩尔。

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

The empirical formula is the simplest whole-number ratio of atoms in a compound. It is determined from percentage composition or mass data.

实验式是化合物中原子的最简整数比。它由百分组成或质量数据确定。

Steps: (i) convert mass (or %) to moles by dividing by Aᵣ; (ii) divide by the smallest number of moles; (iii) obtain the simplest ratio; (iv) write the empirical formula.

步骤：(i) 将质量（或百分比）除以Aᵣ转换为摩尔数；(ii) 除以最小的摩尔数；(iii) 得到最简比；(iv) 写出实验式。

The molecular formula is a multiple of the empirical formula. The multiplier n = (molar mass of compound) / (molar mass of empirical formula). For example, if the empirical formula is CH₂ and Mᵣ is 56, then n = 56/14 = 4, so molecular formula is C₄H₈.

分子式是实验式的整数倍。倍数n = （化合物的摩尔质量）/（实验式的摩尔质量）。例如，如果实验式为CH₂且Mᵣ为56，则n = 56/14 = 4，因此分子式为C₄H₈。

4. Reacting Masses | 反应质量计算

Reacting mass calculations use the balanced equation to relate the moles of reactants and products. The mole ratio comes directly from the coefficients in the equation.

反应质量计算利用配平的方程式将反应物和生成物的摩尔数关联起来。摩尔比直接来自方程式中的系数。

General method: (1) write the balanced equation; (2) convert given mass to moles; (3) use the mole ratio to find moles of the unknown substance; (4) convert moles of unknown to mass.

通用方法：(1) 写出配平方程式；(2) 将给定质量转换为摩尔数；(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质量。

Example: 2Mg + O₂ → 2MgO. What mass of MgO is formed from 6 g of Mg? Moles Mg = 6/24 = 0.25 mol; ratio 2:2 gives 0.25 mol MgO; mass = 0.25 × 40 = 10 g.

示例：2Mg + O₂ → 2MgO。6 g的Mg生成多少MgO？Mg的摩尔数 = 6/24 = 0.25 mol；比例2:2得到0.25 mol MgO；质量 = 0.25 × 40 = 10 g。

5. Molar Volume of Gases | 气体摩尔体积

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

在常温常压（RTP，20 °C和1 atm）下，一摩尔任何气体占据24 dm³（或24,000 cm³）。这就是气体摩尔体积Vₘ。

Volume of gas (dm³) = moles × 24. To find the volume of a gas produced in a reaction, first calculate moles of gas, then multiply by 24.

气体体积（dm³）= 摩尔数 × 24。要计算反应中生成的气体体积，首先计算气体的摩尔数，然后乘以24。

If conditions are not RTP, the ideal gas equation pV = nRT may be used, but IGCSE typically uses the 24 dm³ rule at RTP.

如果条件不是RTP，可以使用理想气体方程pV = nRT，但IGCSE通常使用RTP下的24 dm³规则。

6. Concentration of Solutions | 溶液浓度

Concentration is the amount of solute dissolved in a given volume of solution. Common units: mol/dm³ (molar concentration) and g/dm³ (mass concentration).

浓度是溶解在给定体积溶液中的溶质数量。常用单位：mol/dm³（摩尔浓度）和g/dm³（质量浓度）。

Molar concentration (c) = moles of solute (n) / volume of solution (V in dm³).

摩尔浓度（c）= 溶质摩尔数（n）/ 溶液体积（V，dm³）。

Mass concentration = mass of solute (g) / volume (dm³). To convert between mass concentration and molar concentration, use c (mol/dm³) = mass concentration (g/dm³) / M.

质量浓度 = 溶质质量（g）/ 体积（dm³）。在质量浓度和摩尔浓度之间转换时，使用c（mol/dm³）= 质量浓度（g/dm³）/ M。

A solution of 0.5 mol/dm³ contains 0.5 moles of solute in 1 dm³. If you have 0.2 moles in 250 cm³ (0.25 dm³), c = 0.2/0.25 = 0.8 mol/dm³.

0.5 mol/dm³的溶液表示在1 dm³中含有0.5摩尔溶质。如果您在250 cm³（0.25 dm³）中有0.2摩尔，则c = 0.2/0.25 = 0.8 mol/dm³。

7. Titration Calculations | 滴定计算

Titration is an experimental technique used to determine the concentration of an acid or base by neutralisation. The key formula is: c₁V₁/n₁ = c₂V₂/n₂ for acid-base reactions, where n is the number of moles of H⁺ or OH⁻ per formula unit.

滴定是一种实验技术，通过中和反应确定酸或碱的浓度。关键公式是：c₁V₁/n₁ = c₂V₂/n₂，用于酸碱反应，其中n是每个化学式单元提供H⁺或OH⁻的摩尔数。

Alternatively, use moles of standard solution = c × V (in dm³), then apply the mole ratio from the balanced equation to find moles of unknown, then concentration.

或者，使用标准溶液的摩尔数 = c × V（dm³），然后应用配平方程中的摩尔比求出未知物的摩尔数，再求浓度。

Example: 25.0 cm³ of 0.1 mol/dm³ NaOH neutralises 20.0 cm³ of H₂SO₄. Equation: 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O. Moles NaOH = 0.1 × 0.025 = 0.0025 mol; moles H₂SO₄ = 0.0025/2 = 0.00125 mol; concentration H₂SO₄ = 0.00125/0.020 = 0.0625 mol/dm³.

示例：25.0 cm³的0.1 mol/dm³ NaOH中和20.0 cm³的H₂SO₄。方程式：2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O。NaOH摩尔数 = 0.1 × 0.025 = 0.0025 mol；H₂SO₄摩尔数 = 0.0025/2 = 0.00125 mol；H₂SO₄浓度 = 0.00125/0.020 = 0.0625 mol/dm³。

8. Percentage Yield and Purity | 产率与纯度

Percentage yield compares the actual mass of product obtained from an experiment to the theoretical mass predicted by stoichiometry.

产率百分比将实验中获得的实际产品质量与化学计量预测的理论质量进行比较。

% yield = (actual mass / theoretical mass) × 100%. Yields are often less than 100% due to incomplete reactions, side reactions, or losses during purification.

产率% =（实际质量 / 理论质量）× 100%。由于反应不完全、副反应或提纯过程中的损失，产率通常低于100%。

Percentage purity applies to impure samples. % purity = (mass of pure substance / total mass of sample) × 100%. Use stoichiometry to find the pure mass from a titration or reacting mass, then calculate purity.

纯度百分比适用于不纯的样品。纯度% =（纯物质的质量 / 样品总质量）× 100%。利用化学计量从滴定或反应质量中求出纯物质的质量，然后计算纯度。

9. Limiting Reactants | 限制反应物

The limiting reactant is the reactant that is completely consumed in a reaction, determining the maximum amount of product formed. The other reactants are in excess.

限制反应物是在反应中被完全消耗的反应物，它决定了生成物的最大量。其他反应物是过量的。

To identify the limiting reactant, calculate the moles of each reactant, then divide by its coefficient in the balanced equation. The reactant with the smallest value is limiting.

要确定限制反应物，计算每种反应物的摩尔数，然后除以配平方程中的系数。比值最小的反应物是限制性的。

All subsequent calculations (theoretical mass, volume, etc.) must be based on the limiting reactant.

所有后续计算（理论质量、体积等）都必须基于限制反应物。

10. Water of Crystallisation | 结晶水计算

Many salts contain water molecules as part of their crystal structure, e.g., CuSO₄·5H₂O. The mass of water can be determined by heating to constant mass.

许多盐含有水分子作为其晶体结构的一部分，例如CuSO₄·5H₂O。水的质量可以通过加热至恒重来确定。

Steps: (1) find mass of hydrated salt; (2) heat and find mass of anhydrous salt; (3) mass of water = mass lost; (4) convert masses to moles; (5) find the simplest mole ratio to get x in formula·xH₂O.

步骤：(1) 测定水合盐的质量；(2) 加热并测定无水盐的质量；(3) 水的质量 = 损失的质量；(4) 将质量转换为摩尔数；(5) 求出最简摩尔比以得到化学式中的x·xH₂O。

Example: 3.22 g of Na₂SO₄·xH₂O gave 2.14 g anhydrous Na₂SO₄. Mass H₂O = 1.08 g. Moles Na₂SO₄ = 2.14/142 = 0.0151; moles H₂O = 1.08/18 = 0.06; ratio ≈ 4; thus x = 4.

示例：3.22 g的Na₂SO₄·xH₂O得到2.14 g无水Na₂SO₄。水的质量=1.08 g。Na₂SO₄摩尔数=2.14/142=0.0151；H₂O摩尔数=1.08/18=0.06；比例≈4；因此x=4。

11. Key Equations and Relationships | 关键公式与关系

Master these essential formulas for stoichiometry problems:

掌握这些解决化学计量问题的基本公式：

n = m / M (moles from mass)

n = m / M（由质量求摩尔数）
m = n × M (mass from moles)

m = n × M（由摩尔数求质量）
V (gas, dm³) = n × 24 (at RTP)

V（气体，dm³）= n × 24（在RTP下）
c = n / V (mol/dm³) or n = c × V

c = n / V（mol/dm³）或 n = c × V
% yield = (actual/theoretical) × 100

产率% =（实际/理论）× 100
% purity = (mass of pure/total mass) × 100

纯度% =（纯质量/总质量）× 100

Mole = Mass / Molar Mass | Volume = Moles × 24 | Moles = Concentration × Volume

12. Common Mistakes and Tips | 常见错误与技巧

Always use a correctly balanced chemical equation. An incorrect mole ratio will throw off all calculations.

务必使用正确配平的化学方程式。错误的摩尔比会导致所有计算错误。

Pay attention to units: volumes in cm³ must be divided by 1000 to get dm³ before using in c = n/V. Do not confuse dm³ and cm³.

注意单位：在使用c = n/V之前，必须将cm³体积除以1000转换为dm³。不要混淆dm³和cm³。

For gases, use 24 dm³ per mole only if conditions are RTP. If not specified, assume RTP.

对于气体，仅在RTP条件下使用每摩尔24 dm³。如未指定，假设为RTP。

In limiting reactant problems, always check which reactant runs out first. Do not assume the one with smaller mass is limiting; compare moles after dividing by coefficients.

在限制反应物问题中，务必先检查哪种反应物会先耗尽。不要假定质量较小的就是限制性的；除以系数后比较摩尔数。

When dealing with hydrated salts, ensure the mass of water is found accurately, and use Aᵣ values from the periodic table correctly.

处理水合盐时，确保准确求出水的质量，并正确使用元素周期表中的Aᵣ值。

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