📚 Stoichiometry and the Mole Concept | 化学计量与摩尔概念
Stoichiometry lies at the heart of quantitative chemistry. In Edexcel IGCSE Science (Chemistry), mastering the mole concept and related calculations is essential for tackling everything from reacting masses to titration results. This article will guide you step by step through the key ideas, equations and problem‑solving techniques you need for the exam.
化学计量是定量化学的核心。在Edexcel IGCSE 科学（化学部分）中，掌握摩尔概念及相关计算对于解决从反应质量到滴定结果等各类问题至关重要。本文将逐步带你梳理关键概念、方程式和解题技巧，助你备考无忧。
1. Relative Atomic Mass and Relative Molecular Mass | 相对原子质量与相对分子质量
element is the weighted mean mass of an atom compared with 1/12th of the mass of a carbon‑ It has no units. For chlorine, Aᵣ = 35.5 because of the mixture of ³⁵Cl and ³⁷Cl isotopes.
相对原子质量（Aᵣ）是指一个元素原子的加权平均质量与一个¹²C原子质量的1/12的比值，它没有单位。氯的 Aᵣ = 35.5，正是因为³⁵Cl 和³⁷Cl 同位素共存。
+ 4×16 = 98. For ionic compounds we use relative formula mass, calculated in the same way.
相对分子质量（Mᵣ）是分子中所有原子的 Aᵣ 之和。例如硫酸 H₂SO₄ 的 Mᵣ = 2×1 + 32 + 4×16 = 98。对于离子化合物我们使用相对式量，计算方法相同。
2. The Mole and Avogadro's Constant | 摩尔与阿伏伽德罗常数
elementary entities (atoms, molecules, ions or electrons). This number is called Avogadro' the visible mass of a substance to the unimaginably large number of particles it contains.
一摩尔任何物质恰好含有 6.02 × 10²³ 个基本单元（原子、分子、离子或电子），这个数字称为阿伏伽德罗常数（Nₐ）。摩尔使我们能够将宏观可称量的物质质量与其中包含的天文数字般的粒子数联系起来。
le. Thus 12 g of carbon = 1 mol of carbon atoms; 18 g of water = 1 mol of water molecules.
实践中，以克为单位称取与 Aᵣ 或 Mᵣ 数值相等的质量，即得到一摩尔。因此，12 g 碳 = 1 mol 碳原子；18 g 水 = 1 mol 水分子。
3. Molar Mass, n = m / M | 摩尔质量与计算式
Molar mass (M) is the mass of one mole of a substance, measured in g mol⁻¹. The key equation linking mass, molar mass and amount (n) is:
摩尔质量（M）是一摩尔物质的质量，单位为 g mol⁻¹。连接质量、摩尔质量和摩尔数（n）的核心公式为：
n = m / M
molar mass in g mol⁻¹. Always make sure you use the correct Mᵣ for the substance and convert any masses given in kg to g.
其中 n 为摩尔数（mol），m 为质量（g），M 为摩尔质量（g mol⁻¹）。务必使用正确的 Mᵣ，并将任何以 kg 给出的质量换算为 g。
4. Empirical and Molecular Formulae | 经验式与分子式
The empirical formula gives 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. For glucose, the empirical formula is CH₂O, while the molecular formula is C₆H₁₂O₆.
经验式表示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实际数目。葡萄糖的经验式为 CH₂O，分子式为 C₆H₁₂O₆。
To find the empirical formula from mass or percentage data, convert masses to moles, divide by the smallest number of moles, and write the ratio as whole numbers. If you then know the Mᵣ of the compound, you can multiply the empirical formula by a whole number to obtain the molecular formula.
由质量或百分比数据求经验式时，先将质量转换为摩尔数，除以最小的摩尔数，将比例写为整数。如果进一步知道化合物的 Mᵣ，可将经验式乘以一个整数得到分子式。
5. Reacting Mass Calculations | 反应质量计算
equation to find moles of the desired substance, then convert back to mass with m = n × M.
反应质量计算遵循清晰的步骤：写出配平的化学方程式，利用 n = m / M 计算已知物质的摩尔数，根据方程式中的摩尔比求出目标物质的摩尔数，再通过 m = n × M 转换为质量。
Example: What mass of CO₂ is produced when 10.0 g of CaCO₃ decomposes? / 100 = 0.100 mol. 1:1 mole ratio gives 0.100 mol CO₂. Mass of CO₂ = 0.100 × 44 = 4.40 g.
示例：10.0 g CaCO₃ 分解会产生多少克 CO₂？CaCO₃ → CaO + CO₂。Mᵣ(CaCO₃)=100，Mᵣ(CO₂)=44。CaCO₃ 的摩尔数 = 10.0/100=0.100 mol。1:1 摩尔比得出 0.100 mol CO₂。CO₂ 的质量 = 0.100×44=4.40 g。
6. Concentration and Solution Calculations | 浓度与溶液计算
Concentration is usually expressed in mol dm⁻³. The relationship is:
浓度通常以 mol dm⁻³ 表示。其关系式为：
n = c × V
given in cm³, divide by 1000 to convert to dm³. This equation is essential for titrations.
其中 c 为浓度（mol dm⁻³），V 为体积（dm³）。若体积以 cm³ 给出，需除以 1000 转化为 dm³。该公式是滴定计算的基础。
00500 mol. Rearranging the formula also lets you find concentration or volume when the other quantities are known.
示例：25.0 cm³、0.200 mol dm⁻³ 的 HCl 含 n = 0.200 × (25.0/1000) = 0.00500 mol。变形公式后可求出浓度或体积。
7. Gas Volume Calculations | 气体体积计算
At room temperature and pressure (r.t.p.), one mole of any gas occupies 24 dm³. The molar gas volume is temperature‑ and pressure‑dependent, but the Edexcel IGCSE typically uses 24 dm³ mol⁻¹ at r.t.p. (20 °C, 1 atm).
在常温常压（r.t.p.）下，一摩尔任何气体的体积均为 24 dm³。气体摩尔体积依赖于温度和压强，但 Edexcel IGCSE 通常以 r.t.p.（20 °C, 1 atm）下的 24 dm³ mol⁻¹ 为准。
The relevant equation is:
相关公式为：
n = V(gas) / 24
0 mol of magnesium with excess acid, the hydrogen evolved will occupy 0.500 × 24 = 12 dm³.
其中 V 的单位为 dm³。如果 0.500 mol 镁与过量酸反应，生成的氢气体积为 0.500 × 24 = 12 dm³。
8. Percentage Yield | 产率
Percentage yield compares the actual mass of product obtained in an experiment to the theoretical mass calculated from reacting masses. It is never greater than 100 %; a value above 100 % indicates an error such as an impure product or incomplete drying.
产率将实验中实际获得的产物质量与根据反应质量计算的理论质量进行比较。产率不会超过 100 %；若数值高于 100 % 则表明存在误差，如产物不纯或未充分干燥。
% Yield = (actual yield / theoretical yield) × 100
Always identify the limiting reactant first, then calculate the theoretical yield from that reactant. Knowing why yields are often below 100 % – incomplete reactions, side reactions, losses during purification – is a common exam question.
务必先确定限量反应物，再根据该反应物计算理论产量。产率常低于 100 % 的原因（反应不完全、副反应、纯化过程中的损失）也是常见的考题。
9. Limiting Reactants | 限量反应物
The limiting reactant is the substance that is completely used up in a reaction and determines the amount of product formed. The other reactants are present in excess. To identify the limiting reactant, calculate the number of moles of each reactant and compare the mole ratio with the balanced equation.
限量反应物是在反应中被彻底消耗的物质，它决定了产物的生成量。其他反应物为过量。要找出限量反应物，需计算各反应物的摩尔数，并与配平方程式中的摩尔比进行比较。
he ratio 3.0:1.0 is greater than the 2:1 required, so O₂ is limiting. Only 2.0 mol of H₂ will be used, leaving 1.0 mol unreacted.
示例：3.0 mol H₂ 与 1.0 mol O₂ 反应生成水：2H₂ + O₂ → 2H₂O。实际比为 3.0:1.0，大于需要的 2:1，因此 O₂ 是限量反应物。仅有 2.0 mol H₂ 会被消耗，剩余 1.0 mol。
10. Titration Calculations | 滴定计算
oles of the known solution, use the mole ratio to find moles of the unknown, then find its concentration or volume.
滴定计算将溶液计量与摩尔比相结合。一般步骤为：写出配平方程式，用 n = c × V 求已知溶液的摩尔数，利用摩尔比求未知物的摩尔数，再计算其浓度或体积。
, so moles NaOH = 0.00200 mol. Concentration of NaOH = 0.00200 / 0.0250 = 0.0800 mol dm⁻³.
例如，25.0 cm³ NaOH 被 20.0 cm³、0.100 mol dm⁻³ HCl 中和。NaOH + HCl → NaCl + H₂O。HCl 的摩尔数 = 0.100×0.0200=0.00200 mol。1:1 比，故 NaOH 摩尔数 = 0.00200 mol。NaOH 浓度 = 0.00200/0.0250 = 0.0800 mol dm⁻³。
11. Water of Crystallisation | 结晶水
Many salts contain water of crystallisation, shown as a dot in the formula, e.g. CuSO₄•5H₂O. To determine the number of water molecules, you heat the hydrated salt to remove the water, record the mass loss, and calculate the ratio of moles of water to moles of anhydrous salt.
许多盐含有结晶水，在化学式中用点表示，如 CuSO₄•5H₂O。要确定结晶水的数目，可加热水合盐除去水分，记录质量损失，再计算水与无水盐的摩尔比。
Example: 4.99 g of CuSO₄• 5 = 0.0200 mol. Moles H₂O = 1.80 / 18 = 0.100 mol. Ratio 0.100 : 0.0200 = 5 : 1, so x = 5.
示例：4.99 g CuSO₄•xH₂O 加热后得到 3.19 g 无水 CuSO₄。失水质量 = 1.80 g。CuSO₄ 摩尔数 = 3.19/159.5 = 0.0200 mol。H₂O 摩尔数 = 1.80/18 = 0.100 mol。比例 0.100:0.0200 = 5:1，因此 x = 5。
12. Key Exam Tips | 考试要点
Below are some common mistakes and tips to help you secure full marks in stoichiometry questions.
以下是一些常见错误和应试技巧，帮助你在化学计量题目中拿满分。
-
Always show full working: write the formula, substitute numbers with units, and state the final answer to the appropriate number of significant figures.
-
务必写出完整计算过程：写下公式，代入带单位的数值，最终答案保留恰当的显著数字位数。
-
Double‑check unit conversions – cm³ to dm³ (÷1000), kg to g (×1000). Forgetting these is one of the most frequent errors.
-
仔细检查单位换算 – cm³ 转 dm³（÷1000），kg 转 g（×1000）。忘记换算是频率最高的失误之一。
-
Make sure the chemical equation is correctly balanced before using mole ratios. An unbalanced equation will lead to a completely wrong answer.
-
在利用摩尔比之前，确保化学方程式已准确配平。未配平的方程式会导致答案全盘错误。
-
When calculating empirical formula, always divide by the smallest number of moles. If you obtain a ratio like 1 : 1.5, multiply through by 2 to get whole numbers.
-
计算经验式时，始终除以最小的摩尔数。若得到类似 1:1.5 的比例，请整体乘以 2 以得到整数比。
-
umes to dm³ if all volumes are in cm³ and the same unit is used on both sides of the ratio – but be careful and convert when using n = cV.
-
在滴定问题中，若所有体积单位均为 cm³ 且比例两方单位一致，可无需换算为 dm³，但在使用 n = cV 时务必换算。
-
Percentage yield and atom economy are different concepts; do not confuse them. Atom economy deals with the efficiency of a reaction route, while percentage yield measures how much product was actually recovered.
-
产率与原子经济是不同概念，切勿混淆。原子经济关注反应路线的效率，而产率衡量实际回收产品的多少。
Mastering these calculations requires consistent practice, but once you recognise the structured method – mass to moles, mole ratio, moles to mass – nearly all stoichiometry problems become straightforward steps.
掌握这些计算需要持续练习，但一旦你建立起“质量 → 摩尔数，摩尔比，摩尔数 → 质量”的结构化思路，几乎所有的化学计量问题都将变得条理清晰。
Published by TutorHao | 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