📚 A-Level OCR Chemistry: Stoichiometry Revision | A-Level OCR 化学：化学计量考点精讲
Stoichiometry forms the quantitative backbone of chemistry, enabling you to relate masses, volumes, and concentrations. Mastering this topic is essential for success in OCR A-Level Chemistry, as it appears in both multiple-choice and structured questions.
化学计量学构建了化学定量计算的基础，使你能够将质量、体积和浓度联系起来。掌握这个主题对于OCR化学考试至关重要，因为在选择题和结构题中都会出现。
1. The Mole Concept & Avogadro’s Constant | 摩尔概念与阿伏伽德罗常数
The mole (symbol mol) is the SI base unit for amount of substance. One mole contains exactly 6.02214076 × 10²³ elementary entities – atoms, molecules, ions, electrons or other particles. This number is Avogadro’s constant, Nₐ.
摩尔（符号mol）是物质的量的国际基本单位。1摩尔恰好包含6.02214076×10²³个基本单元——原子、分子、离子、电子或其他粒子。这个数就是阿伏伽德罗常数Nₐ。
The relationship is: amount (n) = number of particles ÷ Nₐ. For example, 3.01 × 10²³ molecules of H₂O is 0.500 mol.
关系式：物质的量(n) = 粒子数 ÷ Nₐ。例如，3.01×10²³个H₂O分子的物质的量为0.500 mol。
2. Molar Mass & Mass-Mole Conversions | 摩尔质量与质量-摩尔换算
Molar mass (M) is the mass of one mole of a substance, expressed in g mol⁻¹. It equals the relative atomic mass (Aᵣ) or relative formula mass (Mᵣ) in grams.
摩尔质量(M)是1摩尔物质的质量，单位为g mol⁻¹。它在数值上等于该物质的相对原子质量(Aᵣ)或相对式量(Mᵣ)。
Key equation: mass (m) = amount (n) × molar mass (M). Always ensure units match: mass in grams, amount in mol.
关键公式：质量(m) = 物质的量(n) × 摩尔质量(M)。务必保证单位一致：质量以克计，物质的量以摩尔计。
Example: The molar mass of CaCO₃ is 100.1 g mol⁻¹. So 0.200 mol of CaCO₃ has a mass of 0.200 × 100.1 = 20.0 g.
示例：CaCO₃的摩尔质量为100.1 g mol⁻¹，则0.200 mol CaCO₃的质量为0.200 × 100.1 = 20.0 g。
3. Empirical & Molecular Formulae | 实验式与分子式
The empirical formula shows the simplest whole‑number ratio of atoms in a compound; the molecular formula gives the actual number of each atom per molecule.
实验式表示化合物中原子最简整数比；分子式给出每个分子中实际的原子数目。
To determine empirical formula from mass data: (1) divide the mass (or %) of each element by its Aᵣ to get moles, (2) divide by the smallest mole value, (3) multiply to obtain whole numbers. For combustion data, use CO₂ and H₂O masses to find C and H moles.
由质量数据求实验式的步骤：(1) 用各元素的质量（或百分数）除以其Aᵣ得到摩尔数，(2) 除以最小的摩尔数，(3) 乘以适当因子得到整数比。对燃烧数据，根据CO₂和H₂O的质量分别计算C和H的摩尔数。
Molecular formula = (empirical formula)ₙ, where n = Mᵣ(compound) ÷ Mᵣ(empirical formula).
分子式 = (实验式)ₙ，其中 n = 化合物的Mᵣ ÷ 实验式的Mᵣ。
4. Water of Crystallisation | 结晶水
Hydrated salts contain water molecules within their crystal lattice, e.g. CuSO₄⋅5H₂O. The water of crystallisation can be determined by heating to constant mass and measuring the mass loss.
水合盐在其晶格中含有水分子，如CuSO₄⋅5H₂O。结晶水可通过加热至恒重并测量质量损失来确定。
From the masses of anhydrous salt and water lost, calculate moles of each, then find the ratio to obtain x in salt⋅xH₂O.
根据无水盐和失去的水的质量，分别计算两者的物质的量，然后求比值以得到盐⋅xH₂O中的x。
5. Balancing Equations & Reaction Ratios | 化学方程式的配平与反应比
A balanced chemical equation respects the conservation of mass and atom economy. The coefficients represent the molar ratios of reactants and products.
配平的化学方程式体现了质量守恒和原子守恒。化学计量数代表反应物与产物之间的物质的量之比。
For example, 2H₂ + O₂ → 2H₂O means 2 mol of H₂ react with 1 mol of O₂ to produce 2 mol of H₂O. Use these ratios as conversion factors.
例如，2H₂ + O₂ → 2H₂O 表示2 mol H₂与1 mol O₂反应生成2 mol H₂O。将这些比值作为换算因子。
Always check that the total mass is the same on both sides. Common OCR pitfalls include forgetting to balance before stoichiometric calculations.
永远检查方程式两边总质量是否相等。OCR常见错误包括在化学计量计算前未配平方程式。
6. Reacting Mass Calculations | 反应质量计算
Using the balanced equation and molar masses, you can calculate the mass of a product formed from a given mass of reactant, or vice versa.
利用配平的方程式和摩尔质量，你可以计算由给定反应物质量生成产物的质量，或反之。
General steps: mass → moles (÷M), then use mole ratio from equation, then moles → mass (×M). Keep intermediate values in your calculator to avoid rounding errors.
通用步骤：质量 → 物质的量（÷M），利用方程式中的摩尔比，再物质的量 → 质量（×M）。计算过程中保留中间值以避免舍入误差。
Example: What mass of CO₂ is produced from 10.0 g of CaCO₃ on heating? (CaCO₃ → CaO + CO₂). M(CaCO₃)=100.1 g mol⁻¹, n=10.0/100.1=0.0999 mol, ratio 1:1, so n(CO₂)=0.0999 mol, mass=0.0999 × 44.0 = 4.40 g.
示例：加热10.0 g CaCO₃会产生多少CO₂？（CaCO₃ → CaO + CO₂）。M(CaCO₃)=100.1 g mol⁻¹，n=10.0/100.1=0.0999 mol，摩尔比1:1，故n(CO₂)=0.0999 mol，质量=0.0999 × 44.0 = 4.40 g。
7. Limiting Reactant | 限制反应物
The limiting reactant is the one that is completely consumed first, determining the maximum amount of product. The other reactant is in excess.
限制反应物是最先被完全消耗的反应物，它决定了产物的最大产量。另一种反应物则为过量。
To identify the limiting reactant, calculate the moles of each reactant and compare with the stoichiometric ratio. The reactant that gives the smallest mole of product (when scaled) is limiting.
要确定限制反应物，计算每种反应物的物质的量，并与化学计量比比较。能够产生最少产物摩尔数（按比例换算后）的反应物即是限制反应物。
OCR questions often ask for the mass of product or mass of excess remaining. Always subtract moles used from moles present.
OCR常考计算产物的质量或剩余过量反应物的质量。始终从现有的物质的量中减去已反应的物质的量。
8. Percentage Yield & Atom Economy | 产率与原子经济性
Percentage yield = (actual yield ÷ theoretical yield) × 100%. It measures the efficiency of a reaction in practice; it is always ≤ 100%.
产率 = (实际产量 ÷ 理论产量) × 100%。它衡量反应的实际效率，始终 ≤ 100%。
Atom economy = (molar mass of desired product ÷ sum of molar masses of all reactants) × 100%. It indicates the proportion of reactant atoms that end up in the desired product, guiding green chemistry decisions.
原子经济性 = (目标产物的摩尔质量 ÷ 所有反应物摩尔质量之和) × 100%。它表明反应物原子进入目标产物的比例，指导绿色化学决策。
High atom economy reduces waste and cost. Addition reactions often have 100% atom economy, while substitution reactions may be lower.
高原子经济性减少废物和成本。加成反应通常具有100%原子经济性，而取代反应可能较低。
9. Solution Concentration | 溶液浓度
Concentration (c) is the amount of solute per unit volume, typically mol dm⁻³. c = n ÷ V, where V is in dm³. To convert cm³ to dm³, divide by 1000.
浓度(c)是单位体积溶液中所含溶质的物质的量，常用单位为mol dm⁻³。
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