📚 Year 11 OCR Chemistry: Quick Reference to Formulas and Theorems | Year 11 OCR 化学：公式定理速查手册
This revision handbook brings together all the essential formulas, relationships, and quantitative principles that you need for Year 11 OCR Chemistry. From relative masses to electrolysis calculations, each section presents the key equation, explains the symbols and units, and gives a typical worked application. Use this as a rapid refresher right before your exams.
这本速查手册汇集了 Year 11 OCR 化学中所有重要的公式、关系和定量原理。从相对质量到电解计算，每一节都给出关键方程，解释符号与单位，并提供典型应用示例。考前用它快速巩固，会非常高效。
1. Relative Atomic Mass and Formula Mass | 相对原子质量与化学式量
The relative atomic mass (Ar) is the weighted average mass of an atom compared to 1/12 of the mass of a carbon-12 atom. It has no units. The relative formula mass (Mr) of a compound is the sum of the Ar values of all atoms in its formula.
相对原子质量（Ar）是一个原子的加权平均质量与一个碳-12 原子质量的 1/12 的比值，没有单位。化合物的相对化学式量（Mr）是其化学式中所有原子 Ar 的总和。
Example: Mr of H₂SO₄ = (2 × 1) + 32 + (4 × 16) = 98.
示例：H₂SO₄ 的 Mr = (2 × 1) + 32 + (4 × 16) = 98。
2. The Mole and Avogadro's Number | 摩尔与阿伏伽德罗常数
02 × 10²³ particles (atoms, molecules, ions, or electrons). This number is called Avogadro' mass of one mole of a substance is its relative atomic or formula mass expressed in grams.
1 摩尔任何物质含有 6.02 × 10²³ 个粒子（原子、分子、离子或电子）。这个数值称为阿伏伽德罗常数。1 摩尔物质的质量就等于以克为单位的相对原子质量或相对化学式量。
The relationship: number of particles = number of moles × 6.02 × 10²³.
关系式：粒子数目 = 摩尔数 × 6.02 × 10²³。
3. Moles from Mass and Molar Mass | 质量、摩尔质量与摩尔数
The central equation linking mass, relative mass, and amount of substance is:
n = m / M
where n is the number of moles (mol), m is the mass in grams (g), and M is the molar mass in g mol⁻¹ (numerically equal to Ar or Mr).
连接质量、相对质量与物质的量的核心方程为：
n = m / M
其中 n 是摩尔数（mol），m 是质量（g），M 是摩尔质量（g mol⁻¹），其数值等于 Ar 或 Mr。
Rearrange to find mass: m = n × M.
变形求质量：m = n × M。
4. Empirical and Molecular Formulas | 经验式与分子式
The empirical formula gives the simplest whole-number ratio of atoms in a compound. To determine it: convert masses (or percentages) to moles, divide by the smallest number of moles, and express as the simplest integer ratio.
经验式表示化合物中原子的最简整数比。确定方法：将质量（或百分比）转换为摩尔数，除以其中最小的摩尔数，得到最简整数比。
of the empirical formula. Multiplier = Mr of molecular formula ÷ Mr of empirical formula.
分子式是经验式的整数倍。倍数 = 分子式的 Mr ÷ 经验式的 Mr。
formula is CH₂O and Mr = 90, then multiplier = 90 ÷ 30 = 3, so molecular formula = C₃H₆O₃.
示例：若经验式为 CH₂O，Mr = 90，则倍数 = 90 ÷ 30 = 3，分子式为 C₃H₆O₃。
5. Reacting Masses and Stoichiometry | 反应质量与化学计量
Balanced chemical equations give mole ratios. To calculate reacting masses: convert the given mass to moles, use the mole ratio to find moles of the target substance, then convert back to mass.
配平的化学方程式给出了摩尔比。计算反应质量时：先将已知质量换算为摩尔，利用摩尔比求出待求物质的摩尔数，再换算回质量。
General steps:
- moles of known = mass of known / molar mass
- moles of unknown = moles of known × (coefficient of unknown / coefficient of known)
- mass of unknown = moles of unknown × molar mass
通用步骤：
- 已知物的摩尔数 = 已知物质量 / 摩尔质量
- 待求物的摩尔数 = 已知物摩尔数 ×（待求物系数 / 已知物系数）
- 待求物质量 = 待求物摩尔数 × 摩尔质量
6. Concentration of Solutions | 溶液浓度
Concentration is defined as the amount of solute per unit volume of solution. The key equation is:
c = n / V
ten units of g dm⁻³ are also used: mass concentration = mass of solute (g) / volume (dm³).
浓度定义为每单位体积溶液中溶质的量。关键公式为：
c = n / V
其中 c 是浓度（mol dm⁻³），n 是摩尔数，V 是体积（dm³）。也常用质量浓度（g dm⁻³）：质量浓度 = 溶质质量（g）/ 体积（dm³）。
f base × (acid coefficient / base coefficient), then concentration = moles / volume (dm³).
滴定计算中：酸的摩尔数 = 碱的摩尔数 ×（酸系数 / 碱系数），然后浓度 = 摩尔数 / 体积（dm³）。
Note: 1 dm³ = 1000 cm³.
注意：1 dm³ = 1000 cm³。
7. Gas Molar Volume | 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 (24 000 cm³).
在室温和常压下（RTP：20 °C，1 atm），1 摩尔任何气体的体积为 24 dm³（24 000 cm³）。
Volume (dm³) = number of moles × 24
This relationship allows you to interconvert volume and moles for gaseous reactants or products.
此关系可用于气体反应物或产物的体积与摩尔数之间的相互换算。
8. Percentage Yield and Atom Economy | 产率与原子经济
Percentage yield measures the efficiency of a reaction in practice:
Percentage yield = (actual mass of product / theoretical mass of product) × 100%
产率衡量实际反应的效率：
产率 = (实际产品质量 / 理论产品质量) × 100%
Atom economy describes how much of the total mass of reactants ends up in the desired product:
Atom economy = (Mr of desired product / sum of Mr of all reactants) × 100%
原子经济描述反应物总质量中有多少进入了目标产物：
原子经济 = (目标产物 Mr / 所有反应物 Mr 总和) × 100%
High atom economy reduces waste and is a key principle of green chemistry.
原子经济高意味着废物少，是绿色化学的重要原则。
9. Rate of Reaction | 反应速率
The average rate of a reaction can be calculated from the change in the amount of a reactant or product per unit time. Common expressions:
Rate = change in mass (g) / time (s)
Rate = change in volume of gas (cm³) / time (s)
平均反应速率可由反应物或产物的量在单位时间内的变化求得。常用表达式：
速率 = 质量变化 (g) / 时间 (s)
速率 = 气体体积变化 (cm³) / 时间 (s)
On a graph of amount against time, the rate at any instant is the slope (gradient) of the tangent to the curve.
在“量-时间”图上，任意时刻的瞬时速率等于该时刻切线的斜率（梯度）。
Concentration, temperature, surface area, and catalysts all affect rate by altering collision frequency or the proportion of successful collisions.
浓度、温度、表面积和催化剂都会通过改变碰撞频率或有效碰撞的比例来影响反应速率。
10. Energy Changes | 能量变化
The heat energy change during a reaction is measured using calorimetry. The basic equation for the heat transferred is:
q = m c ΔT
where q is heat energy (J), m is the mass of the solution (g), c is specific heat capacity (usually 4.18 J g⁻¹ °C⁻¹ for water), and ΔT is the temperature change (°C).
反应放出的热量通过量热法测定。基本热传导方程为：
q = m c ΔT
其中 q 是热能（J），m 是溶液质量（g），c 是比热容（水的 c 通常取 4.18 J g⁻¹ °C⁻¹），ΔT 是温度变化（°C）。
The molar enthalpy change (ΔH) is then found using:
ΔH = – q / n
where n is the moles of the limiting reactant. The negative sign indicates an exothermic reaction (ΔH negative). For endothermic reactions, ΔH is positive.
摩尔焓变（ΔH）由下式求得：
ΔH = – q / n
其中 n 是限制反应物的摩尔数。负号代表放热反应（ΔH 为负），吸热反应则 ΔH 为正。
nergy profile diagrams and to recognise that overall ΔH = energy absorbed in bond breaking – energy released in bond making.
OCR 要求你能解释能量-进程图，并理解总 ΔH = 断键吸收的能量 – 成键释放的能量。
11. Water of Crystallisation | 结晶水
Many salts exist as hydrated crystals containing water molecules built into the crystal lattice. The amount of water of crystallisation can be determined by heating a known mass of the hydrated salt to constant mass, then calculating the loss in mass.
许多盐以水合晶体形式存在，晶格中含有结晶水。通过将已知质量的水合盐加热至恒重，再计算质量损失，即可确定结晶水的含量。
Method:
- Find mass of water lost = mass of hydrated salt – mass of anhydrous salt
- Moles of water = mass of water / 18
- Moles of anhydrous salt = mass of anhydrous salt / Mr (anhydrous)
- The ratio 'moles of water : moles of salt' gives the water of crystallisation number.
方法：
- 失去水的质量 = 水合盐质量 – 无水盐质量
- 水的摩尔数 = 水的质量 / 18
- 无水盐的摩尔数 = 无水盐质量 / 无水盐 Mr
- 水与盐的摩尔比即给出结晶水数。
ter = 12.6/18 = 0.7; moles MgSO₄ = 12.0/120 = 0.1; ratio 0.7:0.1 = 7 → formula MgSO₄·7H₂O.
示例：加热 24.6 g MgSO₄·xH₂O 得到 12.0 g 无水 MgSO₄（Mr = 120）。失去水 12.6 g → 水的摩尔数 = 12.6/18 = 0.7；MgSO₄ 摩尔数 = 0.1；比例 0.7:0.1 = 7 → 化学式为 MgSO₄·7H₂O。
12. Quantitative Electrolysis | 定量电解
During electrolysis, the mass of a substance deposited or liberated at an electrode is directly proportional to the quantity of electricity passed. The quantity of charge (Q) is calculated from current and time:
Q = I × t
where Q is measured in coulombs (C), I is current in amperes (A), and t is time in seconds (s).
电解时，在电极上析出或释放的物质质量与通过的电量成正比。电荷量（Q）由电流和时间计算：
Q = I × t
其中 Q 单位为库仑（C），I 为电流（A），t 为时间（s）。
The mass of metal deposited can then be determined using Faraday's laws:
m = (M × I × t) / (n × F)
Cu), and F is the Faraday constant (96 500 C mol⁻¹). In OCR exams, F is usually provided.
沉积金属的质量可由法拉第定律求出：
m = (M × I × t) / (n × F)
其中 m 为质量（g），M 为物质的摩尔质量（g mol⁻¹），n 为每离子所需电子数（如 Cu²⁺ → Cu，n = 2），F 为法拉第常数（96 500 C mol⁻¹）。OCR 考试中 F 通常会给出。
f electrons = Q / 96 500 → moles of substance = moles of electrons / n → mass = moles × M.
也可以分步推理：Q = It → 电子的摩尔数 = Q / 96 500 → 物质的摩尔数 = 电子的摩尔数 / n → 质量 = 摩尔数 × M。
This topic links electrical measurements to chemical change and is often assessed alongside half-equations at the electrodes.
这一主题将电学测量与化学变化联系起来，常与电极半反应式一同考查。
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