📚 GCSE Chemistry: Formula Summary Handbook | GCSE 化学：公式汇总手册
This handbook brings together all the essential equations and calculations you will meet in GCSE Chemistry. Mastering these formulas is crucial for tackling quantitative chemistry questions, required practicals and examinations. Each section includes the formula, explains what the symbols mean and provides tips on using the correct units. Use this as your go-to reference throughout your revision.
这本手册汇集了 GCSE 化学中所有关键的方程式与计算。掌握这些公式对于解答定量化学题、完成必修实验以及应对考试至关重要。每一节都给出了公式，解释了符号的含义并提供了正确使用单位的提示。请将本文作为你复习过程中的首选参考。
1. Relative Formula Mass (Mr) | 相对分子质量
Relative formula mass is the sum of the relative atomic masses (Ar) of all atoms in the formula unit. It has no units because it compares the mass to the carbon-12 standard. To find Mr, multiply each element’s Ar by the number of atoms present in the formula and add the results.
相对分子质量是化学式单元中所有原子的相对原子质量（Ar）之和。它没有单位，因为它是与碳-12 标准比较的质量比值。计算 Mr 时，将每种元素的 Ar 乘以其原子个数，然后相加即可。
Mr = Σ (Ar × number of atoms)
（Σ 表示求和）
For example, for H₂SO₄: (2 × 1) + 32 + (4 × 16) = 98. Always use the Ar values from the Periodic Table, normally rounded to one decimal place if required by the exam board.
例如，对于 H₂SO₄：(2×1) + 32 + (4×16) = 98。始终使用周期表上的 Ar 值，考试中通常要求保留一位小数。
2. Moles, Mass and Molar Mass | 摩尔、质量与摩尔质量
The mole is the chemical amount unit. One mole of any substance contains 6.02 × 10²³ particles (Avogadro’s number) and has a mass equal to its Mr in grams. The relationship between moles (n), mass (m) and molar mass (Mr) is the most important equation in GCSE quantitative chemistry.
摩尔是物质的量的单位。1 摩尔任何物质含有 6.02×10²³ 个粒子（阿伏伽德罗常数），它的质量以克为单位时等于其 Mr。摩尔（n）、质量（m）和摩尔质量（Mr）之间的关系是 GCSE 定量化学中最重要的公式。
n = m / Mr
n = number of moles (mol), m = mass (g), Mr = relative formula mass (g/mol)
n = 物质的量 (mol)，m = 质量 (g)，Mr = 相对分子质量 (g/mol)
Rearrange the equation to find mass: m = n × Mr. This is used to work out reacting masses, convert grams to moles and vice versa. Remember to use grams, not kilograms, unless the question specifically states otherwise.
可以通过变换公式求质量：m = n × Mr。这用于计算反应质量、转换克与摩尔。除非题目特别说明，否则均使用克而非千克。
3. Concentration and Moles in Solutions | 溶液中的浓度与摩尔
The concentration of a solution tells us how many moles of solute are dissolved in each dm³ of solution. The standard unit is mol/dm³ (moles per cubic decimetre). You need to be familiar with converting cm³ to dm³ – divide by 1000.
溶液浓度表示每 dm³ 溶液中溶解了多少摩尔溶质。标准单位是 mol/dm³（摩尔每立方分米）。你需要熟练掌握 cm³ 与 dm³ 之间的换算——除以 1000。
n = c × V
n = moles (mol), c = concentration (mol/dm³), V = volume (dm³)
n = 物质的量 (mol)，c = 浓度 (mol/dm³)，V = 体积 (dm³)
When the volume is given in cm³, first convert: V(dm³) = V(cm³) / 1000. You can also use concentration in g/dm³, which is equal to mass(g) / volume(dm³). However, the mole-based concentration is essential for titration calculations.
当体积以 cm³ 给出时，先换算：V(dm³) = V(cm³) / 1000。你也可以使用 g/dm³ 作为浓度单位，它等于质量(g)除以体积(dm³)。但基于摩尔的浓度对于滴定计算至关重要。
4. Moles and Gas Volumes | 摩尔与气体体积
At room temperature and pressure (RTP, around 20 °C and 1 atm), one mole of any gas occupies 24 dm³. This molar gas volume simplifies calculations involving gases produced or consumed in reactions. The formula links the amount of gas directly to its volume.
在常温常压下（RTP，约 20 °C 和 1 个标准大气压），1 摩尔任何气体的体积是 24 dm³。这个气体摩尔体积简化了涉及气体生成或消耗反应的有关计算。该公式将气体的量与其体积直接联系起来。
n = V / 24
n = moles of gas (mol), V = volume of gas (dm³) at RTP, 24 = molar volume (dm³/mol)
n = 气体的物质的量 (mol)，V = 气体在 RTP 下的体积 (dm³)，24 = 摩尔体积 (dm³/mol)
If the volume is given in cm³, convert to dm³ before using the equation. Also, note that this relationship is valid only at RTP. If conditions are different you will be given the molar volume. Use the same general form: n = V / molar volume.
如果体积单位是 cm³，在使用公式前先换算成 dm³。另外需注意，此关系仅在常温常压下成立。如果条件不同，题目会给出对应的摩尔体积。通用形式为 n = V / 摩尔体积。
5. Percentage Yield | 产率百分比
The percentage yield compares the mass of product actually obtained in an experiment with the maximum theoretical mass calculated from the balanced equation. It is always less than or equal to 100% due to incomplete reactions, practical losses and side reactions.
产率百分比将实验中实际获得的产品质量与根据化学方程式计算的理论最大质量进行比较。由于反应不完全、操作损失和副反应等原因，产率百分比总是小于或等于 100%。
% Yield = (actual yield / theoretical yield) × 100
Actual yield and theoretical yield must have the same units (usually grams). First calculate the theoretical yield using the mole ratio from the balanced equation and the formula m = n × Mr. Then divide the actual mass by the theoretical mass and multiply by 100.
实际产量和理论产量必须具有相同单位（通常为克）。首先利用化学方程式的摩尔比和公式 m = n × Mr 计算理论产量。然后用实际质量除以理论质量，再乘以 100。
A result of 100% would mean no loss. Yields below 50% often prompt scientists to improve the method. Exam questions frequently ask you to calculate yield and suggest reasons why it is not 100%.
结果为 100% 表示没有损失。产率低于 50% 通常会促使科学家改进方法。考试题中常要求你计算产率并说明为什么不是 100% 的原因。
6. Atom Economy | 原子经济性
Atom economy measures how efficiently reactants are turned into the desired product. It is a key concept in green chemistry because high atom economy means less waste. The equation uses relative formula masses of the product and all reactants as written in the balanced equation.
原子经济性衡量反应物转化为目标产品的效率。它是绿色化学的核心概念，因为高原子经济性意味着更少的废物。该公式使用目标产物和化学方程式中所有反应物的相对分子质量。
% Atom Economy = (Mr of desired product / Σ Mr of all reactants) × 100
The sum of Mr of all reactants means adding the Mr of each reactant exactly as it appears in the balanced equation (do not forget any coefficients). This shows the percentage of total atomic mass that ends up in the useful product.
所有反应物的 Mr 之和是指把化学方程式中每种反应物的 Mr 直接相加（不要遗漏任何化学计量数）。它显示出总原子质量中有百分之多少最终进入了有用的产品。
For example, when iron is extracted from iron oxide using carbon monoxide, some atoms end up in carbon dioxide, which is not the desired product. Atom economy helps compare different routes to making the same product.
例如，用一氧化碳从氧化铁中提取铁时，有些原子最终进入二氧化碳，而后者并非目标产品。原子经济性有助于比较制造同一产品的不同路线。
7. Energy Changes (Calorimetry) | 能量变化（量热法）
The heat energy change in a reaction is often measured by the temperature change of water or solution. The equation Q = mcΔT lets you calculate the heat energy transferred. In GCSE Chemistry, the substance being heated is usually water, so c = 4.18 J/g°C.
反应中的热量变化通常通过测量水或溶液的温度变化来得到。公式 Q = mcΔT 能让你计算出传递的热量。在 GCSE 化学中，被加热的物质通常是水，因此 c = 4.18 J/g°C。
Q = m × c × ΔT
Q = heat energy (J), m = mass of water or solution (g), c = specific heat capacity (4.18 J/g°C for water), ΔT = temperature change (°C)
Q = 热量 (J)，m = 水或溶液的质量 (g)，c = 比热容（水为 4.18 J/g°C），ΔT = 温度变化 (°C)
After finding Q, you can calculate the molar enthalpy change by dividing Q by the number of moles that reacted. Remember that an increase in temperature means an exothermic reaction (negative ΔH), and a decrease means endothermic (positive ΔH). Some specifications use 4.2 J/g°C, but 4.18 is more precise.
求出 Q 后，你可以通过 Q 除以反应的物质的量来计算摩尔焓变。记住，温度升高表示放热反应（ΔH 为负），温度降低表示吸热反应（ΔH 为正）。有些教材使用 4.2 J/g°C，但 4.18 更精确。
8. Rate of Reaction | 反应速率
The mean rate of a chemical reaction can be determined by measuring how quickly a reactant is used up or a product is formed. The rate is expressed as the change in quantity per unit time. Quantities can be mass, volume of gas, or moles.
化学反应的速率可以通过测量反应物消耗或产物生成的速度来确定。速率表示为单位时间内某量的变化。量可以是质量、气体体积或物质的量。
Mean rate = quantity of reactant used or product formed / time
平均速率 = 反应物消耗量或产物生成量 / 时间
For example, mean rate (g/s) = mass lost (g) / time (s). When a gas is produced, you might use volume per second (cm³/s). On a graph, the rate at a specific point can be found from the gradient of a tangent. The steeper the gradient, the faster the rate.
例如，平均速率 (g/s) = 质量损失 (g) / 时间 (s)。如果产生气体，你可以使用每秒产生的体积 (cm³/s)。在图表上，某一点的速率可以通过切线的斜率求得。斜率越大，速率越快。
Rate is affected by temperature, concentration, surface area and catalysts. Collision theory explains these effects; you are expected to link the formula to experimental data.
速率受温度、浓度、表面积和催化剂影响。碰撞理论能解释这些影响；考试中要求你能够将公式与实验数据联系起来。
9. Titration Calculations | 滴定计算
Titrations are used to find the unknown concentration of an acid or alkali. The calculation relies on the equation n = cV, combined with the mole ratio from the balanced neutralisation reaction. Careful unit conversion is critical here.
滴定用于找出酸或碱的未知浓度。计算依赖于公式 n = cV，并结合中和反应方程式中的摩尔比。此处仔细的单位换算至关重要。
Step 1: n = c × V (unknown solution)
Step 2: Use mole ratio to find n of other solution
Step 3: c = n / V (to find concentration)
For instance, in the reaction H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O, the mole ratio is 1:2. Convert all volumes from cm³ to dm³. Take the average titre volume (concordant results) and avoid using rough titres. Remember to multiply moles by the appropriate factor.
例如，在反应 H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O 中，摩尔比是 1:2。将所有体积从 cm³ 换算成 dm³。用平均滴定体积（一致结果），避免使用粗测数据。记住将物质的量乘以合适的倍数。
A typical question: 25.0 cm³ of NaOH requires 23.40 cm³ of 0.100 mol/dm³ HCl. Find the concentration of NaOH. You must show a clear step-by-step method, which is a common 6-mark question.
一个典型问题：25.0 cm³ 的 NaOH 需要 23.40 cm³ 的 0.100 mol/dm³ HCl 来中和。求 NaOH 的浓度。你必须展示清晰的逐步解法，这通常是 6 分的题目。
10. Chromatography: Rf Value | 色谱法：Rf 值
Paper chromatography separates mixtures of soluble substances. Each component has a retention factor (Rf) that is constant under the same conditions. It is used to identify substances by comparing with known Rf values.
纸上色谱法可分离可溶性混合物。在相同条件下每种组分都有一个固定的比移值（Rf）。通过比较已知 Rf 值可以鉴定物质。
Rf = distance moved by substance / distance moved by solvent
Measure both distances from the origin (baseline) to the centre of the spot and to the solvent front respectively. Rf has no units because it is a ratio. It is always less than 1 because a substance cannot travel further than the solvent.
两个距离均从原点（基线）量起，分别是斑点中心到原点的距离与溶剂前沿到原点的距离。Rf 是一个比值，没有单位。它总是小于 1，因为物质不可能比溶剂移动得更远。
Use a pencil to draw the baseline, as ink would dissolve. The solvent must be below the baseline. Rf values for pure substances can be looked up to identify components in a mixture.
用铅笔绘制基线，因为墨水会溶解。液面必须低于基线。纯物质的 Rf 值可查表比对，用来确定混合物中的成分。
11. Converting Units and Key Constants | 单位换算与关键常数
Many mistakes in GCSE calculations come from using the wrong units. The table below summarises the common conversions and essential numbers you must know.
GCSE 计算中的许多错误都源于使用了错误的单位。以下表格总结了常见的换算和你必须掌握的关键数值。
| Conversion / Constant | Value |
| cm³ to dm³ | ÷ 1000 |
| dm³ to cm³ | × 1000 |
| m³ to dm³ | × 1000 |
| tonne to g | × 10⁶ |
| Avogadro’s number | 6.02 × 10²³ |
| Molar gas volume (RTP) | 24 dm³/mol |
| Specific heat capacity of water | 4.18 J/g°C |
Always check the units given in the question. If mass is in kilograms, convert to grams before using n = m/Mr. For gas volumes, ensure you are at RTP unless a different molar volume is provided. Understanding these constants will save you from avoidable errors.
始终检查题目给出的单位。若质量以千克给出，在使用 n = m/Mr 前先换算成克。对于气体体积，除非给出了不同的摩尔体积，否则要确认是否处于常温常压。掌握这些常数能帮你避免本可避免的错误。
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