📚 Essential AS Chemistry Formula Handbook | AS 化学：公式汇总手册
This handbook collects the essential equations and formulas needed for AS-level Chemistry. It covers mole calculations, empirical and molecular formulae, stoichiometry, concentration and dilution, the ideal gas equation, percentage yield and atom economy, energetics and Hess's law, equilibrium constants, acids and bases with pH and Ka, titration and redox calculations, and electrolysis with Faraday's laws. Keep it beside your notes for quick revision before papers and practicals.
这份手册汇总了 AS 化学所需的关键方程式和公式，涵盖摩尔计算、经验式与分子式、化学计量学、浓度与稀释、理想气体方程、产率与原子经济性、能量学与盖斯定律、平衡常数、酸碱与 pH/Ka、滴定与氧化还原计算，以及电解与法拉第定律等内容。建议放在笔记旁，备考与做实验前快速查阅。
1. The Mole & Avogadro's Number | 摩尔与阿伏加德罗常数
The mole is the chemist' articles - atoms, molecules, ions or electrons. This number is the Avogadro constant, L, and it links the number of particles N to the amount of substance n in moles.
摩尔是化学家的"计数单位"。任何物质 1 摩尔都含有 6.02 × 10²³ 个指定微粒——原子、分子、离子或电子。这个数就是阿伏加德罗常数 L，它把微粒数 N 与物质的量 n 联系起来。
N = n × L (L = 6.02 × 10²³ mol⁻¹)
Mass and moles are linked by the molar mass M, which is numerically equal to the relative molecular mass in g mol⁻¹. Rearranging the definition gives three tidy forms you should be able to write instantly.
质量与物质的量通过摩尔质量 M 联系起来，M 的数值等于相对分子质量，单位为 g mol⁻¹。把定义式变形，就得到三个必须能立刻写出的式子。
m = n × M n = m / M M = m / n
For gases at room temperature and pressure (RTP, 20 °C and 1 atm), one mole occupies 24 dm³, that is 24,000 cm³. This single fact converts any gas volume at RTP into moles without using the ideal gas equation.
对于室温常压（RTP，20 °C、1 atm）下的气体，1 摩尔占据 24 dm³，即 24,000 cm³。记住这一条，就能不用理想气体方程直接把 RTP 下的气体体积换算成物质的量。
V(gas, RTP) = n × 24 dm³ mol⁻¹
- Particles: N = n × 6.02 × 10²³ | 微粒数：N = n × 6.02 × 10²³
- Mass: n = m / M - keep mass in g and M in g mol⁻¹ | 质量：n = m / M，质量用 g，摩尔质量用 g mol⁻¹
- Gas volume: n = V / 24 (V in dm³) at RTP | 气体体积：n = V / 24（V 单位为 dm³，RTP 下）
- Solution: n = c × V (V in dm³) | 溶液：n = c × V（V 单位为 dm³）
Unit consistency is the single biggest source of lost marks. Convert cm³ to dm³ by dividing by 1000, and always check that volumes are in dm³ before multiplying by a concentration in mol dm⁻³.
单位不一致是失分最多的原因。把 cm³ 除以 1000 换成 dm³；在用 mol dm⁻³ 的浓度相乘之前，务必确认体积已经是 dm³。
2. Empirical & Molecular Formulae | 经验式与分子式
The empirical formula shows the simplest whole-number ratio of atoms in a compound, while the molecular formula shows the actual number of atoms in one molecule. To find the empirical formula, convert each mass to moles, divide all the values by the smallest, and scale to whole numbers.
经验式表示化合物中原子最简整数比，分子式表示一个分子中实际的原子数目。求经验式的方法：把各元素的质量换算成物质的量，全部除以最小值，再按比例化为整数。
moles of element = mass of element / Aᵣ
Once the empirical formula is known, compare the relative molecular mass Mᵣ with the empirical formula mass. The ratio n tells you how many empirical units are present in the molecule.
得到经验式后，把相对分子质量 Mᵣ 与经验式质量相比，比值 n 就是分子中含有多少个经验式单元。
molecular formula = (empirical formula)ₙ n = Mᵣ / empirical formula mass
elative molecular mass is 180, then n = 180 / 30 = 6, so the molecular formula is C₆H₁₂O₆.
示例：经验式 CH₂O 的经验式质量为 30 g mol⁻¹。若相对分子质量为 180，则 n = 180 / 30 = 6，分子式为 C₆H₁₂O₆。
- Percentages can be used directly: treat % as mass in 100 g | 百分组成可直接使用：把百分数当作 100 g 中的质量
- Water of crystallisation: find moles of anhydrous salt and moles of H₂O, then write the ratio | 结晶水：分别求出无水盐与 H₂O 的物质的量，再写比值
- Round only at the final step; values such as 1.5 and 2.5 must be doubled | 只在最后一步取整；出现 1.5、2.5 时必须整体乘以 2
3. Stoichiometry & Reacting Masses | 化学计量学与反应质量
A balanced equation gives the mole ratio in which substances react and are formed. The standard method is: convert the known mass to moles, apply the mole ratio from the equation, then convert the answer back to mass, volume or concentration.
配平的化学方程式给出反应物与生成物之间的摩尔比。标准解题流程是：把已知质量换算成物质的量，套用方程式中的摩尔比，再把结果换算回质量、体积或浓度。
mass of product = (moles of limiting reagent) × (mole ratio) × M(product)
The limiting reagent is the reactant that runs out first; it controls the maximum amount of product. Compare the actual mole ratio with the ratio required by the equation - the substance present in the smaller proportion relative to that requirement is limiting.
限量试剂是先被消耗完的反应物，它决定产物的最大量。把实际的物质的量之比与方程式要求的比例相比较，相对于要求而言比例更小的那种物质即为限量试剂。
es of hydrogen react with one mole of oxygen. If you start with 3 mol H₂ and 1 mol O₂, hydrogen is in excess and oxygen is limiting, giving a maximum of 2 mol H₂O.
例如，在 2H₂ + O₂ → 2H₂O 中，2 mol 氢气与 1 mol 氧气反应。若起始为 3 mol H₂ 和 1 mol O₂，则氢气过量、氧气限量，最多生成 2 mol H₂O。
- Gas volumes at RTP can replace moles directly, because V is proportional to n | RTP 下气体体积可直接替代物质的量，因为 V 与 n 成正比
- In titration calculations, moles of the known reactant = c × V (V in dm³) | 滴定计算中，已知反应物的物质的量 = c × V（V 以 dm³ 计）
- Then use the equation ratio to find moles of the unknown, and divide by its volume to obtain its concentration | 再用方程式比例求出未知物的物质的量，除以其体积即得浓度
n(unknown) = n(known) × (ratio from equation)
4. Concentration & Dilution | 浓度与稀释
Concentration c in mol dm⁻³ is the amount of solute per unit volume of solution. The three rearrangements below cover almost every AS calculation involving solutions.
浓度 c（mol dm⁻³）是单位体积溶液中溶质的物质的量。下面三个变形式几乎可以解决 AS 中所有涉及溶液的計算。
n / V n = c × V V = n / c (V in dm³)
When a solution is diluted, water is added but the number of moles of solute does not change. This conservation of moles gives the dilution equation, which is extremely useful in practical work.
稀释溶液时只是加水，溶质的物质的量不变。由物质的量守恒可得稀释公式，在实验操作中非常实用。
c₁V₁ = c₂V₂
c = n / V, however, the volume must be in dm³: 1 dm³ = 1000 cm³, so 25.0 cm³ = 0.0250 dm³.
在稀释公式中，两边体积只要单位一致即可，因为单位会约去。但使用 c = n / V 时，体积必须是 dm³：1 dm³ = 1000 cm³，所以 25.0 cm³ = 0.0250 dm³。
- Concentration in g dm⁻³ = concentration in mol dm⁻³ × M | 以 g dm⁻³ 表示的浓度 = 以 mol dm⁻³ 表示的浓度 × M
- Dilution factor = V₂ / V₁; if you dilute tenfold, the concentration falls to one tenth | 稀释倍数 = V₂ / V₁；稀释 10 倍，浓度降为原来的 1/10
- Standard solution preparation: moles needed = c × V, then mass = n × M | 配制标准溶液：所需物质的量 = c × V，再算质量 = n × M
5. The Ideal Gas Equation & Molar Volume | 理想气体方程与摩尔体积
The ideal gas equation relates pressure, volume, temperature and amount of gas in a single expression. It is used when conditions are not RTP, or when you need to find the relative molecular mass of a gas from its measured volume.
理想气体方程把压强、体积、温度和气体物质的量统一在同一个表达式中。当条件不是 RTP，或需要由测得的气体体积求相对分子质量时就要用它。
pV = nRT
- p = pressure in Pa (1 atm = 101,000 Pa; 1 kPa = 1000 Pa) | p 为压强，单位 Pa（1 atm = 101,000 Pa；1 kPa = 1000 Pa）
- V = volume in m³ (1 m³ = 1000 dm³ = 10⁶ cm³) | V 为体积，单位 m³（1 m³ = 1000 dm³ = 10⁶ cm³）
- n = amount in mol | n 为物质的量，单位 mol
- R = gas constant = 8.31 J K⁻¹ mol⁻¹ | R 为气体常数 = 8.31 J K⁻¹ mol⁻¹
- T = temperature in K; T = θ in °C + 273 | T 为热力学温度，单位 K；T = θ（°C）+ 273
The molar volume of any gas depends only on the temperature and pressure, not on the identity of the gas. A useful pair of values to remember is 24 dm³ mol⁻¹ at RTP and 22.4 dm³ mol⁻¹ at standard temperature and pressure (STP, 0 °C and 1 atm).
任何气体的摩尔体积只取决于温度和压强，与气体种类无关。值得记住的两个数值是：RTP 下 24 dm³ mol⁻¹，标准状况（STP，0 °C、1 atm）下 22.4 dm³ mol⁻¹。
n = V / 24 (RTP) Mᵣ = mRT / pV
The combined expression on the right is the standard route for finding Mᵣ from a gas experiment: measure m, V, p and T, then substitute.
右式是气体实验测定 Mᵣ 的标准方法：测出 m、V、p、T 后代入计算。
6. Percentage Yield & Atom Economy | 产率与原子经济性
Percentage yield measures how much product you actually obtained compared with the theoretical maximum predicted by the equation. Atom economy measures how much of the reactant mass ends up in the desired product - a green chemistry measure of efficiency.
产率衡量的是实际得到的产物量与方程式预测的理论最大量之比。原子经济性衡量的是反应物质量中有多少最终进入目标产物，是绿色化学的衡量指标。
percentage yield = (actual yield / theoretical yield) × 100%
atom economy = (M of desired product / Σ M of all reactants) × 100%
Yields fall below 100% because reactions are reversible, side reactions occur, some product is lost during transfer, filtration or purification, and the reaction may not reach completion in the time allowed.
产率低于 100% 的原因包括：反应可逆、发生副反应、转移/过滤/提纯过程中损失部分产物，以及在给定时间内反应未进行完全。
Note the difference: yield depends on how you ran the experiment, whereas atom economy is fixed by the equation itself. An addition reaction, such as the hydration of ethene, has 100% atom economy because all reactant atoms appear in the single product.
注意区别：产率取决于你怎么做实验，而原子经济性由方程式本身决定。加成反应（例如乙烯水合）原子经济性为 100%，因为反应物所有原子都进入唯一产物。
- Choose routes with high atom economy to reduce waste | 选择原子经济性高的路线以减少废料
- Report yield to a sensible number of significant figures | 产率保留合理的有效数字
- Useful practical yields are often 60-90%, rarely 100% | 实际实验产率常为 60-90%，几乎不可能 100%
7. Enthalpy Changes & Calorimetry | 焓变与量热法
Enthalpy change ΔH is the heat energy transferred at constant pressure. Exothermic reactions release heat and have negative ΔH; endothermic reactions absorb heat and have positive ΔH. In solution calorimetry, the heat released or absorbed is calculated from the temperature change of the surrounding water.
焓变 ΔH 是恒压条件下传递的热能。放热反应放出热量，ΔH 为负；吸热反应吸收热量，ΔH 为正。在溶液量热实验中，热量由周围水的温度变化计算。
q = m × c × ΔT
- q = heat energy transferred, in J | q 为传递的热能，单位 J
- m = mass of water (or solution) heated, in g; assume 1.00 cm³ = 1.00 g | m 为被加热的水（或溶液）质量，单位 g；假设 1.00 cm³ = 1.00 g
- c = specific heat capacity = 4.18 J g⁻¹ K⁻¹ for water | c 为比热容，水为 4.18 J g⁻¹ K⁻¹
- ΔT = temperature change in K or °C (the two scales have the same size) | ΔT 为温度变化，单位 K 或 °C（两种刻度间隔相同）
To convert a measured heat into a molar enthalpy change, divide q by the number of moles that reacted, and attach the correct sign.
把测得的热量换算成摩尔焓变时，用 q 除以发生反应的物质的量，并加上正确的符号。
/ n (negative sign because the water gains the heat the reaction loses)
Common standard enthalpy definitions you should quote precisely: standard enthalpy of combustion (1 mol of substance burned completely in excess oxygen), of formation (1 mol of compound formed from its elements in their standard states), of neutralisation (1 mol of water formed from acid and alkali), and of solution (1 mol of substance dissolved in a large excess of water).
需要准确表述的常见标准焓变定义：标准燃烧焓（1 mol 物质在过量氧气中完全燃烧）、标准生成焓（由元素在标准状态下生成 1 mol 化合物）、标准中和焓（酸碱反应生成 1 mol 水）、标准溶解焓（1 mol 物质溶于大量过量水）。
8. Hess's Law & Bond Enthalpies | 盖斯定律与键焓
Hess's law states that the total enthalpy change of a reaction is independent of the route taken from reactants to products. This lets you calculate an unknown enthalpy change by building a cycle from known values, usually enthalpies of formation or combustion.
盖斯定律指出：反应的总焓变与从反应物到产物所经过的路径无关。由此可借助已知数据（通常是生成焓或燃烧焓）构建循环，求出未知焓变。
ΔH(reaction) = Σ ΔHf(products) - Σ ΔHf(reactants)
ΔH(reaction) = Σ ΔHc(reactants) - Σ ΔHc(products)
For reactions involving gases, mean bond enthalpies can be used instead. Bond breaking is endothermic and bond making is exothermic, so the total enthalpy change equals the energy absorbed minus the energy released.
对于涉及气体的反应，也可以用平均键焓。断键吸热、成键放热，所以总焓变等于吸收的能量减去放出的能量。
ΔH = Σ(bonds broken) - Σ(bonds formed)
Mean bond enthalpies are averages over many different compounds, so values calculated this way are approximate and will usually differ from the accepted value. Calculations from standard enthalpy of formation data are more reliable because those values are defined precisely.
平均键焓是多种化合物中的平均值，所以由它算出的结果只是近似值，通常与公认值有偏差。用标准生成焓数据计算更可靠，因为这些值有精确定义。
- Draw the cycle clearly and label every arrow with a value and a sign | 清楚画出循环图，给每个箭头标注数值与符号
- Reverse an arrow when you travel against its direction and change the sign of ΔH | 逆箭头方向走时要反向并改变 ΔH 的符号
- Multiply ΔH values by the number of moles in the equation | 按方程式中的摩尔数乘以相应的 ΔH 值
9. Equilibrium & the Equilibrium Constant Kc | 平衡与平衡常数
When a reversible reaction reaches equilibrium, the forward and reverse rates are equal and the concentrations of reactants and products remain constant. The equilibrium constant Kc expresses the position of that equilibrium at a given temperature.
可逆反应达到平衡时，正、逆反应速率相等，反应物与产物的浓度保持不变。平衡常数 Kc 表示在给定温度下该平衡所处的位置。
aA + bB ⇌ cC + dD Kc = [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ
- Square brackets mean equilibrium concentration in mol dm⁻³ | 方括号表示平衡浓度，单位 mol dm⁻³
- Pure solids and pure liquids never appear in the expression | 纯固体和纯液体不出现在表达式中
- The value of Kc depends only on temperature; changing concentration, pressure or adding a catalyst does not change Kc | Kc 只随温度改变；改变浓度、压强或
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