📚 Year 10 OCR Chemistry: Formula & Theorem Quick Reference Handbook | Year 10 OCR 化学公式定理速查手册
This quick reference handbook brings together all the essential formulas, equations and quantitative relationships you will meet in Year 10 OCR Chemistry. It is designed to help you revise calculations involving moles, concentrations, energy changes and reaction yields, giving you a solid foundation for the GCSE course.
本速查手册汇集了你在 Year 10 OCR 化学中将遇到的所有基本公式、方程式和定量关系。它旨在帮助你复习涉及摩尔、浓度、能量变化和反应产率的计算，为你打好 GCSE 课程的坚实基础。
1. Relative Atomic Mass (Aᵣ) and Relative Formula Mass (Mᵣ) | 相对原子质量 (Aᵣ) 与相对式量 (Mᵣ)
The relative atomic mass (Aᵣ) is the average mass of an atom of an element compared to 1/12 of the mass of a carbon‑12 atom, taking account of the abundance of its natural isotopes. The relative formula mass (Mᵣ) is the sum of the Aᵣ values of all the atoms in the formula unit of a compound. For molecular substances, Mᵣ is often called the relative molecular mass.
相对原子质量 (Aᵣ) 是元素的一个原子相对于一个碳‑12 原子质量的 1/12 的平均质量，已经考虑了其天然同位素的丰度。相对式量 (Mᵣ) 是化合物化学式中所有原子 Aᵣ 值的总和。对于分子物质，Mᵣ 通常也称为相对分子质量。
Calculating Mᵣ: add the relative atomic masses of every atom shown by the formula.
计算 Mᵣ: 将化学式所示的每个原子的相对原子质量相加。
Mᵣ = Σ (number of atoms × Aᵣ)
Example: Mᵣ of CaCO₃ = 40.1 + 12.0 + (3 × 16.0) = 100.1
示例: CaCO₃ 的 Mᵣ = 40.1 + 12.0 + (3 × 16.0) = 100.1
Example: Mᵣ of Mg(OH)₂ = 24.3 + (2 × 16.0) + (2 × 1.0) = 58.3
示例: Mg(OH)₂ 的 Mᵣ = 24.3 + (2 × 16.0) + (2 × 1.0) = 58.3
2. The Mole and Mass Calculations | 摩尔与质量计算
ro constant). The mass of one mole of a substance is its formula mass in grams. The key relationship linking moles, mass and relative formula mass is:
摩尔是物质的量的单位。一摩尔任何物质都含有 6.02 × 10²³ 个粒子（阿伏伽德罗常数）。一摩尔物质的质量就是其以克为单位的式量。联系摩尔、质量和相对式量的关键关系是：
n = m / Mᵣ
n = number of moles (mol)
m = mass (g)
Mᵣ = molar mass (g mol⁻¹), numerically equal to the relative formula mass
n = 摩尔数 (mol)
m = 质量 (g)
Mᵣ = 摩尔质量 (g mol⁻¹)，数值上等于相对式量
This equation can be rearranged to m = n × Mᵣ and Mᵣ = m / n. Always check that the mass unit is grams and the molar mass uses g mol⁻¹.
该公式可变形为 m = n × Mᵣ 和 Mᵣ = m / n。请始终确认质量单位是克，摩尔质量使用 g mol⁻¹。
- Example: How many moles are there in 8.0 g of NaOH? (Mᵣ = 40.0) n = 8.0 / 40.0 = 0.20 mol
- 示例: 8.0 g NaOH 有多少摩尔？(Mᵣ = 40.0) n = 8.0 / 40.0 = 0.20 mol
- Example: What is the mass of 0.50 mol of CO₂? (Mᵣ = 44.0) m = 0.50 × 44.0 = 22.0 g
- 示例: 0.50 mol CO₂ 的质量是多少？(Mᵣ = 44.0) m = 0.50 × 44.0 = 22.0 g
3. Moles and Gas Volumes | 摩尔与气体体积
At room temperature and pressure (RTP, approximately 20 °C and 1 atm), one mole of any gas occupies the same volume. In OCR GCSE Chemistry this molar gas volume is taken as:
在室温常压下 (RTP，约 20 °C 和 1 atm)，一摩尔任何气体占据相同的体积。在 OCR GCSE 化学中，该气体摩尔体积取为：
Vₘ = 24 dm³ mol⁻¹ (or 24000 cm³ mol⁻¹)
Therefore, the volume of a gas can be found directly from moles:
因此，气体的体积可以直接由摩尔数求得：
V (dm³) = n × 24
V (cm³) = n × 24000
m³) / 24 or n = V (cm³) / 24000. Remember: these relationships apply only to gases at RTP.
变形可得 n = V (dm³) / 24 或 n = V (cm³) / 24000。记住：这些关系仅适用于 RTP 下的气体。
- Example: What volume does 0.750 mol of O₂ occupy at RTP? V = 0.750 × 24 = 18.0 dm³
- 示例: 0.750 mol O₂ 在 RTP 下占据多大体积？V = 0.750 × 24 = 18.0 dm³
- Example: How many moles are in 4.80 dm³ of H₂ at RTP? n = 4.80 / 24 = 0.200 mol
- 示例: 4.80 dm³ H₂ 在 RTP 下有多少摩尔？n = 4.80 / 24 = 0.200 mol
4. Concentration of Solutions | 溶液的浓度
Concentration can be expressed in two main ways: either as molar concentration (mol dm⁻³) or as mass concentration (g dm⁻³). The mole‑based concentration formula is the most important for titration calculations.
浓度主要有两种表示方法：摩尔浓度 (mol dm⁻³) 或质量浓度 (g dm⁻³)。基于摩尔的浓度公式对滴定计算最重要。
c = n / V
c = concentration in mol dm⁻³
n = number of moles of solute
V = volume of solution in dm³
c = 浓度，单位为 mol dm⁻³
n = 溶质的摩尔数
V = 溶液的体积，单位为 dm³
If the volume is given in cm³, convert to dm³ by dividing by 1000. The same formula can be used with mass: mass concentration (g dm⁻³) = mass of solute (g) / volume (dm³).
如果体积以 cm³ 给出，先除以 1000 转化为 dm³。同样的公式也可用于质量：质量浓度 (g dm⁻³) = 溶质质量 (g) / 体积 (dm³)。
- .0) is dissolved in water to make 250 cm³ of solution. Find the concentration in mol dm⁻³.
0 / 40.0 = 0.100 mol; volume = 250 / 1000 = 0.250 dm³; c = 0.100 / 0.250 = 0.400 mol dm⁻³. - 示例: 将 4.00 g NaOH (Mᵣ = 40.0) 溶于水配成 250 cm³ 溶液。求摩尔浓度。
n(NaOH) = 4.00/40.0 = 0.100 mol; 体积 = 250/1000 = 0.250 dm³; c = 0.100/0.250 = 0.400 mol dm⁻³。
5. Atom Economy and Percentage Yield | 原子经济性与百分比产率
These two green chemistry quantities allow chemists to assess the efficiency of a reaction. Percentage yield measures how much product is actually obtained compared to the theoretical maximum. Atom economy measures how much of the starting materials ends up in the desired product.
这两个绿色化学量允许化学家评估反应的效率。百分比产率衡量实际获得产物与理论最大产量的比较。原子经济性衡量起始原料中有多少最终进入了目标产物。
Percentage yield = (actual yield / theoretical yield) × 100
Atom economy = (Mᵣ of desired product / Σ Mᵣ of all reactants) × 100
In atom economy calculations, use only the reactants that take part in the reaction (the balanced equation shows the mole ratios). Both values are given as percentages. High atom economy means less waste, which is more sustainable.
在原子经济性计算中，只使用参与反应的物质（配平方程式给出摩尔比）。两者都以百分数表示。原子经济性高意味着废弃物更少，更可持续。
- duct is 15.0 g but only 11.7 g is collected, percentage yield = (11.7 / 15.0) × 100 = 78.0%.
- 示例: 某产物理论产量为 15.0 g，但只收集到 11.7 g，百分比产率 = (11.7/15.0) × 100 = 78.0%。
- / [(2 × 55.8 + 3 × 16.0) + (3 × (12.0+16.0))] × 100 = 111.6 / (159.6 + 84.0) × 100 ≈ 45.8%.
- 示例: 对于反应 Fe₂O₃ + 3CO → 2Fe + 3CO₂，铁的原子经济性 = (2×55.8) / [(2×55.8+3×16.0)+(3×(12.0+16.0))] × 100 = 111.6/(159.6+84.0)×100 ≈ 45.8%。
6. Energy Changes and Enthalpy | 能量变化与焓变
In reactions carried out in solution, the heat energy change can be calculated using the specific heat capacity of water (or of very dilute solutions). The temperature change is measured, then the heat energy change q is found:
在溶液中进行的反应里，可利用水的比热容（或极稀溶液的比热容）计算热量变化。先测量温度变化，然后求算热量变化 q：
q = m c ΔT
q = heat energy change (J)
m = mass of solution or water (g)
c = specific heat capacity (for water, c = 4.18 J g⁻¹ °C⁻¹)
ΔT = temperature change (°C)
q = 热量变化 (J)
m = 溶液或水的质量 (g)
c = 比热容（对水而言，c = 4.18 J g⁻¹ °C⁻¹）
ΔT = 温度变化 (°C)
The enthalpy change (ΔH) per mole can then be calculated. For exothermic reactions heat is released, so ΔH is negative; for endothermic reactions ΔH is positive.
然后可计算出每摩尔的焓变 (ΔH)。对放热反应，热量释放，因此 ΔH 为负；对吸热反应，ΔH 为正。
ΔH = -q / n
where n is the number of moles of the limiting reactant or the specified substance. The negative sign ensures that an exothermic reaction (temperature rise, +q) gives a negative ΔH.
其中 n 是限制反应物或指定物质的摩尔数。负号确保放热反应（温度升高，+q）给出负的 ΔH。
- 3 J. If 0.0500 mol of fuel reacted, ΔH = -5643 / 0.0500 = -112860 J mol⁻¹ ≈ -113 kJ mol⁻¹.
- 示例: 100 cm³ 水（质量 100 g）由 21.0 °C 升至 34.5 °C，燃料燃烧对其加热。q = 100 × 4.18 × 13.5 = 5643 J。若燃烧了 0.0500 mol 燃料，ΔH = -5643/0.0500 = -112860 J mol⁻¹ ≈ -113 kJ mol⁻¹。
7. Rate of Reaction | 反应速率
The rate of a reaction tells us how quickly reactants are used up or products are formed. Average rate can be expressed as a change in some measurable quantity (mass, volume of gas, concentration, etc.) divided by the time taken.
反应速率告诉我们反应物消耗或产物生成的快慢。平均速率可以表示为某个可测量量（质量、气体体积、浓度等）的变化除以所用时间。
Average rate = change in measurable quantity / change in time
Common units are g s⁻¹, cm³ s⁻¹, mol dm⁻³ s⁻¹. For reactions that produce a gas, rate is often followed by measuring the volume of gas evolved per unit time.
常用单位有 g s⁻¹、cm³ s⁻¹、mol dm⁻³ s⁻¹。对产生气体的反应，常通过测量单位时间内释放的气体体积来跟踪速率。
- dissolves in 120 seconds. Average rate = 0.45 / 120 = 0.00375 g s⁻¹ or 3.75 × 10⁻³ g s⁻¹.
- 示例: 某反应中，0.45 g 镁在 120 秒内溶解。平均速率 = 0.45/120 = 0.00375 g s⁻¹ 或 3.75 × 10⁻³ g s⁻¹。
- ample: A reaction produces 96 cm³ of gas in 160 s. Average rate = 96 / 160 = 0.60 cm³ s⁻¹.
- 示例: 某反应在 160 s 内产生 96 cm³ 气体。平均速率 = 96/160 = 0.60 cm³ s⁻¹。
Using a graph of quantity vs time, the rate at any instant can be found from the gradient of the tangent at that time.
利用“量–时间”图像，任意时刻的瞬时速率可通过该点切线的斜率求得。
8. Mole Ratio from Balanced Equations | 化学方程式中的摩尔比
The coefficients in a balanced chemical equation show the ratio of the amounts of reactants and products in moles. This ratio is the foundation for all quantitative predictions in chemistry.
配平化学方程式中的化学计量数（系数）给出了反应物和产物之间的摩尔比。这个摩尔比是化学中所有定量预测的基础。
aA + bB → cC + dD
This means a moles of A react with b moles of B to form c moles of C and d moles of D. Once the moles of one substance are known, mole ratio can be used to calculate the moles of any other substance.
这意味着 a 摩尔 A 与 b 摩尔 B 反应生成 c 摩尔 C 和 d 摩尔 D。一旦知道某物质的摩尔数，就可以利用摩尔比计算出任何其他物质的摩尔数。
- ce 2 mol MgO, so the mole ratio is 1:1 between Mg and MgO. If we have 0.80 mol Mg, we can make 0.80 mol MgO.
- 示例: 2Mg + O₂ → 2MgO。2 mol Mg 生成 2 mol MgO，因此 Mg 与 MgO 的摩尔比为 1:1。若有 0.80 mol Mg，可制得 0.80 mol MgO。
- th 3 mol H₂ to give 2 mol NH₃. If 0.30 mol N₂ reacts, it produces 2 × 0.30 = 0.60 mol NH₃.
- 示例: N₂ + 3H₂ → 2NH₃。1 mol N₂ 与 3 mol H₂ 反应生成 2 mol NH₃。若 0.30 mol N₂ 反应，则生成 2 × 0.30 = 0.60 mol NH₃。
c = n/V or V = n × 24 to convert moles into mass, concentration or gas volume as required.
然后可根据需要，使用 n = m/Mᵣ、c = n/V 或 V = n × 24 将摩尔转化为质量、浓度或气体体积。
9. Isotopic Abundance and Relative Atomic Mass | 同位素丰度与相对原子质量
Most elements exist as a mixture of isotopes. The Aᵣ shown in the periodic table is a weighted average that reflects the natural abundances of these isotopes.
大多数元素以同位素混合物的形式存在。周期表中显示的 Aᵣ 是反映这些同位素天然丰度的加权平均值。
Aᵣ = Σ (isotopic mass × % abundance) / 100
decimals (fractions of 1), simply sum the products: Aᵣ = Σ (mass × fractional abundance).
如果丰度以小数（分数）给出，则直接求和乘积：Aᵣ = Σ（同位素质量 × 百分比丰度）。
- Example: Boron has two isotopes: ¹⁰B (mass 10.0, 20.0%) and ¹¹B (mass 11.0, 80.0%). Aᵣ = (10.0 × 20.0 + 11.0 × 80.0) / 100 = (200 + 880) / 100 = 1080 / 100 = 10.8.
- 示例: 硼有两种同位素：¹⁰B（质量 10.0，占 20.0%）和 ¹¹B（质量 11.0，占 80.0%）。Aᵣ = (10.0×20.0 + 11.0×80.0) / 100 = (200 + 880) / 100 = 1080/100 = 10.8。
- Example: Chlorine has ⁷⁵Cl (mass 35.0, 75.0%) and ³⁷Cl (mass 37.0, 25.0%). Aᵣ = (35.0 × 0.750) + (37.0 × 0.250) = 26.25 + 9.25 = 35.5.
- 示例: 氯有 ⁷⁵Cl（质量 35.0，丰度 75.0%）和 ³⁷Cl（质量 37.0，丰度 25.0%）。Aᵣ = (35.0×0.750)+(37.0×0.250)=26.25+9.25=35.5。
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