Edexcel A Level Chemistry Topic 5: Formulae, Equations and Amount of Substance | 爱德思A Level化学第5单元：公式、方程式与物质的量

📚 Edexcel A Level Chemistry Topic 5: Formulae, Equations and Amount of Substance | 爱德思A Level化学第5单元：公式、方程式与物质的量

Amount of substance is one of the most calculation-heavy areas in Edexcel A Level Chemistry. Topic 5 covers the mole, balanced equations, reacting masses, titrations and gas calculations. Mastering these skills is essential for both AS and A2 papers, as quantitative ideas appear in physical, inorganic and organic chemistry.

物质的量是爱德思A Level化学中计算最密集的板块之一。第5单元涵盖摩尔、配平方程式、反应质量、滴定和气体计算。掌握这些技能对AS和A2试卷都至关重要，因为定量思想会出现在物理化学、无机化学和有机化学中。

1. The Mole and Avogadro Constant | 摩尔与阿伏伽德罗常数

The mole is the SI unit for amount of substance. One mole contains exactly 6.02 × 10²³ particles, known as the Avogadro constant, L or Nₐ.

摩尔是物质的量的国际单位。1摩尔包含6.02 × 10²³个粒子，称为阿伏伽德罗常数，记作L或Nₐ。

n = m ÷ M

Here n is the amount of substance in mol, m is the mass in g, and M is the molar mass in g mol⁻¹. The molar mass is numerically equal to the relative formula mass.

其中n为物质的量（mol），m为质量（g），M为摩尔质量（g mol⁻¹）。摩尔质量在数值上等于相对式量。

2. Empirical and Molecular Formulae | 实验式与分子式

The empirical formula gives the simplest whole-number ratio of atoms in a compound, while the molecular formula gives the actual number of atoms of each element. To find an empirical formula, convert percentage composition to moles by dividing each mass percentage by the relative atomic mass, then divide all results by the smallest number of moles.

实验式给出化合物中原子的最简整数比，而分子式给出每种元素的原子的实际数目。求实验式时，将质量分数除以相对原子质量得到摩尔数，再将所有结果除以最小的摩尔数。

For example, a compound with 40.0% carbon, 6.7% hydrogen and 53.3% oxygen gives a mole ratio of 1 : 2 : 1, so the empirical formula is CH₂O. The molecular formula is a whole-number multiple of the empirical formula.

例如，含40.0%碳、6.7%氢和53.3%氧的化合物得到摩尔比为1 : 2 : 1，因此实验式为CH₂O。分子式是实验式的整数倍。

3. Balancing Chemical Equations | 化学方程式的配平

Balanced equations obey the law of conservation of mass and charge. Coefficients must be whole numbers arranged so that the number of atoms of each element is equal on both sides.

配平方程式遵守质量守恒和电荷守恒。系数必须为整数，使每种元素的原子数在两边相等。

For example, the combustion of methane is balanced as:

例如，甲烷的燃烧配平如下：

CH₄ + 2O₂ → CO₂ + 2H₂O

This shows one mole of methane reacts with two moles of oxygen to produce one mole of carbon dioxide and two moles of water.

这表示1摩尔甲烷与2摩尔氧气反应，生成1摩尔二氧化碳和2摩尔水。

4. Reacting Mass Calculations | 反应质量计算

Reacting masses are calculated using stoichiometric ratios from a balanced equation. First convert the given mass to moles, apply the mole ratio, then convert the resulting moles back to mass.

反应质量利用配平方程式中的化学计量比进行计算。先将已知质量换算为摩尔，应用摩尔比，再将所得摩尔换算回质量。

Using the combustion of methane, 16 g of CH₄ has 16 ÷ 16 = 1 mol. The CH₄ : CO₂ ratio is 1 : 1, so 1 mol CO₂ forms, which has a mass of 1 × 44 = 44 g.

以甲烷燃烧为例，16 g CH₄ 为16 ÷ 16 = 1 mol。CH₄ : CO₂ 摩尔比为1 : 1，因此生成1 mol CO₂，质量为1 × 44 = 44 g。

5. Limiting Reactants and Excess | 限制性反应物与过量

The limiting reactant is the one that is completely consumed first and therefore determines the maximum amount of product formed. The other reactants are present in excess.

限制性反应物是最先被完全消耗的反应物，因此决定产物的最大生成量。其他反应物过量存在。

For the reaction 2H₂ + O₂ → 2H₂O, if 4 mol of H₂ reacts with 1 mol of O₂, oxygen is limiting because 4 mol H₂ requires 2 mol O₂. Only 2 mol H₂ react, so 2 mol H₂ remain in excess.

对于反应 2H₂ + O₂ → 2H₂O，如果4 mol H₂ 与1 mol O₂ 反应，氧气是限制性反应物，因为4 mol H₂ 需要2 mol O₂。只有2 mol H₂ 反应，因此2 mol H₂ 过量。

6. Concentration and Molarity | 浓度与摩尔浓度

Concentration is the amount of solute dissolved in a given volume of solution. The standard unit is mol dm⁻³.

浓度是溶解在给定体积溶液中的溶质的量。标准单位为 mol dm⁻³。

c = n ÷ V

Here c is concentration in mol dm⁻³, n is amount in mol, and V is volume in dm³. To convert from cm³ to dm³, divide by 1000.

其中c为浓度（mol dm⁻³），n为物质的量（mol），V为体积（dm³）。将cm³换算为dm³需除以1000。

7. Titration Calculations | 滴定计算

Titrations use the reaction between a solution of known concentration and a solution of unknown concentration. The balanced equation gives the mole ratio between the two reactants.

滴定利用已知浓度溶液与未知浓度溶液之间的反应。配平方程式给出两种反应物之间的摩尔比。

For the neutralisation HCl + NaOH → NaCl + H₂O, n(HCl) = c(HCl) × V(HCl) and n(NaOH) = c(NaOH) × V(NaOH). Because the ratio is 1 : 1, the two amounts are equal, so the unknown concentration can be calculated.

对于中和反应 HCl + NaOH → NaCl + H₂O，n(HCl) = c(HCl) × V(HCl)，n(NaOH) = c(NaOH) × V(NaOH)。由于摩尔比为1 : 1，两者的物质的量相等，因此可计算未知浓度。

8. Ideal Gas Equation | 理想气体状态方程

The ideal gas equation links pressure, volume, temperature and amount of gas.

理想气体状态方程将气体的压强、体积、温度与物质的量联系起来。

pV = nRT

Here p is pressure in Pa, V is volume in m³, n is amount in mol, T is absolute temperature in K, and R is the gas constant, 8.31 J K⁻¹ mol⁻¹. At room temperature and pressure, one mole of gas occupies 24.0 dm³.

其中p为压强（Pa），V为体积（m³），n为物质的量（mol），T为绝对温度（K），R为气体常数 8.31 J K⁻¹ mol⁻¹。在室温常压下，1摩尔气体体积为24.0 dm³。

9. Percentage Yield and Atom Economy | 产率与原子经济性

Percentage yield shows how much product is obtained compared with the theoretical maximum.

产率表示实际得到的产物与理论最大产量的比值。

% yield = (actual yield ÷ theoretical yield) × 100%

Atom economy measures the efficiency of a reaction in incorporating reactant atoms into the desired product.

原子经济性衡量反应将反应物原子并入目标产物的效率。

% atom economy = (molar mass of desired product ÷ sum of molar masses of all reactants) × 100%

A higher atom economy means fewer waste by-products and a more sustainable process.

原子经济性越高，意味着副产物废物越少，过程越可持续。

10. Common Pitfalls in Amount of Substance Calculations | 物质的量计算常见误区

Always convert units to the standard SI units before substituting into equations. Volumes must be in dm³ for concentration calculations and in m³ for the ideal gas equation.

代入方程前务必将单位换算为国际标准单位。浓度计算中体积必须为dm³，理想气体方程中体积必须为m³。

Do not confuse RTP molar volume 24.0 dm³ with STP molar volume 22.4 dm³. Edexcel uses 24.0 dm³ at room temperature and pressure unless stated otherwise.

不要将室温常压下的摩尔体积24.0 dm³与标准状况下的22.4 dm³混淆。除非另有说明，爱德思使用室温常压下的24.0 dm³。

Balance the equation before using mole ratios, and never round numbers until the final step to avoid losing accuracy.

在使用摩尔比之前一定要配平方程式，并且不要在最后一步之前取整，以免损失准确性。

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Edexcel A Level Chemistry Topic 5: Formulae, Equations and Amount of Substance | 爱德思A Level化学第5单元：公式、方程式与物质的量