Amount of Substance | 物质的量

📚 Amount of Substance | 物质的量

Welcome to the cornerstone of quantitative chemistry – the concept of amount of substance. In A-Level Chemistry, mastering the mole is the key to unlocking everything from reacting mass calculations to volumetric analysis and gas equations. This article walks you through every essential topic, linking definitions with worked examples so that you can tackle any stoichiometry problem with confidence.

欢迎来到定量化学的基石——物质的量的概念。在 A-Level 化学中，掌握摩尔是解决从反应质量计算到容量分析、气体方程等一系列问题的关键。本文带你逐一梳理每个核心主题，将定义与实例相结合，让你能自信应对任何化学计量难题。

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

The mole (symbol: mol) is the SI unit for amount of substance. Since 2019, one mole has been defined as containing exactly 6.02214076×10²³ elementary entities (atoms, molecules, ions, electrons, etc.). This number is called Avogadro’s constant, Nₐ.

摩尔（符号：mol）是物质的量的国际单位。自 2019 年起，一摩尔被定义为恰好包含 6.02214076×10²³ 个基本单元（原子、分子、离子、电子等）。这个数被称为阿伏伽德罗常数，记作 Nₐ。

In practice, we often use the approximation Nₐ = 6.02×10²³ mol⁻¹. The beauty of the mole is that it allows chemists to count particles by weighing. For example, one mole of carbon‑12 atoms has a mass of exactly 12 g.

实际使用中我们常取近似值 Nₐ = 6.02×10²³ mol⁻¹。摩尔的妙处在于它让化学家能够通过称量来计数微粒。例如，一摩尔碳‑12 原子的质量恰好为 12 g。

The relationship between number of particles (N), amount of substance (n), and Avogadro’s constant is given by:

粒子数（N）、物质的量（n）与阿伏伽德罗常数的关系为：

N = n × Nₐ

This simple equation underpins all quantitative chemistry and must be memorised.

这个简单的方程是所有定量化学的基础，必须牢记。

2. Molar Mass and the Mole | 摩尔质量与摩尔

Molar mass (M) is the mass of one mole of a substance, expressed in grams per mole (g mol⁻¹). It is numerically equal to the relative atomic mass (Aᵣ) or relative formula mass (Mᵣ) of the substance.

摩尔质量（M）是一摩尔物质的质量，单位为克每摩尔（g mol⁻¹）。它在数值上等于该物质的相对原子质量（Aᵣ）或相对式量（Mᵣ）。

Substance	Formula	Mᵣ	Molar mass / g mol⁻¹
Water	H₂O	18.0	18.0
Sodium chloride	NaCl	58.5	58.5
Sulfuric acid	H₂SO₄	98.1	98.1

The core equation linking mass (m), molar mass (M), and amount (n) is:

连接质量（m）、摩尔质量（M）和物质的量（n）的核心方程是：

n = m ÷ M

Be careful with units: mass must be in grams. If a problem gives mass in kilograms or milligrams, convert to grams first. Always show your working, especially in structured exam questions, to avoid losing marks on unit errors.

要注意单位：质量必须以克为单位。如果题目给出的质量是千克或毫克，要先换算成克。在结构化考题中，务必展示运算过程，避免因单位错误而失分。

3. Empirical and Molecular Formulae | 经验式与分子式

The empirical formula gives the simplest whole‑number ratio of atoms of each element in a compound. The molecular formula shows the actual number of atoms of each element in one molecule.

经验式表示化合物中各元素原子最简整数比。分子式则表示一个分子中各元素原子的实际数目。

To determine an empirical formula from experimental data, convert the mass (or percentage) of each element into moles, then divide by the smallest number of moles to obtain a simple ratio. For example, a compound containing 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass gives the following:

由实验数据确定经验式时，先将各元素的质量（或百分比）转化为物质的量，再除以最小的物质的量，得到简单整数比。例如，某化合物中碳、氢、氧的质量分数分别为 40.0%、6.7% 和 53.3%，计算如下：

n(C) = 40.0 / 12.0 = 3.33 mol; n(H) = 6.7 / 1.0 = 6.7 mol; n(O) = 53.3 / 16.0 = 3.33 mol. Divide by 3.33 → ratio C:H:O = 1:2:1, so the empirical formula is CH₂O.

n(C) = 40.0 / 12.0 = 3.33 mol；n(H) = 6.7 / 1.0 = 6.7 mol；n(O) = 53.3 / 16.0 = 3.33 mol。除以 3.33 得到比例 C:H:O = 1:2:1，故经验式为 CH₂O。

The molecular formula is a whole‑number multiple of the empirical formula. If the relative molecular mass of the compound above is 180.0, then the multiplier is 180.0 / 30.0 = 6, giving the molecular formula C₆H₁₂O₆ (glucose).

分子式是经验式的整数倍。若上述化合物的相对分子质量为 180.0，则倍数为 180.0 / 30.0 = 6，分子式即为 C₆H₁₂O₆（葡萄糖）。

4. Reacting Mass Calculations | 反应质量计算

Stoichiometry allows us to relate masses of reactants and products in a chemical reaction. The balanced equation gives the mole ratio, which acts as a conversion factor. Follow this general method:

化学计量学使我们可以通过化学反应关联反应物与产物的质量。配平后的化学方程式给出了摩尔比，作为换算因子。一般采用以下步骤：

Write the balanced equation.
Convert known mass to moles using n = m / M.
Use the mole ratio to find moles of the target substance.
Convert moles back to mass with m = n × M.

写出配平的化学方程式。
用 n = m / M 将已知质量转化为物质的量。
利用摩尔比求出目标物质的物质的量。
用 m = n × M 将物质的量转化回质量。

Example: What mass of water is produced when 4.0 g of hydrogen gas reacts completely with oxygen? 2H₂ + O₂ → 2H₂O. n(H₂) = 4.0 / 2.0 = 2.0 mol. Mole ratio H₂ : H₂O = 1 : 1, so n(H₂O) = 2.0 mol. m(H₂O) = 2.0 × 18.0 = 36.0 g.

例题：4.0 g 氢气与氧气完全反应生成多少克水？2H₂ + O₂ → 2H₂O。n(H₂) = 4.0 / 2.0 = 2.0 mol。摩尔比 H₂ : H₂O = 1 : 1，故 n(H₂O) = 2.0 mol。m(H₂O) = 2.0 × 18.0 = 36.0 g。

Always check for limiting reactants if masses of two or more reactants are given. The reactant that runs out first defines the maximum amount of product that can form.

如果给出两种或多种反应物的质量，一定要检查限量反应物。首先耗尽的那一种反应物决定了产物的最大生成量。

5. Gas Volumes and the Molar Volume | 气体体积与摩尔体积

At room temperature and pressure (RTP, about 20 °C and 1 atm), one mole of any ideal gas occupies approximately 24.0 dm³ (or 24,000 cm³). This is called the molar gas volume, Vₘ.

在常温常压（RTP，约 20 °C、1 atm）下，一摩尔任何理想气体的体积约为 24.0 dm³（即 24,000 cm³）。这称为气体摩尔体积，记作 Vₘ。

Use the relationship between volume, amount, and molar volume:

体积、物质的量与摩尔体积的关系为：

n = V / Vₘ

Be mindful of unit conversions: 1 dm³ = 1000 cm³. When a reaction produces gases, the stoichiometric coefficients in the balanced equation give the volume ratio directly, provided all gases are measured at the same temperature and pressure (Avogadro’s law).

注意单位换算：1 dm³ = 1000 cm³。当反应中有气体生成时，只要所有气体都在相同温度和压力下测量，配平方程中的化学计量系数直接给出体积比（阿伏伽德罗定律）。

Example: N₂(g) + 3H₂(g) → 2NH₃(g). 10 dm³ of nitrogen reacts completely with hydrogen, how much ammonia is produced? Volume ratio N₂ : NH₃ = 1 : 2, so volume of NH₃ = 20 dm³.

例题：N₂(g) + 3H₂(g) → 2NH₃(g)。10 dm³ 氮气与氢气完全反应，生成多少氨气？体积比 N₂ : NH₃ = 1 : 2，所以 NH₃ 体积为 20 dm³。

6. The Ideal Gas Equation | 理想气体状态方程

When temperature and pressure vary, the molar volume is not constant. Instead, use the ideal gas equation:

当温度和压力变化时，摩尔体积不再恒定，应使用理想气体状态方程：

pV = nRT

Where p = pressure (Pa), V = volume (m³), n = amount (mol), R = 8.31 J K⁻¹ mol⁻¹, T = temperature (K). Convert all units carefully: 1 atm = 1.01325×10⁵ Pa; °C to K by adding 273.15.

其中 p = 压强（Pa），V = 体积（m³），n = 物质的量（mol），R = 8.31 J K⁻¹ mol⁻¹，T = 温度（K）。务必小心换算单位：1 atm = 1.01325×10⁵ Pa；℃ 转 K 需加 273.15。

This equation allows you to calculate molar mass of a volatile liquid, determine the formula of a gas, or find the amount of gas in a reaction where mass is hard to measure. Simply plug in the known values and solve for the unknown.

该方程可用于计算挥发性液体的摩尔质量、确定气体分子式，或在难以测量质量的反应中求气体的物质的量。只需代入已知量并求解未知数。

7. Concentrations and Titrations | 浓度与滴定

Solution stoichiometry revolves around concentration, defined as amount of solute per unit volume of solution:

溶液中的化学计量围绕浓度展开，定义为每单位体积溶液中溶质的物质的量：

c = n / V

Common units are mol dm⁻³. In titration experiments, the balanced equation gives the reacting ratio of acid and base (or other reactants). The key formula for titration calculations is:

常用单位是 mol dm⁻³。在滴定实验中，配平方程式给出了酸和碱（或其他反应物）的反应比。滴定计算的关键公式是：

nₐ = cₐ × Vₐ, and use mole ratio nₐ / n_b from the equation.

Example: 25.0 cm³ of 0.10 mol dm⁻³ HCl neutralises 20.0 cm³ of NaOH solution. Find the concentration of NaOH. HCl + NaOH → NaCl + H₂O. n(HCl) = 0.10 × 0.0250 = 0.00250 mol. Mole ratio 1:1, so n(NaOH) = 0.00250 mol. c(NaOH) = 0.00250 / 0.0200 = 0.125 mol dm⁻³.

例题：25.0 cm³ 0.10 mol dm⁻³ HCl 恰好中和 20.0 cm³ NaOH 溶液，求 NaOH 的浓度。HCl + NaOH → NaCl + H₂O。n(HCl) = 0.10 × 0.0250 = 0.00250 mol。摩尔比 1:1，所以 n(NaOH) = 0.00250 mol。c(NaOH) = 0.00250 / 0.0200 = 0.125 mol dm⁻³。

Volumes in cm³ must be converted to dm³ by dividing by 1000. Always present calculated concentrations to an appropriate number of significant figures, reflecting the precision of the apparatus used.

以 cm³ 为单位的体积必须除以 1000 转换为 dm³。计算出的浓度应根据所用仪器的精度保留合适的有效数字位数。

8. Limiting Reactant and Excess | 限量反应物与过量

In most reactions, one reactant is completely consumed before the others. This is the limiting reactant – it determines the theoretical yield. The other reactants are present in excess.

在大多数反应中，有一种反应物会先于其他反应物耗尽。这就是限量反应物——它决定了理论产量。其他反应物则过量存在。

To identify the limiting reactant, calculate the moles of each reactant. Then compare the available mole ratio with the stoichiometric ratio from the balanced equation. The reactant that gives the smaller amount of product is limiting.

辨别限量反应物时，先计算各反应物的物质的量，再比较实际摩尔比与配平方程中的化学计量比。生成产物较少的那个反应物即为限量反应物。

Example: 2.0 mol of hydrogen reacts with 1.0 mol of oxygen: 2H₂ + O₂ → 2H₂O. From the equation, 2 mol H₂ needs 1 mol O₂. Both are exactly matched, so neither is limiting – they react completely. If we had 1.5 mol H₂ and 1.0 mol O₂, H₂ would be limiting because it requires 0.75 mol O₂ and would run out first.

例题：2.0 mol 氢气与 1.0 mol 氧气反应：2H₂ + O₂ → 2H₂O。由方程式可知，2 mol H₂ 需要 1 mol O₂，两者恰好匹配，均非限量——完全反应。若为 1.5 mol H₂ 与 1.0 mol O₂，则 H₂ 是限量反应物，因为其完全反应仅需 0.75 mol O₂，会先耗尽。

Remember to use the limiting reactant for all yield calculations, as the excess reactant simply remains unreacted.

所有产率计算都要使用限量反应物，因为过量反应物会剩余未反应。

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

The percentage yield compares the actual yield (experimentally obtained) to the theoretical yield (calculated from stoichiometry):

产率（百分产率）将实际产量（实验获得）与理论产量（按化学计量计算）进行比较：

Percentage yield = (actual yield / theoretical yield) × 100%

A yield less than 100% may result from incomplete reaction, side reactions, or product losses during purification. Yields can never exceed 100% in a genuine chemical process.

产率低于 100% 可能是因为反应不完全、副反应发生或纯化过程中产物损失。在真正的化学过程中，产率绝不会超过 100%。

Atom economy, on the other hand, measures the efficiency of a reaction in incorporating reactant atoms into the desired product:

另一方面，原子经济性衡量反应将反应物原子纳入目标产物的效率：

Atom economy = (molar mass of desired product / sum of molar masses of all products) × 100%

Unlike yield, atom economy is a theoretical number derived from the balanced equation. It is crucial in green chemistry for designing processes that minimise waste. A high atom economy means fewer by‑products and a more sustainable reaction.

与产率不同，原子经济性是由配平方程式得出的理论数值。它在绿色化学中至关重要，用于设计减少废弃物的工艺。原子经济性高意味着副产物更少、反应更可持续。

10. Water of Crystallisation | 结晶水

Many ionic compounds crystallise from aqueous solution with a fixed number of water molecules incorporated into the crystal lattice. These are called hydrates, e.g. CuSO₄·5H₂O, Na₂CO₃·10H₂O.

许多离子化合物从水溶液中结晶时，会在晶格中结合固定数目的水分子。这些称为水合物，如 CuSO₄·5H₂O，Na₂CO₃·10H₂O。

To determine the number of water molecules (x) in a hydrated salt, you can heat a known mass to drive off the water and weigh the anhydrous residue. From the mass loss, calculate the amount of water removed and the amount of anhydrous salt remaining. The value of x is the ratio n(H₂O) : n(anhydrous salt).

要测定水合盐中水分子数目（x），可加热已知质量使其失去结晶水，称量所得无水残留物的质量。根据质量损失可计算出失去的水的物质的量及剩余无水盐的物质的量。x 值即为 n(H₂O) : n(无水盐) 的比值。

Example: 2.50 g of hydrated magnesium sulfate (MgSO₄·xH₂O) is heated to constant mass, leaving 1.22 g of anhydrous MgSO₄. Mass of water = 2.50 – 1.22 = 1.28 g. n(MgSO₄) = 1.22 / 120.4 = 0.0101 mol; n(H₂O) = 1.28 / 18.0 = 0.0711 mol. Ratio ≈ 7, so x = 7, giving MgSO₄·7H₂O.

例题：2.50 g 水合硫酸镁（MgSO₄·xH₂O）加热至恒重，剩余 1.22 g 无水 MgSO₄。水的质量 = 2.50 – 1.22 = 1.28 g。n(MgSO₄) = 1.22 / 120.4 = 0.0101 mol；n(H₂O) = 1.28 / 18.0 = 0.0711 mol。比值约为 7，故 x = 7，得到 MgSO₄·7H₂O。

11. Back Titration and More Complex Stoichiometry | 返滴定与复杂化学计量

In a back titration, an excess of a standard reagent is added to the analyte, and the unreacted excess is then titrated. This approach is used when the reaction is slow, when the endpoint is difficult to detect, or when the analyte is impure or insoluble.

在返滴定中，先将过量标准试剂加入待测物中，然后再滴定未反应的过量部分。当反应较慢、终点难以判断，或待测物不纯、难溶时，常采用此方法。

The steps are: (1) React the sample with a known amount of reagent A. (2) Titrate the remaining unreacted A with reagent B. (3) Calculate the amount of A that reacted with the sample by subtraction. (4) Use stoichiometry to find the amount or purity of the analyte.

步骤为：(1) 用已知量的试剂 A 与样品反应。(2) 用试剂 B 滴定剩余的未反应 A。(3) 通过减法算出与样品反应的 A 的量。(4) 根据化学计量求出待测物的量或纯度。

Example: A sample of impure limestone (CaCO₃) is treated with 50.0 cm³ of 1.00 mol dm⁻³ HCl. After reaction, the excess HCl requires 25.0 cm³ of 1.00 mol dm⁻³ NaOH for neutralisation. n(HCl) initial = 0.0500 mol; n(NaOH) = 0.0250 mol → excess HCl = 0.0250 mol. Therefore HCl that reacted with CaCO₃ = 0.0500 – 0.0250 = 0.0250 mol. From CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O, n(CaCO₃) = 0.0125 mol, mass of pure CaCO₃ = 0.0125 × 100.1 = 1.25 g. This is then used to calculate purity by comparing with the original sample mass.

例题：一个不纯的石灰石（CaCO₃）样品用 50.0 cm³ 1.00 mol dm⁻³ HCl 处理。反应后，过量 HCl 需 25.0 cm³ 1.00 mol dm⁻³ NaOH 中和。初始 n(HCl) = 0.0500 mol；n(NaOH) = 0.0250 mol → 过量 HCl = 0.0250 mol。故与 CaCO₃ 反应的 HCl = 0.0500 – 0.0250 = 0.0250 mol。由 CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O 得 n(CaCO₃) = 0.0125 mol，纯 CaCO₃ 质量 = 0.0125 × 100.1 = 1.25 g。再与原始样品质量比较即可算出纯度。

All the calculations you need for A‑Level stoichiometry arise from the simple law of conservation of mass and the mole concept. Regular practice with multi‑step problems, unit conversions, and careful reading of the question will make these skills second nature.

A‑Level 化学计量中的所有计算都源于质量守恒定律和摩尔概念。多多练习多步问题、注意单位换算并仔细审题，这些技能将变得得心应手。

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Amount of Substance | 物质的量