Amount of Substance | 物质的量

📚 Amount of Substance | 物质的量

The concept of “amount of substance” is one of the most fundamental ideas in A-Level chemistry. It provides the vital bridge between the microscopic world of atoms and molecules and the macroscopic world of laboratory measurements. This revision guide covers all the essential definitions, equations and worked calculations you need for AQA A-Level Chemistry.

“物质的量”这一概念是A-Level化学中最基础的思想之一。它架起了原子、分子的微观世界与实验室测量的宏观世界之间的重要桥梁。本复习指南涵盖了AQA A-Level化学所需的全部关键定义、方程和计算实例。

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

The mole is the SI unit for amount of substance, abbreviated as “mol”. One mole of any substance contains exactly 6.022 × 10²³ particles (atoms, molecules, ions, or electrons). This number is known as the Avogadro constant, Nₐ, and has units of mol⁻¹. The mole allows chemists to count particles by weighing them, since individual atoms are far too small to count directly.

摩尔是物质的量的SI单位，缩写为“mol”。任何物质的一摩尔恰好包含6.022 × 10²³个粒子（原子、分子、离子或电子）。这个数目称为阿伏伽德罗常数，用Nₐ表示，单位为mol⁻¹。由于单个原子太小而无法直接计数，摩尔使化学家能够通过称量来“数”粒子。

The number of particles, N, is related to the amount of substance, n, by the following equation:

粒子数N与物质的量n的关系通过以下方程表示：

N = n × Nₐ

N = number of particles (no units)
n = amount of substance (mol)
Nₐ = 6.022 × 10²³ mol⁻¹

N = 粒子数（无单位）
n = 物质的量（mol）
Nₐ = 6.022 × 10²³ mol⁻¹

For example, 0.50 mol of carbon dioxide contains 0.50 × 6.022 × 10²³ = 3.01 × 10²³ molecules of CO₂. Each molecule contains three atoms, so there are 9.03 × 10²³ atoms in total.

例如，0.50 mol二氧化碳含有0.50 × 6.022 × 10²³ = 3.01 × 10²³个CO₂分子。每个分子含有三个原子，因此总共有9.03 × 10²³个原子。

2. Molar Mass | 摩尔质量

Molar mass (M) is the mass of one mole of a substance, expressed in grams per mole (g mol⁻¹). For an element, the molar mass equals its relative atomic mass in grams; for a compound, it equals the sum of the relative atomic masses of all atoms in the formula. For example, the molar mass of H₂SO₄ is (2 × 1.0) + 32.1 + (4 × 16.0) = 98.1 g mol⁻¹.

摩尔质量（M）是一摩尔物质的质量，单位为克每摩尔（g mol⁻¹）。对于元素，摩尔质量等于以其克数表示相对原子质量；对于化合物，它等于化学式中所有原子的相对原子质量之和。例如，H₂SO₄的摩尔质量为(2 × 1.0) + 32.1 + (4 × 16.0) = 98.1 g mol⁻¹。

The relationship between mass, molar mass and amount of substance is given by:

质量、摩尔质量与物质的量之间的关系为：

n = m / M

where m is the mass in grams (g) and M is the molar mass (g mol⁻¹), giving n in moles (mol).

其中m为质量（克，g），M为摩尔质量（g mol⁻¹），计算得到的n为物质的量（mol）。

Substance	Formula	Molar mass (g mol⁻¹)
Sodium chloride	NaCl	23.0 + 35.5 = 58.5
Calcium carbonate	CaCO₃	40.1 + 12.0 + (3 × 16.0) = 100.1
Ammonium sulfate	(NH₄)₂SO₄	(2 × 14.0) + (8 × 1.0) + 32.1 + (4 × 16.0) = 132.1

3. The Ideal Gas Equation | 理想气体方程

Gases can be described by the ideal gas equation, which links pressure, volume, temperature and the amount of substance. This equation applies to ideal gases, which are assumed to have negligible particle volume and no intermolecular forces. Real gases approximate this behaviour well at low pressure and high temperature.

气体可以用理想气体方程描述，该方程将压强、体积、温度和物质的量联系起来。该方程适用于理想气体，理想气体被假设为粒子体积可忽略且无分子间作用力。真实气体在低压和高温下能很好地近似这种行为。

pV = nRT

p = pressure (Pa)
V = volume (m³)
n = amount of substance (mol)
R = molar gas constant = 8.31 J K⁻¹ mol⁻¹
T = temperature (K)

p = 压强（Pa）
V = 体积（m³）
n = 物质的量（mol）
R = 摩尔气体常数 = 8.31 J K⁻¹ mol⁻¹
T = 温度（K）

When using this equation, you must convert all quantities to the correct SI units. A common trap in exams is using cm³ or dm³ for volume instead of m³: 1 m³ = 1000 dm³ = 1 000 000 cm³. Temperature in °C must be converted to kelvin by adding 273.15.

使用此方程时，必须将所有量转换为正确的SI单位。考试中常见的陷阱是用cm³或dm³作为体积单位而不用m³：1 m³ = 1000 dm³ = 1 000 000 cm³。摄氏温度必须加上273.15转换为开尔文。

4. Gas Volumes at Room Temperature and Pressure | 室温常压下的气体体积

At room temperature and pressure (RTP, defined as 298 K and 1 atm ≈ 101 kPa), one mole of any ideal gas occupies a volume of 24.0 dm³. This is a convenient shortcut for gas calculations in A-Level questions, as you can rearrange n = V / 24.0 to find the amount of gas directly when conditions are at RTP.

在室温常压（RTP，定义为298 K和1 atm ≈ 101 kPa）下，任何理想气体的摩尔体积为24.0 dm³。这是A-Level题目中气体计算的便捷捷径，因为当条件为RTP时，可以直接使用n = V / 24.0求出气体的物质的量。

n = V / 24.0 dm³ mol⁻¹

Worked Example: What volume, in dm³, is occupied by 4.40 g of carbon dioxide at RTP? The molar mass of CO₂ is 44.0 g mol⁻¹. First, n(CO₂) = 4.40 / 44.0 = 0.100 mol. Then V = n × 24.0 = 0.100 × 24.0 = 2.40 dm³.

例题：在RTP下，4.40 g二氧化碳占据的体积为多少dm³？CO₂的摩尔质量为44.0 g mol⁻¹。首先，n(CO₂) = 4.40 / 44.0 = 0.100 mol。然后V = n × 24.0 = 0.100 × 24.0 = 2.40 dm³。

5. Concentration of Solutions | 溶液的浓度

Concentration is defined as the amount of solute dissolved in a given volume of solution. The unit of concentration in A-Level chemistry is mol dm⁻³ (often written as mol/L). A 1 mol dm⁻³ solution contains one mole of solute in every cubic decimetre (1 dm³ = 1000 cm³) of solution.

浓度定义为溶质溶解在给定体积溶液中的物质的量。A-Level化学中浓度的单位是mol dm⁻³（常写作mol/L）。1 mol dm⁻³的溶液在每立方分米（1 dm³ = 1000 cm³）溶液中含有溶质一摩尔。

c = n / V

where c is concentration (mol dm⁻³), n is amount (mol) and V is volume (dm³). If the volume is given in cm³, the equation becomes n = c × V / 1000.

其中c为浓度（mol dm⁻³），n为物质的量（mol），V为体积（dm³）。如果体积以cm³给出，方程变为n = c × V / 1000。

Worked Example: Calculate the amount of NaOH in 25.0 cm³ of a 0.150 mol dm⁻³ solution. Using n = c × V / 1000 = 0.150 × 25.0 / 1000 = 3.75 × 10⁻³ mol.

例题：计算25.0 cm³的0.150 mol dm⁻³ NaOH溶液中所含NaOH的物质的量。使用n = c × V / 1000 = 0.150 × 25.0 / 1000 = 3.75 × 10⁻³ mol。

6. 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. To find the empirical formula from experimental data, divide the mass of each element by its relative atomic mass to get the mole ratio, then divide through by the smallest number.

经验式给出了化合物中各元素原子的最简整数比。分子式则显示一个分子中每种元素的实际原子数目。要从实验数据求经验式，将每种元素的质量除以其相对原子质量得到摩尔比，然后同时除以最小的数。

Worked Example: A compound contains 1.20 g of carbon, 0.30 g of hydrogen and 1.60 g of oxygen. Find its empirical formula.

例题：一种化合物含碳1.20 g、氢0.30 g、氧1.60 g。求其经验式。

C: 1.20 / 12.0 = 0.100 mol
H: 0.30 / 1.0 = 0.300 mol
O: 1.60 / 16.0 = 0.100 mol

Dividing each by 0.100 gives a ratio of C : H : O = 1 : 3 : 1, so the empirical formula is CH₃O. If the molar mass of the compound is 62.0 g mol⁻¹, then since the empirical formula mass is 31.0 g mol⁻¹, the molecular formula is (CH₃O)₂ = C₂H₆O₂.

每一项除以0.100得到C : H : O = 1 : 3 : 1的比值，因此经验式为CH₃O。如果该化合物的摩尔质量为62.0 g mol⁻¹，经验式量为31.0 g mol⁻¹，则分子式为(CH₃O)₂ = C₂H₆O₂。

7. Balanced Equations and Stoichiometry | 配平方程式与化学计量学

A balanced chemical equation shows the stoichiometric relationships between reactants and products. The coefficients in a balanced equation represent the number of moles of each substance, not the number of molecules directly. This allows us to convert between quantities of different substances using mole ratios.

配平的化学方程式显示了反应物与产物之间的化学计量关系。配平方程中的系数代表各物质的摩尔数，而不仅仅是分子个数。这使我们能够利用摩尔比在不同物质之间进行换算。

Worked Example: Consider the combustion of propane: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O. How many moles of carbon dioxide are produced when 2.20 g of propane burns completely?

例题：考虑丙烷燃烧：C₃H₈ + 5O₂ → 3CO₂ + 4H₂O。当2.20 g丙烷完全燃烧时，产生多少摩尔二氧化碳？

Molar mass of C₃H₈ = (3 × 12.0) + (8 × 1.0) = 44.0 g mol⁻¹. Therefore n(C₃H₈) = 2.20 / 44.0 = 0.0500 mol. From the equation, 1 mol C₃H₈ produces 3 mol CO₂, so n(CO₂) = 3 × 0.0500 = 0.150 mol.

C₃H₈的摩尔质量 = (3 × 12.0) + (8 × 1.0) = 44.0 g mol⁻¹。因此n(C₃H₈) = 2.20 / 44.0 = 0.0500 mol。由方程式可知，1 mol C₃H₈产生3 mol CO₂，所以n(CO₂) = 3 × 0.0500 = 0.150 mol。

8. Limiting Reagents | 限制试剂

When two reactants are mixed, the limiting reagent is the reactant that is completely used up first, and it determines the maximum amount of product that can be formed. The other reactant is said to be in excess. To identify the limiting reagent, compare the actual mole ratio of the reactants with the stoichiometric ratio in the balanced equation.

当两种反应物混合时，限制试剂是首先完全消耗的反应物，它决定了可以生成的最大产物量。另一种反应物则称是过量的。要识别限制试剂，需要将反应物的实际摩尔比与配平方程中的化学计量比进行比较。

Worked Example: Zinc reacts with hydrochloric acid: Zn + 2HCl → ZnCl₂ + H₂. If 6.50 g of zinc is added to 0.100 dm³ of 2.00 mol dm⁻³ HCl, which is the limiting reagent?

例题：锌与盐酸反应：Zn + 2HCl → ZnCl₂ + H₂。如果将6.50 g锌加入0.100 dm³的2.00 mol dm⁻³ HCl中，哪个是限制试剂？

n(Zn) = 6.50 / 65.4 = 0.0994 mol. n(HCl) = 2.00 × 0.100 = 0.200 mol. From the equation, 1 mol Zn requires 2 mol HCl, so 0.0994 mol Zn requires 0.1988 mol HCl. Since 0.200 mol HCl is available (slightly more than 0.1988 mol), HCl is in excess and zinc is the limiting reagent.

n(Zn) = 6.50 / 65.4 = 0.0994 mol。n(HCl) = 2.00 × 0.100 = 0.200 mol。由方程式可知，1 mol Zn需要2 mol HCl，因此0.0994 mol Zn需要0.1988 mol HCl。由于实际有0.200 mol HCl（略多于0.1988 mol），HCl过量，锌是限制试剂。

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

Percentage yield measures the efficiency of a reaction in terms of how much product was actually obtained compared to the theoretical maximum. In AQA, you must be able to calculate percentage yield from experimental data and explain why yields are often less than 100% (e.g., side reactions, incomplete reactions, or loss during separation).

产率衡量反应在产物的实际获得量与理论最大量之间的效率。在AQA中，你必须能够根据实验数据计算产率，并解释产率常低于100%的原因（例如副反应、反应不完全或分离过程中的损失）。

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

Atom economy is a measure of how many of the atoms in the reactants end up in the desired product. It is calculated using the molar masses of the desired product and all reactants, without using actual masses of chemicals. A high atom economy means less waste is produced.

原子经济性是衡量反应物中有多少原子进入目标产物的指标。它使用目标产物和所有反应物的摩尔质量计算，不涉及实际质量。高原子经济性意味着产生更少的废物。

Atom economy = (molar mass of desired product / total molar mass of all reactants) × 100%

10. Titration Calculations | 滴定计算

Titration is a core practical skill in A-Level chemistry. In a titration, a solution of known concentration (the titrant) is added from a burette to a known volume of the analyte until the endpoint is reached. Titration calculations involve using n = c × V (with V in dm³) alongside the stoichiometric ratio from the balanced equation.

滴定是A-Level化学的核心实验技能。在滴定中，将已知浓度的溶液（滴定剂）从滴定管加入已知体积的待测溶液，直到达到终点。滴定计算使用n = c × V（V以dm³为单位），并结合配平方程中的化学计量比。

Worked Example: In a titration, 20.0 cm³ of sulfuric acid is neutralised by 25.0 cm³ of 0.100 mol dm⁻³ NaOH. The equation is H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Calculate the concentration of the sulfuric acid.

例题：在滴定中，20.0 cm³硫酸被25.0 cm³的0.100 mol dm⁻³ NaOH中和。方程式为H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O。计算硫酸的浓度。

n(NaOH) = 0.100 × 25.0 / 1000 = 2.50 × 10⁻³ mol. From the equation, 1 mol H₂SO₄ reacts with 2 mol NaOH, so n(H₂SO₄) = 2.50 × 10⁻³ / 2 = 1.25 × 10⁻³ mol. Therefore c(H₂SO₄) = n / V = 1.25 × 10⁻³ / (20.0 / 1000) = 0.0625 mol dm⁻³.

n(NaOH) = 0.100 × 25.0 / 1000 = 2.50 × 10⁻³ mol。由方程式可知，1 mol H₂SO₄与2 mol NaOH反应，因此n(H₂SO₄) = 2.50 × 10⁻³ / 2 = 1.25 × 10⁻³ mol。因此c(H₂SO₄) = n / V = 1.25 × 10⁻³ / (20.0 / 1000) = 0.0625 mol dm⁻³。

11. Water of Crystallisation | 结晶水

Many ionic compounds form hydrated salts containing water molecules in a fixed ratio, known as water of crystallisation. The formula of a hydrated salt is written, for example, as CuSO₄·5H₂O. In AQA calculations, you may be asked to determine the value of x in a hydrated salt formula by heating a known mass of the hydrated salt to drive off water and calculating the mole ratio.

许多离子化合物形成含有固定比例水分子的水合盐，这些水分子称为结晶水。水合盐的化学式写作如CuSO₄·5H₂O。在AQA计算题中，通过加热已知质量的水合盐以除去水分，并计算摩尔比，可能要求你确定水合盐化学式中的x值。

Worked Example: 2.50 g of hydrated sodium carbonate, Na₂CO₃·xH₂O, is heated until constant mass. The anhydrous salt remaining has a mass of 0.93 g. Find x.

例题：将2.50 g水合碳酸钠Na₂CO₃·xH₂O加热至恒重。剩余无水盐的质量为0.93 g。求x。

Mass of water lost = 2.50 − 0.93 = 1.57 g. n(Na₂CO₃) = 0.93 / 106.0 = 8.77 × 10⁻³ mol. n(H₂O) = 1.57 / 18.0 = 0.0872 mol. Ratio H₂O : Na₂CO₃ = 0.0872 / 8.77 × 10⁻³ = 9.94, which rounds to 10. Therefore x = 10, and the formula is Na₂CO₃·10H₂O.

失去水的质量 = 2.50 − 0.93 = 1.57 g。n(Na₂CO₃) = 0.93 / 106.0 = 8.77 × 10⁻³ mol。n(H₂O) = 1.57 / 18.0 = 0.0872 mol。H₂O与Na₂CO₃的摩尔比 = 0.0872 / 8.77 × 10⁻³ = 9.94，四舍五入为10。因此x = 10，化学式为Na₂CO₃·10H₂O。

12. Summary of Key Equations | 关键方程总结

The table below summarises the essential equations from this topic. You should be able to recall and rearrange each one confidently in exam conditions without a calculator guide.

下表总结了本主题的核心方程。在考试中，你应该能够自信地回忆并变换每一个方程，无需计算器指南。

Relationship	Equation	Units
Amount from mass	n = m / M	mol, g, g mol⁻¹
Number of particles	N = n × Nₐ	particles, mol, mol⁻¹
Ideal gas law	pV = nRT	Pa, m³, mol, J K⁻¹ mol⁻¹, K
Gas volume at RTP	n = V / 24.0	mol, dm³
Concentration	c = n / V	mol dm⁻³, mol, dm³

Mastering “Amount of Substance” is essential because these calculations appear throughout the AQA A-Level Chemistry specification: in physical chemistry, inorganic analysis and organic synthesis questions. Practice rearranging the equations in multiple forms, carry out the unit conversions carefully and always check that your final answer has sensible units.

掌握“物质的量”至关重要，因为这些计算贯穿AQA A-Level化学大纲的始终：无论是物理化学、无机分析还是有机合成题目中都会出现。练习以多种形式重排方程，仔细进行单位换算，并始终检查最终答案是否具有合理的单位。

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Amount of Substance | 物质的量