Mole Calculations and Reacting Masses | 摩尔计算与反应质量

📚 Mole Calculations and Reacting Masses | 摩尔计算与反应质量

Mole calculations are a core part of Edexcel IGCSE Science (Chemistry). This revision guide covers every type of calculation you need to master, from the mole concept to limiting reactants, with worked examples in the exact style of Edexcel exam questions.

摩尔计算是 Edexcel IGCSE 科学（化学）的核心内容。本复习指南讲解你需要掌握的全部计算类型，从摩尔概念到限量反应物，并配有完全符合 Edexcel 考试风格的例题。

1. The Mole Concept | 摩尔概念

One mole of any substance contains the same number of particles as 12 g of carbon-12. This number, 6.02 × 10²³, is called the Avogadro constant (L). The particles can be atoms, molecules, ions or electrons depending on the substance.

任何物质的一摩尔含有的粒子数与 12 g 碳-12 中含有的原子数相同。这个数目为 6.02 × 10²³，称为阿伏加德罗常数（L）。粒子可以是原子、分子、离子或电子，取决于物质种类。

number of particles = amount in mol × Avogadro constant
粒子数 = 物质的量（mol）× 阿伏加德罗常数

2. Molar Mass | 摩尔质量

The molar mass (M) of a substance is the mass of one mole of that substance, measured in g/mol. It is numerically equal to the relative atomic mass (Ar) or the relative formula mass (Mr), which you find by adding the Ar values of all atoms in the formula.

物质的摩尔质量（M）是指 1 mol 该物质的质量，单位为 g/mol。它在数值上等于相对原子质量（Ar）或相对分子质量（Mr），后者通过将化学式中所有原子的 Ar 值相加得到。

Example: Find the molar mass of sulfuric acid, H₂SO₄.

例题：求硫酸 H₂SO₄ 的摩尔质量。

M = (2 × 1) + 32 + (4 × 16) = 98 g/mol

3. Converting Mass, Moles and Particles | 质量、摩尔与粒子数的换算

These three quantities are linked by simple formulas. Always show your working and your units in the exam, because method marks are often awarded even for incorrect final answers.

这三个量由简单公式相互联系。考试中务必写出计算过程和单位，因为即使最终答案错误，步骤分也常常会被保留。

n = m / M (moles = mass ÷ molar mass)

n = m / M（物质的量 = 质量 ÷ 摩尔质量）
m = n × M (mass = moles × molar mass)

m = n × M（质量 = 物质的量 × 摩尔质量）

Worked example: Calculate the number of moles in 4.4 g of carbon dioxide, CO₂.

例题：计算 4.4 g 二氧化碳 CO₂ 的物质的量。

M(CO₂) = 12 + (2 × 16) = 44 g/mol
n = 4.4 ÷ 44 = 0.10 mol

The number of molecules = 0.10 × 6.02 × 10²³ = 6.02 × 10²² molecules.

分子数 = 0.10 × 6.02 × 10²³ = 6.02 × 10²² 个分子。

4. Empirical and Molecular Formula | 实验式与分子式

To find the empirical formula, divide each element’s mass (or percentage) by its relative atomic mass. Then divide every result by the smallest value to obtain a whole-number ratio.

求实验式的方法：将各元素的质量（或百分比）除以各自的相对原子质量，再用所有结果除以最小值，得到整数比。

Worked example: A compound contains 2.4 g of carbon and 0.8 g of hydrogen. Find its empirical formula.

例题：某化合物含 2.4 g 碳和 0.8 g 氢，求其实验式。

C: 2.4 ÷ 12 = 0.20
H: 0.8 ÷ 1 = 0.80
Ratio = 0.20 : 0.80 = 1 : 4, so the empirical formula is CH₄

The molecular formula is a whole-number multiple of the empirical formula. Given Mr = 16, the molecular formula is CH₄.

分子式是实验式的整数倍。若 Mr = 16，则分子式也是 CH₄。

5. Reacting Mass Calculations | 反应质量计算

Use a balanced symbol equation. Convert the known mass to moles, use the mole ratio from the equation, then convert the required moles back to mass.

使用配平的符号方程式。将已知质量换算成物质的量，利用方程式中的摩尔比，再将所需物质的量换算回质量。

Worked example: When 5.6 g of calcium oxide is formed by heating limestone, what mass of calcium carbonate was used? Equation: CaCO₃ → CaO + CO₂.

例题：加热石灰石生成 5.6 g 氧化钙时，消耗了多少质量的碳酸钙？方程式：CaCO₃ → CaO + CO₂。

n(CaO) = 5.6 ÷ 56 = 0.10 mol
Mole ratio CaCO₃ : CaO = 1 : 1
m(CaCO₃) = 0.10 × 100 = 10 g

6. Concentration of Solutions | 溶液浓度

Concentration in mol dm⁻³ is the amount of solute in moles divided by the volume in dm³. Remember that 1 dm³ = 1000 cm³, so convert volumes before calculating.

以 mol dm⁻³ 为单位的浓度等于溶质的物质的量除以体积（dm³）。注意 1 dm³ = 1000 cm³，计算前务必先将体积换算为 dm³。

c = n / V

Worked example: How many moles of sodium hydroxide are in 250 cm³ of a 0.40 mol dm⁻³ solution?

例题：250 cm³ 的 0.40 mol dm⁻³ 氢氧化钠溶液中含有多少摩尔 NaOH？

V = 250 ÷ 1000 = 0.250 dm³
n = c × V = 0.40 × 0.250 = 0.10 mol

7. Gas Volumes | 气体体积

At room temperature and pressure (r.t.p.), one mole of any gas occupies exactly 24 dm³. This value is called the molar volume of a gas.

在室温常压（r.t.p.）下，1 mol 任何气体的体积恰好为 24 dm³。这个值称为气体的摩尔体积。

gas volume (dm³) = amount (mol) × 24

Worked example: Calculate the volume of 0.20 mol of carbon dioxide at r.t.p.

例题：计算 0.20 mol 二氧化碳在室温常压下的体积。

V = 0.20 × 24 = 4.8 dm³

8. Limiting Reactants | 限量反应物

The limiting reactant is the substance that is completely used up first. It determines the maximum amount of product that can be formed; the other reactants are said to be in excess.

限量反应物是最先被完全消耗的物质，它决定了产物的最大生成量；其他反应物则称为过量。

Worked example

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Mole Calculations and Reacting Masses | 摩尔计算与反应质量