GCSE OCR Chemistry: Mole Calculations - Key Points | GCSE OCR 化学：摩尔计算考点精讲

📚 GCSE OCR Chemistry: Mole Calculations – Key Points | GCSE OCR 化学：摩尔计算考点精讲

The mole concept is the gateway to quantitative chemistry. It allows us to move easily between the number of particles, mass, and volume of substances. For OCR GCSE Chemistry, you must be confident in using moles to perform calculations involving reacting masses, gas volumes, solution concentrations, titrations, and yields. This revision guide walks you through each essential skill.

摩尔概念是定量化学的大门。它让我们能够轻松地在粒子的数目、质量和体积之间转换。对于 OCR GCSE 化学，你必须自信地运用摩尔进行反应质量、气体体积、溶液浓度、滴定和产率等计算。本复习指南带你逐一攻克每个关键技能。

1. What is a Mole? | 什么是摩尔？

A mole (symbol: mol) is the SI unit for ‘amount of substance’. It provides a bridge between the atomic scale and the laboratory scale. One mole of any substance contains exactly 6.02 × 10²³ discrete particles – atoms, molecules, ions, or electrons – just as a dozen means 12 items.

摩尔（符号：mol）是“物质的量”的 SI 单位。它在原子尺度和实验室尺度之间架起桥梁。1 摩尔任何物质恰好包含 6.02 × 10²³ 个独立粒子——原子、分子、离子或电子——就像一打代表 12 个物品。

The number 6.02 × 10²³ is called the Avogadro constant. Knowing this number allows chemists to count atoms by simply weighing a sample.

数字 6.02 × 10²³ 被称为阿伏伽德罗常数。知道这个常数，化学家就可以通过称量样品来计数原子。

On the OCR specification, you will often see the mole used to interpret chemical equations: coefficients in a balanced equation represent mole ratios, not directly masses.

在 OCR 考纲中，你常会看到摩尔被用于解释化学方程式：配平方程中的系数代表的是摩尔比，而非直接的质量比。

2. The Avogadro Constant | 阿伏伽德罗常数

The Avogadro constant has a numerical value of 6.02 × 10²³ and the unit mol&supmin;¹. It links the number of formula units to the amount in moles. For example, 1 mol of carbon atoms contains 6.02 × 10²³ carbon atoms; 1 mol of water molecules contains 6.02 × 10²³ H&sub2;O molecules.

阿伏伽德罗常数的数值是 6.02 × 10²³，单位是 mol&supmin;¹。它将粒子的数量与摩尔数联系起来。例如，1 mol 碳原子含有 6.02 × 10²³ 个碳原子；1 mol 水分子含有 6.02 × 10²³ 个 H&sub2;O 分子。

If you are given the number of particles in a sample, you can use the relationship: number of moles = number of particles / (6.02 × 10²³). This is a common GCSE calculation type, particularly when asked about electrons or ions formed during electrolysis.

如果给出样品中的粒子数，你可以使用关系式：摩尔数 = 粒子数 / (6.02 × 10²³)。这是 GCSE 中常见的计算类型，尤其在涉及电解中形成的电子或离子数时。

3. Molar Mass (M_r) and Relative Formula Mass | 摩尔质量与相对式量

Relative formula mass (M_r) is the sum of the relative atomic masses (A_r) of all atoms in a formula unit. It has no unit. When we express this quantity in grams per mole, we call it the molar mass. For example, the M_r of H&sub2;O is (2×1) + 16 = 18, so its molar mass is 18 g mol&supmin;¹.

相对式量 (M_r) 是化学式中所有原子相对原子质量 (A_r) 的总和。它没有单位。当我们以克每摩尔来表达这个量时，称其为摩尔质量。例如，H&sub2;O 的 M_r = (2×1) + 16 = 18，所以其摩尔质量是 18 g mol&supmin;¹。

For ionic compounds like CaCO&sub3;, the M_r = 40 + 12 + (3×16) = 100. For diatomic elements such as Cl&sub2;, you must remember to double the A_r of chlorine (35.5 × 2 = 71).

对于离子化合物如 CaCO&sub3;，M_r = 40 + 12 + (3×16) = 100。对于双原子分子如 Cl&sub2;，你必须记得将氯的 A_r 加倍 (35.5 × 2 = 71)。

Always use the A_r values given on your OCR data sheet. Common pitfalls include using wrong A_r values for chlorine (35.5 not 35) or forgetting to count all atoms in brackets, e.g. in Mg(OH)&sub2;.

务必使用 OCR 数据表中给出的 A_r 值。常见的易错点包括使用错误的氯原子量（应为 35.5 而非 35），或者忘记数括号内的所有原子，例如 Mg(OH)&sub2;。

4. Calculating Moles from Mass | 由质量计算摩尔数

The central equation is:

moles = mass / molar mass n = m / M

where mass is in grams (g) and molar mass in grams per mole (g mol&supmin;¹). You can rearrange this to mass = moles × molar mass.

核心公式是：

摩尔数 = 质量 / 摩尔质量 n = m / M

其中质量以克 (g) 为单位，摩尔质量以克每摩尔 (g mol&supmin;¹) 为单位。你可以将公式变形为 质量 = 摩尔数 × 摩尔质量。

Worked example: How many moles are in 30 g of calcium carbonate?
M_r of CaCO&sub3; = 100
n = 30 / 100 = 0.30 mol

例题：30 g 碳酸钙是多少摩尔？
CaCO&sub3; 的 M_r = 100
n = 30 / 100 = 0.30 mol

In OCR exams, you must show all working: write the formula, substitute the numbers, and give the final answer with the correct unit (mol). Marks are often awarded for correct use of the equation even if the arithmetic contains a slip.

在 OCR 考试中，你必须展示所有步骤：写出公式，代入数字，并给出带正确单位 (mol) 的最终答案。即使计算有误，正确使用公式通常也能得分。

5. Moles and Gas Volumes | 摩尔与气体体积

At room temperature and pressure (RTP), one mole of any gas occupies 24 dm³ (24 000 cm³). This is called the molar gas volume. The relationship is:

volume (dm³) = moles × 24

在常温常压 (RTP) 下，1 摩尔任何气体占据 24 dm³ (24 000 cm³)。这被称为气体摩尔体积。关系式为：

体积 (dm³) = 摩尔数 × 24

You must be careful with units. If a volume is given in cm³, convert it to dm³ by dividing by 1000 before using the formula. For example, 48 cm³ of CO&sub2; at RTP is (48/1000) / 24 = 0.0020 mol.

你必须注意单位。如果体积以 cm³ 给出，先除以 1000 换算为 dm³ 再代入公式。例如，RTP 下 48 cm³ CO&sub2; 的摩尔数为 (48/1000) / 24 = 0.0020 mol。

Questions may ask you to find the volume of a gas produced in a reaction. Use the balanced equation to find the moles of gas, then multiply by 24. For instance, if 0.050 mol of magnesium reacts with excess acid to produce H&sub2;, the volume of H&sub2; released at RTP is 0.050 × 24 = 1.2 dm³.

考题可能要求你找出反应中生成的气体体积。使用配平的方程式求出气体的摩尔数，然后乘以 24。例如，若 0.050 mol 镁与过量酸反应产生 H&sub2;，则 RTP 下释放的 H&sub2; 体积为 0.050 × 24 = 1.2 dm³。

6. Moles in Solution: Concentration | 溶液中的摩尔：浓度

The concentration of a solution tells you how many moles of solute are dissolved in 1 dm³ of solution. The key equation is:

concentration (mol dm&supmin;³) = moles / volume (dm³) c = n / V

溶液的浓度告诉你每 1 dm³ 溶液中含有多少摩尔溶质。核心方程为：

浓度 (mol dm&supmin;³) = 摩尔数 / 体积 (dm³) c = n / V

Again, volume must be in dm³. If you are given a volume in cm³, divide by 1000. For example, a solution containing 0.10 mol of NaCl in 500 cm³ has a concentration of 0.10 / (500/1000) = 0.20 mol dm&supmin;³.

同样，体积必须以 dm³ 为单位。若给出的体积为 cm³，需除以 1000。例如，将 0.10 mol NaCl 溶解在 500 cm³ 水中，溶液的浓度为 0.10 / (500/1000) = 0.20 mol dm&supmin;³。

You can also calculate the mass of solute needed to make a solution of a given volume and concentration. First calculate n = c × V, then mass = n × M_r.

你也可以计算配制一定体积和浓度的溶液所需的溶质质量。首先计算 n = c × V，然后质量 = n × M_r。

7. Reacting Mass Calculations | 反应质量计算

Reacting mass questions ask you to find the mass of a product formed from a given mass of reactant, or vice versa. Follow these steps:

1. Write the balanced chemical equation.

1. 写出配平的化学方程式。
2. Convert the given mass to moles using n = m / M.

2. 利用 n = m / M 将已知质量转化为摩尔数。
3. Use the mole ratio from the equation to find moles of the unknown substance.

3. 利用方程式中的摩尔比求出未知物的摩尔数。
4. Convert moles back to mass using m = n × M.

4. 利用 m = n × M 将摩尔数转回质量。

Example: What mass of CO&sub2; is produced when 50 g of CaCO&sub3; is completely decomposed?
CaCO&sub3; → CaO + CO&sub2;
Moles of CaCO&sub3; = 50 / 100 = 0.50 mol
Mole ratio CaCO&sub3; : CO&sub2; = 1

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GCSE OCR Chemistry: Mole Calculations – Key Points | GCSE OCR 化学：摩尔计算考点精讲