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AQA A-Level Chemistry 1.2 Amount of Substance | AQA A-Level 化学 1.2 物质的量

2026-08-21 19:35 · 本文部分内容由AI辅助创作

📚 AQA A-Level Chemistry 1.2 Amount of Substance | AQA A-Level 化学 1.2 物质的量

Amount of substance is a central quantitative idea in AQA A-Level Chemistry. This topic links the microscopic scale of atoms and molecules to measurable laboratory quantities. You will use the mole to convert between mass, concentration, gas volume and particle number, and to analyse chemical equations. Mastery of these calculations is essential for practical questions, synoptic papers and university preparation.

物质的量是 AQA A-Level 化学中核心的定量概念。它把原子和分子的微观尺度与实验室可测量的量联系起来。你将使用摩尔在质量、浓度、气体体积和粒子数之间进行换算,并分析化学方程式。掌握这些计算对实验题、综合题和大学预备都至关重要。


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

that contains exactly 6.022 × 10²³ particles. This number is called the Avogadro constant.

在 AQA 化学中,物质的量以摩尔(mol)为单位。1 摩尔恰好含有 6.022 × 10²³ 个粒子。这个数称为阿伏伽德罗常数。

tains 6.022 × 10²³ H₂O molecules, while 1 mol of NaCl contains 6.022 × 10²³ formula units.

粒子可以是原子、分子、离子、电子或化学式单元。例如,1 mol H₂O 分子含有 6.022 × 10²³ 个 H₂O 分子,而 1 mol NaCl 含有 6.022 × 10²³ 个化学式单元。

To convert between the number of particles N and amount n, use:

要在粒子数 N 和物质的量 n 之间换算,使用:

N = n × Nₐ

the number of atoms in 0.250 mol of magnesium is 0.250 × 6.022 × 10²³ = 1.51 × 10²³ atoms.

例如,0.250 mol 镁中的原子数为 0.250 × 6.022 × 10²³ = 1.51 × 10²³ 个原子。


2. Molar Mass and Mass-to-Mole Conversion | 摩尔质量与质量-物质的量换算

Molar mass, symbol M, is the mass per mole of a substance. It has units g mol⁻¹ and is numerically equal to the relative atomic mass Aᵣ or relative formula mass Mᵣ.

摩尔质量,符号 M,是物质每摩尔的质量。单位为 g mol⁻¹,数值上等于相对原子质量 Aᵣ 或相对化学式质量 Mᵣ。

The key equation connecting mass, molar mass and amount is:

连接质量、摩尔质量和物质的量的关键公式为:

n = m ÷ M

of CO₂ in 4.40 g. The Mᵣ of CO₂ is 12.0 + 2 × 16.0 = 44.0, so n = 4.40 ÷ 44.0 = 0.100 mol.

例如,计算 4.40 g CO₂ 的物质的量。CO₂ 的 Mᵣ 为 12.0 + 2 × 16.0 = 44.0,因此 n = 4.40 ÷ 44.0 = 0.100 mol。

or 0.150 mol of CaCO₃, Mᵣ = 40.1 + 12.0 + 3 × 16.0 = 100.1, so m = 0.150 × 100.1 = 15.0 g.

重新整理公式可计算质量:m = n × M。对于 0.150 mol CaCO₃,Mᵣ = 40.1 + 12.0 + 3 × 16.0 = 100.1,因此 m = 0.150 × 100.1 = 15.0 g。


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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 gives the actual number of atoms of each element in one molecule.

经验式给出化合物中各元素原子的最简整数比。分子式给出一个分子中各元素的实际原子数。

To determine an empirical formula from percentage composition, convert each percentage to mass in grams, divide by Aᵣ, then divide each answer by the smallest value. If necessary, multiply all ratios by the same integer to obtain whole numbers.

要从质量百分组成确定经验式,将每个百分数换算为以克为单位的质量,除以 Aᵣ,再除以最小值。如有必要,将所有比值乘以相同整数以获得整数。

Example: a compound contains 40.0% C, 6.7% H and 53.3% O by mass.

示例:某化合物含 40.0% C、6.7% H 和 53.3% O(质量分数)。

Element% by mass÷ Aᵣ÷ smallestRatio
C40.040.0 ÷ 12.0 = 3.333.33 ÷ 3.33 = 11
H6.7
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