Edexcel A-Level Chemistry: Quantitative Chemistry Combined 110 | 爱德思 A-Level 化学：定量化学综合 110

📚 Edexcel A-Level Chemistry: Quantitative Chemistry Combined 110 | 爱德思 A-Level 化学：定量化学综合 110

Quantitative chemistry is the backbone of A-Level calculations. It links the submicroscopic world of atoms and molecules to measurable laboratory quantities such as mass, volume, concentration and gas pressure. Mastering these skills is essential for Edexcel practical-based questions and synoptic papers.

定量化学是 A-Level 计算的核心。它将原子与分子的微观世界与质量、体积、浓度和气体压力等可测量的实验室量联系起来。掌握这些技能对爱德思实验题和综合试卷至关重要。

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

The mole is the SI unit for amount of substance. One mole contains exactly 6.022 × 10²³ elementary particles, a number known as the Avogadro constant.

摩尔是国际单位制中物质的量的单位。1 摩尔恰好包含 6.022 × 10²³ 个基本单元，这个数称为阿伏伽德罗常数。

You can convert between number of particles N and amount n using N = nL, where L = 6.022 × 10²³ mol⁻¹. In most calculations, this relationship is used to find the number of ions, atoms or molecules produced or consumed.

可以使用 N = nL 在粒子数 N 与物质的量 n 之间转换，其中 L = 6.022 × 10²³ mol⁻¹。在大多数计算中，这个关系用于求生成或消耗的离子、原子或分子数目。

n = N ÷ L

The symbol n is measured in moles (mol), while molar mass M has units g mol⁻¹. The central equation for mass-mole conversions is n = m ÷ M.

符号 n 的单位是摩尔（mol），摩尔质量 M 的单位是 g mol⁻¹。质量与摩尔转换的核心公式是 n = m ÷ M。

n = m ÷ M

When using this equation, always give m in grams and M in g mol⁻¹. If the mass is given in kilograms or milligrams, convert it to grams before substituting.

使用此公式时，质量 m 必须以克为单位，摩尔质量 M 以 g mol⁻¹ 为单位。如果给出的质量是千克或毫克，代入前必须先换算成克。

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2. Empirical and Molecular Formulae | 实验式与分子式

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

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

To determine an empirical formula, convert the percentage or mass of each element into moles by dividing by its relative atomic mass, Aᵣ. Then divide each mole value by the smallest mole value to obtain the simplest ratio.

要确定实验式，先将各元素的质量或百分比除以相对原子质量 Aᵣ 得到摩尔数。然后将每个摩尔数除以最小的摩尔数，得到最简整数比。

Worked example: A compound contains 40.0% carbon, 6.70% hydrogen and 53.3% oxygen by mass. Moles of C = 40.0 ÷ 12.0 = 3.33, moles of H = 6.70 ÷ 1.00 = 6.70, moles of O = 53.3 ÷ 16.0 = 3.33. Dividing by 3.33 gives C₁H₂O₁, so the empirical formula is CH₂O.

示例：某化合物按质量含碳 40.0%、氢 6.70%、氧 53.3%。碳的摩尔数 = 40.0 ÷ 12.0 = 3.33，氢的摩尔数 = 6.70 ÷ 1.00 = 6.70，氧的摩尔数 = 53.3 ÷ 16.0 = 3.33。除以 3.33 得到 C₁H₂O₁，因此实验式为 CH₂O。

If the relative molecular mass Mᵣ of the compound is known, the molecular formula can be found. Divide Mᵣ by the mass of the empirical formula unit, then multiply the empirical formula by this factor.

如果已知化合物的相对分子质量 Mᵣ，就能求出分子式。将 Mᵣ 除以实验式单元的质量，再将实验式乘以该倍数。

For CH₂O, the empirical formula mass is 12.0 + 2.00 + 16.0 = 30.0. If Mᵣ = 180, the factor is 180 ÷ 30.0 = 6, giving the molecular formula C₆H₁₂O₆.

对于 CH₂O，实验式质量为 12.0 + 2.00 + 16.0 = 30.0。如果 Mᵣ = 180，倍数为 180 ÷ 30.0 = 6，因此分子式为 C₆H₁₂O₆。

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3. Reacting Mass Calculations | 反应质量计算

Reacting mass calculations connect the masses of reactants and products through the mole ratio in the balanced equation. The three steps are: convert mass to moles, use the mole ratio, then convert moles back to mass.

反应质量计算通过配平方程式中的摩尔比连接反应物和生成物的质量。三个步骤是：将质量转换为摩尔，使用摩尔比，再将摩尔转换回质量。

Worked example: Calculate the mass of magnesium oxide produced when 5.00 g of magnesium burns fully in oxygen. The equation is 2Mg + O₂ → 2MgO.

示例：计算 5.00 g 镁完全燃烧后生成的氧化镁质量。方程式为 2Mg + O₂ → 2MgO。

Moles of Mg = 5.00 ÷ 24.3 = 0.206 mol. The mole ratio Mg : MgO is 1 : 1, so moles of MgO = 0.206 mol. Molar mass of MgO = 24.3 + 16.0 = 40.3 g mol⁻¹. Mass of MgO = 0.206 × 40.3 = 8.30 g.

镁的摩尔数 = 5.00 ÷ 24.3 = 0.206 mol。Mg : MgO 的摩尔比为 1 : 1，因此 MgO 的摩尔数 = 0.206 mol。MgO 的摩尔质量 = 24.3 + 16.0 = 40.3 g mol⁻¹。MgO 的质量 = 0.206 × 40.3 = 8.30 g。

Always check that the equation is balanced before using mole ratios. If the ratio is not 1 : 1, scale the moles accordingly.

在使用摩尔比之前务必检查方程式是否配平。如果比例不是 1 : 1，则相应地换算摩尔数。

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4. Concentration and Molarity | 浓度与摩尔浓度

Concentration measures the amount of solute dissolved in a given volume of solution. In A-Level chemistry, concentration is usually expressed in mol dm⁻³, often written as M.

浓度表示溶解在一定体积溶液中的溶质的量。在 A-Level 化学中，浓度通常以 mol dm⁻³ 表示，常写作 M。

The key equation is c = n ÷ V, where c is concentration, n is amount in mol, and V is volume in dm³.

关键公式是 c = n ÷ V，其中 c 为浓度，n 为物质的量（mol），V 为体积（dm³）。

c = n ÷ V

Many questions give volumes in cm³. Convert to dm³ by dividing by 1000 before substituting into the equation.

很多题目给出的体积单位是 cm³。代入公式前先除以 1000 转换成 dm³。

Worked example: Calculate the concentration of a solution made by dissolving 0.250 mol of sodium hydroxide in 500 cm³ of water. V = 500 ÷ 1000 = 0.500 dm³, so c = 0.250 ÷ 0.500 = 0.500 mol dm⁻³.

示例：计算将 0.250 mol 氢氧化钠溶解在 500 cm³ 水中所得溶液的浓度。V = 500 ÷ 1000 = 0.500 dm³，因此 c = 0.250 ÷ 0.500 = 0.500 mol dm⁻³。

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5. Standard Solutions and Dilutions | 标准溶液与稀释

A standard solution is a solution whose concentration is accurately known. It is usually prepared by dissolving a known mass of a primary standard in deionised water and making up to a known volume in a volumetric flask.

标准溶液是浓度精确已知的溶液。通常将已知质量的一级标准物溶解在去离子水中，并在容量瓶中定容到已知体积来配制。

The solute is weighed, transferred to the volumetric flask, rinsed with water, dissolved fully, and then filled to the graduation mark. The flask is inverted several times to mix thoroughly.

称量溶质后，转移至容量瓶中，用水冲洗，完全溶解，然后定容至刻度线。反复倒转容量瓶使其充分混匀。

Dilution calculations use the relationship c₁V₁ = c₂V₂, where 1 refers to the original solution and 2 refers to the diluted solution.

稀释计算使用关系式 c₁V₁ = c₂V₂，其中下标 1 表示原溶液，下标 2 表示稀释后的溶液。

c₁V₁ = c₂V₂

Worked example: What volume of 2.00 mol dm⁻³ hydrochloric acid is needed to prepare 250 cm³ of 0.100 mol dm⁻³ acid? V₁ = (c₂V₂) ÷ c₁ = (0.100 × 0.250) ÷ 2.00 = 0.0125 dm³ = 12.5 cm³.

示例：配制 250 cm³ 的 0.100 mol dm⁻³ 盐酸需要多少体积的 2.00 mol dm⁻³ 盐酸？V₁ = (c₂V₂) ÷ c₁ = (0.100 × 0.250) ÷ 2.00 = 0.0125 dm³ = 12.5 cm³。

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6. Titration Calculations | 滴定计算

Acid-base titrations are used to find the concentration of an unknown solution. The endpoint is reached when the indicator changes colour, showing that the acid and base have reacted in the exact mole ratio given by the equation.

酸碱滴定用于测定未知溶液的浓度。当指示剂变色时达到滴定终点，表明酸和碱已按方程式给出的精确摩尔比完全反应。

For a 1 : 1 reaction such as HCl + NaOH → NaCl + H₂O, the moles of acid equal the moles of base at the endpoint: cₐVₐ = c_bV_b.

对于 HCl + NaOH → NaCl + H₂O 这样的 1 : 1 反应，终点时酸的摩尔数等于碱的摩尔数：cₐVₐ = c_bV_b。

Worked example: 25.0 cm³ of 0.100 mol dm⁻³ NaOH is neutralised by 23.5 cm³ of hydrochloric acid. Calculate the concentration of the acid.

示例：25.0 cm³ 的 0.100 mol dm⁻³ NaOH 被 23.5 cm³ 盐酸中和。计算盐酸的浓度。

Moles of NaOH = 0.100 × 0.0250 = 0.00250 mol. Since the mole ratio is 1 : 1, moles of HCl = 0.00250 mol. Concentration of HCl = 0.00250 ÷ 0.0235 = 0.106 mol dm⁻³.

NaOH 的摩尔数 = 0.100 × 0.0250 = 0.00250 mol。由于摩尔比为 1 : 1，HCl 的摩尔数 = 0.00250 mol。HCl 的浓度 = 0.00250 ÷ 0.0235 = 0.106 mol dm⁻³。

For reactions with a ratio other than 1 : 1, such as H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O, use the mole ratio from the balanced equation to convert moles of one substance to moles of the other.

对于比例不是 1 : 1 的反应，如 H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O，使用配平方程式中的摩尔比将一种物质的摩尔数转换为另一种物质的摩尔数。

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7. Ideal Gas Equation | 理想气体状态方程

The ideal gas equation links pressure, volume, temperature and amount of gas: pV = nRT.

理想气体状态方程将气体的压力、体积、温度和物质的量联系起来：pV = nRT。

pV = nRT

In this equation, p is pressure in pascals (Pa), V is volume in cubic metres (m³), n is amount in mol, T is temperature in kelvin (K), and R = 8.31 J mol⁻¹ K⁻¹.

在公式中，p 为压力，单位帕斯卡（Pa）；V 为体积，单位立方米（m³）；n 为物质的量（mol）；T 为温度（开尔文，K）；R = 8.31 J mol⁻¹ K⁻¹。

Temperature in Celsius must be converted to kelvin by adding 273. Pressure in kPa must be multiplied by 1000 to give Pa, and volume in cm³ must be divided by 1,000,000 to give m³.

摄氏温度必须加 273 转换为开尔文。压力单位 kPa 必须乘以 1000 换算成 Pa，体积单位 cm³ 必须除以 1,000,000 换算成 m³。

Worked example: Calculate the volume of 0.500 mol of an ideal gas at 100 kPa and 298 K. First p = 100,000 Pa. Then V = nRT ÷ p = (0.500 ×

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