📚 Year 12 CCEA Chemistry: Case Study in Action | Year 12 CCEA 化学：案例分析实战演练
Case studies are a key component of the CCEA AS Chemistry examination. They test your ability to apply knowledge to unfamiliar contexts, interpret experimental data, perform calculations, and evaluate procedures. This article provides a hands-on drill through a realistic case study involving the synthesis and analysis of aspirin, equipping you with the skills to tackle similar questions confidently.
案例分析是 CCEA AS 化学考试的重要组成部分。它们考察你将知识应用于陌生情境、解释实验数据、进行计算以及评价实验步骤的能力。本文将通过一个涉及阿司匹林合成与分析的现实案例进行实战演练，帮助你掌握应对类似问题的技能。
1. Understanding Case Study Requirements | 理解案例分析题的要求
CCEA case study questions often present a multi-step practical scenario with data tables, graphs, or spectra. You are expected to extract relevant information, link it to chemical principles, perform quantitative reasoning, and critically assess the method. Marks are allocated for accurate calculations, correct units, logical explanations, and sensible improvements.
CCEA 的案例分析题通常会给出一个包含多步实验的情境，并附带数据表、图表或谱图。你需要从中提取相关信息，与化学原理联系起来，进行定量推理，并对方法做出批判性评价。准确的计算、正确的单位、逻辑清晰的解释以及合理的改进建议都是得分点。
2. Sample Case: Synthesis of Aspirin | 示例案例：阿司匹林的合成
In a typical experiment, a student reacted 2.00 g of salicylic acid (C₇H₆O₃) with 4.00 cm³ of ethanoic anhydride (density 1.08 g cm⁻³, excess) in the presence of a few drops of concentrated sulfuric acid. The mixture was heated on a water bath for 20 minutes, then cautiously added to cold water. The crude aspirin was filtered, washed, and dried, yielding 2.35 g. After recrystallisation from hot ethanol/water, the pure aspirin (acetylsalicylic acid, C₉H₈O₄) weighed 1.80 g. Subsequent analyses were performed: a back titration, thin-layer chromatography, IR, and mass spectrometry.
在一次典型实验中，学生将 2.00 g 水杨酸 (C₇H₆O₃) 与 4.00 cm³ 乙酸酐（密度 1.08 g cm⁻³，过量）在几滴浓硫酸催化下反应。混合物在水浴上加热 20 分钟，然后小心地倒入冷水中。粗品阿司匹林经过滤、洗涤、干燥后，得到 2.35 g。经热乙醇/水重结晶后，纯阿司匹林（乙酰水杨酸，C₉H₈O₄）的质量为 1.80 g。随后进行了返滴定、薄层色谱、红外光谱和质谱分析。
3. Calculating Moles and Identifying the Limiting Reagent | 计算摩尔并确定限制反应物
First, calculate the amount of salicylic acid: Mᵣ = 138.0 g mol⁻¹ → moles = 2.00 / 138.0 = 0.0145 mol. For ethanoic anhydride, mass = 4.00 cm³ × 1.08 g cm⁻³ = 4.32 g; Mᵣ of (CH₃CO)₂O = 102.0 g mol⁻¹ → moles = 4.32 / 102.0 = 0.0424 mol. The reaction is 1:1, so salicylic acid is the limiting reactant.
首先计算水杨酸的物质的量：Mᵣ = 138.0 g mol⁻¹ → 物质的量 = 2.00 / 138.0 = 0.0145 mol。乙酸酐的质量 = 4.00 cm³ × 1.08 g cm⁻³ = 4.32 g；(CH₃CO)₂O 的 Mᵣ = 102.0 g mol⁻¹ → 物质的量 = 4.32 / 102.0 = 0.0424 mol。该反应为 1:1，因此水杨酸是限制反应物。
The balanced equation using molecular formulae is:
使用分子式的配平方程式如下：
C₇H₆O₃ + C₄H₆O₃ → C₉H₈O₄ + C₂H₄O₂
Theoretical moles of aspirin = 0.0145 mol. Mᵣ of aspirin = 180.0 g mol⁻¹, so theoretical mass = 0.0145 × 180.0 = 2.61 g.
阿司匹林的理论物质的量 = 0.0145 mol。阿司匹林的 Mᵣ = 180.0 g mol⁻¹，因此理论质量 = 0.0145 × 180.0 = 2.61 g。
4. Calculating Percentage Yield and Atom Economy | 计算百分产率与原子经济性
The crude yield = (2.35 g / 2.61 g) × 100 = 90.0%. The yield after recrystallisation = (1.80 g / 2.61 g) × 100 = 69.0%. A drop during purification is normal due to mechanical losses and solubility.
粗产率 = (2.35 g / 2.61 g) × 100 = 90.0%。重结晶后的产率 = (1.80 g / 2.61 g) × 100 = 69.0%。在提纯过程中产率下降是正常的，这是由于操作损失以及产品在溶剂中的溶解度所致。
Atom economy is calculated from the reaction: [Mᵣ(C₉H₈O₄) / (Mᵣ(C₇H₆O₃) + Mᵣ(C₄H₆O₃))] × 100 = [180.0 / (138.0+102.0)] × 100 = 75.0%. The 25% loss in atom economy represents the mass of the co-product ethanoic acid.
原子经济性由反应式计算：[Mᵣ(C₉H₈O₄) / (Mᵣ(C₇H₆O₃) + Mᵣ(C₄H₆O₃))] × 100 = [180.0 / (138.0+102.0)] × 100 = 75.0%。25% 的原子经济性损失对应副产物乙酸的质量。
5. Purification and Recrystallisation | 提纯与重结晶
Recrystallisation exploits differences in solubility between the desired product and impurities in a chosen solvent. Aspirin is dissolved in a minimum volume of hot ethanol/water, then cooled slowly. Pure crystals form, while soluble impurities stay in solution. Rinsing with cold solvent minimises loss.
重结晶利用了目标产物与杂质在所选溶剂中的溶解度差异。阿司匹林被溶解在最少量的热乙醇/水中，然后缓慢冷却。纯晶体析出，而可溶性杂质留在溶液中。用冷溶剂淋洗可减少产品损失。
Reasons for a yield below 100% include incomplete transfer, crystals sticking to glassware, and slight solubility even in cold solvent. In an exam, always suggest practical improvements such as using a chilled solvent and filtering quickly under reduced pressure.
产率低于 100% 的原因包括转移不完全、晶体粘附在玻璃仪器上，以及即使在冷溶剂中也有少量溶解。在考试中，要提出实际的改进措施，例如使用冷冻过的溶剂和快速减压过滤。
6. Determining Purity by Back Titration | 通过返滴定测定纯度
In the case study, 0.50 g of the purified aspirin was hydrolysed by heating with 25.0 cm³ of 0.200 mol dm⁻³ NaOH (excess). The equation shows 1 mol aspirin consumes 2 mol NaOH: C₉H₈O₄ + 2NaOH → C₇H₄O₃Na₂ + CH₃COONa + H₂O. After cooling, the unreacted NaOH was titrated with 0.100 mol dm⁻³ HCl, requiring an average of 10.2 cm³ to reach the endpoint using phenolphthalein.
在本案例中，0.50 g 纯化后的阿司匹林与 25.0 cm³ 0.200 mol dm⁻³ NaOH（过量）一起加热水解。反应方程式显示 1 mol 阿司匹林消耗 2 mol NaOH：C₉H₈O₄ + 2NaOH → C₇H₄O₃Na₂ + CH₃COONa + H₂O。冷却后，用 0.100 mol dm⁻³ HCl 滴定未反应的 NaOH，以酚酞为指示剂，平均消耗 10.2 cm³ 到达终点。
Calculation steps:
Total moles of NaOH added = 25.0 × 10⁻³ dm³ × 0.200 mol dm⁻³ = 0.00500 mol.
Moles of HCl used = 10.2 × 10⁻³ dm³ × 0.100 mol dm⁻³ = 0.00102 mol = moles of unreacted NaOH.
Moles of NaOH that reacted with aspirin = 0.00500 – 0.00102 = 0.00398 mol.
Moles of aspirin = 0.00398 / 2 = 0.00199 mol.
Mass of aspirin = 0.00199 mol × 180.0 g mol⁻¹ = 0.358 g.
Purity = (0.358 / 0.50) × 100 = 71.6%.
计算步骤：
加入的 NaOH 总物质的量 = 25.0 × 10⁻³ dm³ × 0.200 mol dm⁻³ = 0.00500 mol。
消耗 HCl 的物质的量 = 10.2 × 10⁻³ dm³ × 0.100 mol dm⁻³ = 0.00102 mol = 未反应 NaOH 的物质的量。
与阿司匹林反应的 NaOH 的物质的量 = 0.00500 – 0.00102 = 0.00398 mol。
阿司匹林的物质的量 = 0.00398 / 2 = 0.00199 mol。
阿司匹林的质量 = 0.00199 mol × 180.0 g mol⁻¹ = 0.358 g。
纯度 = (0.358 / 0.50) × 100 = 71.6%。
The result indicates the recrystallised product still contains about 28% impurity, possibly unreacted salicylic acid or co-crystallised solvent. In an exam, you would comment on how to improve purity, such as repeating the recrystallisation.
该结果表明重结晶后的产物仍含有约 28% 的杂质，可能是未反应的水杨酸或共结晶的溶剂。在考试中，你需要评论如何提高纯度，例如重复重结晶。
7. Thin-Layer Chromatography (TLC) Analysis | 薄层色谱分析
TLC was run on a silica gel plate using ethyl ethanoate/hexane as the mobile phase. The Rf value of a salicylic acid reference was 0.70, while pure aspirin gave Rf = 0.56. The crude product showed two spots: one at Rf 0.56 (aspirin) and one at 0.70 (unreacted salicylic acid). The recrystallised product gave a single spot at Rf 0.56, confirming enhanced purity.
使用硅胶板、以乙酸乙酯/己烷为流动相进行薄层色谱分析。水杨酸对照品的 Rf 值为 0.70，纯阿司匹林的 Rf 值为 0.56。粗产物显示两个斑点：一个在 Rf 0.56（阿司匹林），另一个在 0.70（未反应的水杨酸）。重结晶产物仅显示一个在 Rf 0.56 的斑点，证实纯度提高。
Rf is calculated as distance moved by substance / distance moved by solvent front. When commenting on TLC results, always compare the number of spots and their Rf values to reference compounds, and state what the comparison reveals about purity.
Rf 值的计算是物质移动距离除以溶剂前沿移动距离。在评论 TLC 结果时，要始终将斑点的数量及其 Rf 值与对照品进行比较，并说明该比较揭示了哪些关于纯度的信息。
8. Infrared Spectroscopy Interpretation | 红外光谱解析
The IR spectrum of the recrystallised product showed a broad absorption from 2500–3300 cm⁻¹ (O–H stretch in carboxylic acid), a strong peak at 1750 cm⁻¹ (C=O stretch of ester), and a peak at 1700 cm⁻¹ (C=O stretch of carboxylic acid). Aromatic C–C stretches appeared around 1600 and 1500 cm⁻¹, while strong C–O stretches were found near 1200 cm⁻¹ and 1050 cm⁻¹. The absence of a broad –OH peak around 3200 cm⁻¹ from phenol (as would be seen in salicylic acid) confirms the conversion.
重结晶产物的红外光谱显示在 2500–3300 cm⁻¹ 的宽吸收峰（羧酸中 O–H 伸缩振动）、1750 cm⁻¹ 处的强峰（酯中 C=O 伸缩振动）以及 1700 cm⁻¹ 处的峰（羧酸中 C=O 伸缩振动）。芳香族 C–C 伸缩振动出现在约 1600 和 1500 cm⁻¹，而在 1200 cm⁻¹ 和 1050 cm⁻¹ 附近有较强的 C–O 伸缩振动。没有出现酚羟基在 3200 cm⁻¹ 附近的宽峰（水杨酸中会
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