📚 CCEA Year 13 Chemistry: Case Study Practical Drill | CCEA Year 13 化学：案例分析实战演练
This article takes you through a highly realistic analytical chemistry case study – identifying an unknown white solid and determining its purity using a combination of combustion analysis, mass spectrometry, IR and NMR spectroscopy, and a back-titration. By following the step‑by‑step reasoning, you will sharpen the skills required for CCEA Year 13 exam questions that demand integrated problem‑solving.
本文将带你完成一个高度真实的化学分析案例——通过燃烧分析、质谱、红外和核磁共振波谱以及返滴定法，鉴定一种未知白色固体并测定其纯度。跟随一步步的推理，你将磨炼应对CCEA Year 13考试中需要综合解题能力的技能。
1. Case Background | 案例背景
A pharmaceutical company sends a white crystalline solid to the laboratory. The label claims it is aspirin (acetylsalicylic acid), but the batch is suspected to be impure. Your task is to confirm the identity of the organic compound and to quantify the percentage of aspirin in the sample using the data provided.
某制药公司将一种白色结晶固体送至实验室。标签声称其为阿司匹林（乙酰水杨酸），但该批次被怀疑含有杂质。你的任务是确认该有机化合物的身份，并利用所给数据定量测定样品中阿司匹林的百分含量。
The investigation is split into two parts: (I) structural elucidation by elemental composition and spectroscopic methods; (II) purity determination by a back‑titration that exploits the chemical properties of aspirin. Record all working, because in CCEA examinations marks are awarded for clear logical steps.
该研究分为两部分：(I) 通过元素组成和光谱方法进行结构解析；(II) 利用阿司匹林的化学性质进行返滴定，测定纯度。请记录所有计算过程，因为在CCEA考试中，清晰的逻辑步骤可获得分数。
2. Combustion Analysis | 燃烧分析
A 0.2000 g portion of the solid is completely burnt in excess oxygen. The products are 0.4400 g of carbon dioxide and 0.0800 g of water. No other elements besides C, H and O are detected.
取0.2000 g固体在过量氧气中完全燃烧。产物为0.4400 g二氧化碳和0.0800 g水。除C、H、O外未检测到其他元素。
Calculate the mass of carbon: (12.01 ÷ 44.01) × 0.4400 g = 0.1200 g. Calculate the mass of hydrogen: (2.016 ÷ 18.015) × 0.0800 g = 0.00895 g. The mass of oxygen is obtained by difference: 0.2000 – (0.1200 + 0.00895) = 0.07105 g.
计算碳的质量：(12.01 ÷ 44.01) × 0.4400 g = 0.1200 g。计算氢的质量：(2.016 ÷ 18.015) × 0.0800 g = 0.00895 g。氧的质量通过差值求得：0.2000 – (0.1200 + 0.00895) = 0.07105 g。
Now convert masses to moles: C: 0.1200 ÷ 12.01 = 0.00999 mol; H: 0.00895 ÷ 1.008 = 0.00888 mol; O: 0.07105 ÷ 16.00 = 0.00444 mol. Divide by the smallest value (0.00444) gives C 2.25, H 2.00, O 1.00. Multiplying by 4 to obtain whole numbers yields C9H8O4. The empirical formula is therefore C9H8O4.
现在将质量换算为物质的量：C：0.1200 ÷ 12.01 = 0.00999 mol；H：0.00895 ÷ 1.008 = 0.00888 mol；O：0.07105 ÷ 16.00 = 0.00444 mol。除以最小值（0.00444）得到比例 C 2.25，H 2.00，O 1.00。乘以4以获得整数比，得到C9H8O4。因此经验式为C9H8O4。
3. Mass Spectrometry | 质谱分析
The mass spectrum of the solid displays a molecular ion peak at m/z = 180, with a very small M+1 peak consistent with the natural abundance of ¹³C. The base peak appears at m/z = 120, corresponding to the loss of a CH₃COOH fragment.
该固体的质谱显示分子离子峰位于m/z = 180，并伴有与¹³C自然丰度相符的极小的M+1峰。基峰出现在m/z = 120，对应于丢失一个CH₃COOH碎片。
Since the molecular ion mass is 180, the relative molecular mass is 180. The empirical formula C9H8O4 has a formula mass of (9×12.01 + 8×1.008 + 4×16.00) = 180.16. This matches the empirical formula exactly, confirming the molecular formula is C9H8O4.
由于分子离子质量为180，相对分子质量为180。经验式C9H8O4的式量为(9×12.01 + 8×1.008 + 4×16.00) = 180.16。这与经验式完全吻合，证实分子式为C9H8O4。
4. Infrared Spectroscopy | 红外光谱
The infrared spectrum shows a broad absorption centred near 3000 cm⁻¹, typical of a hydrogen‑bonded O–H stretch in a carboxylic acid. A strong, sharp peak at 1755 cm⁻¹ is assigned to the C=O stretch of an ester, while a second intense peak at 1690 cm⁻¹ indicates the C=O stretch of a carboxylic acid. Peaks at 1605 cm⁻¹ and 1460 cm⁻¹ are characteristic of aromatic C=C bonds.
红外光谱显示一个中心约在3000 cm⁻¹的宽吸收峰，这是羧酸中氢键结合的O–H伸缩振动的典型特征。一个位于1755 cm⁻¹的强而尖的峰归属于酯的C=O伸缩振动，而另一个位于1690 cm⁻¹的强峰表明存在羧酸的C=O伸缩振动。1605 cm⁻¹和1460 cm⁻¹的峰则是芳香族C=C键的特征。
The presence of an ester carbonyl (1755 cm⁻¹) together with a carboxylic acid carbonyl (1690 cm⁻¹) strongly suggests the molecule is acetylsalicylic acid, which contains an acetyl ester group attached to a benzoic acid framework.
酯羰基（1755 cm⁻¹）和羧酸羰基（1690 cm⁻¹）同时存在，强烈暗示该分子是乙酰水杨酸，它含有一个连接在苯甲酸骨架上的乙酰酯基团。
5. NMR Spectroscopy | 核磁共振谱
The ¹H NMR spectrum (400 MHz, CDCl₃) shows three distinct signals: a singlet at δ 2.35 integrating for 3H, a complex multiplet between δ 7.10 and δ 8.22 integrating for 4H, and a broad singlet at δ 11.5 integrating for 1H. The broad peak disappears upon shaking with D₂O, confirming it is an exchangeable O–H proton.
¹H核磁共振谱（400 MHz，CDCl₃）显示三个不同的信号：δ 2.35处的单峰，积分对应3个H；δ 7.10到8.22之间的复杂多重峰，积分对应4个H；以及δ 11.5处的宽单峰，积分对应1个H。该宽峰在与D₂O振摇后消失，证实其为可交换的O–H质子。
The singlet at δ 2.35 is typical of a methyl group adjacent to a carbonyl (CH₃CO–). The four aromatic protons indicate a disubstituted benzene ring, and the broad signal at δ 11.5 is the carboxylic acid –OH. No other signals are present, which is entirely consistent with the structure of 2‑acetyloxybenzoic acid (aspirin).
δ 2.35处的单峰是甲基与羰基相邻（CH₃CO–）的典型信号。四个芳香质子表明存在二取代苯环，而δ 11.5处的宽信号是羧酸的–OH。图谱中没有其他信号，这与2‑乙酰氧基苯甲酸（阿司匹林）的结构完全一致。
6. Putting It All Together: Structure Elucidation | 综合解析确定结构
Collectively, the data converge on a single structure: a benzene ring with two substituents – a carboxylic acid group (–COOH) and an acetyl ester group (–OCOCH₃). The NMR splitting pattern and the mass spectral fragmentation both support the ortho‑arrangement, hence the compound is 2‑(acetyloxy)benzoic acid, commonly known as aspirin.
综合所有数据，可以确定唯一的结构：一个苯环上带有两个取代基——一个羧基（–COOH）和一个乙酰酯基（–OCOCH₃）。核磁共振的裂分模式和质谱碎裂方式都支持邻位取代，因此该化合物是2‑（乙酰氧基）苯甲酸，即通常所说的阿司匹林。
Molecular structure: C₆H₄(OCOCH₃)COOH (C₉H₈O₄)
分子结构：C₆H₄(OCOCH₃)COOH (C₉H₈O₄)
7. Back‑Titration for Purity | 返滴定测纯度
Having identified the compound, the laboratory now determines its purity. Aspirin reacts with sodium hydroxide in a 1:2 molar ratio because both the carboxylic acid group and the ester linkage consume one mole of OH⁻ when the ester hydrolyses. The reaction scheme is:
确认化合物身份后，实验室现在测定其纯度。阿司匹林与氢氧化钠以1:2的摩尔比反应，因为在酯水解时，羧酸基团和酯键各消耗一摩尔OH⁻。反应式为：
C₉H₈O₄ + 2NaOH → CH₃COONa + HOC₆H₄COONa + H₂O
A 0.5000 g sample of the impure aspirin is placed in a conical flask, and 50.0 cm³ of 0.100 mol dm⁻³ NaOH solution is added. The mixture is heated gently under reflux for 15 minutes to ensure complete hydrolysis. After cooling, the excess NaOH is titrated with 0.100 mol dm⁻³ hydrochloric acid using phenolphthalein as indicator. The titre is 25.0 cm³.
将0.5000 g不纯阿司匹林样品置于锥形瓶中，加入50.0 cm³ 0.100 mol dm⁻³ NaOH溶液。混合物在回流下温和加热15分钟以确保完全水解。冷却后，以酚酞为指示剂，用0.100 mol dm⁻³盐酸滴定过量的NaOH。滴定体积为25.0 cm³。
8. Calculations and Results | 计算与结果
Calculate the amount of NaOH initially added: 0.0500 dm³ × 0.100 mol dm⁻³ = 0.00500 mol. The amount of HCl used for the back‑titration: 0.0250 dm³ × 0.100 mol dm⁻³ = 0.00250 mol, which equals the moles of NaOH in excess. Therefore, the amount of NaOH that reacted with aspirin is 0.00500 – 0.00250 = 0.00250 mol.
计算最初加入的NaOH的物质的量：0.0500 dm³ × 0.100 mol dm⁻³ = 0.00500 mol。返滴定所用HCl的物质的量：0.0250 dm³ × 0.100 mol dm⁻³ = 0.00250 mol，这等于过量的NaOH的物质的量。因此，与阿司匹林反应的NaOH的物质的量为0.00500 – 0.00250 = 0.00250 mol。
Since 1 mol aspirin consumes 2 mol NaOH, the amount of aspirin present = 0.00250 ÷ 2 = 0.00125 mol. Mass of pure aspirin = 0.00125 mol × 180.16 g mol⁻¹ = 0.2252 g. The percentage purity is therefore (0.2252 g ÷ 0.5000 g) × 100% = 45.0%.
由于1 mol阿司匹林消耗2 mol NaOH，实际存在的阿司匹林的物质的量 = 0.00250 ÷ 2 = 0.00125 mol。纯阿司匹林的质量 = 0.00125 mol × 180.16 g mol⁻¹ = 0.2252 g。因此百分纯度为(0.2252 g ÷ 0.5000 g) × 100% = 45.0%。
The result shows the sample is only 45% aspirin, with the remainder likely being inert tablet excipients such as starch or lactose, which do not interfere with the titration.
结果表明该样品仅含45%的阿司匹林，其余部分可能是惰性片剂辅料，如淀粉或乳糖，这些物质不会干扰滴定。
9. Discussion of Errors | 误差讨论
Several sources of uncertainty must be considered: incomplete hydrolysis of the ester would leave some aspirin unreacted, leading to an overestimation of the excess NaOH and thus an underestimation of purity. To minimise this, the reagent was heated under reflux for a generous period. The use of a calibrated burette and pipette reduces volumetric errors. Furthermore, if the sample had absorbed moisture, the apparent purity would be lowered.
必须考虑几个不确定度的来源：酯的水解不完全将使部分阿司匹林未反应，导致高估过量NaOH，从而低估纯度。为尽量减少这一影响，试剂在回流下加热了足够长的时间。使用校准过的滴定管和移液管可减少体积误差。此外，如果样品吸了潮，则表观纯度会偏低。
In an exam context, you might be asked to explain why a back‑titration is preferred over a direct titration for aspirin. The answer is that aspirin is a weak acid and its endpoint with NaOH is difficult to detect sharply; moreover, the hydrolysis step ensures complete reaction of the acetyl group.
在考试情境中，你可能需要解释为什么测定阿司匹林时优先采用返滴定而非直接滴定。答案是阿司匹林为弱酸，其与NaOH的滴定终点难以敏锐检测；而且水解步骤确保了乙酰基的完全反应。
10. Conclusion and Exam Tips | 结论与应试技巧
This case study demonstrates how CCEA Year 13 chemistry integrates organic analysis, spectroscopy, and quantitative reasoning. The key is to approach problems systematically: first use empirical evidence to identify the molecule, then design or interpret a volumetric method to quantify it. Always show all working, include units, and check that your answer makes chemical sense.
本案例研究展示了CCEA Year 13化学如何将有机分析、光谱学和定量推理融为一体。关键在于系统地处理问题：首先利用实验证据鉴定分子，然后设计或解释滴定方法来定量分析。一定要展示所有计算过程，注明单位，并检查答案在化学上是否合理。
When revising, practise constructing logical sequences – from combustion figures to empirical formula, from mass spectra to molecular formula, and from NMR/IR to functional groups. Then connect the volumetric data through balanced equations. In the CCEA examination, structured questions often mirror this exact workflow, so familiarity with the whole chain of reasoning will save time and earn marks.
在复习时，练习构建逻辑顺序——从燃烧数据到经验式，从质谱到分子式，从核磁/红外到官能团。然后通过配平的方程式将容量分析数据联系起来。在CCEA考试中，结构化试题常常恰好反映出这样的工作流程，因此熟悉整个推理链条将节省时间并取得分数。
Published by TutorHao | Chemistry Revision Series | aleveler.com
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