📚 Mastering Combustion Analysis: From Data to Empirical Formula | 燃烧分析：从实验数据求实验式与分子式
Combustion analysis is one of the most powerful experimental techniques in organic chemistry. By burning a known mass of compound in excess oxygen and carefully measuring the masses of carbon dioxide and water produced, you can determine the percentages of carbon and hydrogen in the sample. This information, together with the molar mass, allows you to work out the empirical and molecular formulae of the unknown substance. In Edexcel A-Level Chemistry Topic 5.8, you are expected to handle combustion data with confidence, converting mass readings into mole ratios and solving for the complete formula of a compound containing carbon, hydrogen, and often oxygen.
燃烧分析是有机化学中最强有力的实验技术之一。在过量氧气中燃烧已知质量的化合物，并精确测量生成的二氧化碳和水的质量，就能求得样品中碳和氢的质量分数。结合化合物的摩尔质量，这些数据可以让你推导出未知物的实验式和分子式。在Edexcel A-Level化学第5.8节中，要求你能熟练处理燃烧数据，将质量读数转化为摩尔比，最终求出含碳、氢甚至含氧化合物的完整化学式。
1. The Core Idea of Combustion Analysis | 燃烧分析的核心思路
When an organic compound containing carbon, hydrogen and possibly oxygen is completely burned in a stream of dry oxygen, all the carbon is converted to carbon dioxide (CO₂) and all the hydrogen is converted to water (H₂O). By passing the products through pre‑weighed absorption tubes – one containing a strong desiccant to trap water and another containing a base (such as soda‑lime or sodium hydroxide) to absorb carbon dioxide – the increase in mass of each tube gives the mass of H₂O and CO₂ formed. From these masses you can calculate the mass of carbon and the mass of hydrogen that were present in the original sample.
含有碳、氢以及可能含氧的有机化合物在干燥氧气流中完全燃烧时，所有的碳都转化为二氧化碳(CO₂)，所有的氢都转化为水(H₂O)。让产物依次通过预先称重的吸收管——一根装有强干燥剂以捕获水，另一根装有碱（如碱石灰或氢氧化钠）以吸收二氧化碳——每根管子增加的质量就是生成的H₂O和CO₂的质量。由这些质量即可计算出原来样品中碳的质量和氢的质量。
If the sum of the masses of carbon and hydrogen is less than the original sample mass, the missing mass is assumed to be oxygen. This ‘oxygen by difference’ approach is used for compounds that contain C, H and O only, which is the most common case in Edexcel exam questions.
如果碳与氢的质量之和小于原始样品的质量，缺失的质量就被认为是氧。这种“差额法求氧”的思路适用于只含C、H和O的化合物，这也是Edexcel考题中最常见的情况。
2. The Reaction Stoichiometry | 反应计量关系
The general equation for complete combustion of a compound with the formula CₓHᵧOᵪ (where the subscript letters are used generically) in excess oxygen is:
通式为CₓHᵧOᵪ的化合物在过量氧气中完全燃烧的通用方程式为：
CₓHᵧOᵪ + (x + y/4 – z/2) O₂ → x CO₂ + (y/2) H₂O
From this, you can see that every mole of CO₂ produced contains one mole of carbon atoms, and every mole of H₂O produced contains two moles of hydrogen atoms. These simple ratios are the key to unlocking the composition of the original compound.
由此可以看出，每生成1 mol CO₂就含有1 mol 碳原子，每生成1 mol H₂O就含有2 mol 氢原子。这些简单的比例关系正是解开原始化合物组成之谜的关键。
3. Step‑by‑Step Calculation Strategy | 分步计算策略
Follow this systematic procedure when you are given combustion data:
当题目给出燃烧数据时，请按照以下系统步骤进行：
Step 1. Calculate the mass of carbon from the mass of CO₂. Use the fact that 44.0 g of CO₂ contains 12.0 g of carbon.
步骤1. 从CO₂的质量计算碳的质量。已知44.0 g CO₂含有12.0 g 碳。
Mass of C = (12.0 / 44.0) × mass of CO₂
Step 2. Calculate the mass of hydrogen from the mass of H₂O. Remember that 18.0 g of H₂O contains 2.0 g of hydrogen.
步骤2. 从H₂O的质量计算氢的质量。记住18.0 g H₂O含有2.0 g 氢。
Mass of H = (2.0 / 18.0) × mass of H₂O
Step 3. If the compound contains oxygen, calculate its mass by subtracting the masses of carbon and hydrogen from the original sample mass.
步骤3. 如果化合物含氧，则用原始样品质量减去碳和氢的质量，得到氧的质量。
Step 4. h mass to moles by dividing by the relative atomic mass (Aᵣ): C = 12.0, H = 1.0, O = 16.0.
步骤4. 将每种元素的质量除以相对原子质量(Aᵣ)转化为摩尔数：C = 12.0, H = 1.0, O = 16.0。
Step 5. Divide all mole values by the smallest number of moles to obtain the simplest whole‑number ratio. This gives the empirical formula.
步骤5. 将所有摩尔数除以最小的摩尔数，得到最简整数比，即实验式。
Step 6. If the relative molecular mass (Mᵣ) is known, find the molecular formula by comparing the mass of the empirical formula unit with Mᵣ.
步骤6. 如果已知相对分子质量(Mᵣ)，将实验式单元的质量与Mᵣ进行比较，求出分子式。
4. Worked Example 1 – A Hydrocarbon | 例题1 – 烃
A 0.400 g sample of a hydrocarbon is burned completely in oxygen. The mass of CO₂ produced is 1.257 g and the mass of H₂O produced is 0.514 g. Determine the empirical formula of the hydrocarbon.
0.400 g烃样品在氧气中完全燃烧，生成CO₂ 1.257 g，H₂O 0.514 g。求该烃的实验式。
Mass of carbon: (12.0/44.0) × 1.257 = 0.343 g
碳的质量： (12.0/44.0) × 1.257 = 0.343 g
Mass of hydrogen: (2.0/18.0) × 0.514 = 0.0571 g
氢的质量： (2.0/18.0) × 0.514 = 0.0571 g
k total mass: 0.343 + 0.0571 = 0.4001 g ≈ 0.400 g, confirming no other element is present.
检查总质量：0.343 + 0.0571 = 0.4001 g ≈ 0.400 g，确认无其他元素。
Moles of C = 0.343 / 12.0 = 0.0286 mol; Moles of H = 0.0571 / 1.0 = 0.0571 mol.
碳的摩尔数 = 0.343 / 12.0 = 0.0286 mol；氢的摩尔数 = 0.0571 / 1.0 = 0.0571 mol。
Divide by the smaller value (0.0286): C: 1.00, H: 2.00. Empirical formula is CH₂.
除以较小值(0.0286)：C: 1.00, H: 2.00。实验式为CH₂。
| Element 元素 | Mass / g 质量/克 | Aᵣ | Moles / mol 摩尔 | Ratio 比 |
|---|---|---|---|---|
| C | 0.343 | 12.0 | 0.0286 | 1 |
| H | 0.0571 | 1.0 | 0.0571 | 2 |
Thus, the empirical formula is CH₂.
因此，实验式为CH₂。
en as 56, then the molecular formula is C₄H₈ (since the mass of CH₂ is 14, and 56/14 = 4).
如果后来给出该烃的Mᵣ是56，则分子式为C₄H₈（CH₂的式量为14，56/14 = 4）。
5. Worked Example 2 – Compound Containing C, H and O | 例题2 – 含C、H和O的化合物
0.250 g of a liquid organic compound containing only carbon, hydrogen and oxygen is completely burnt. 0.550 g of CO₂ and 0.225 g of H₂O are formed. Determine the empirical formula of the compound.
某液态有机化合物只含碳、氢、氧，取0.250 g完全燃烧，生成0.550 g CO₂和0.225 g H₂O。求该化合物的实验式。
Mass of C from CO₂: (12.0/44.0) × 0.550 = 0.150 g
从CO₂求碳的质量：(12.0/44.0) × 0.550 = 0.150 g
Mass of H from H₂O: (2.0/18.0) × 0.225 = 0.0250 g
从H₂O求氢的质量：(2.0/18.0) × 0.225 = 0.0250 g
Mass of oxygen by difference: 0.250 – (0.150 + 0.0250) = 0.0750 g
氧的质量（差额法）：0.250 – (0.150 + 0.0250) = 0.0750 g
ert to moles: C = 0.150/12.0 = 0.0125; H = 0.0250/1.0 = 0.0250; O = 0.0750/16.0 = 0.00469.
转化为摩尔数：C = 0.150/12.0 = 0.0125；H = 0.0250/1.0 = 0.0250；O = 0.0750/16.0 = 0.00469。
469): C ≈ 2.67, H ≈ 5.33, O = 1.00. Multiply by 3 to clear fractions: C₈H₁₆O₃ (empirical).
除以最小值0.00469：C ≈ 2.67，H ≈ 5.33，O = 1.00。乘以3消除分数：C₈H₁₆O₃（实验式）。
6. Dealing with Fractional Ratios | 处理非整数比
When you divide by the smallest number of moles, you may obtain ratios such as 1.5, 2.5 or 1.33. These are not acceptable as subscripts in a chemical formula. Multiply all the ratios by the smallest integer that will turn them all into whole numbers. Common multipliers are 2, 3 or 4. For instance, a ratio of 1 : 1.5 becomes 2 : 3, and a ratio of 1 : 1.33 : 1 becomes 3 : 4 : 3.
当除以最小摩尔数时，可能得到1.5、2.5或1.33这样的比值。这些不能用作化学式中的下标。需要将所有比值乘以能使其全部变为整数的最小整数。常用的乘数是2、3或4。例如，1 : 1.5乘以2得2 : 3；1 : 1.33 : 1乘以3得3 : 4 : 3。
Be careful when rounding – have 2.33, recognise it as 7/3 and multiply by 3. The precision of the experimental data determines how much rounding is acceptable.
取整时要小心——只有当数值非常接近整数时（如2.98 ≈ 3.0）才能取整。如果得到2.33，应识别为7/3并乘以3。实验数据的精度决定了可以接受的取整程度。
7. The Role of Molar Mass in Finding the Molecular Formula | 摩尔质量在求分子式中的作用
The empirical formula tells you the simplest ratio of atoms. The molecular formula is a whole‑number multiple of the empirical formula. Once you have the empirical formula mass (EMM), you can use the given Mᵣ to find the multiplier, n.
实验式给出的是原子个数的最简整数比，分子式则是实验式的整数倍。一旦求得实验式质量(EMM)，就可以利用题给的Mᵣ算出倍数n。
n = Mᵣ / EMM
l formula is CH₂O (EMM = 30) and Mᵣ = 60, then n = 2, giving the molecular formula C₂H₄O₂.
例如，实验式为CH₂O (EMM = 30)，Mᵣ = 60，则n = 2，分子式为C₂H₄O₂。
In some exam questions, the Mᵣ is obtained from mass spectrometry or the ideal gas equation. You must be ready to combine these pieces of information.
在某些考题中，Mᵣ来源于质谱或理想气体状态方程，你必须能够综合运用这些信息。
8. Combustion Analysis of Compounds Containing Other Elements | 含有其他元素的化合物的燃烧分析
If a compound contains elements such as nitrogen, sulfur or halogens, the combustion products are more complex. Nitrogen is often converted to N₂ gas, sulfur to SO₂ or SO₃, and halogens can form hydrogen halides. Edexcel questions at this level usually restrict themselves to C, H and O, but you should be aware that the ‘oxygen by difference’ method only works if oxygen is the only element besides carbon and hydrogen whose mass is not directly measured.
如果化合物含有氮、硫或卤素，燃烧产物会更复杂。氮通常转化为N₂，硫转化为SO₂或SO₃，卤素可形成卤化氢。在目前的学习阶段，Edexcel考题通常只涉及C、H和O，但你应该明白，“差额法求氧”仅当氧是除碳和氢外唯一没有直接称量的元素时才有效。
9. Experimental Considerations and Sources of Error | 实验操作要点与误差来源
To obtain reliable results, the combustion apparatus must be carefully dried before use, and the sample must be weighed accurately. The oxygen supply must be dry and free of CO₂. The absorption tubes are weighed before and after combustion, and the mass increases correspond exactly to the H₂O and CO₂ produced. Common errors include incomplete combustion (producing CO or soot), insufficient trapping of water, or leakage. Incomplete combustion causes the mass of CO₂ to be lower than expected, leading to an underestimation of carbon content.
为获得可靠结果，燃烧装置在使用前必须彻底干燥，样品必须准确称量。氧气必须干燥且不含CO₂。吸收管在燃烧前后分别称重，增加的质量即为生成的H₂O和CO₂。常见误差包括燃烧不完全（生成CO或炭黑）、水吸收不完全或系统泄漏。不完全燃烧会使CO₂的质量偏低，导致碳含量被低估。
10. Common Pitfalls and How to Avoid Them | 常见易错点及应对策略
Using atomic masses incorrectly is a frequent mistake – always use 12.0 for C, 1.0 for H and 16.0 for O unless the question states otherwise. Another pitfall is forgetting to account for the two hydrogen atoms in each H₂O molecule; many students mistakenly use a factor of 1.0/18.0 instead of 2.0/18.0. Also, do not forget to check whether the compound contains oxygen. If the sum of the masses of C and H is less than the original mass, there must be oxygen, and you must calculate its mass before finding moles.
使用错误的原子量是常见错误——除非题目另有说明，通常C取12.0，H取1.0，O取16.0。另一个易错点是忘记H₂O分子中有两个氢原子；不少学生错误地使用了1.0/18.0的系数，而不是2.0/18.0。此外，别忘了检查化合物是否含氧。如果碳和氢的质量之和小于原始质量，必然含有氧，在计算摩尔数之前必须先求出氧的质量。
11. Edexcel Exam-Style Questions | Edexcel考试题型示例
Edexcel often sets questions where you must deduce the empirical formula from combustion data and then use the Mᵣ determined from a mass spectrum or an ideal gas calculation to find the molecular formula. A typical multi‑step question may ask you to:
Edexcel常出的题型是先让你从燃烧数据推导实验式，再结合由质谱或理想气体计算得出的Mᵣ求出分子式。一道典型的多步题可能要求：
- Calculate the mass of C and H from CO₂ and H₂O data.
- 从CO₂和H₂O数据计算碳和氢的质量。
- Determine the mass of oxygen, if present.
- 确定氧的质量（若存在）。
- Work out the empirical formula.
- 求出实验式。
- Use PV = nRT to find Mᵣ from gas density data.
- 利用PV = nRT从气体密度数据求出Mᵣ。
- Scale the empirical formula to the molecular formula.
- 将实验式放大得到分子式。
Practising these linked problems will build your confidence for the Unit 2 or Unit 4 exams.
练习这类连锁题能帮助你在Unit 2或Unit 4考试中树立信心。
12. Summary and Key Takeaways | 总结与核心要点
la. Master the three mass factors (12/44 for C, 2/18 for H, and difference for O) and always express your final formula with whole‑ inally, exam success depends on systematic working, careful rounding and plenty of practice with past Edexcel questions.
燃烧分析是定量化学的直接应用：质量 → 摩尔 → 比 → 实验式。掌握三个质量系数（碳12/44、氢2/18、氧用差额法），并始终用整数下标写出最终化学式。将这一技术与摩尔质量数据结合，即可从实验式过渡到分子式。考试成功取决于系统化的解题步骤、谨慎的取整以及大量Edexcel真题练习。
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