CU-103 / L09 · lecture

Combustion Analysis

~8 min read · Tier toggle changes depth (T1 plain · T2 standard · T3 rigour)

BEFORE YOU START

You must be comfortable converting mass to moles using molar mass and determining empirical formulae from percentage composition data, as covered in the previous lecture on molecular formulae.

Why this matters

Combustion analysis is the primary experimental method for determining the elemental composition of organic compounds.

In organic chemistry, we rarely know the exact structure of a new molecule immediately. Instead, we isolate a pure sample and need to know exactly which atoms it contains and in what proportions. Combustion analysis provides this data by completely burning a known mass of the compound in excess oxygen. The resulting carbon dioxide and water are collected and weighed. By measuring the mass of these products, we can back-calculate the mass of carbon and hydrogen in the original sample. If the compound contains other elements like nitrogen, sulfur, or halogens, specific adaptations to this method allow their detection as well. This technique bridges the gap between a physical sample and its chemical identity, forming the foundation for structural elucidation.

The Core Idea: Burning to Count

Imagine you have a sealed box containing a mixture of carbon and hydrogen atoms, but you cannot see inside. If you burn the contents in pure oxygen, every single carbon atom will end up in a molecule of carbon dioxide, and every single hydrogen atom will end up in a molecule of water. This is the fundamental principle of combustion analysis: complete oxidation ensures a one-to-one correspondence between the atoms in the original compound and the atoms in the products. Carbon does not disappear; it transforms into CO₂. Hydrogen does not vanish; it transforms into H₂O. Oxygen from the air joins the reaction, but since we use excess oxygen, we do not need to measure how much oxygen was consumed. We only need to measure the products. By weighing the CO₂ and H₂O produced, we can determine exactly how much carbon and hydrogen were present in the original sample. This allows us to calculate the empirical formula, which tells us the simplest whole-number ratio of atoms in the compound. For example, if we find that a sample contains 40% carbon, 6.7% hydrogen, and 53.3% oxygen by mass, we can convert these percentages to moles and find the ratio. This method is robust because it relies on the law of conservation of mass and the fixed stoichiometry of combustion reactions. It is a direct, physical measurement that does not rely on theoretical assumptions about the compound's structure, making it an indispensable tool in analytical chemistry for identifying unknown organic substances.

Calculating Carbon and Hydrogen

The first step in combustion analysis is to determine the mass of carbon and hydrogen in the original sample. When a hydrocarbon or organic compound burns, the carbon atoms are converted to carbon dioxide. The molar mass of CO₂ is 44.01 g/mol, and the molar mass of carbon is 12.01 g/mol. Therefore, the mass fraction of carbon in CO₂ is \( \frac{12.01}{44.01} \). If we measure the mass of CO₂ produced, we can find the mass of carbon by multiplying the mass of CO₂ by this fraction. Similarly, hydrogen is converted to water. The molar mass of H₂O is 18.02 g/mol, and the molar mass of hydrogen is 1.008 g/mol. However, since there are two hydrogen atoms in each water molecule, the mass of hydrogen in the water is \( \frac{2 \times 1.008}{18.02} \) of the total mass of water. By applying these factors to the measured masses of the combustion products, we obtain the precise masses of carbon and hydrogen in the original sample. This step is purely arithmetic but requires careful attention to significant figures and unit consistency. For instance, 1.00 g of hexane (\(\mathrm{C_6H_{14}}\)) burns to 3.064 g of CO₂ and 1.464 g of H₂O. The mass of carbon is \( 3.064 \times \frac{12.01}{44.01} = 0.836 \) g and the mass of hydrogen is \( 1.464 \times \frac{2.016}{18.02} = 0.164 \) g. These values are the actual masses of those elements in the starting material.

\[ m_{\mathrm{C}} = m_{\mathrm{CO_2}} \times \frac{M_{\mathrm{C}}}{M_{\mathrm{CO_2}}}, \qquad m_{\mathrm{H}} = m_{\mathrm{H_2O}} \times \frac{2 M_{\mathrm{H}}}{M_{\mathrm{H_2O}}} \]

Determining Oxygen and Other Elements

Once the masses of carbon and hydrogen are known, we can determine the mass of any other elements present in the compound. In many organic compounds, oxygen is the third element. Since oxygen is also present in the combustion products (CO₂ and H₂O), we cannot measure it directly from the products. Instead, we use the law of conservation of mass. The total mass of the original sample must equal the sum of the masses of all elements in it. If the compound contains only carbon, hydrogen, and oxygen, the mass of oxygen is simply the total mass of the sample minus the masses of carbon and hydrogen. For the hexane example above, \( 1.00 - 0.836 - 0.164 = 0.000 \) g, so there is no oxygen, as expected for a hydrocarbon. A clearly negative difference would signal an experimental error. In practice, if the sum of carbon and hydrogen masses is less than the sample mass, the difference is attributed to oxygen. If the sum is greater than the sample mass, it suggests impurities or measurement errors. For compounds containing nitrogen, sulfur, or halogens, specific detection methods are used. Nitrogen is often converted to N₂ or NO_x, sulfur to SO₂, and halogens to their respective ions. These elements are then quantified using specific chemical tests or instrumental methods, and their masses are subtracted from the total sample mass along with carbon and hydrogen to find the mass of oxygen, if present.

\[ m_{\mathrm{O}} = m_{\text{sample}} - m_{\mathrm{C}} - m_{\mathrm{H}} - m_{\text{other elements}} \]

From Masses to Empirical Formula

With the masses of all elements in the compound determined, we can now calculate the empirical formula. This involves converting each mass to moles using the atomic molar masses. For carbon, divide the mass by 12.01 g/mol. For hydrogen, divide by 1.008 g/mol. For oxygen, divide by 16.00 g/mol. Once we have the number of moles of each element, we divide each value by the smallest number of moles among them to get the simplest whole-number ratio. This ratio gives the subscripts in the empirical formula. If the ratios are not close to whole numbers, we may need to multiply all ratios by a small integer to achieve whole numbers. This process is identical to determining empirical formulae from percentage composition, as cited in the previous lecture. The empirical formula tells us the simplest ratio of atoms, but not the actual number of atoms in a molecule. To find the molecular formula, we need the molar mass of the compound, which can be determined by other methods such as mass spectrometry or boiling point elevation. The molecular formula is always a whole-number multiple of the empirical formula. Combustion analysis thus provides the essential elemental composition needed to identify the compound's structure.

\[ n_{\mathrm{C}} = \frac{m_{\mathrm{C}}}{12.01}, \quad n_{\mathrm{H}} = \frac{m_{\mathrm{H}}}{1.008}, \quad n_{\mathrm{O}} = \frac{m_{\mathrm{O}}}{16.00} \]

The result

◆ RESULT · T-014 · full treatment on its own page

Empirical and molecular formulae

Given a compound's percentage composition by mass, converting each element's mass to moles (\(n=m/M\), mole-avogadro) and dividing by the smallest mole value gives the simplest whole-number mole ratio between elements — the empirical formula. The true molecular formula is always a whole-number multiple of the empirical formula, with the multiplier found by comparing the compound's actual (experimentally measured) molar mass to the empirical formula's mass.

\[ r_i = \frac{n_i}{\min_j n_j}\ \xrightarrow{\text{clear fractions, round}}\ \text{empirical formula}, \qquad k=\frac{M_{\text{actual}}}{M_{\text{empirical}}}\ \Rightarrow\ \text{molecular formula} \]

Open the full result →

Tier 3 · watch out

A common misconception is that the oxygen collected in the CO₂ and H₂O measures the oxygen in the compound. It does not: product oxygen comes partly from the compound and partly from the O₂ supplied, so the compound\'s oxygen can only be found by difference, assuming no other elements are present. If the compound also contains nitrogen, sulfur or a halogen, those must be determined separately before taking the difference; otherwise the oxygen mass, and the empirical formula, will be wrong. C

Worked example

Problem. A 0.500 g sample of an organic compound containing only carbon, hydrogen, and oxygen is combusted. The products are 1.10 g of CO₂ and 0.453 g of H₂O. Determine the empirical formula of the compound.

Calculate the mass of carbon in the CO₂: \( m_{\mathrm{C}} = 1.10 \times \frac{12.01}{44.01} = 0.300 \) g.
Calculate the mass of hydrogen in the H₂O: \( m_{\mathrm{H}} = 0.453 \times \frac{2.016}{18.02} = 0.0507 \) g.
Calculate the mass of oxygen by difference: \( m_{\mathrm{O}} = 0.500 - 0.300 - 0.0507 = 0.149 \) g.
Convert masses to moles: \( n_{\mathrm{C}} = \frac{0.300}{12.01} = 0.0250 \) mol, \( n_{\mathrm{H}} = \frac{0.0507}{1.008} = 0.0503 \) mol, \( n_{\mathrm{O}} = \frac{0.149}{16.00} = 0.00931 \) mol.
Determine the simplest ratio by dividing by the smallest value (0.00931): C: \( \frac{0.0250}{0.00931} = 2.68 \), H: \( \frac{0.0503}{0.00931} = 5.40 \), O: \( \frac{0.00931}{0.00931} = 1.00 \). Multiply by 3 to get whole numbers: C: 8.04, H: 16.2, O: 3.00. Round to nearest whole numbers: C₈H₁₆O₃.

\[ \boxed{\mathrm{C_8H_{16}O_3}} \]

Units check. The masses are in grams, molar masses in g/mol, so moles are dimensionless ratios, yielding a valid empirical formula.

CHECK YOURSELF

Three quick tests

What is the mass fraction of carbon in CO₂?
Answer
0.273
If a sample contains 40% C, 6.7% H, and 53.3% O, what is the empirical formula?
Answer
CH₂O
Why do we use excess oxygen in combustion analysis?
Answer
To ensure complete combustion and simplify calculations by not needing to measure oxygen consumption.

UNIT

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NOTES

Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.

Results cited: T-014 · Threads: structure