Combustion analysis is a foundational technique in analytical chemistry used to determine the elemental composition of an unknown organic compound. This method focuses primarily on finding the percentage composition of carbon and hydrogen, which are common to most organic molecules. By precisely measuring the amounts of carbon dioxide (\(\text{CO}_2\)) and water (\(\text{H}_2\text{O}\)) produced when a sample is burned, chemists deduce the simplest whole-number ratio of atoms in the compound. The goal of this process is to establish the empirical formula, which provides the atomic ratio of the elements present.
The Essential Combustion Apparatus
The specialized equipment used for this analysis is often referred to as a combustion train, designed to ensure complete combustion and quantitative collection of the products. At the heart of the apparatus is a furnace or combustion chamber, which is heated to a high temperature, typically between 800 and 1000 degrees Celsius, to ensure the sample is fully oxidized. A continuous stream of pure oxygen is supplied to the furnace, acting as the oxidizing agent to convert all carbon to carbon dioxide (\(\text{CO}_2\)) and all hydrogen to water vapor (\(\text{H}_2\text{O}\)).
The combustion products are then directed through a series of absorption traps, which are a distinctive feature of the apparatus. The first trap in the flow path contains a highly hygroscopic material, such as magnesium perchlorate (\(\text{Mg}(\text{ClO}_4)_2\)) or Drierite (anhydrous calcium sulfate, \(\text{CaSO}_4\)). This trap is specifically designed to absorb all the water vapor produced during the reaction.
Following the water trap, the gas stream passes into a second trap containing a strong base, such as Ascarite (sodium hydroxide-coated silica). This chemical reacts with and captures all the carbon dioxide gas that was generated. The entire system is meticulously sealed to ensure that the only change in mass for each trap is due to the mass of the absorbed combustion product. By weighing these traps before and after the experiment, the mass of water and the mass of carbon dioxide produced are determined with high precision.
Executing the Combustion Analysis Procedure
The procedure begins with the careful preparation of the sample, which involves accurately weighing a small amount of the unknown compound using an analytical balance. This initial mass of the sample is a fundamental data point required for the final calculations. The pre-weighed sample is then placed into a boat or capsule and introduced into the high-temperature combustion chamber.
Before the sample is combusted, the absorption traps are also weighed precisely and connected in sequence to the outlet of the furnace. Once the sample is in the chamber, the oxygen flow is initiated, and the heat is applied to start the combustion reaction. The sample burns rapidly, and the gaseous products are swept by the oxygen stream through the remainder of the combustion train.
As the products pass through the train, the water vapor is retained by the first trap, and the carbon dioxide is captured by the second trap. After the combustion is complete, the system is flushed with oxygen to ensure all products have reached the traps. The two absorption traps are carefully removed and weighed again. The difference between the initial and final mass of the water trap is the exact mass of \(\text{H}_2\text{O}\) produced, and the mass increase of the carbon dioxide trap is the exact mass of \(\text{CO}_2\) produced. These two mass measurements represent the collected data from the physical experiment.
Translating Mass Measurements to Empirical Formulas
The transition from physical measurements to the empirical formula requires a series of stoichiometric calculations. The first step involves converting the mass of each combustion product into the mass of the original element it contains.
Since every carbon atom in the original compound is converted into one \(\text{CO}_2\) molecule, the mass of carbon in the sample is calculated using the molar mass ratio of carbon to carbon dioxide (\(12.01 \text{ g}/\text{mol}\) divided by \(44.01 \text{ g}/\text{mol}\)). Similarly, the mass of hydrogen is found using the molar mass ratio of two hydrogen atoms to one water molecule (\(2 \times 1.008 \text{ g}/\text{mol}\) divided by \(18.02 \text{ g}/\text{mol}\)). This accounts for the two hydrogen atoms present in every \(\text{H}_2\text{O}\) molecule.
For a hypothetical example, if the experiment yielded \(4.401 \text{ g}\) of \(\text{CO}_2\), this would contain \(1.201 \text{ g}\) of carbon. If \(1.802 \text{ g}\) of \(\text{H}_2\text{O}\) was collected, this would contain \(0.202 \text{ g}\) of hydrogen.
Determining Mass of Oxygen (By Difference)
If the original compound contained only carbon and hydrogen, the analysis would be complete. However, many organic compounds also contain a third element, most commonly oxygen. The mass of this third element is determined using the ‘mass by difference’ method. The calculated masses of carbon and hydrogen are subtracted from the initial mass of the unknown sample.
If the initial sample mass was \(2.000 \text{ g}\) and the calculated masses of C and H were \(1.201 \text{ g}\) and \(0.202 \text{ g}\) respectively, the mass of oxygen would be \(2.000 \text{ g} – 1.201 \text{ g} – 0.202 \text{ g}\), resulting in \(0.597 \text{ g}\) of oxygen.
Finding the Empirical Formula
Once the mass of every element (\(\text{C}\), \(\text{H}\), and \(\text{O}\)) in the original sample has been determined, the next step is to convert each mass into the number of moles by dividing by the element’s atomic mass. These mole values represent the true mole ratio of the elements in the compound.
Continuing the example:
- \(1.201 \text{ g}\) of \(\text{C}\) is \(0.100 \text{ moles}\).
- \(0.202 \text{ g}\) of \(\text{H}\) is \(0.200 \text{ moles}\).
- \(0.597 \text{ g}\) of \(\text{O}\) is approximately \(0.0373 \text{ moles}\).
The final stage is to find the simplest whole-number ratio. This is accomplished by dividing the mole amount of every element by the smallest mole value calculated. In the example, the smallest value is \(0.0373 \text{ moles}\) (oxygen). The resulting ratios are \(\text{C}\) (\(0.100/0.0373 \approx 2.68\)), \(\text{H}\) (\(0.200/0.0373 \approx 5.36\)), and \(\text{O}\) (\(0.0373/0.0373 = 1.00\)). Since these values are not whole numbers, the entire set must be multiplied by a small integer (in this case, three) to yield the simplest whole-number ratio, giving \(\text{C}_8\text{H}_{16}\text{O}_3\) as the empirical formula.