4.6 Combustion Analysis
Combustion Analysis:
Combustion analysis is a method used in both organic chemistry and analytical chemistry to determine the elemental composition (i.e. the empirical formula) of a pure organic compound by combustion reaction of the sample and the products of this combustion will be quantitative analyzed.
sample
Figure 4.31 Combustion Analysis Set up
Reference:http://www.chem.wilkes.edu/~mencer/combustion/combust_app.htm
| The combustion reaction is carried out under the excess of oxygen. Oxygen is passed through a closed tube which is heated. The carbon and hydrogen present in the sample are converted into carbon dioxide CO2 and water H2O. The excess O2 helps to push through the products through the tube. Inside the tube, the products pass through copper-II-oxide, CuO solid. CuO is acting as a catalyst (similar to Nick, Platinum and Palladium compounds act as catalysts in the catalytic converter in the vehicles). CuO catalyzes the conversion of any leftover traces of carbon and hydrogen and into CO2 and H2O accordingly. |
| H2O is collected in an absorber filled with magnesium perchlorate, Mg(ClO4)2. The change in mass of this absorber is followed and observed. The mass change is equal to the mass of water produced during the combustion reaction.CO2 is collected in a separate absorber filled with sodium hydroxide, NaOH. The change in mass of this absorber is equal to the mass of carbon dioxide produced during the combustion process.Nitrogen and oxygen (N or O) masses if present can be determined by the difference between the mass of combusted sample and the sum of the masses of the carbon in the carbon dioxide and hydrogen in the water (the masses of the C and H are found using stoichiometric calculations). |
The video below illustrates the elemental analysis of a compound:
Example:
Elemental analysis of an organic compound using the combustion reaction reveals the results below:
2.6406 g CO2
0.5400 g H2O
The molecular formula is to be determined if the molar mass of this organic compound is 78.11 g/mol
The combustion reaction is given below:
CxHy + (excess) O2 à (x) CO2 + (y/2) H2O
CO2 grams are converted into mol C (x)
H2O grams are converted into mol H (y)
The empirical formula is CH
The empirical formula has the molar mass = CH = 12.0 g/mole + 1.00 g/mol = 13.0 g/mole
The molar mass of the organic compound = 78.11 g/mole
The empirical formula unit = [molar mass of the compound] / [molar mass of empirical formula]
[The empirical formula unit] = [78.11 g/mole] / [13.0 g/mole] = 6 units
The molecular formula = 6 x CH = C6H6
One can confirm the result by calculating the molar mass of C6H6 which is then calculated to be 78.0 g/ mole.