Combustion Analysis Combustion analysis determines the empirical formula of molecules by burning them in the presence of excess oxygen. The process involves placing a sample of know weight in the analyzer and recording the weight of both CO2 and H2O that are emitted from the machine. From this information you can calculate the amount of carbon and hydrogen in the sample. However, since oxygen is in excess, you must find oxygen through indirect means (the mass comes from what is not accounted for by carbon and hydrogen, in a sample that only contains CHO). The information you get from this analysis provides you with the empirical formula of the molecule. If there are other elements present in the molecule, such nitrogen or chlorine, you may get information by decomposition analysis (the same process as described for combustion, however a different term is used when it is not carbon or hydrogen).

Combustion Analysis of Vitamin C If 0.200 g of Vitamin C are subject to combustion analysis, the mass of CO2 and water found is 0.2889 g and 0.0819 g respectively. What is the Empirical Formula? Step 1: First determine the mass of carbon in 0.2998g of CO2 by converting to mol C. then to grams of C.

Determining H in Vitamin C Similar fashion for finding H2O.

Determining O in Vitamin C & Percent Composition of Elements Weight of original sample minus the weight of carbon and hydrogen (0.200g –(0.08182 g C + 0.00916 g H) = 0.1090 g O

Combustion and Decomposition While combustion analysis only gives information about carbon, hydrogen, and oxygen indirectly, you can get information about other elements from decomposition analysis. In these types of problems, you will have combustion analysis data from one sample with a specific weight, and decomposition data from another sample with a different weight.

Lysine Example Lysine is an amino acid which has the following elemental composition: C, H, O, N. In one experiment, 2.175 g of lysine was combusted to produce 3.94 g of CO2 and 1.89 of H2O. In a separate experiment, 1.873 g of lysine was burned to produce 0.436 g of NH3. The molar mass of lysine is about 150 g/mol. Determine the empirical and molecular formula of lysine. Step One: determine the mass of each element present. Carbon: 3.94 g x (12.011 / 44.0098) = 1.0753 g Hydrogen: 1.89 g x (2.016 / 18.0152) = 0.2115 g Nitrogen: 0.436 g x (14.007 / 17.031) = 0.3585 g Oxygen: cannot yet be done Why can't the oxygen be determined yet? Because our C, H, and N data come from TWO different sources.