Combustion analysis is a way of determining empirical/molecular formulas of unknown substances. It is based on the following chemical reaction:
Unknown substance (i.e.) CxHyOz + O2(g) =>CO2 (g) + H20(g)
If you remember, the key to finding out the empirical formula is to know the actual mass of each of atoms in the unknown sample. To do this we use two tricks: 1) Conservation of Mass-the mass of each atom into the chemical reaction (from reactants) must equal the mass of that atom out (products) and 2) Law of Difference- if you know the mass of two things, you can find out third if you know total.
Here is how to do it.
Example:
The characteristic odor of pineapple is due to ethyl butyrate. Combustion of 2.78 mg of ethyl butyrate yields 6.32 mg of CO2 and 2.58 mg of H20. Determine the empirical formula of ethyl butyrate.
Using the Conservation of Matter, you can figure out the mass of carbon and hydrogen in the sample since the carbon in the CO2 is solely from the unknown sample and the hydrogen in the water is from the unknown sample.
A trick that works for these problems (as long as all masses are in the same units) is to simply assume the milligrams (mg) are grams (g).
Now convert grams of CO2 to moles and do a molar ratio from moles of CO2 to moles of carbon atoms. Then convert to mass of carbon as follows:
C: (6.32 g CO2)(1 mol CO2/44g CO2)((1 mol C)/(1 mol CO2))(12g C/1 mol C) = 1.72g C
Similarly for hydrogen:
H: (2.58g H20)(1 mol H20/18g H20)((2 mol H)/(1 mol H20))(1g H/1 mol H) = 0.29g H
Now i can use the law of difference, since I know the total sample mass and two of the components:
Total mass - gC - gH = g0
2.78g - 1.72g - 0.29g = 0.77gO
Now that we have the masses of each atom in the unknown sample, we can determine the empirical formula as follows:
Convert actual masses to moles:
C: (1.72g)(1mol/12g) = 0.143 moles C
H: (0.29g)(1mol/1g) = 0.29 moles H
O: (0.77g)(1mol/16g) = 0.0481 moles O
divide the lowest number into all
C: (0.143mol)/(0.0481mol) = 2.9 = 3.0
H: (0.29mol)/(0.0481mol) = 6.0
O: (0.0481mol)/(0.0481mol) = 1.0
Therefore, the empirical formula for ethyl butyrate is C3H6O.
NOW FOR SOMETHING COMPLETELY DIFFERENT ABOUT COMBUSTION REACTIONS