Frequently Asked Questions
How does combustion analysis determine a molecular formula?
Burning an organic compound in excess oxygen converts all carbon to CO2 and all hydrogen to H2O. By weighing the absorbed CO2 (in a CO2 trap) and H2O (in a desiccant tube), the masses of C and H in the sample are calculated. Any remaining mass is assumed to be oxygen (or other elements analyzed separately). The mole ratios give the empirical formula.
What is the empirical formula and how is it different from the molecular formula?
The empirical formula gives the simplest whole-number ratio of atoms. The molecular formula gives the actual count of atoms per molecule. Combustion analysis gives the empirical formula; you need the molar mass (from mass spectrometry or the ideal gas law) to find the molecular formula. Example: empirical CH2O (MW 30) could be formaldehyde (CH2O), acetic acid (C2H4O2), glucose (C6H12O6) etc.
Why is oxygen calculated by difference in combustion analysis?
Oxygen from the sample combines with carbon and hydrogen during combustion, so it ends up split between CO2 and H2O and cannot be separately collected. Instead: mass O = sample mass - mass C - mass H - mass N (from separate Dumas analysis). This subtraction works because all other elements are accounted for, and the remainder must be oxygen.
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