we must write empirical formula of $\ce{CH3COOH}$ as $\ce{CH2O}$ or $\ce{COH2}$ ? Alike ? Any nomenclature?

Somebody told me it was because of bonding, different way of writing the chemical formula mean different way or bonding, but I'm not sure if it is correct or not,


The reason that acetic acid's chemical formula is often written as $\ce{CH3COOH}$ is that the $\ce{COOH}$ portion of the formula indicates that the compound is a carboxylic acid. You'll also see it written as $\ce{HC2H3O2}$ sometimes, again to indicate an acid, this time through the leading $\ce{H}$.

An empirical formula is only intended to give the relative number of atoms in a compound; writing the formula to indicate the type of compound is not usually done. I would typically follow something like the Hill System for this purpose: carbon, hydrogen, then other elements in alphabetical order, so $\ce{CH2O}$ here. Empirical formulas are never named.

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    $\begingroup$ Even though your answer is right, I find the idea of using the empirical formula rather ugly. The normal thing to do for an organic compound would be to express the molecular formula in the Hill System: C2H4O2. There is much to be lost and little to be gained from dividing everything by 2 to get the empirical formula CH2O, though this was indeed what the OP asked for. On the other hand, for inorganic substances it's more normal to see, for example, the empirical formula (P2O5 or S) than the molecular formula (P4O10 or S8.) $\endgroup$ – Level River St Jun 8 '15 at 12:27
  • $\begingroup$ @steveverrill I agree that finding an empirical formula from a molecular formula is very nearly pointless, particularly for organic compounds. The empirical formula's usefulness is primarily in identifying compounds, for example from combustion analysis. Working from empirical to molecular makes sense, but I know from experience that many chemistry instructors do not emphasize that aspect of it and instead teach it as an analogue of reducing fractions to simplest form. I considered addressing that in my answer, but the question really doesn't ask about why the empirical formula exists. $\endgroup$ – Jason Patterson Jun 8 '15 at 13:49

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