I am making vinegar by mixing vinegar (containing the mother) and wine, and then allowing fermentation to occur. I am wondering how to calculate the final % acidity.


  1. I know the initial concentration (by volume) of acetic acid, X, alcohol Y
  2. All alcohol will be converted to acetic acid
  3. No sugar is converted to acetic acid

How do I calculate the final acidity?

  • Which kind of percentage are we doing here? % by Volume, % by Weight, or % by moles? The answer will be different for each, and in some cases depend on temperature (volumes change with temperature). – BobMcGee Jun 16 '12 at 2:25
  • As stated in the question, % by volume. I'm of course only looking at an approximation – Marc-André Lafortune Jun 16 '12 at 3:25
  • I think my confusion comes from having a chemistry background, where the terminology is used somewhat differently. This also means that if I can find a good half hour to look up densities, it won't be hard to answer. In any case, I have taken the liberty of rewriting it to clearly state the assumptions. – BobMcGee Jun 16 '12 at 4:10
  • @BobMcGee Percent by volume is by no means uncommon in everyday usage. For example, it's pretty common for alcohol. I'm not sure "concentration (by volume)" is really more understandable, even though scientists wouldn't use percentages. – Cascabel Jun 16 '12 at 4:24
  • @Jefromi: I know what volume percent is, but "% per volume" just seems awkward, rather than "concentration (by volume)". Many concentrations are per volume... moles of solute per volume solvent, grams of solute per volume solvent. Even if you don't think the point is significant, I think you can agree the rewritten question is clearer. – BobMcGee Jun 16 '12 at 4:30

This is actually a great chemistry question! First off, you need the density and molecular weight of the acetic acid (1.039 g/mL, 60.05 g/mol) and alcohol (which is ethanol — 0.709 g/mL, 46.07 g/mol). Assuming 100% conversion of ethanol (y) to acetic acid (x), you will end up with the same number of moles of acetic acid as the amount of ethanol you started with.

So if you started with y mL ethanol, you would have 0.709/(46.07 * y) moles ethanol.

Since we are assuming 100% conversion to acetic acid, we end with the same number of moles acetic acid, which we can then convert back to mL. mL acetic acid = 60.05/(moles acetic acid/ethanol * 1.049).

If we condense all that into one calculation, you end up with: mL acetic acid = 1.753 * mL ethanol.

If you add the volume of acetic acid you made in the fermentation process and the amount you started with, you have the total volume of acetic acid in your vinegar. Simply divide this by the total volume of vinegar to get the % acetic acid!

  • Doesn't the reaction that produces acetic acid also produce water? – Cascabel Jun 18 '12 at 2:48
  • Yes it does, the reaction produces 1 mole of acetic acid and 1 mole of water for each mole of alcohol though. C2H5OH + O2 → CH3COOH + H2O – scientifics Jun 18 '12 at 3:08
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    I get the impression that the OP wants to be able to figure this out before making the vinegar, so calculations that don't include that (i.e. depend on measuring the resulting volume) aren't a complete answer. – Cascabel Jun 18 '12 at 4:13
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    I should add then how to convert initial concentration by volume to volume. Since % by volume is just volume alcohol or acetic acid/total volume, it is easy to convert to mL. – scientifics Jun 18 '12 at 13:21

Go to a home winemaking supply shop or www.countrywines.com. Buy an Acid Test Kit. Dilute homemade vinegar: 1 ounce vinegar with 9 ounces water (distilled, preferably). Follow directions in the acid test kit multiplying neutralizer used by 0.075 as indicated in the test kit instructions.
Multiply this result by 8 to account for the dilution to get your end result: Total Acid expresed as percentage tartaric acid.

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