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!