brad
Active member
Everyone says use 5 times as much vinegar as vodka, but no one says whether this is by weight, volume, mole or other unit. So I went through the math:
1 liter of 80 proof vodka, assuming that is 31.571% ethanol by weight and the rest is water = 315.71 g ethanol = 6.85283264597351855871 moles ethanol = 13.70566529194703711743 moles Carbon
1 liter of distilled white vinegar = 52.45 grams acetic acid (assuming 1.049 density of acetic acid and that the vinegar is 5% acetic acid by volume*) = 0.8734388009991673605 moles acetic acid = 1.746877601998334721 moles Carbon.
So 1 liter of vodka is closer to 7-8 liters of vinegar, 7.84580744310218855036 to be exact.
1 liter of vodka is 914.8 grams (assuming a specific gravity of ethanol of .787) and 1 liter of vinegar is 1002.45 grams. So 1 gram of vodka is 8 or 9 grams of vinegar, 8.5975400867269227288 to be exact.
By mole is easy, with 1 mole of acetic acid adding 2 moles of Carbon, which is the same as from 1 mole of ethanol.
Can someone verify my math is correct (or close)?
*I don't know, but am guessing vinegar is actually 5% by weight. The weight of acetic acid and water are close enough the result is almost the same, but the math is much easier doing it by volume.
1 liter of 80 proof vodka, assuming that is 31.571% ethanol by weight and the rest is water = 315.71 g ethanol = 6.85283264597351855871 moles ethanol = 13.70566529194703711743 moles Carbon
1 liter of distilled white vinegar = 52.45 grams acetic acid (assuming 1.049 density of acetic acid and that the vinegar is 5% acetic acid by volume*) = 0.8734388009991673605 moles acetic acid = 1.746877601998334721 moles Carbon.
So 1 liter of vodka is closer to 7-8 liters of vinegar, 7.84580744310218855036 to be exact.
1 liter of vodka is 914.8 grams (assuming a specific gravity of ethanol of .787) and 1 liter of vinegar is 1002.45 grams. So 1 gram of vodka is 8 or 9 grams of vinegar, 8.5975400867269227288 to be exact.
By mole is easy, with 1 mole of acetic acid adding 2 moles of Carbon, which is the same as from 1 mole of ethanol.
Can someone verify my math is correct (or close)?
*I don't know, but am guessing vinegar is actually 5% by weight. The weight of acetic acid and water are close enough the result is almost the same, but the math is much easier doing it by volume.
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