by-volume/by-weight conversion formula

Question

Okay, because I have nothing better to do on a Friday night (i.e., my carboys are all currently fermenting), I've decided to fiddle a little bit with math.

I know that 0% alcohol by weight (ABW) is also 0% alcohol by volume (ABV), and I know that 100% ABW is 100% ABV. I also know that, in between those two extremes, the relationship is not linear (e.g., 3.2% ABW is approximately 4.0% ABV). However, any web-site calculator that I see pretty much performs the calculation as linear.

I've been fiddling with my TI-89 from high school, and I think that I have it figured out. Assuming ethyl alcohol has a density of .789 kg/l and water has a density of 1 kg/l and assuming that water and alcohol are the only substances in an alcoholic beverage (which I know isn't true), the formula is:

ABV = ABW ÷ (.211 · ABW + .789)

The problem is that I can't confirm my formula. I'm searching Google and can't find much on the topic. Well… I can find plenty of calculators, but they all seem to be using a linear formula.

So… is this formula correct?

---

EDIT: So, I've done a little more tweaking on this (after I already got the accepted answer). I already made the assumption that the density of water is 1 kg/l. The density that I used for ethyl alcohol was for 20°C. At that same temperature, water has a density of approximately .99823 kg/l and not 1 kg/l. Using this, I arrived at a different formula that's probably more accurate (at 20°C, that is):

ABV = (99823 · ABW) ÷ (20923 · ABW + 78945)

EDIT: Please see AlkonMikko's comment.

Answer 1

I'm a Chemical Engineer and it frustrates me that everywhere on the internet appears to be using this linear approximation. Yeah it's probably okay for hobby brewers, but for anyone actually interested in solving the problem for example use in software design or higher ABV products, this approximation is really not good enough.

This is actually a ridiculously complicated subject and it all boils down to the fact that at a molecular level, water and ethanol interact with one another with what are known as hydrogen bonds. Now you can just use an empirical lookup table or regress a tonne of data, but if you're interested in the root of this problem you should do some research into what are known as equations of state. In particular UNIQUAC for ethanol-water mixtures. It get's even worse when you realise that these equations of state describe gaseous behaviour remarkably well, but require additional thought to be applied to liquids.

People dedicate their entire lives into perfecting a generalised equation of state to describe the interactions between all chemical components. ANYWAY I digress...

I've done some number crunching myself using UNIQUAC and have come up with the following simple formula. Like yours above, it's only applicable to ethanol-water mixtures, and I've only validated this equation up to an ABV of 50% so be weary of going higher. This equation has an R-Squared value of 1.

ABW = 0.1893*ABV*ABV + 0.7918*ABV + 0.0002

Try it yourself :)

Answer 2

It is easier to reason about how to derive the alcohol-by-weight value from the alcohol-by-volume rather than the reverse.

If abv is the alcohol-by-volume value expressed as a number between 0 and 1, then for a unit volume of the liquid, the weight of the alcohol will be 0.789 * abv. Similarly, the weight of the non-alcohol component will be 1 - abv (assuming it has a density of 1). So the total weight of the liquid will be:

  0.789 * abv + (1 - abv)
= 1 - 0.211 * abv

Using the weight of alcohol and the total weight, we can easily determine the alcohol-by-weight (also expressed as a number between 0 and 1):

abw = 0.789 * abv / (1 - 0.211 * abv)

So this is clearly not linear (it is hyperbolic), and maintains identity for the 0% and 100% cases as expected. We can invert the equation in a few steps:

                    0.789 * abv = abw * (1 - 0.211 * abv)
                    0.789 * abv = abw - 0.211 * abv * abw
0.789 * abv + 0.211 * abv * abw = abw
    (0.789 + 0.211 * abw) * abv = abw
                            abv = abw / (0.789 + 0.211 * abw)

So that confirms the formula you derived. Wolfram Alpha seems to agree.

Answer 3

According to Alcohol by volume (Wikipedia):

ABV * 0.78924 = ABW * SpecificGravity(at 20°C in g/ml)

Thus ABV = ABW * SpecificGravity / 0.78924

This formula is only correct for a mixture of ethanol and pure water. You can't plug in the s.g. of your beer and have it work.

I have found tables that relate ABW to specific gravity for ethanol/water solutions, one can be found here

I dumped the table into a spreadsheet, and used the formula to create a conversion table for ABV and ABW.

Please note that the formula is not completely accurate against the table, 100% ABW calculates to 100.01% ABV. 0.01% difference is not really significant for our purposes. Not sure if the table or formula is incorrect.

Answer 4

The table prduced by Evil Genius should produce the correct answer. The error reported on 0.01% for 100% ABW is because the siource table given density of 100% alcohol as 0.78934 and he divided by 0.78924. Either the table or his divisor must be wrong.