Ethanol weight percentage -> volume percentage

[CaptainPanic]

I'm in need of a table or formula (empirical fit) of the conversion of ethanol/water mixtures weight percentage to volume percentage, and back.

 

I know how to calculate the theoretical conversion (ideal mixtures), but I doubt that ethanol/water can be considered ideal. I also found an online conversion tool, but I'd really prefer to have all data neatly in a table. Does anyone have a reference for me?

 

p.s. If the volume percentage is as simple as taking the volume of the pure ethanol and the volume of the pure water, and then dividing the ethanol volume by the sum of the water and ethanol, then I am just a nut. (My results for that don't seem to correspond with the little data I have).

 

[math]Ethanol\ volume\ percentage = \frac{Pure\ Ethanol\ volume}{Pure\ Ethanol\ volume\ +\ Pure\ Water\ Volume}*100\% [/math]

 

Wikipedia says on volume percentage:

[math] volume\ percent=\frac{volume\ of\ solute}{volume\ of\ solution}*100\% [/math]

which indicates that you do have to take into account the excess volume (the volume change upon mixing)... then again, I am not certain that brewers also use this definition. It's a whole industry, and they might as well have their own conventions.

 

Thanks in advance.

 

p.s. the tool I linked to seems crappy... (Or I completely fail to understand something here).

IMHO, 100% ethanol is 100% vol. and also 100% weight... this converter seems to disagree with me.

Edited July 2, 2008 by CaptainPanic

[ecoli]

it shouldn't be too hard, if you know the density of water/ethanol (which depends on temperature, but that should be known and given in a table somewhere, perhaps even on wikipedia). From there you could convert the volume to mass.

[CharonY]

I think I see what confused you. The formula you put forth is the same as wikipedias, the volume of the solution is equal to ethanol and water together. For most intents and purposes a percent solution is simply calculated against a given total volume (e.g. 1l).

 

What you have to take into account is the density of ethanol which is 0.789 kg/l (at 25°).

What the calculator does wrongly is probably calculating %(w/v) against a kg instead of an liter.

In that case 1l of Ethanol would have a weight of 0.789 kg or 78.9%(w/w) against the 1 kg.

 

But using the density, however, the correct calculation is fairly easy.

[Bignose]

Part of the problem is that the water and ethanol interacts, so it isn't going to be linear.

 

A reference like the CRC Handbook of Chemistry and Physics should have some water/ethanol mixtures.

[CaptainPanic]

Let's use an example - because I am still confused.

 

 

If you add 1 liter of ethanol and 1 liter of water, and you mix, the resulting solution will not be exactly 2 liters.

 

1 liter of water is 0.998 kg.

1 liter of ethanol is 0.789 kg.

 

Mixing would mean that there is 44.2%wt ethanol in the mixture.

 

The density of the resulting solution of 2 liters (if this was linear) would be 0.894 kg/l.

 

However, using Perry's handhook (7th edition), I find that the actual value is 0.92642 kg/l.

 

This means that the 2 liters, upon mixing, were "compacted" to only 1.93 liters.

 

Knowing all this, what is the volume% of the ethanol? (I am just confused about the damned definition - if the answer in this case is 50%, then it means that the definition of %vol. is "the volume percentage after separating all compounds into pure species - if not, then I also don't know what it should be...)

[katmar]

Captain,

 

You are on the correct track with your calculations based on the Perry data. I will come back to this.

 

The problem is that mixing ethanol with water is the microscopic equivalent of a builder mixing a bucket of sand with a bucket of stone and not finishing up with 2 buckets of mixture. In a similar way as the sand fits between the stones, the ethanol and water molecules are able to pack together more tightly.

 

Let's look at the definition of volume %. It is not the volume after separating the components as you suggested. A 40% ethanol by volume solution is made up by (say) taking 40 ml of ethanol and then adding however much water is required to give 100 ml of mixture. If you did this at 20 deg C you would find that you needed to add 63.34 ml of water to bring the total volume to 100 ml. So we have the weird situation where a solution can have 40 vol% of one component and 63.34 vol% of the other. It's simply a function of the way vol % is defined.

 

Getting back to your calculation - As you correctly calculated when you added 1 liter of ethanol to 1 liter of water, you get 1.9289 liters of mix with a mass of 1.787 kg and the mass % is 44.15%. Now that we have the definition of vol% we can work this out to be (100 * 1/1.9289 ) = 51.84 vol%. This is one point on your conversion table and you could work all the way through the Perry data and build up the complete table you need.

 

In fact, it is even more complicated than this. A 50 vol% mixture at 20 deg C would have a strength of 50.91 vol% at 80 deg C because ethanol has a higher coefficient of thermal expansion than water. If you do find a conversion table just be aware of the temperature to which it applies.

 

How do I know all this useless info? I have worked a lot in ethanol distillery design where it is much easier to do all the calculations based on mass%, but for historical reasons the clients always want the answers in vol%. I do not know the site rules here about whether I am allowed to say it here, but I have written and I sell a computer program that does these conversions. The program will work in evaluation mode for 30 days for free, so if you want to download it and draw up the table for yourself it is available from http://katmarsoftware.com Follow the links to AlcoDens.

 

Alternatively you could type the Perry data into a spreadsheet and do it that way.

[CaptainPanic]

Let's look at the definition of volume %. It is not the volume after separating the components as you suggested. A 40% ethanol by volume solution is made up by (say) taking 40 ml of ethanol and then adding however much water is required to give 100 ml of mixture. If you did this at 20 deg C you would find that you needed to add 63.34 ml of water to bring the total volume to 100 ml. So we have the weird situation where a solution can have 40 vol% of one component and 63.34 vol% of the other. It's simply a function of the way vol % is defined.

 

Great post. Exactly what I needed Thanks katmar! (And welcome to the forum!)