I am making Limoncello, I have added 8 lemon zests to 1 liter of 96% alcohol and put everything in a big glass container.

After 1 month I started making the real liqueur by diluting the alcohol down with a sugar syrup. The sugar syrup consists of a dilution of 800 grams sugar in 1 liter of water.

I've put the big container on a scale, used the tare button to put it to zero. In all my silliness I've decided to use the scale/a weight factor in order to 'measure' my dilution - so I went on and added 1 kilogram of my sugar syrup.

I think that the next time it would be better to measure an amount of 1 liter of sugar syrup instead of 1 kilogram - but I am still thinking about this.

Now I am calculating like crazy, but I cannot really figure it out. The most simple and erroneous calculation would be to say:

1 L alcohol + 1 Liter/Kilogram sugar syrup = 2 Liter/Kilograms
1 L alcohol = 96%
2 L mixture = 96 / 2 = 48%

But I know this is wrong. I however cannot figure out how and why this is wrong. Diluting alcohol with water will work this way, I guess. But the added parts sugar to the water makes it hard. So the questions are:

  1. How much alcohol do I have in my mixture?
  2. How do I calculate alcohol percentages after diluting with a certain sugar syrup concentration?

Because alcohol concentration is measured in terms of volume, the key question in this is:

What is the volume of 1kg of sugar syrup?

Assuming dissolving the sugar in water doesn't change the total volume much, your liter of sugar syrup weighs 1kg + 800g = 1.8kg. Assuming it was perfectly mixed, the kilogram you added was only 1/1.8 = 0.555... = 55.6% water.

So your total volume should be 1.556 liters.

That said, if I'm wrong you can also measure the volume you actually have and use that amount.

Now in terms of alcohol, one liter of that total 1.556 liters was 96% alcohol, so you total ABV should be

   1L * 0.96ABV
------------------ = 0.6169 ABV = 61.69%
1.556L (or actual)

Again, adjust for your total volume if necessary. But it should be close to 62%.

Edit: it should be noted, per the discussion in the comments, that the volume may change with the ABV. However, this should be fairly minimal and likely offset/limited by the sugar in the syrup you added.

There doesn't seem to be a lot of data on mixing alcoholic substances with anything but pure water for dilution, so giving an exact answer without running an experiment would be difficult. That said, the method above should be a fairly good estimate given everything I've seen.

  • You forgot important factor - dissolving alcohol changes volume, depending on concentration. 2 liters of 30% ABV will contain different amount of alcohol than one liter of 60% ABV. – Mołot Nov 14 '15 at 19:20
  • @Molot - It likely does to some degree, but dividing volume of alcohol (0.96L) by the total volume afterward, which is directly measurable, will still yield the ABV. – thesquaregroot Nov 14 '15 at 19:22
  • @Molot Looking into it more, it seems this is definitely more complicated. From Wikipedia: " Mixing two solutions of alcohol of different strengths usually causes a change in volume. Mixing pure water with a solution less than 24% by mass causes a slight increase in total volume..." The sugar makes this especially complicated though. It's unclear whether this would apply given how saturated the water is with other molecules. Given this, I still think dividing 0.96 by the actual volume would give a decent approximation, but it's difficult to determine what the error would be. – thesquaregroot Nov 14 '15 at 21:42
  • 1
    @Molot - I found this chart alcbevtesting.com/wp-content/uploads/2009/05/… and if we use the first column as a percentage increase in volume we can find an approximate final ABV by row. Assuming the difference in volume is similar to what I calculated I think it puts us in pretty much the same range. The sugar is still a tough factor but I would think the potential extra volume from sugar would reasonably counteract the decrease in volume due to the alcohol. Regardless, the differences seem to be fairly minimal, maybe with an error of about 2-3%. – thesquaregroot Nov 14 '15 at 22:31
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    @Molot - I've edited my answer to include some caveats based on our discussion here; thanks for keeping me honest! – thesquaregroot Nov 15 '15 at 16:06

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