I wanted to use my standard hydrometer for measuring the sugar content of a number of drinks.
But alcohol is well lighter than water, ~20% less weight.
It should have an impact of the SG reading.
So I imagine that if I have a commercial drink that is supposed to have 10% ABV, and a commercial drink that is supposed to have 32% ABV, and if I measure SG=1.050 for both of them, in reality the sugar content is not the same.
I searched a bit but couldn’t find a reference, a formula or a chart.
Any advise about how to correct gravity readings for drinks with high alcohol content?

You can’t calculate the ABV using a hydrometer with one measurement. You need to measure pre and post fermentation I’m afraid.

Originally posted by Danko
You can’t calculate the ABV using a hydrometer with one measurement. You need to measure pre and post fermentation I’m afraid.
but that’s not my question...

Hydrometers measure gravity, I believe, which I think is more like density of the liquid. I don’t think you can use a hydrometer without a ton of calculations to just figure out the amount of sugar in a given drink. When we use it to measure alcohol in homebrew, we do a before and after because we know the variable changing will be sugar to alcohol, etc., and even that isn’t exact.


Originally posted by drowland
Hydrometers measure gravity, I believe, which I think is more like density of the liquid. I don’t think you can use a hydrometer without a ton of calculations to just figure out the amount of sugar in a given drink. When we use it to measure alcohol in homebrew, we do a before and after because we know the variable changing will be sugar to alcohol, etc., and even that isn’t exact.
No he is right. This is doable. The hydrometer will measure the density of the liquid, but the amount of alcohol in the final product will have an effect on the percentage of that reading that is created by the actual sugar content. Unfortunately, I don’t really know an appropriate correction. Let me look around for a minute and see if I can find anything.

Ok, I derived an equation. Math in the morning isn’t always the best, so someone check this for accuracy, but this should work.
Convert your measured gravity to deg Plato, which is more directly related to sugar concentration (essentially 1 deg Plato = 1 g/mL of sugar in water).
Sugar concentration (in g/mL) = (Plato) / [(1  %abv) + %abv(.79)]
So, using your examples:
1.050 = 12.4 P
the 10%abv would be 12.4/[(1.1) +.1(.79)] = 12.67 g/mL of sugar
the 32%abv would be 12.4/[(1.32) + .32(.79)] = 13.3 g/mL of sugar
Of course, this ignores anything else in the solution (proteins, esters, whatever), but those should be in small enough concentrations that you won’t have to worry about it unless you want more precision than above.

Originally posted by tronraner Ok, I derived an equation. Math in the morning isn’t always the best, so someone check this for accuracy, but this should work. Convert your measured gravity to deg Plato, which is more directly related to sugar concentration (essentially 1 deg Plato = 1 g/mL of sugar in water). Sugar concentration (in g/mL) = (Plato) / [(1  %abv) + %abv(.79)] So, using your examples: 1.050 = 12.4 P the 10%abv would be 12.4/[(1.1) +.1(.79)] = 12.67 g/mL of sugar the 32%abv would be 12.4/[(1.32) + .32(.79)] = 13.3 g/mL of sugar Of course, this ignores anything else in the solution (proteins, esters, whatever), but those should be in small enough concentrations that you won’t have to worry about it unless you want more precision than above. Aren’t you ignoring the fact that the liquid consists of more than just alcohol and water? It seems so considering this expression: [(1  %abv) + %abv(.79)]. It should be doable if you know or can approximate the gravity/density (kg/m^3) of the sugars. Otherwise both the percentage of sugar in the drink and the gravity of those sugars will be unknown and the only thing you could solve out of the equation would then be the product of the two. Inb4 anybody correcting me: I’m considerably hung over.

Originally posted by VsXsV
Aren’t you ignoring the fact that the liquid consists of more than just alcohol and water? It seems so considering this expression: [(1  %abv) + %abv(.79)]. It should be doable if you know or can approximate the gravity/density (kg/m^3) of the sugars. Otherwise both the percentage of sugar in the drink and the gravity of those sugars will be unknown and the only thing you could solve out of the equation would then be the product of the two. Inb4 anybody correcting me: I’m considerably hung over.
Yes, I intentionally ignored the contribution of the sugar to the volume. I didn’t calculate it, but I’m pretty sure that the volume contribution of 12g of sugar to 100 mL of liquid is not significant enough to warrant making the equation much more cumbersome. Maybe not even within the margin of error of Plato measurements.


Now that I have worked it through, it seems like you have to make too large of an assumption (weight of sugar being negligible) to get a useful answer. I’m not positive though, so here it goes. Ok, best info/guess is that ethanol is 0.789 g/cm3 at 20C, while water is 1.000 g/cm3 at 20C and table sugar is 1.590 g/cm3. You should probably use ABW rather than ABV for this because we’re talking about varying the percentage weights of the solution (water/sugar/alcohol). So let’s take your example numbers with one modification, two solutions of SG 1.010 (1.050 is a crazy high FG), one with 32% ABV and the other with 10% ABV. First convert to ABW by dividing by 1.25 gives ABWs of 25.6% and 8% respectively. We do have to assume that the weight of the sugar is negligible I guess. So you want to take the percentage of the total made up by alcohol and water and multiply the percentage by their respective densities and add them. Then you want to subtract that total from the total gravity of 1.010 to see what portion must be made up from sugar. Lastly divide this gravity portion from sugar by the density of sugar to get the % by weight of the total. I think... So 0.256x0.789 + 0.744x1.000 = 0.946 And 0.080x0.789 + 0.920x1.000 = 0.983 (1.0100.946)/1.590 = ~0.040 or a 4.0% sugar solution by weight in the 32% beer. (1.0100.983)/1.590 = ~0.017 or a 1.7% sugar solution by weight in the 10% beer. So really, the assumption that the sugar is negligible is probably not a good one, but I guess this would be a rough way to work it out. If you did have a starting gravity, you could get a better idea. In reality, I imagine that a 32% beer is going to have a much, much larger FG than a 10% beer just because how the yeast work.

Originally posted by tronraner
Yes, I intentionally ignored the contribution of the sugar to the volume. I didn’t calculate it, but I’m pretty sure that the volume contribution of 12g of sugar to 100 mL of liquid is not significant enough to warrant making the equation much more cumbersome. Maybe not even within the margin of error of Plato measurements.
Ok, that makes sense.
If you would approximate the sugars as i.e. maltose (with the density 1540 kg/m^3) you could use the following expression to approximate the sugar content:
x = 1/rho_maltose*(rho  ABV*rho_ethanol  (1  ABV)*rho_water)
where:
x is the sugar content
rho_maltose is the density of maltose
rho_ethanol is the density of ethanol
rho_water is the density of water
ABV is the alcohol content of the liquid.
With you two examples you get:
x_10 = 1/1540*(1050  0.1*789  (10.1)*1000) = 4,6 % sugar
x_32 = ... = 7,6 % sugar
All densities are the values at 25 degrees centigrade. I’m not entirely sure how well these equations work. I’ll try it out on some booze that I have at home that I can find the sugar content of on the internetz.

Originally posted by tronraner
Convert your measured gravity to deg Plato, which is more directly related to sugar concentration (essentially 1 deg Plato = 1 g/mL of sugar in water).
Sugar concentration (in g/mL) = (Plato) / [(1  %abv) + %abv(.79)]
So, using your examples:
1.050 = 12.4 P
the 10%abv would be 12.4/[(1.1) +.1(.79)] = 12.67 g/mL of sugar
the 32%abv would be 12.4/[(1.32) + .32(.79)] = 13.3 g/mL of sugar
One thing I screwed up: Plato is g sugar / 100 mL water, so the decimal points need moved back two spaces. 0.127 and 0.133 should be the results for g/mL.
Yeah, someone just needs to test this experimentally and come up with a chart.
