How much pure alcohol will 100g of white sugar produce ?

Homebrew Talk - Beer, Wine, Mead, & Cider Brewing Discussion Forum

Help Support Homebrew Talk - Beer, Wine, Mead, & Cider Brewing Discussion Forum:

This site may earn a commission from merchant affiliate links, including eBay, Amazon, and others.

andy-10

Member
Joined
Jul 9, 2010
Messages
6
Reaction score
0
Location
england
If I were to put 100g of pure white sugar into 1 lt of water, with some alcohol-resistant yeast (eg Champagne yeast), how many mililitres of pure alcohol would be produced when all of the sugar had been converted ?

Is there a table somewhere which displays this ?

Thanks.
 
None.

To become pure alcohol you would have to distill out the water. Otherwise, you'd just have watery ethanol.

There are softwares, even free ones, that will calculate the percentage alcohol given the fermentables you choose.
 
100g sucrose in 1000mL of water yields a solution with an original specific gravity of 1.039. Assuming you do something about the lack of free amino nitrogen during fermentation, an apparent attenuation of 75% - 80% is reasonable. That should yield a finishing specific gravity of 1.007 - 1.009. From there, the math is easy.
 
One, how would you get an attenuation of OVER 100% and two, it's yeast dependent on attenuation. Like Gila said, you need to distill to get pure alcohol, and we don't do that here.
 
Huh? Sucrose is going to be far more fermentable than that, isn't it? i would expect an apparent attenuation of well over 100%.

+1, I've done pure sucrose fermentations and got FGs substantially below 1.000. IrregularPulse: that's *apparent* attenuation. 100% attenuation is about 122% apparent attenuation. A 1.040 solution would ferment down to about 0.991.

To answer the question as directly as possible, 100 g of sucrose is 0.292 mol, and stoichiometrically 1 mol sucrose yields 4 mol EtOH, so you'd expect to get 54.0 g, which is about 68.3 mL.
 
One, how would you get an attenuation of OVER 100% and two, it's yeast dependent on attenuation. Like Gila said, you need to distill to get pure alcohol, and we don't do that here.

One, as a10t2 already pointed out, there's a big difference between apparent attenuation and real attenuation. Don't forget that ethanol has a lower specific gravity than water.

Two, it is dependent on the yeast to a small extent, but the apparent attenuation numbers the labs give are based on an average wort. A sucrose solution is going to be far, far more fermentable without all of the long chain sugars.
 
Thanks.

After posting I found a recipe for fermented spirit.

It said that 8kg of sugar added to 21 lt of water would produce 25lt of 20% spirit.

So that's 5lt of pure alcohol.

1.6kg of sugar would make 1lt, and so 100g would make 62.5 ml.

(Yes, realise that distillation would be required, but I was asking about how much pure ethanol would be suspended in the solution once 100% of the sugar had been converted.)
 
That concoction is almost sure to be foul. Unless you're making fuel, you need to rethink your gameplan. Look around this site for mead and cider recipes.
 
Thanks.

After posting I found a recipe for fermented spirit.

It said that 8kg of sugar added to 21 lt of water would produce 25lt of 20% spirit.

So that's 5lt of pure alcohol.

1.6kg of sugar would make 1lt, and so 100g would make 62.5 ml.

(Yes, realise that distillation would be required, but I was asking about how much pure ethanol would be suspended in the solution once 100% of the sugar had been converted.)

1. 20% spirit: ABV or ABW?
2. 20% of 25 l is 5.2 l
3. But water and alcohol mix funny; if you add 1l of water to 1l of ethanol, the total volume of the mixture is only 1.92 l. Trying to find the volume of fractional parts given the ABV and volume of the mixture requires a little math.
 
1. 20% spirit: ABV or ABW?
2. 20% of 25 l is 5.2 l
3. But water and alcohol mix funny; if you add 1l of water to 1l of ethanol, the total volume of the mixture is only 1.92 l. Trying to find the volume of fractional parts given the ABV and volume of the mixture requires a little math.

Number 3 made my head explode! So mixing ethonal and water changes the density of the mixture (not just an average of the 2 densities) thats wack :fro:.
Also look at page 3 of this paper dealing with fermentation products Fermentation lecture
The theoretical yeild is 51.1% so 100g = 51.1 g ethonal but as it states some of the carbon(?) is used for reproduction/oother cell functions so the actual observed yeild is more around 47%.
 
Balling determined way back when that 2.0665 g of extract were required to produce 1 gram of ethanol, 0.9565 g of CO2 and 0.11 g of yeast biomass so 100 grams sucrose (extract) would produce 100/2.0665 = 48.4 grams of alcohol. Since the density of ethanol is 0.789 g/mL this would be 61.3 mL if the alcohol were removed from the solution. Assuming no water loss (to evaporation or to the yeast cells) you would then have 1000 + 48.4 grams = 1048.4 grams of solution with 48.4 grams of alcohol which is 48.4/1048.4 = 4.616% ABW which corresponds to a density of approximately 0.988 g/cc. Thus the volume of the solution would be about 1050.5 mL. Note that 100 grams of sucrose in 1000 cc of water yields 1100 grams of solution at 100/1100 = 9.09 °P solution which has a density of 1.0344 so it's initial volume would be 1064.4 cc. So the net result of the fermentation would be a 14 cc decrease in the volume --- if I did my math right.
 
It's a perfectly reasonable question. If no one had ever asked it there woudn't be a Balling formula and you wouldn't be able to use Balling's formula to estimate the alcohol content of your beer from initial and final gravity readings (and you do use Balling's formula whether you are aware of it or not).
 
Just spend the 2 dollars and find out the answer. Take the OG, ferment, take the FG, and you shall have the data required to calculate the MLs of alcohol present in the booze water.
 
It's a perfectly reasonable question. If no one had ever asked it there woudn't be a Balling formula and you wouldn't be able to use Balling's formula to estimate the alcohol content of your beer from initial and final gravity readings (and you do use Balling's formula whether you are aware of it or not).


...check out his 3rd post. Not picking on the guy but just find it odd. Whatever!
 
Back
Top