A pound of sugar per five gallons of beer will add nine points to the original gravity and reduce the final gravity by two points.

We’ve all heard it. I’m ashamed to say that I’ve even parroted it myself in the past. But it’s only half true.

The OG contribution of simple sugars is certainly easy enough to calculate. Most brewers are probably aware that potential extract is typically given as a percentage of the potential extract of pure sucrose (46.21 point-gal/lb, 96.39 °P-L/kg). Let’s assume a volume of 20 L (5.28 gal) just to make the math easy. For a 5.0-5.5 gal batch volume, that will get us within 5%, which I think most brewers would concede is “close enough”.

(96.39 °P-L/kg)(0.454 kg)/(20 L) = 2.19 °P = 1.00856

So if your batch volume is 5.5 gal, eight points might be a better rule of thumb, but the oft-quoted value is essentially correct.

Determining how sugars will affect the FG gets a little more complicated, because, let’s face it, for most of us it’s been a while since high school chemistry. There also has to be an assumption made about how much of the sugar is consumed, and how much is fermented. I think it’s reasonable to assume that a healthy, active population of yeast will consume nearly all of the simple sugars available. I’m going to further assume that the sugar is being added during the anaerobic fermentation phase, so that nearly all of it will be fermented, as opposed to being used for aerobic respiration. The amount of ethanol generated then becomes a simple question of stoichiometry. The relevant reaction is:

C_{12}H_{22}O_{11}
+ H_{2}O → 4C_{2}H_{5}OH + 4CO_{2}

So four moles of EtOH will be produced per mole of sucrose fermented. On a volumetric basis:

4((454 g)/(342.3 g/mol))(46.07 g/mol)/(0.789 g/mL) = 309.8 mL

Which is 1.525% ABV, incidentally. The reduction in density is a two-term weighted average:

(20000*1.0000 + 309.8*0.789)/(20000 + 309.8) = 0.9968

So when added to 20 L of water (or beer), a pound of fully-fermented sugar will actually reduce the SG by about 3.2 points, or 0.8°P. Maybe I’m splitting hairs here, but saying “two points” is off by about 60%.

Some suggestions:

1. You neglected that only a fraction of the sugar consumed will be used by yeast for energy (which is where the EtOH comes from). This fraction is about 0.9, according to Balling. In the net, about 10% of the sugar will go to form more yeast cells. So, right out of the chute, you’ve over-estimated mL of EtOH produced by 100*(1/0.9) = 11%.

2. EtOH weighs 46 g/mol, not 44 g/mol.

3. In your final equation, the added mass contributed (in the end) from having added the 1 lb of sucrose is not the mass of the EtOH formed, but rather the difference between the 1 lb of sucrose minus the masses of CO2 evolved and settled yeast cells. And the denominator’s volume is not the sum of the original 20 L + volume of EtOH, since the volume of their mixture is less than their respective, added volumes (i.e, 1 liter of water + 1 liter of EtOH gives less than 2 liters, when intermixed).

Jim,

Thanks for the feedback.

1. As I mentioned, this makes the assumption that the sugar is added during active anaerobic fermentation, and that the sugar going to cellular synthesis is therefore negligible.

2. Well damn. Apparently I’d been working too much with CO

_{2}.3. While your point about combining volumes is certainly true, when dealing with a 1.5% ABV solution the effect is so slight that it won’t affect the results at this level of precision.

Sean

Sean,

The fraction of sucrose used in synthesis is about 10% in anaerobic fermentation. It’s more like 50% in aerobic respiration. Oh… and I was too hasty in my comment #3 about your numerator. The added mass contributed to the numerator *is* equal to the EtOH mass formed because that *does* equal the difference between the 1 lb of sucrose added minus the masses of CO2 evolved and settled yeast cells. For a fuller discussion of stoichiometry, visit my page at http://www.ithacoin.com/brewing/stoichiometry.htm

Best, Jim