How Much CO2 is Produced from Brewing?

Off-week bonus post!  Ever wonder how much CO2 is actually produced when you ferment your favorite bubbly beverage?  I’ve often found myself staring at my air-lock or blow-off bucket thinking it must be a lot.  I finally decided that I should just go ahead and calculate it; as turns out, I was right.

The results of my calculations are presented here, in both English and metric units.  Below I give a bit more detail on the actual calculations.  If you look at the plots, you’ll notice that not only is the volume of gas large (left axis), the weight of the gas (right axis) is also quite appreciable.  Yes, your fermentor looses weight over the course of the brew!  In fact, some people have proposed monitoring the attenuation of a batch of beer in real time by monitoring the weight of the fermentor.  I always assumed you would need an extremely precise scale, but it turns out the difference is so large, you actually wouldn’t.

co2 production, english

CO2 production for various specific gravities and apparent attenuations.

co2 production, metric

CO2 production for various specific gravities and apparent attenuations.

For anyone interested in the math behind the plots, read on.

The first thing I did was to convert specific gravity (SG in 1.xxx format) to degrees Plato (°P) with the equation ^{\circ}P=135.997SG^{3}-630.272SG^{2}+1111.14SG-616.868.  This is a third order polynomial fit derived from the American Society of Brewing Chemists gravity to Plato tables, given on en.wikipedia.org/wiki/brix.  This is a convenient unit to work in because every 1°P is 1% by weight sugar, so a 1.060 SG wort would be 14.741°P, or 14.741% sugar by weight.

To find the mass of sugar in each wort, I then found the weight of a 5 gal/ 18.9 L batch by multiplying this volume by its density (if working in metric, this amounts to simply multiplying by the specific gravity since water is 1 kg/L).  Then, its a simple matter of multiplying by the sugar weight percent.  The example 1.060 wort contains (44.23 lb wort)*(0.14741)= 6.52 lb or 2.96 kg sugar.  Note that I actually did all of my calculations in metric, then converted to English because the units work out so nicely.

Next, we must find how much sugar is actually consumed.  For this, remember that the “attenuation” used most often is “apparent attenuation” as measured by a hydrometer.  Hydrometers actually measure the density of the solution, so when alcohol is created (its density is lower than water), the hydrometer is tricked into thinking more of the sugar has been consumed than really was.  Actual attenuation can by found by multiplying the apparent attenuation by 0.814, as given in Greg Noonan’s New Brewing Lager Beer.  So the actual amount of sugar consumed in a 1.060 75% apparent attenuation beer is (6.52 lb)(0.75*0.814)= 3.98 lb or 1.81 kg.

Now we must bring in some very basic chemistry.  Lets assume this sugar is all (or could become) glucose.  A single mole (in units of mol) of glucose weighs 180.156 g or  6.35 oz.  A mole is simply a way to keep track of how many molecules of something there is so you can predict chemical reactions.  Basically, there are a different number of molecules of acid in vinegar in a gram than the number of baking soda molecules in a gram, but reactions happen to molecules.  So to have a complete reaction, you need to add the same number of molecules of acid in vinegar and baking soda, not the same number of grams of each.  The upshot is we have (1.81 kg)/(0.180156 kg/mol)=10.02 mol glucose consumed in this example wort.

For every 1 mol glucose consumed, 2 mol of ethanol and 2 mol of CO2 are produced.  Thus we have 10.02*2=20.04 mol of total CO2 produced from fermenting this 1.060 wort to 75% apparent attenuation.

Now we can put this in terms of volume or mass of gas.  For any “ideal” gas, its volume can be predicted by the ideal gas law at standard temperature and pressure, PV=nRT, where P is pressure (1 atm or atmospheres), V is volume (liters or L), n is the number of moles of gas (1 mol), R is the ideal gas constant (0.0821 (atm*L)/(mol*K)), and T is the temperature (273 Kelvin or K).  Solving for volume and plugging in the numbers, we get V=\frac{nRT}{P}=\frac{(1 mol)(0.0821 \frac{atm*L}{mol*K})(273 K)}{1 atm}=22.41 L.  Thus every mol of gas occupies 22.41 L or 0.791 cubic feet.  This would be (20.04 mol)*(22.41 L/mol)=449.1 L or 15.86 ft3 of CO2 produced for our example beer.  Put another way, from 5 gallons of 1.060 SG wort, you could fill the glass carboy its in almost 24 times with the CO2!  In the most extreme case shown on the graph (1.110 wort, 85% attenuation), this more than doubles.

To find the mass of the gas, we simply use the mass of a mole of CO2, 0.04401 kg/mol, to convert back.  We have (20.04 mol)*(0.04401 kg/mol)=0.88 kg or 1.94 lb of CO2 produced.

– Dennis,
Life, Fermented

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About Dennis
Home brewer, home chef, garage tinkerer. Author of Life Fermented blog.

23 Responses to How Much CO2 is Produced from Brewing?

  1. Rob says:

    Neat story – and an amazing amount of gas – that explains a lot…
    I am having a little trouble with the 44.23 lbs – and am getting 44.17 (not a big difference, but why?) 1.06kg/L x 18.9L = 20.034 kg. 20.034 kg too labs = 44.1674096.?

    • Dennis says:

      Hey Rob,
      I must say I am pleasantly surprised anyone went through the math. Its just a rounding error. I reported using a value of 18.9L, but used the slightly more precise 18.927 L in my calculations, yielding in 20.0626… kg of wort. I then used a factor of exactly 2.20462 to convert back to lb. I also did everything in a spreadsheet, so any rounding throughout the process might be different if you are taking the result of each step and typing it back into a calculator, as you would likely be carrying less digits to the next step.
      – Dennis

  2. Bernardo says:

    Hi Dennis,
    I was wondering whether the relation between the CO2 produced and the volume of the batch is linear. From my understanding, a 18.9 L batch of a wort starting at 1.070 with a 85% attenuation, will produce 600 Liters of CO2. Let’s say that I want to estimate the CO2 produced by 2000 liters of the same wort with the very same fermentation. Can I just proceed as in the following: 600*(2000/18.9)?
    Thanks for your help
    Bernardo

    • Dennis says:

      Bernardo,
      That’s right. You would still have the same amount of sugar per liter of wort no matter how big the batch, thus you would have the same amount of gas produced per liter of wort. So, you can linearly scale the estimate, just as you suggested. I used 18.927 L in the actual calculations though, so 600*(2000/18.927) will yield a slightly more accurate estimate.
      – Dennis

  3. Kerry says:

    When it comes to Ideal Gas Law calculation, you chose to use Standard Temperature and Pressure, 273K and 1 atm. What if the beer is fermented in a temperature control and pressure controlled environment? Would you use the pressure and temperature measured at the top of the tank?

    • Dennis says:

      Hi Kerry,
      Thanks for the question, and sorry for the delay in answering. In this case, I am calculating the CO2 released from the fermentor into the room. So, ultimately, I am calculating how much there is at standard temperature/ pressure (STP), which is usually pretty close to average room conditions. Ultimately, though, the volume calculation was just to give people some context; the real answer is the number of moles produced, but it much harder to picture a mole of something. If you want to know how much CO2 is in the head space of the tank, that’s a whole different problem, which I think you could estimate from the temperature/ pressure/ volume of the head space.
      – Dennis

  4. Michael says:

    I just printed this and took it to class. You are my new best friend. Savior. I will be sharing your blog!

  5. Jennifer says:

    Thank you for the interesting article. Conversely, how much CO2 is USED in brewing? I have seen info on CO2 capture and was wondering if after capturing would any be left over that could be sold or used in other applications.

    • Dennis says:

      First, my apologies for taking so long to respond!

      I’m not sure I understand the question. If by used you mean in the entire process—plants uptaking CO2, harvesting and transporting creating CO2 etc—not sure, but I’ve always been curious. Otherwise, if you mean during kegging and force carbonation, one volume is one liter of CO2 in one liter of liquid at room temperature, so you could use the procedure above to find out how many grams that is.

      A fermentation will throw off far far more than what is ultimately used to carbonate the beer: 3 volumes is on the high end for most styles. There are some breweries that will cap their beer near the end of ferment and carbonate their beer that way (not advisable at home!), and they need only a plato or few remaining to do so. This is generally done to stay within the German beer purity law and is not do much outside of Germany. Fewer still, there are some breweries who do have recapture equipment to save all or most of the CO2 generated during ferment, but I don’t recall what they do with it all.
      – Dennis

      • david neale says:

        Hi. I don’t know how universal it is but I got some numbers for a brewery here in Colorado that has CO2 provided to it each month. From the barrels of beer they are producing each month and the total co2 they are provided, the math boils down to 5.84lbs per barrel. The same brewery gave numbers that indicated they produce 14.01lbs per barrel through the fermentation process. So there is a significant amount left over. I too am interested in co2 recapture and use so would love to talk more about that.

  6. Brian says:

    Great article. I have been brewing beer in the 2 liter soda bottles using the pressure release caps that limit to 30-60 ps I use the wt to determine when it’s done and the ABV.

    Startgm = 2004 (where I enter the wt)
    Finishgm = 1899

    TowLitergm= 50
    EthanolPerCO2=46/44
    DensityEthanol = .789
    S=Startgm-TwoLitergm
    F=Finishgm=TwoLitergm
    Dif=S-F
    PercentEthanolMass=Dif*100*EthanolPerCO2/F
    PercentEthanolVol=PercentEthanolMass/DensityEthanol

    PercentEthanolMass = 5.937 (for this example)
    PercentEthanolVol=7.5 (for this example)

    This batch used 1.4 lbs of DME per gallon of water.
    Every beer batch with this amount of DME produced nearly identical final calculated ABV.
    The perceived strength seems to match the calculated and I have checked it with some other folks.

    The beers can loss 50 gm per 1l bottle the first days then slow down to 1 in about a week and finally not change for days. The rate of loss also seems to match the visible yeast activity.

    I keep meaning to include the absorbed CO2 given the PSI.

    • Dennis says:

      Always nice to see science works in real life! If I understand correctly, you are keeping your brew under a constant 30-60 psi while fermenting? This much CO2 is quite toxic to the yeast: 30 psi is on the high side for final CO2 after bottle conditioning or kegging and pressurizing. 60 psi I would be wary of blowing up the soda bottle… not sure what they rate those at, but that’s probably double what they normally sell soda at.
      – Dennis

      • Justin H says:

        Dennis, to calculate weight of co2 released, could you take the (SG x 8.35) – (FG x 8.35) = weight difference = weight CO2 produced?

  7. Justin Henderson says:

    Oops, sorry I meant :

    (8.35 lbs/gal x gal wort x SG)- (8.35 lbs/gal x gal beer x FG) = weight difference = weight CO2 produced?

    So in a 1.065 SG wort fermented with bacteria and S Cerv. Down to 1.008 (95% attenuation) you would produce 2.36 pounds of CO2 using only the hydrometer reading and a quick calculation in a 5 gallon batch.

    • Dennis says:

      Great question. I’d have to think about it more, but my first inclination would be no, in general you could not use this formula. This would assume you end with the exact same volume you started with, and I’m not sure you can make this assumption. Perhaps it would be close enough, though.

      As for bacteria, I know that bacteria produce CO2 at a much different rate than S. Cerv. for a given mass of sugar. I assume this is because other byproducts besides ethanol and CO2 are produced in larger quantities, but that’s outside my realm of expertise. The upshot is I would be willing to bet that even if your formula is “close enough” for a 100% S. Cerv. ferment, its probably way off for bacteria. Mike Tonsmiere would be the reference for this. I know I’ve seen something of his dealing with CO2 from bacteria, but its been a while and I don’t recall where.
      – Dennis

  8. Kris says:

    Good stuff. Normal temperature and pressure (NTP) would be a little better reference than STP. Normal temp is 293 K (20 C). Standard temp (0 C) is a cold room — some would say it is freezing.

  9. Zachriel says:

    Fermentation is carbon neutral. The carbon is captured from the atmosphere by the growing hops and barley.

  10. George Songulashvili says:

    Can someone tell me, what is the pressure (bar) produced by 1 L wort fermentation per Plato (P°)?

  11. Doug says:

    …or you could simply use the Balling Equation, as used by the EBC: every 2.0665 gms of extract consumed produces 1 gm of alcohol, 0.9565 gm CO2 and 0.11 gm yeast mass. By this method your 5 gallon fermenter of 1060 wort with an AE of 75% would produce 1.74 lb of CO₂.
    Cheers!

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  13. Paul hamer says:

    Not to mention CO2 produced by the tractors and trucks to produce the barley hops,ect

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