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Hi fuelfarmer, from information others have given me via the forum, when titrating with KOH the FFA % is @ 60% of the titration. So a titration of 10 = @ 6% FFA's.  

Member 
Hi Fuelfarmer, To calculate the % FFA from a standard homebrew titration: FFA% = 0.766 x t for titration with 1g NaOH per litre where t is the volume of solution in ml for 1ml oil FFA% = 0.546 x t for titration with 1g KOH per litre. This assumes a chemical purity of 100%. If your chemicals {typically KOH) are lower purity you need to adjust for lack of purity If you are using KOH with a purity of 90% that comes out to 0.4914 For mental calculations I usually round this off to every 1 point of KOH titration is 1/2% FFA. So a titration of 2KOH is about 1% FFA and a titration of 4KOH is about 2% FFA. Close enough.  

member 
I calculated the average molecular weight of free fatty acids in corn oil was about 279 grams per mole. KOH has 56.11 grams per mole. NaOH has 40. grams per mole. Hexadecanoic acid (palmitic acid) has a density of 0.8527 grams per millilitre. So, one gram KOH divided by 56.11 grams per mole equals 0.017822135 moles. 0.017822135 moles multiplied by 279 grams per mole equals 4.972375691 grams. 4.972375691 grams divided by 0.8527 grams per milliliter equals 5.831330703 ml of free fatty acids per litre of vegetable oil with a titration of 1 KOH. That's assuming the density of palmitic acid is close to the average density of free fatty acids present in the vegetable oil. For NaOH titration of one gram per litre of vegetable oil. 1 gram divided by 40 grams per mole equals 0.025 moles. 0.025 moles times 279 grams per mole equals 6.975 grams (of free fatty acids), 6.975 grams divided by 0.8527 grams per millilitre equals 8.179899144 millilitres of free fatty acids per litre of vegetable oil with a titration of 1 NaOH. So for a titration of 1 gram KOH that's about 5.8 millilitres free fatty acids in 1 litre of oil, or rendered animal fat. For a titration of 1 NaOH that's about 8.2 millitres of free fatty acids in one litre of vegetable oil or rendered animal fat. For higher titrations just multiply it out. Like a titration of 5 NaOH would be 5 multiplied by 8.2 millilitres which equals about 41 millilitres free fatty acids per litre of vegetable oil. 41 millilitres of free fatty acids in 1000 millilitres of vegetable oil would be about 4.1% ffa in vegetable oil. You could say ffa percentage by weight (mass) or volume. This 4.1% ffa is by volume.  

Member 
Hi Wesley, This is Neutral's explanation for the numbers I posted above. "The formula used in industry for calculation of free fatty acid percentage is: FFA% = (v  b) x N x 28.2 / w where v is the volume in ml of titration solution b is the volume in ml of the blank N is the normality of the titration solution w is the weight of the sample of oil in grams With NaOH the molecular weight is 40 so a 0.1N solution contains 4g per litre. That is the usual concentration used in industry. Note that the formula contains 28.2 which is the molecular weight of oleic acid divided by ten. Oils are not made of only oleic acid hence this formula results in small errors, normally accepted. The range of molecular weights encountered in the used oil trade is from 270 for palm to 281 for canola. For the homebrew titration the concentration of the solution is usually 1/4 of the above, 1g NaOH per litre. Sometimes it is 1g per litre of KOH. To calculate the FFA from a homebrew titration use the following with a 1ml sample of oil: FFA% = 0.766 x t for titration with 1g NaOH per litre where t is the volume of solution in ml for 1ml oil FFA% = 0.546 x t for titration with 1g KOH per litre Acid Value = 1.99 x FFA% \" http://biodiesel.infopop.cc/ev...29605551/m/846105159 EDIT On Page two of this discussion Neutral further posts: "Re conversion factors. If you look at the formula for the Industrial method at the beginning of this thread you will see that it boils down to: FFA% = 28.2vN/w where v is the titration volume using N strength solution with w grams of oil. In the Homebrew method we use 1g NaOH per litre of water. The molecular weight of NaOH is 40 so this is a 0.025N solution. We also use 1ml of oil which is 0.92g oil. So in this case: FFA% = 28.2v x 0.025 / 0.92 = 0.766v On page three neutral says I have been asked on another thread how to titrate in such a way as to obtain the FFA%. The post that starts this thread sets out how it is done in industry. Here is an alternative method: To calculate the FFA% from a homebrew titration multiply the number of ml by 0.5463 where you have 1g KOH dissolved in 1 litre of water and titrate against 1ml of oil. So if you dissolved 1 / 0.5463 = 1.83 g KOH in 1 litre of water and titrated against 1ml of oil you would get the FFA% direct. Note that this assumes the KOH is 100% pure. Allowance must be made if it is not. PS on the make biodiesel website at the bottom it further explains: "Note that since we estimated both the density of our oil sample and the molecular weight of the average fatty acid in our oil, the result will also be an estimate. Even though it is an estimate, it is close enough for our use." http://www.makebiodiesel.org/...ffaconversions.htmlThis message has been edited. Last edited by: Tilly,  

member 
I suppose you can show my explanation of this to Neutral, I think he was in Australia. I wonder if he agrees with my method of calculation. My answers in % of ffa are slightly different than yours. I have no clue how to contact Neutral.  

Member 
Thanks for all the info. I just might have to average all of the results. My daughter is working on a small enzymatic biodiesel project.  

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