Re: [Biofuel] %FFA of vegetable oil
Thank to both Keith and Tom, For clarifying the use of the FFA formula regarding KOH or NaOH.. Since I use KOH, my titration solution is also based on .1% KOH... Never use NaOH for tirtrating or otherwise. I many time I have been approached to sell my oil by brokers and homebrewers, brokers are always talking %FFA and homebrewers are ml of titration.. See this formula drew my interest as an easy means to convert one to the other with a fair amount of accuracy... Thanks again.. Shawn - Original Message - From: Keith Addison [EMAIL PROTECTED] To: biofuel@sustainablelists.org Sent: Friday, April 10, 2009 6:25:49 AM GMT -06:00 US/Canada Central Subject: Re: [Biofuel] %FFA of vegetable oil I guess I am not clear on a few of the following points: * Are we to assume 1mole of KOH nuetralizes 1mole of OA as does NaOH * What %KOH is to be used as with this assumption, is there some compensation factor for vary %KOH Yes, the same conversion as always, depending on the concentration of your KOH. If you're not clear on that, then how do you know how much KOH to use for your biodiesel? Are you assuming that the Biofuel Systems calculation is correct? I don't know if it's correct or not. It seems you are assuming that. How will knowing the %FFA be useful to you? Just interested to know. Best Keith Shawn - Original Message - From: Keith Addison [EMAIL PROTECTED] To: biofuel@sustainablelists.org Sent: Thursday, April 9, 2009 2:17:42 AM GMT -06:00 US/Canada Central Subject: Re: [Biofuel] %FFA of vegetable oil Found this email in my list of unopened and I am curious... For those of us who use KOH... How would this formula translate... The same way as usual. What's the difficulty? Everybody uses KOH, or they should, but titration levels are usually given for NaOH. A good reason for that is that NaOH is always the same concentration, 99%+, and KOH varies. Are you assuming that the Biofuel Systems calculation is correct? I don't know if it's correct or not. Keith Shawn - Original Message - From: Keith Addison [EMAIL PROTECTED] To: biofuel@sustainablelists.org Sent: Monday, March 9, 2009 6:01:34 AM GMT -06:00 US/Canada Central Subject: [Biofuel] %FFA of vegetable oil Hi all I've seen several conflicting ratios for converting homebrewer titration results of # ml 0.1% NaOH solution to % FFA. This below is from Biofuel Systems in the UK http://www.biofuelsystems.com/. Can anyone confirm whether it's correct or not? How to determine percentage free fatty acid (%FFA) of vegetable oil / used cooking oil Add 10ml of oil to 100ml of isopropyl alchohol (propan-2-ol) Mix thoroughly until oil dissolves. Add a few drops of Universal Indicator solution (UI) Measure how much 0.025M sodium hydroxide solution (1g NaOH / 1 litre water) is required to neutralise the oil solution - ie. raise pH to 8 / turn UI blue/green While continually stirring this mixture, add the NaOH solution drop by drop until the mixture turns and remains green / blue. Note the number of millilitres (ml) of NaOH solution required to do this. In stoichiometric terms, 1 mole NaOH will neutralise 1 mole of oleic acid (OA) Firstly, determine how many moles of NaOH have been used· Moles = molarity (mol/l) x volume (l) = [ molarity x volume (ml) ÷ 1000 ] Example If 10ml of NaOH solution was used Moles of NaOH = (molarity of solution x volume in ml) ÷ 1000 = (0.025 x 10) ÷ 1000 = 0.00025 moles of NaOH Therefore, the equivalent to 0.00025 moles of OA have been neutralised mass = moles x molecular weight (molecular weight of OA is 282.52 Daltons) mass of OA in 10ml sample = 0.00025 x 282.52 = 0.07063 = 0.7063g per 100ml of oil rounding to 2 decimal places, %FFA = 0.71 Put simply... %FFA = number ml 0.025M NaOH solution used x 0.07063 - Since most homebrewers use 1 ml of oil in 10 ml of isopropanol, that would be: %FFA = number ml 0.025M NaOH solution used x 0.7063 I guess blue/green Universal Indicator solution is the same as magenta phenolphthalein, no? Thanks! Best Keith ___ Biofuel mailing list Biofuel@sustainablelists.org http://sustainablelists.org/mailman/listinfo/sustainablelorgbiofuel Biofuel at Journey to Forever: http://journeytoforever.org/biofuel.html Search the combined Biofuel and Biofuels-biz list archives (70,000 messages): http://www.mail-archive.com/biofuel@sustainablelists.org/ -- next part -- An HTML attachment was scrubbed... URL: /pipermail/attachments/20090413/0115d8f9/attachment.html ___ Biofuel mailing list Biofuel@sustainablelists.org http://sustainablelists.org/mailman/listinfo/sustainablelorgbiofuel Biofuel at Journey to Forever: http://journeytoforever.org/biofuel.html Search
Re: [Biofuel] %FFA of vegetable oil
Hi Tom Thankyou, nice! Keith, I was hoping to hear from someone with greater expertise in the area, but will pass along my thoughts. I'm not sure that the calculations provided from Biofuels Systems are accurate for the following reasons. 1. The % FFA calculations were based on the neutralization of Oleic Acid (molecular mass = 283). The idea, as I see it, is that calculations can be made to determine the mass of substances consumed in reactions, if we know the mass of one of the consumed reactants and know the ratio(s) of the involved reactants. Ex. If we know the mass of NaOH needed to neutralize a known volume of a specific acid, and we know that there exists a 1-to-1 ratio for the reaction, we can calculate the mass of the acid that is neutralized. -However many moles of NaOH were consumed will equal the moles of acid neutralized. -The mass of the acid neutralized = moles neutralized X mol. mass of the acid Problem: Used vegetable oils have a mix of free fatty acids, with different molecular masses, and hence different masses per mole. To do an accurate calculation of % FFA in veg oil, I think we would need to know the specific FFAs present, in what concentrations, and the molecular mass of each. I'm sure you're right. According to this table, the molecular weight of common fatty acids varies from 88 to 324, and oleic acid is 282: http://journeytoforever.org/biodiesel_meth.html After all, We wouldn't use a titration solution with an unknown mix of NaOH and KOH to do a quantitative analysis. We wouldn't, no. Although I've heard it referred to, I don't know why one would want to know %FFA in veg oil. It seems to be the way chemists do it, where we give a titration result, they talk of %FFA. Maybe it's just a comparative measure rather than a real-world one: if this WVO were pure oleic acid it would have this much FFA in it, as opposed to that WVO, if it were also pure oleic acid. It reminds me a bit of the way they test the protein content of grain crops and so on: N x 6.25, which is the ratio of N in protein. I think that presumes a lot, but it's easy to measure N and very difficult to measure protein. You see. Or something like that. Quite often chemists write to me about this or that %FFA oil, I encounter it in various reports too, and it would be nice to know what that means when it comes to processing it. Personally I'll stick with titration results. 2. To determine %, one divides the total mass by the mass of the individual component and then multiply by 100 Ex. 1 g of NaOH + enough distilled water to equal 1 L The total mass of the solution = 1000g 1g NaOH divided by 1000g = .001X 100 = 0.1% The Biofuel Systems calculations state that there is 0.7063 g of FFA in 100ml of oil. It then divides mass by volume (). It assumes that 100ml of the oil has a mass of 100g. So he does, you're right. This would be true if it was water instead of oil. The veg oils I am familiar with float on water indicating a lower density ... I think along the order of 0.9g/ml Close to that. At 0.9g/ml, the 100ml sample of veg oil would have a mass of 90g 0.7063g divided by 90g X 100 = 0.785% vs The 0.71% calculated in the example provided by Biofuel systems. If the oil being tested is very high in oleic acid, and we use the density of the oil when calculating %, I suspect we could get a ballpark figure for % FFAs in a sample of veg oil. According to the chart I reffed above it works quite well for some oils, especially soy. But not rapeseed. Hm. Interesting. We might just as well use 0.75 - 0.80 as a coefficient (constant) and multiply it times the NaOH titration result rather than going through the laborious calculations described. Good! That's much better. And divide by 0.785% to convert %FFA to a ballpark titration result. Many thanks Tom, very good answer, IMHO. Regards Keith Although I've heard it referred to, I don't know why one would want to know %FFA in veg oil. Best to You, Tom - Original Message - From: Keith Addison [EMAIL PROTECTED] To: biofuel@sustainablelists.org Sent: Monday, March 09, 2009 7:01 AM Subject: [Biofuel] %FFA of vegetable oil Hi all I've seen several conflicting ratios for converting homebrewer titration results of # ml 0.1% NaOH solution to % FFA. This below is from Biofuel Systems in the UK http://www.biofuelsystems.com/. Can anyone confirm whether it's correct or not? How to determine percentage free fatty acid (%FFA) of vegetable oil / used cooking oil Add 10ml of oil to 100ml of isopropyl alchohol (propan-2-ol) Mix thoroughly until oil dissolves. Add a few drops of Universal Indicator solution (UI) Measure how much 0.025M sodium hydroxide solution (1g NaOH / 1 litre water) is required to neutralise the oil solution - ie. raise pH
Re: [Biofuel] %FFA of vegetable oil
I guess I am not clear on a few of the following points: * Are we to assume 1mole of KOH nuetralizes 1mole of OA as does NaOH * What %KOH is to be used as with this assumption, is there some compensation factor for vary %KOH Yes, the same conversion as always, depending on the concentration of your KOH. If you're not clear on that, then how do you know how much KOH to use for your biodiesel? Are you assuming that the Biofuel Systems calculation is correct? I don't know if it's correct or not. It seems you are assuming that. How will knowing the %FFA be useful to you? Just interested to know. Best Keith Shawn - Original Message - From: Keith Addison [EMAIL PROTECTED] To: biofuel@sustainablelists.org Sent: Thursday, April 9, 2009 2:17:42 AM GMT -06:00 US/Canada Central Subject: Re: [Biofuel] %FFA of vegetable oil Found this email in my list of unopened and I am curious... For those of us who use KOH... How would this formula translate... The same way as usual. What's the difficulty? Everybody uses KOH, or they should, but titration levels are usually given for NaOH. A good reason for that is that NaOH is always the same concentration, 99%+, and KOH varies. Are you assuming that the Biofuel Systems calculation is correct? I don't know if it's correct or not. Keith Shawn - Original Message - From: Keith Addison [EMAIL PROTECTED] To: biofuel@sustainablelists.org Sent: Monday, March 9, 2009 6:01:34 AM GMT -06:00 US/Canada Central Subject: [Biofuel] %FFA of vegetable oil Hi all I've seen several conflicting ratios for converting homebrewer titration results of # ml 0.1% NaOH solution to % FFA. This below is from Biofuel Systems in the UK http://www.biofuelsystems.com/. Can anyone confirm whether it's correct or not? How to determine percentage free fatty acid (%FFA) of vegetable oil / used cooking oil Add 10ml of oil to 100ml of isopropyl alchohol (propan-2-ol) Mix thoroughly until oil dissolves. Add a few drops of Universal Indicator solution (UI) Measure how much 0.025M sodium hydroxide solution (1g NaOH / 1 litre water) is required to neutralise the oil solution - ie. raise pH to 8 / turn UI blue/green While continually stirring this mixture, add the NaOH solution drop by drop until the mixture turns and remains green / blue. Note the number of millilitres (ml) of NaOH solution required to do this. In stoichiometric terms, 1 mole NaOH will neutralise 1 mole of oleic acid (OA) Firstly, determine how many moles of NaOH have been used· Moles = molarity (mol/l) x volume (l) = [ molarity x volume (ml) ÷ 1000 ] Example If 10ml of NaOH solution was used Moles of NaOH = (molarity of solution x volume in ml) ÷ 1000 = (0.025 x 10) ÷ 1000 = 0.00025 moles of NaOH Therefore, the equivalent to 0.00025 moles of OA have been neutralised mass = moles x molecular weight (molecular weight of OA is 282.52 Daltons) mass of OA in 10ml sample = 0.00025 x 282.52 = 0.07063 = 0.7063g per 100ml of oil rounding to 2 decimal places, %FFA = 0.71 Put simply... %FFA = number ml 0.025M NaOH solution used x 0.07063 - Since most homebrewers use 1 ml of oil in 10 ml of isopropanol, that would be: %FFA = number ml 0.025M NaOH solution used x 0.7063 I guess blue/green Universal Indicator solution is the same as magenta phenolphthalein, no? Thanks! Best Keith ___ Biofuel mailing list Biofuel@sustainablelists.org http://sustainablelists.org/mailman/listinfo/sustainablelorgbiofuel Biofuel at Journey to Forever: http://journeytoforever.org/biofuel.html Search the combined Biofuel and Biofuels-biz list archives (70,000 messages): http://www.mail-archive.com/biofuel@sustainablelists.org/
Re: [Biofuel] %FFA of vegetable oil
Shawn, You ask: • Are we to assume 1mole of KOH nuetralizes 1mole of OA as does NaOH • What %KOH is to be used as with this assumption, is there some compensation factor for vary %KOH I will give it a try. The following is based on information provided at JtF. Any mistakes I make are mine alone. 1. • Are we to assume 1mole of KOH nuetralizes 1mole of OA as does NaOH Yes, I would think that 1 mole of KOH would neutralize 1 mole of Oleic Acid as does 1 mole of NaOH. A mole of a substance contains the same number of molecules as a mole of any other substance. If 1 NaOH molecule neutralizes 1 O.A. molecule, then 1 KOH will also neutralize 1 O.A. 2. • What %KOH is to be used as with this assumption, is there some compensation factor for vary %KOH It is important to know the concentration (purity) of the KOH being used in order to calculate the molarity of the KOH titration solution. That is, how do you know how much mass constitutes a mole of a substance if you don't know the purity of the substance you are measuring? You may want to skip to* To be sure we are talking about the same thing: A mole of a substance is equal to the molecular mass of the substance in grams. A mole of Sodium Hydroxide (NaOH) = 40 g i.e. The sum of the atomic masses ofNa (23) O (16) and H (1) A mole of Potassium Hydroxide (KOH) = 56 g K (39) + O (16) + H (1) Therefore in order to achieve equal molarity, we must increase KOH over NaOH by a factor of 1.4 (56 divided by 40) To do the calculations described by Biofuel Systems using a KOH titration solution rather than a 0.1% NaOH titration solution we must do two things. 1. Either a) make KOH solution of equal molarity (0.025 M) to the NaOH solution by dissolving 1.4g KOH to get a 1L titration solution (0.025moles/L X 56 g KOH/mole = 1.4g KOH/L) Orb) make a 0.1% solution of pure KOH and multiply titration results by 1.4 Orc) make a 0.1% solution of pure KOH realizing that it is a 0.0179 molar solution rather than the 0.025 molar solution used in the calculations and then use 0.0179 moles in the calculation of Oleic Acid mass 2. As was pointed out by Keith, NaOH is typically used as a titration solution because it is easily obtained at purities approaching 100%. KOH is typically 90% pure or even 85% pure. To compensate for this one must divide the amount of KOH by its purity to determine the number of grams needed to produce a mole of KOH Ex: 56 g of 100% pure KOH = 1 mole of KOH If the KOH is 90% pure:divide 56g by .9- 62.2g of the KOH = 1 mole *** To compensate for titration results obtained from a 0.1% KOH titration solution (To get the results that would be produced on the same veg oil using a 0.1% NaOH titration solution.) a. Multiply KOH titration results by the purity b. Then divide by 1.4 Ex If 3ml of 0.1% KOH (made from 90% KOH) neutralized 1ml of veg oil: a) 3ml (hence 3mg KOH of 90% KOH)) X .9 = 2.7ml (made from pure KOH) b) 2.7ml divided by 1.4 = 1.93ml 1.93 ml of 0.1% NaOH solution would have neutralized the same volume of the same veg oil as 3ml of 0.1% KOH solution made from 90% KOH Use the 1.93ml in the calculation provided by Biofuel Systems. Tom - Original Message - From: [EMAIL PROTECTED] To: sustainablelorgbiofuel@sustainablelists.org Sent: Thursday, April 09, 2009 10:26 AM Subject: Re: [Biofuel] %FFA of vegetable oil I guess I am not clear on a few of the following points: • Are we to assume 1mole of KOH nuetralizes 1mole of OA as does NaOH • What %KOH is to be used as with this assumption, is there some compensation factor for vary %KOH Shawn - Original Message - From: Keith Addison [EMAIL PROTECTED] To: biofuel@sustainablelists.org Sent: Thursday, April 9, 2009 2:17:42 AM GMT -06:00 US/Canada Central Subject: Re: [Biofuel] %FFA of vegetable oil Found this email in my list of unopened and I am curious... For those of us who use KOH... How would this formula translate... The same way as usual. What's the difficulty? Everybody uses KOH, or they should, but titration levels are usually given for NaOH. A good reason for that is that NaOH is always the same concentration, 99%+, and KOH varies. Are you assuming that the Biofuel Systems calculation is correct? I don't know if it's correct or not. Keith Shawn - Original Message - From: Keith Addison [EMAIL PROTECTED] To: biofuel@sustainablelists.org Sent: Monday, March 9, 2009 6:01:34 AM GMT -06:00 US/Canada Central Subject: [Biofuel] %FFA of vegetable oil Hi all I've seen several conflicting ratios for converting homebrewer titration results of # ml 0.1% NaOH solution to % FFA. This below is from Biofuel Systems in the UK http
Re: [Biofuel] %FFA of vegetable oil
Keith, I was hoping to hear from someone with greater expertise in the area, but will pass along my thoughts. I'm not sure that the calculations provided from Biofuels Systems are accurate for the following reasons. 1. The % FFA calculations were based on the neutralization of Oleic Acid (molecular mass = 283). The idea, as I see it, is that calculations can be made to determine the mass of substances consumed in reactions, if we know the mass of one of the consumed reactants and know the ratio(s) of the involved reactants. Ex. If we know the mass of NaOH needed to neutralize a known volume of a specific acid, and we know that there exists a 1-to-1 ratio for the reaction, we can calculate the mass of the acid that is neutralized. -However many moles of NaOH were consumed will equal the moles of acid neutralized. -The mass of the acid neutralized = moles neutralized X mol. mass of the acid Problem: Used vegetable oils have a mix of free fatty acids, with different molecular masses, and hence different masses per mole. To do an accurate calculation of % FFA in veg oil, I think we would need to know the specific FFAs present, in what concentrations, and the molecular mass of each. After all, We wouldn't use a titration solution with an unknown mix of NaOH and KOH to do a quantitative analysis. 2. To determine %, one divides the total mass by the mass of the individual component and then multiply by 100 Ex. 1 g of NaOH + enough distilled water to equal 1 L The total mass of the solution = 1000g 1g NaOH divided by 1000g = .001X 100 = 0.1% The Biofuel Systems calculations state that there is 0.7063 g of FFA in 100ml of oil. It then divides mass by volume (). It assumes that 100ml of the oil has a mass of 100g. This would be true if it was water instead of oil. The veg oils I am familiar with float on water indicating a lower density ... I think along the order of 0.9g/ml At 0.9g/ml, the 100ml sample of veg oil would have a mass of 90g 0.7063g divided by 90g X 100 = 0.785% vs The 0.71% calculated in the example provided by Biofuel systems. If the oil being tested is very high in oleic acid, and we use the density of the oil when calculating %, I suspect we could get a ballpark figure for % FFAs in a sample of veg oil. We might just as well use 0.75 - 0.80 as a coefficient (constant) and multiply it times the NaOH titration result rather than going through the laborious calculations described. Although I've heard it referred to, I don't know why one would want to know %FFA in veg oil. Best to You, Tom - Original Message - From: Keith Addison [EMAIL PROTECTED] To: biofuel@sustainablelists.org Sent: Monday, March 09, 2009 7:01 AM Subject: [Biofuel] %FFA of vegetable oil Hi all I've seen several conflicting ratios for converting homebrewer titration results of # ml 0.1% NaOH solution to % FFA. This below is from Biofuel Systems in the UK http://www.biofuelsystems.com/. Can anyone confirm whether it's correct or not? How to determine percentage free fatty acid (%FFA) of vegetable oil / used cooking oil Add 10ml of oil to 100ml of isopropyl alchohol (propan-2-ol) Mix thoroughly until oil dissolves. Add a few drops of Universal Indicator solution (UI) Measure how much 0.025M sodium hydroxide solution (1g NaOH / 1 litre water) is required to neutralise the oil solution - ie. raise pH to 8 / turn UI blue/green While continually stirring this mixture, add the NaOH solution drop by drop until the mixture turns and remains green / blue. Note the number of millilitres (ml) of NaOH solution required to do this. In stoichiometric terms, 1 mole NaOH will neutralise 1 mole of oleic acid (OA) Firstly, determine how many moles of NaOH have been usedS Moles = molarity (mol/l) x volume (l) = [ molarity x volume (ml) ÷ 1000 ] Example If 10ml of NaOH solution was used Moles of NaOH = (molarity of solution x volume in ml) ÷ 1000 = (0.025 x 10) ÷ 1000 = 0.00025 moles of NaOH Therefore, the equivalent to 0.00025 moles of OA have been neutralised mass = moles x molecular weight (molecular weight of OA is 282.52 Daltons) mass of OA in 10ml sample = 0.00025 x 282.52 = 0.07063 = 0.7063g per 100ml of oil rounding to 2 decimal places, %FFA = 0.71 Put simply... %FFA = number ml 0.025M NaOH solution used x 0.07063 - Since most homebrewers use 1 ml of oil in 10 ml of isopropanol, that would be: %FFA = number ml 0.025M NaOH solution used x 0.7063 I guess blue/green Universal Indicator solution is the same as magenta phenolphthalein, no? Thanks! Best Keith ___ Biofuel mailing list Biofuel@sustainablelists.org http://sustainablelists.org/mailman/listinfo/sustainablelorgbiofuel Biofuel at Journey to Forever: http://journeytoforever.org
Re: [Biofuel] %FFA of vegetable oil
Found this email in my list of unopened and I am curious... For those of us who use KOH... How would this formula translate... The same way as usual. What's the difficulty? Everybody uses KOH, or they should, but titration levels are usually given for NaOH. A good reason for that is that NaOH is always the same concentration, 99%+, and KOH varies. Are you assuming that the Biofuel Systems calculation is correct? I don't know if it's correct or not. Keith Shawn - Original Message - From: Keith Addison [EMAIL PROTECTED] To: biofuel@sustainablelists.org Sent: Monday, March 9, 2009 6:01:34 AM GMT -06:00 US/Canada Central Subject: [Biofuel] %FFA of vegetable oil Hi all I've seen several conflicting ratios for converting homebrewer titration results of # ml 0.1% NaOH solution to % FFA. This below is from Biofuel Systems in the UK http://www.biofuelsystems.com/. Can anyone confirm whether it's correct or not? How to determine percentage free fatty acid (%FFA) of vegetable oil / used cooking oil Add 10ml of oil to 100ml of isopropyl alchohol (propan-2-ol) Mix thoroughly until oil dissolves. Add a few drops of Universal Indicator solution (UI) Measure how much 0.025M sodium hydroxide solution (1g NaOH / 1 litre water) is required to neutralise the oil solution - ie. raise pH to 8 / turn UI blue/green While continually stirring this mixture, add the NaOH solution drop by drop until the mixture turns and remains green / blue. Note the number of millilitres (ml) of NaOH solution required to do this. In stoichiometric terms, 1 mole NaOH will neutralise 1 mole of oleic acid (OA) Firstly, determine how many moles of NaOH have been used· Moles = molarity (mol/l) x volume (l) = [ molarity x volume (ml) ÷ 1000 ] Example If 10ml of NaOH solution was used Moles of NaOH = (molarity of solution x volume in ml) ÷ 1000 = (0.025 x 10) ÷ 1000 = 0.00025 moles of NaOH Therefore, the equivalent to 0.00025 moles of OA have been neutralised mass = moles x molecular weight (molecular weight of OA is 282.52 Daltons) mass of OA in 10ml sample = 0.00025 x 282.52 = 0.07063 = 0.7063g per 100ml of oil rounding to 2 decimal places, %FFA = 0.71 Put simply... %FFA = number ml 0.025M NaOH solution used x 0.07063 - Since most homebrewers use 1 ml of oil in 10 ml of isopropanol, that would be: %FFA = number ml 0.025M NaOH solution used x 0.7063 I guess blue/green Universal Indicator solution is the same as magenta phenolphthalein, no? Thanks! Best Keith ___ Biofuel mailing list Biofuel@sustainablelists.org http://sustainablelists.org/mailman/listinfo/sustainablelorgbiofuel Biofuel at Journey to Forever: http://journeytoforever.org/biofuel.html Search the combined Biofuel and Biofuels-biz list archives (70,000 messages): http://www.mail-archive.com/biofuel@sustainablelists.org/
Re: [Biofuel] %FFA of vegetable oil
I guess I am not clear on a few of the following points: • Are we to assume 1mole of KOH nuetralizes 1mole of OA as does NaOH • What %KOH is to be used as with this assumption, is there some compensation factor for vary %KOH Shawn - Original Message - From: Keith Addison [EMAIL PROTECTED] To: biofuel@sustainablelists.org Sent: Thursday, April 9, 2009 2:17:42 AM GMT -06:00 US/Canada Central Subject: Re: [Biofuel] %FFA of vegetable oil Found this email in my list of unopened and I am curious... For those of us who use KOH... How would this formula translate... The same way as usual. What's the difficulty? Everybody uses KOH, or they should, but titration levels are usually given for NaOH. A good reason for that is that NaOH is always the same concentration, 99%+, and KOH varies. Are you assuming that the Biofuel Systems calculation is correct? I don't know if it's correct or not. Keith Shawn - Original Message - From: Keith Addison [EMAIL PROTECTED] To: biofuel@sustainablelists.org Sent: Monday, March 9, 2009 6:01:34 AM GMT -06:00 US/Canada Central Subject: [Biofuel] %FFA of vegetable oil Hi all I've seen several conflicting ratios for converting homebrewer titration results of # ml 0.1% NaOH solution to % FFA. This below is from Biofuel Systems in the UK http://www.biofuelsystems.com/. Can anyone confirm whether it's correct or not? How to determine percentage free fatty acid (%FFA) of vegetable oil / used cooking oil Add 10ml of oil to 100ml of isopropyl alchohol (propan-2-ol) Mix thoroughly until oil dissolves. Add a few drops of Universal Indicator solution (UI) Measure how much 0.025M sodium hydroxide solution (1g NaOH / 1 litre water) is required to neutralise the oil solution - ie. raise pH to 8 / turn UI blue/green While continually stirring this mixture, add the NaOH solution drop by drop until the mixture turns and remains green / blue. Note the number of millilitres (ml) of NaOH solution required to do this. In stoichiometric terms, 1 mole NaOH will neutralise 1 mole of oleic acid (OA) Firstly, determine how many moles of NaOH have been used· Moles = molarity (mol/l) x volume (l) = [ molarity x volume (ml) ÷ 1000 ] Example If 10ml of NaOH solution was used Moles of NaOH = (molarity of solution x volume in ml) ÷ 1000 = (0.025 x 10) ÷ 1000 = 0.00025 moles of NaOH Therefore, the equivalent to 0.00025 moles of OA have been neutralised mass = moles x molecular weight (molecular weight of OA is 282.52 Daltons) mass of OA in 10ml sample = 0.00025 x 282.52 = 0.07063 = 0.7063g per 100ml of oil rounding to 2 decimal places, %FFA = 0.71 Put simply... %FFA = number ml 0.025M NaOH solution used x 0.07063 - Since most homebrewers use 1 ml of oil in 10 ml of isopropanol, that would be: %FFA = number ml 0.025M NaOH solution used x 0.7063 I guess blue/green Universal Indicator solution is the same as magenta phenolphthalein, no? Thanks! Best Keith ___ Biofuel mailing list Biofuel@sustainablelists.org http://sustainablelists.org/mailman/listinfo/sustainablelorgbiofuel Biofuel at Journey to Forever: http://journeytoforever.org/biofuel.html Search the combined Biofuel and Biofuels-biz list archives (70,000 messages): http://www.mail-archive.com/biofuel@sustainablelists.org/ -- next part -- An HTML attachment was scrubbed... URL: /pipermail/attachments/20090409/30e51014/attachment.html ___ Biofuel mailing list Biofuel@sustainablelists.org http://sustainablelists.org/mailman/listinfo/sustainablelorgbiofuel Biofuel at Journey to Forever: http://journeytoforever.org/biofuel.html Search the combined Biofuel and Biofuels-biz list archives (70,000 messages): http://www.mail-archive.com/biofuel@sustainablelists.org/
Re: [Biofuel] %FFA of vegetable oil
Found this email in my list of unopened and I am curious... For those of us who use KOH... How would this formula translate... Shawn - Original Message - From: Keith Addison [EMAIL PROTECTED] To: biofuel@sustainablelists.org Sent: Monday, March 9, 2009 6:01:34 AM GMT -06:00 US/Canada Central Subject: [Biofuel] %FFA of vegetable oil Hi all I've seen several conflicting ratios for converting homebrewer titration results of # ml 0.1% NaOH solution to % FFA. This below is from Biofuel Systems in the UK http://www.biofuelsystems.com/. Can anyone confirm whether it's correct or not? How to determine percentage free fatty acid (%FFA) of vegetable oil / used cooking oil Add 10ml of oil to 100ml of isopropyl alchohol (propan-2-ol) Mix thoroughly until oil dissolves. Add a few drops of Universal Indicator solution (UI) Measure how much 0.025M sodium hydroxide solution (1g NaOH / 1 litre water) is required to neutralise the oil solution - ie. raise pH to 8 / turn UI blue/green While continually stirring this mixture, add the NaOH solution drop by drop until the mixture turns and remains green / blue. Note the number of millilitres (ml) of NaOH solution required to do this. In stoichiometric terms, 1 mole NaOH will neutralise 1 mole of oleic acid (OA) Firstly, determine how many moles of NaOH have been usedŠ Moles = molarity (mol/l) x volume (l) = [ molarity x volume (ml) ÷ 1000 ] Example If 10ml of NaOH solution was used Moles of NaOH = (molarity of solution x volume in ml) ÷ 1000 = (0.025 x 10) ÷ 1000 = 0.00025 moles of NaOH Therefore, the equivalent to 0.00025 moles of OA have been neutralised mass = moles x molecular weight (molecular weight of OA is 282.52 Daltons) mass of OA in 10ml sample = 0.00025 x 282.52 = 0.07063 = 0.7063g per 100ml of oil rounding to 2 decimal places, %FFA = 0.71 Put simply... %FFA = number ml 0.025M NaOH solution used x 0.07063 - Since most homebrewers use 1 ml of oil in 10 ml of isopropanol, that would be: %FFA = number ml 0.025M NaOH solution used x 0.7063 I guess blue/green Universal Indicator solution is the same as magenta phenolphthalein, no? Thanks! Best Keith ___ Biofuel mailing list Biofuel@sustainablelists.org http://sustainablelists.org/mailman/listinfo/sustainablelorgbiofuel Biofuel at Journey to Forever: http://journeytoforever.org/biofuel.html Search the combined Biofuel and Biofuels-biz list archives (70,000 messages): http://www.mail-archive.com/biofuel@sustainablelists.org/ -- next part -- An HTML attachment was scrubbed... URL: /pipermail/attachments/20090408/6ba17b37/attachment.html ___ Biofuel mailing list Biofuel@sustainablelists.org http://sustainablelists.org/mailman/listinfo/sustainablelorgbiofuel Biofuel at Journey to Forever: http://journeytoforever.org/biofuel.html Search the combined Biofuel and Biofuels-biz list archives (70,000 messages): http://www.mail-archive.com/biofuel@sustainablelists.org/
[Biofuel] %FFA of vegetable oil
Hi all I've seen several conflicting ratios for converting homebrewer titration results of # ml 0.1% NaOH solution to % FFA. This below is from Biofuel Systems in the UK http://www.biofuelsystems.com/. Can anyone confirm whether it's correct or not? How to determine percentage free fatty acid (%FFA) of vegetable oil / used cooking oil Add 10ml of oil to 100ml of isopropyl alchohol (propan-2-ol) Mix thoroughly until oil dissolves. Add a few drops of Universal Indicator solution (UI) Measure how much 0.025M sodium hydroxide solution (1g NaOH / 1 litre water) is required to neutralise the oil solution - ie. raise pH to 8 / turn UI blue/green While continually stirring this mixture, add the NaOH solution drop by drop until the mixture turns and remains green / blue. Note the number of millilitres (ml) of NaOH solution required to do this. In stoichiometric terms, 1 mole NaOH will neutralise 1 mole of oleic acid (OA) Firstly, determine how many moles of NaOH have been used Moles = molarity (mol/l) x volume (l) = [ molarity x volume (ml) ÷ 1000 ] Example If 10ml of NaOH solution was used Moles of NaOH = (molarity of solution x volume in ml) ÷ 1000 = (0.025 x 10) ÷ 1000 = 0.00025 moles of NaOH Therefore, the equivalent to 0.00025 moles of OA have been neutralised mass = moles x molecular weight (molecular weight of OA is 282.52 Daltons) mass of OA in 10ml sample = 0.00025 x 282.52 = 0.07063 = 0.7063g per 100ml of oil rounding to 2 decimal places, %FFA = 0.71 Put simply... %FFA = number ml 0.025M NaOH solution used x 0.07063 - Since most homebrewers use 1 ml of oil in 10 ml of isopropanol, that would be: %FFA = number ml 0.025M NaOH solution used x 0.7063 I guess blue/green Universal Indicator solution is the same as magenta phenolphthalein, no? Thanks! Best Keith ___ Biofuel mailing list Biofuel@sustainablelists.org http://sustainablelists.org/mailman/listinfo/sustainablelorgbiofuel Biofuel at Journey to Forever: http://journeytoforever.org/biofuel.html Search the combined Biofuel and Biofuels-biz list archives (70,000 messages): http://www.mail-archive.com/biofuel@sustainablelists.org/
