Based on stoichiometry ratio, you'll need 3 moles of MeOH and 3 moles of K/Na-OH per mole of triglyceride ["oil"]. The outputs for this reaction is 1 mole of glycerol and 3 moles of biodiesel. The reaction, furthermore, is two step, where the first reaction cleaves the ester bond and the second methylates the carboxylate. Excess hydroxide should be used to ensure full conversion and prior to methylation a separation of the glycerol layer should be done to prevent any contamination with an esterified glycerol in the fuel. Chemically speaking, your moles of hydroxide and methanol are entirely dependent on the type of fatty acids attached to the glycerol because that will affect the moles you need, which also means the actual amount of material you should use for full conversion changes per oil type used. Lets say one manufacturer uses a mix of fatty acids - coconut,, olive, vegetable, etc. - you'll need to adjust stoichiometry based on this. Let us assume you are gathering 55 gallons of vegetable oil waste from a commercial business partner. This is strictly vegetable oil; Based on the following research (doi: 10.3390/ijms160612871), you can expect 4.6-20 % palmitic acid derivative (256.4 g/mol) , 6.2-71.1% oleic acid (282.47 g/mol), and 1.6-79% linoleic acid (280.45 g/mol). These units are bound to glycerol (92.09 g/mol). There are three units bound to glycerol. It would be fair to assume about 15.22% palmitic, 41.57% oleic, and 43.22% linoleic acids. All of these fatty acids are bound to glycerol, so the combined molar mass would then be 3 fatty acid average weights + glycerol weight - 3(18.02) [water abstraction to bind FAs to glycerol]. Thus, 3[(256.4(0.1522) + 282.47(0.4157) + 280.45(0.4322)] + 92.09 - 3(18.02) = 871 g/mol avg per triglyceride molecule Density of vegetable oil is about 0.92 g/mL so you'll roughly have 208.20 L of solution, which corresponds to about 192 kg of material (around 422 pounds!) per 55 gallons of oil. An easy way to phrase the calculation would be that you need 9.47 mL of oil per mole of oil, then from there the rest is simple. Because you need to cleave the ester bond at 3 different locations, 3 moles of hydroxide will be used. If using KOH, you need 168.32 g per 9.47 mL oil. Because excess is recommended, 180g KOH/9.5 mL oil is suitable. For methanol, you'll also need 3 moles so that'll be 96.12 g MeOH/9.5 mL oil; at a density of 0.791g/mL, this can also be phrased as about 125 mL MeOH/9.5 mL oil (slight excess of MeOH here). To ensure the above is correct, realize the average is 871 g/mol for weight of vegetable oil. 1 gal = 3785.41 mL --> 208197.55 mL total of oil == 191541.75 g == 220 mol. Now we can finally plug in numbers to calculate the proper amount of material to use for 55 gallons of vegetable oil: 192 kg material == 55 gallons material == 208.20 L material == 221 mol. material 221 moles triglyceride per 3(221) moles KOH + 3(221) moles MeOH and use excess KOH = 663 moles KOH or 37.20 kg KOH [about 82 pounds] = 663 moles MeOH or 21.24 kg MeOH [about 47 pounds or 26.84 L] Assume Std. cond (25C/1ATM)
First off thanks for this excellent explanation/tutorial! This cleared up a lot of the clutter in my head surrounding the titration process when using WVO. I think he was asking why the 0.6ml was not included in the purity calculation. To my limited understanding, the titration solution amount is already corrected for purity since the higher the purity of the catalyst, the less amount of solution it will take to titrate WVO at a given PH level and vice versa. Am I in the right ballpark?
Hello I studies chemistry and polymer sciences and I have done many titrations and reports thereof, you basically want to see where the equivalence point is with regards to pH. When the solution goes from acidic to basic (under pH 7 and over pH 7 respectively). What I want to know is why you cant use a pH meter instead?
Thank you once again @UtahBiodieselSupply Your videos are highly effective and efficient. Is there a way to contact you for advice on PME related projects please?
Chemists several years back reacted several batches of Biodiesel with different bases (5 through about 10) and found the 7.0 was the sweet spot for KOH. Be sure and account for purity (7.0/.90 = 7.8). Nope. Value stays the same no matter which country your in. For Biodiesel we work in grams per liter so it'll work where-ever you ar. Just do the purity calculation for your particular KOH you're using.
Is there a reason you only calculated your purity against the base and not the titration amount? I am guessing you figured it in with the round up after titration but you didn't really specify. Great video!
so if you start with 190l of used cooking oil, 38l of methanol, 1,6kg of KOH, how many liters pure biodiesel we can get? What about glycerine waste? Thank you (great videos)
What is the pink color change indicate? A certain pH value? If so what is that value, it seems more appropriate to use a pH meter instead of color indicator
You'll need: 50 gallons of oil, 1596 grams Potassium Hydroxide and 10 gallons of methanol. Then heat to 135-140, mix thoroughly for 2 hours, let it settle for about 8-10 hours, drain off the glycerin and start washing.
yes but i am new to this and in the video you dont say how much alcohol why have to add (you calculate 50 cal of oil 1596 xydroxide potassium but you dont say what alse why need??)
"We know that when we're using brand new oil, it'll take 7.0g of KOH per litre of oil to make it" - how is this base value calculated/how do you know you need 7.0g? On a titration kit I recently bought, I was given a base value of 3.5g for NaOH instead of 5.5g, as it says on here. Also, I am unsure as to how to convert between the two - can you just use the good old moles = mass/Mr equation and find the ratio, or is it more complex? Thanks!
I got a sample of WVO...i titrated it..i did the exact same thing you did..but the colour change in my sample is red instead of pink...Can you please tell me the reason behind it???..also i need your email address for further questionnaire on my project....
All you needed to do is divide 1444 by 90% as 1444g is the amount of sodium hydroxide you needed for the reaction if they were pure, now since the one you have is only 90% pure, divide 1444 by 90%, you will get the more accurate result of 1604.4, I have absolutely no idea you went back to the beginning and ditched all your previous efforts, I can only think your math skills are at elementary levels (compared to the rest of the world, in the US you are already above average adults).
for my project I really want to know How To determination of total glycerol in biodiesel by not use Gas chromatography but simple test i can do ... How To Bio-Diesel Titration of Total Glycerin? Can you please tell me Can you send text data on e-mail?
Morale of the story, do not let Americans do math! Funny how he went the extra mile and rounded up during the calculation to get to 1596 when all he had to do was dividing 1444 by 90%, that way he would have gotten the more accurate result of 1604.4 without extra calculation needed!
KOH & NaOH don't come to us 100% pure. KOH is generally 90 to 93% pure & NaOH is usually 95 to 98% pure. So, with NaOH, we use 5.5 as our base assuming 100% purity. But to be accurate, you correct it for purity by dividing the base by the chemical purity. ie. 5.5 / 95% [5.5/0.95=5.8]. Purity corrected base for NaOH of 95% purity is 5.8. Titrate, add the value to 5.8 and times it by the liters of oil you have and you've identified how much catalyst to use with that batch.
We use 7.0 as a base because it's the amount of KOH that would be needed to react 1 liter of oil into Biodiesel if the titration was 0. Yes. It's applicable to all types of organic oils.
If you visit the Biodiesel Discussion forum (google it), you can look up the discussions on the research that was done in regards to the two base numbers.
3.5 is a terribly old, outdated number. I'm not sure why people don't update it. Use 5.5 and then adjust for purity. There was a lot of research done by chemists that established the 5.5 for NaOH & 7.0 for KOH. Just correct the figures for purity. Don't convert between the two. Yes. It can be done, but it'll never equal out right because of purity differences. Just use the same catalyst for titrating that you do for reacting & you'll be set.
Based on stoichiometry ratio, you'll need 3 moles of MeOH and 3 moles of K/Na-OH per mole of triglyceride ["oil"]. The outputs for this reaction is 1 mole of glycerol and 3 moles of biodiesel. The reaction, furthermore, is two step, where the first reaction cleaves the ester bond and the second methylates the carboxylate. Excess hydroxide should be used to ensure full conversion and prior to methylation a separation of the glycerol layer should be done to prevent any contamination with an esterified glycerol in the fuel. Chemically speaking, your moles of hydroxide and methanol are entirely dependent on the type of fatty acids attached to the glycerol because that will affect the moles you need, which also means the actual amount of material you should use for full conversion changes per oil type used. Lets say one manufacturer uses a mix of fatty acids - coconut,, olive, vegetable, etc. - you'll need to adjust stoichiometry based on this.
Let us assume you are gathering 55 gallons of vegetable oil waste from a commercial business partner. This is strictly vegetable oil;
Based on the following research (doi: 10.3390/ijms160612871), you can expect 4.6-20 % palmitic acid derivative (256.4 g/mol) , 6.2-71.1% oleic acid (282.47 g/mol), and 1.6-79% linoleic acid (280.45 g/mol). These units are bound to glycerol (92.09 g/mol). There are three units bound to glycerol. It would be fair to assume about 15.22% palmitic, 41.57% oleic, and 43.22% linoleic acids. All of these fatty acids are bound to glycerol, so the combined molar mass would then be 3 fatty acid average weights + glycerol weight - 3(18.02) [water abstraction to bind FAs to glycerol].
Thus, 3[(256.4(0.1522) + 282.47(0.4157) + 280.45(0.4322)] + 92.09 - 3(18.02) = 871 g/mol avg per triglyceride molecule
Density of vegetable oil is about 0.92 g/mL so you'll roughly have 208.20 L of solution, which corresponds to about 192 kg of material (around 422 pounds!) per 55 gallons of oil.
An easy way to phrase the calculation would be that you need 9.47 mL of oil per mole of oil, then from there the rest is simple.
Because you need to cleave the ester bond at 3 different locations, 3 moles of hydroxide will be used. If using KOH, you need 168.32 g per 9.47 mL oil. Because excess is recommended, 180g KOH/9.5 mL oil is suitable. For methanol, you'll also need 3 moles so that'll be 96.12 g MeOH/9.5 mL oil; at a density of 0.791g/mL, this can also be phrased as about 125 mL MeOH/9.5 mL oil (slight excess of MeOH here).
To ensure the above is correct, realize the average is 871 g/mol for weight of vegetable oil. 1 gal = 3785.41 mL --> 208197.55 mL total of oil == 191541.75 g == 220 mol.
Now we can finally plug in numbers to calculate the proper amount of material to use for 55 gallons of vegetable oil:
192 kg material == 55 gallons material == 208.20 L material == 221 mol. material
221 moles triglyceride per 3(221) moles KOH + 3(221) moles MeOH and use excess KOH
= 663 moles KOH or 37.20 kg KOH [about 82 pounds]
= 663 moles MeOH or 21.24 kg MeOH [about 47 pounds or 26.84 L]
Assume Std. cond (25C/1ATM)
Excellent point. It's always a good idea to adjust for purity. I forgot to show it in the video. My bad.
First off thanks for this excellent explanation/tutorial! This cleared up a lot of the clutter in my head surrounding the titration process when using WVO.
I think he was asking why the 0.6ml was not included in the purity calculation. To my limited understanding, the titration solution amount is already corrected for purity since the higher the purity of the catalyst, the less amount of solution it will take to titrate WVO at a given PH level and vice versa. Am I in the right ballpark?
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Hello
I studies chemistry and polymer sciences and I have done many titrations and reports thereof, you basically want to see where the equivalence point is with regards to pH. When the solution goes from acidic to basic (under pH 7 and over pH 7 respectively). What I want to know is why you cant use a pH meter instead?
Thank you once again @UtahBiodieselSupply Your videos are highly effective and efficient. Is there a way to contact you for advice on PME related projects please?
Super lesson! I have a problem:în my contry is not în market metanol. I can use alcool etilic from biodiesel? I have palm oil very much.
Chemists several years back reacted several batches of Biodiesel with different bases (5 through about 10) and found the 7.0 was the sweet spot for KOH. Be sure and account for purity (7.0/.90 = 7.8). Nope. Value stays the same no matter which country your in.
For Biodiesel we work in grams per liter so it'll work where-ever you ar. Just do the purity calculation for your particular KOH you're using.
Is there a reason you only calculated your purity against the base and not the titration amount? I am guessing you figured it in with the round up after titration but you didn't really specify. Great video!
Thanks for the reply. What exactly is correcting for purity, and how do you go about it? Apologies - I'm a little new to this!
Thank you!
so if you start with 190l of used cooking oil, 38l of methanol, 1,6kg of KOH, how many liters pure biodiesel we can get? What about glycerine waste? Thank you (great videos)
Gr8 vid. But you didn't tell us how much methanol you need for 50 gallons cheers cheers
Great video. When using Turmeric as a PH indicator, would I expect an orange color (turmeric color) as it neutralizes or the pink color?
do you do a titration test on new vegetable oil , or only on waste used oil.
Why do you not adjust for purity on the 7.6 grams (total per litre amount)? It means an extra 8 grams on the 190lts?
This was a very good video. Thank you from England. And wtf America why use imperial and metric? lol
What is the pink color change indicate? A certain pH value? If so what is that value, it seems more appropriate to use a pH meter instead of color indicator
One question. If im going to use an electronic pH meter, what pH value should I look for instead of "pink"?
You'll need:
50 gallons of oil, 1596 grams Potassium Hydroxide and 10 gallons of methanol.
Then heat to 135-140, mix thoroughly for 2 hours, let it settle for about 8-10 hours, drain off the glycerin and start washing.
How will the turmeric power look when its the correct PH will it also be pink like that?
It took 50 mils to titrate my oil something must be wrong i followec the intructions perfectly did it 5 times could my oil really be that bad?
how to find out the right amount of alcohol?
So, how much tumuric would you use for the titration test?
Wait so where does Methanol come in then? How much Methanol do you mix with Potassium Hydroxide?
thanks
Thanks very much!
yes but i am new to this and in the video you dont say how much alcohol why have to add
(you calculate 50 cal of oil 1596 xydroxide potassium but you dont say what alse why need??)
could you explain why you used 7.0 as base?? is it applicable to all types of oil?
"We know that when we're using brand new oil, it'll take 7.0g of KOH per litre of oil to make it" - how is this base value calculated/how do you know you need 7.0g? On a titration kit I recently bought, I was given a base value of 3.5g for NaOH instead of 5.5g, as it says on here.
Also, I am unsure as to how to convert between the two - can you just use the good old moles = mass/Mr equation and find the ratio, or is it more complex? Thanks!
s/o to the Mr. Ehnes squad
thank you so much
Thank you
But i want ask you a question
Before titration is it important to heat oil?!!
Nope, the process will turn out different if you heat oil
how you determined that you need 7 grams of KOH? If i am from a other country that value's change?
do you spell litre different in America? you spell it as liter ... just asking
90%x=1444, solve x and you get the answer, it’s that simple.
I got a sample of WVO...i titrated it..i did the exact same thing you did..but the colour change in my sample is red instead of pink...Can you please tell me the reason behind it???..also i need your email address for further questionnaire on my project....
Abi Thapa Red is fine too. You're probably just using Phenol Red.
Could you explain that a little bit more?
how do i know how much catalyst to add if it is a solid acid catalyst?
use water to make a solution
All you needed to do is divide 1444 by 90% as 1444g is the amount of sodium hydroxide you needed for the reaction if they were pure, now since the one you have is only 90% pure, divide 1444 by 90%, you will get the more accurate result of 1604.4, I have absolutely no idea you went back to the beginning and ditched all your previous efforts, I can only think your math skills are at elementary levels (compared to the rest of the world, in the US you are already above average adults).
for my project I really want to know How To determination of total glycerol in biodiesel by not use Gas chromatography but simple test i can do ... How To Bio-Diesel Titration of Total Glycerin? Can you please tell me Can you send text data on e-mail?
how to make Phenolphthalein solution 1% from Phenolphthalein powder, what I need to add and how much.
100 ml of water
1g of Phenolphthalein powder
Why use isopropyl alcohol instead of methanol? I can get methanol at 99% purity but the purest isopropyl alcohol I can get is 85% in my country
For titration, you can use methanol. It's just acting as a solvent.
Morale of the story, do not let Americans do math! Funny how he went the extra mile and rounded up during the calculation to get to 1596 when all he had to do was dividing 1444 by 90%, that way he would have gotten the more accurate result of 1604.4 without extra calculation needed!
KOH & NaOH don't come to us 100% pure. KOH is generally 90 to 93% pure & NaOH is usually 95 to 98% pure. So, with NaOH, we use 5.5 as our base assuming 100% purity. But to be accurate, you correct it for purity by dividing the base by the chemical purity. ie. 5.5 / 95% [5.5/0.95=5.8]. Purity corrected base for NaOH of 95% purity is 5.8. Titrate, add the value to 5.8 and times it by the liters of oil you have and you've identified how much catalyst to use with that batch.
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We use 7.0 as a base because it's the amount of KOH that would be needed to react 1 liter of oil into Biodiesel if the titration was 0. Yes. It's applicable to all types of organic oils.
If you visit the Biodiesel Discussion forum (google it), you can look up the discussions on the research that was done in regards to the two base numbers.
No problem. Happy brewing!
3.5 is a terribly old, outdated number. I'm not sure why people don't update it. Use 5.5 and then adjust for purity. There was a lot of research done by chemists that established the 5.5 for NaOH & 7.0 for KOH. Just correct the figures for purity. Don't convert between the two. Yes. It can be done, but it'll never equal out right because of purity differences. Just use the same catalyst for titrating that you do for reacting & you'll be set.
It's always 20 percent of the oil you start with. For a 50 gallon batch, that would be 10 gallons.
About 8 to 8.5.
Thanks a lot!