The chemistry of the glycerol (and soap) phase

[Dr Mark Senior Member Find all posts View Profile Close Dr Mark]

When making biodiesel, we're mostly concerned with the biodiesel phase, and the thick black syrup is just this annoying stuff that we have to dispose of somehow.

My method overturns this, however, as it produces a glycerol phase that is so thin that it can be tipped down the sink like water. See the videos in my method.

So I thought I'd post up an explanation of this phase, and the location of the various components in the system.

Let's start at the beginning. The state of a substance (gas, liquid, solid) is determined by the intermolecular forces - that is, the degree and type of interaction between molecules of the same type. For our purposes, the two relevant ones are hydrogen bonding and Van der Waals forces. These are the two weakest types of forces between molecules, with VDW being the weakest. Hydrogen bonding occurs when an organic molecule contains and oxhgen-rich atom such as oxygen, nitrogen or sulfor (but mostly O), with whose electron clouds tha naked hydrogen proton can interact.

Consequently, the first four carbon hydrocarbons are gases - methane, ethane, propane and butane. It's not until you get to C5 - pentane, that we get a liquid at RTP. This is simply because the weak VDW forces are not strong enough to overcome the thermal energy at RTP.

Let's now look at propane. C3H8. Let's now add a single oxygen, so we have C3H8O. This is propanol. This is now a liquid at RTP, simply because the oxygen provides a site at which hydrogen-bonding can occur. Add a further two oxygens and we have glycerol. Now with three oxygens, we get three sites per molecule at which hydrogen bonding can occur, and we now have a thick, viscous liquid.

This stuff is not easy to dispose of. It's so thick that it can clog plumbing if you tip it down the sink or toilet, and in any case the foam can cause problems.

So how is my method different? How is it that it is so thin that it can be tipped down the sink, and doesn't foam?

Let's look at the chemistry of the transesterification process:

1. Methoxide solution:

Eq 1: MeOH + KOH <-> KMeO + H2O

2. Transesterification reaction

Eq 2: C6O6H5R3 + 3KMeO + 3H2O -> C3O3H5K3 (potassium salt of glycerol) + 3(MeOCOR) + 3H2O

3. Regeneration of catalyst

Eq 3: C3O3H5K3 + 3H2O -> C3H8O3 (glycerol) + 3KOH

So the final step in the process results in the regeneration of the catalyst and protonation of the potassium salt of the glycerol to form glycerol.

My method differs in that it removes water from the process, which means that the final state of the glycerol is the potassium salt of the glycerol. Since the hydrogen bonding in glycerol is almost entirely due to the electron-poor hydroxy proton, its absence means that there is less interaction between the molecules. What bonding there is, is almost entirely due to the protons on the carbon backbone, somewhat ameliorated by the fact that the highly electronegative potassium cation will have a substantial electron withdrawing effect on the nonbonding electron pair on the oxygen.

Here is my process:

Methoxide solution:

Eq 1: MeOH + KOH <-> KMeO + H2O

Drying of methoxide solution:

Eq 2: H2O + CaO -> Ca(OH)2

Adding Eq 1 and Eq 2:

Eq 4: MeOH + KOH + CaO -> KMeO + Ca(OH)2

Transesterification reaction:

Eq 5: C6O6H5R3 + 3KMeO + 3H2O -> C3O3H5K3 (potassium salt of glycerol) + 3(MeOCOR) + 3H2O

And this is the final reaction. As the water has been removed, the KOH is not regenerated, and the final state of the glycerol is the potassium salt, which is much thinner than the glycerol and therefore more easily disposed of.

One more issue - what happens to the soap?

When the highly alkaline methoxide is added to the WVO it will obviously first react with the Free Fatty Acids:

KMeO + RCOOH -> RCOOK (soap) + MeOH

In other words, the methoxide reacts with the FFA to form soap + methanol.

So what happens to the soap?

Soap, of course, is a surfactant. It will then, obviously, look for interfaces. In this case, it will be the interface between the hydrophilic glycerol phase and the hydrophobic phase.

So the upper phase will be the biodiesel. Below this will be the soap, and underneath that will be the glycerol phase.

And this certainly aligns with my observations. When I leave the raw material to clarify (by bubbling air), when it has done so (by removing all the excess MeOH) I see a light brown layer settling on top of the dark brown glycerol. And, as I would expect, this light brown material is somewhat gelatinous and clumpy in nature, settling out on top of the liquid glycerol.

So that's it. Happy to answer any genuine questions from any genuine people. As with all my posts, I will ignore all contributions from a certain individual, a bloke who sees himself as an expert on biodiesel but who has no qualifications in chemistry, whose understanding of chemistry could be written on the back of a postage stamp, and whose posts are nothing more than noise.

[WesleyB Donating Member Find all posts View Profile Close WesleyB]

Re: The chemistry of the glycerol (and soap) phase

"it will obviously react first with the Free Fatty Acids" . There is reaction rate to consider. I made soap from fairly pure hot/soft octadecanoic acid (present in vegetable oils and animal fats). The soap cures for days or weeks. It's not a fast inorganic reaction like muriatic acid and sodium hydroxide. I expect the transesterification reaction rate might exceed the neutralisation reaction rate especially in very dry conditions. So biodiesel might form as fast as soap forms. Tell me where that's wrong.

[Dr Mark Senior Member Find all posts View Profile Close Dr Mark]

Re: The chemistry of the glycerol (and soap) phase

"it will obviously react first with the Free Fatty Acids" . There is reaction rate to consider. I made soap from fairly pure hot/soft octadecanoic acid (present in vegetable oils and animal fats). The soap cures for days or weeks. It's not a fast inorganic reaction like muriatic acid and sodium hydroxide. I expect the transesterification reaction rate might exceed the neutralisation reaction rate especially in very dry conditions. So biodiesel might form as fast as soap forms. Tell me where that's wrong.

The reaction with the FFA is a simple acid-base reaction and is all but instantaneous. In years of teaching and doing analytical chemistry I've not encountered one occasion where an acid-base process wasn't instantaneous. Kinetic data for acid-base reactions simply doesn't exist. And of course we know that because if it wasn't instantaneous, you wouldn't be able to titrate the FFAs with an instant result, as many people on this forum do.

The transesterification process, OTOH, is a Nucleophilic Substitution reaction with an SN2 mechanism (I think) and is therefore subject to kinetic considerations. We know, for example, that the Free Energy chart has an Activation Energy, which is why it needs a catalyst.

But even if it wasn't instantaneous it's a moot point, as both reactions go to completion anyway

[tillyfromparadise Senior Member Find all posts View Profile Close tillyfromparadise]

Re: The chemistry of the glycerol (and soap) phase

Hi Mark,

When making biodiesel, we're mostly concerned with the biodiesel phase, and the thick black syrup is just this annoying stuff that we have to dispose of somehow.

My method overturns this, however, as it produces a glycerol phase that is so thin that it can be tipped down the sink like water. See the videos in my method.

So I thought I'd post up an explanation of this phase, and the location of the various components in the system.

Interesting.
However, the more likely reason that your by-product (which you call glycerol) is so thin is that because the largest component of your by-product is probably Biodiesel instead of glycerol.

You have posted that you have a large by-product layer.
By using such a huge excess of KOH (20g! KOH per litre of WVO) in the reaction instead of performing a simple 5 minute titration to accurately determine the amount of KOH required, you are likely to be producing an excess of soap.
The soap, along with about twice it's volume of biodiesel, becomes part of the by-product phase
My calculations suggest that your by-product composition is probably about 36% biodiesel and only about 30% glycerol.

I do urge that in the future you perform a simple titration to determine the amount of KOH actually required for the reaction to prevent pouring so much biodiesel down the drain.

[WesleyB Donating Member Find all posts View Profile Close WesleyB]

Re: The chemistry of the glycerol (and soap) phase

Dr Mark; I believe in the formulas you have not accounted for wetness in waste vegetable oil. You described the neutralization reaction (acid+base--> salt + water) but you did not describe saponification reaction where trifatty acid glyceryl + caustic in the presence of water --> soap + glycerol +regenerated water (saponification). Some of the error in making biodioesel stems from wetness of vegetable oil. Your no titration fool proof method doesn't consider wet oil or very high titration oil. For Example a titration of 20 oil. add the suggested amount of caustic, the instantaneous neutralization reaction consumes the caustic and there's no caustic to form methoxide plus the water formed by the neutralization reaction. Beautiful weather here today sir.

[Dr Mark Senior Member Find all posts View Profile Close Dr Mark]

Re: The chemistry of the glycerol (and soap) phase

Dr Mark; I believe in the formulas you have not accounted for wetness in waste vegetable oil. You described the neutralization reaction (acid+base--> salt + water) but you did not describe saponification reaction where trifatty acid glyceryl + caustic in the presence of water --> soap + glycerol +regenerated water (saponification). Some of the error in making biodioesel stems from wetness of vegetable oil. Your no titration fool proof method doesn't consider wet oil or very high titration oil. For Example a titration of 20 oil. add the suggested amount of caustic, the instantaneous neutralization reaction consumes the caustic and there's no caustic to form methoxide plus the water formed by the neutralization reaction. Beautiful weather here today sir.

Correct. I didn't bother with the saponification process. I'm going to do that in a subsequent post devoted to the generation and fate of soap.

The amount of water in clear, free-phase oil is negligible. For about the millionth time, those of you that use the dr Pepper method (or any method without the drying step) add water to your mixture from the methoxide solution, and given the 1:1 stoichiometric ratio with the methoxide, the amount of water will easily swamp whatever small amount may be present in the oil:

KOH + MeOH <-> KMeO + H2O

And as for "high titration" oil, yes, in principle a high concentration of FFAs will neutralise the methoxide solution. But I have been using this method now for about ten years with not a single problem. And if I wasn't going to have any problems at a SVO/MeOH ratio of 100:15 then I'm not going to have any problems with a ratio of 100:20

Also, unless I'm very much mistaken, the amount of caustic I use in my method is much higher than is used by other methods, so if the conc of FFAs do not inhibit the catalysis in these methods, it certainly isn't going to in mine.

By way of example, the equivalent concentration of KOH in my mix is about 0.3M.

But even if this is wrong - even if the amount of caustic in my mixture is less than the Dr Pepper method, my method doesn't have the interfering saponification reaction to consume catalyst.

The simplicity of my method means that I can toss in as much KOH as I want, and there are never any interfering reactions to get in the way.

And the water sure makes a difference. I know this as I once had a batch where I was lazy when transferring the WVO and some free-phase water transferred across. She'll be right, I thought and went ahead anyway. When I started the reaction, it wasn't going - not a nice situation to be in when you make 1000L at a time. It was the middle of winter and it was about 14 degrees I think. So I pit my heater in and kept it recirculating overnight. When it got to about 23 the reaction kicked off.

So the great key to my method is that it is anhydrous and the interfering saponification reaction is eliminated. I'm composing a post about soap at the moment, and this just occurred to me, so I thought I'd better update it

[tillyfromparadise Senior Member Find all posts View Profile Close tillyfromparadise]

Re: The chemistry of the glycerol (and soap) phase

Hi Mark,

For about the millionth time, those of you that use the dr Pepper method (or any method without the drying step) add water to your mixture from the methoxide solution, and given the 1:1 stoichiometric ratio with the methoxide, the amount of water will easily swamp whatever small amount may be present in the oil:

KOH + MeOH <-> KMeO + H2O

You seem to keep forgetting, so for about the millionth time I will remind you that water is only produced when you actually Produce methoxide.
You claim that without your drying procedure, there is very little methoxide produced. That means VERY LITTLE WATER is produced.

The Chemist Neutral used to point out that when you add KOH to Methanol you are mostly just dissolving KOH in methanol.
Dissolving KOH in methanol does not produce water.
It is only when you actually produce a tiny bit of methoxide that an equivalent tiny bit of water is produced.
That means that those of us who use the Dr Pepper Method (or any method without the drying step) are putting Very Little Water in the reaction.

[Dr Mark Senior Member Find all posts View Profile Close Dr Mark]

Re: The chemistry of the glycerol (and soap) phase

Dissolving KOH in methanol does not produce water.

Wrong, as usual. Here is the equation:

KOH + MeOH <-> MeOK + H2O

In case you don't know, H2O is water. You may look at it and think it is "dihydrogen monoxide" or something, but it's actually water.

MeOK is the potassium salt of the methoxide. Unlike the methanol (MeOH), in which the hydrogen is covalently bonded to the oxygen, the potassium ion is bound as an ion pair, so it's more accurate to write it as K+MeO-

This was all explained in my post on Le Chateleir's principle, and it is now clear that you didn't understand a word of it, despite my attempts to put it in layman's terms.

It is now plain that if you don't understand such a simple equilibrium as this, you are in no position to advise others on anything pertaining to the chemistry of any part of the biodiesel process. Please leave such advice to the professionals. In other words, people with tertiary qualifications in chemistry

Note: replying in shouty bold text does not increase your credibility. It just makes you look desperate, and very, very much out of your depth.

Both of which, of course, you are.

[Tony From West Oz Vice Chairperson of WARFA Find all posts View Profile Close Tony From West Oz]

Re: The chemistry of the glycerol (and soap) phase

[quote]Here is the equation: KOH + MeOH <-> MeOK + H2O[/]quote
As you have stated previously, Le Chateleir's principle applies, but you do not give the proportions at STP for this reaction.

Tilly's statement could be true if the reaction is weighted in favour of the solution over the reaction products.
So here are my questions:
1. What is the formula for calculating the ratios?
2, So, if I have 1Kg of KOH and 20 litres of 'MeOH', How much 'MeOK' and H2o would be produced and how much KOH dissolved in H20 would be present at any one point in time?

As a Chemist you should be able to specify the ratios of each side of the reaction at STP.
I await your reply with the answers to my questions.

[tillyfromparadise Senior Member Find all posts View Profile Close tillyfromparadise]

Re: The chemistry of the glycerol (and soap) phase

Hi Mark,
Wrong again, as usual.
You do seem to have a problem understanding what is written.

H2O is only produced when you actually produce MeOK.
As you continually point out, when enough water is produced, the Production of MeOK stops. You may have forgotten, but this is an equilibrium reaction. At the point where the production of MeOK stops because of the presence of too much H2O, the production of H2O also stops. No more MeOK or H2O/ water is produced.
The remaining unreacted KOH is just dissolved in the methanol without producing H2O or MeOK.
A real chemist going by the name Neutral who had over 30 years R&D in the alternate fuel field including actually making "biodiesel" in the 70's {before it was called biodiesel) pointed that out to me years ago.
At the same time he told me that he thought that not much MeOK was actually produced when you mixed KOH with methanol so not much H2O was produced. He thought it was mostly just KOH dissolved in methanol

PS. THIS IS SHOUTING. This is emphasizing an important point.
If you have any further confusion about any of this please do not hesitate to ask.

[Dr Mark Senior Member Find all posts View Profile Close Dr Mark]

Re: The chemistry of the glycerol (and soap) phase

[QUOTE=Tony From West Oz;66558]

Here is the equation: KOH + MeOH <-> MeOK + H2O[/]quote
As you have stated previously, Le Chateleir's principle applies, but you do not give the proportions at STP for this reaction.

Tilly's statement could be true if the reaction is weighted in favour of the solution over the reaction products.
So here are my questions:
1. What is the formula for calculating the ratios?
2, So, if I have 1Kg of KOH and 20 litres of 'MeOH', How much 'MeOK' and H2o would be produced and how much KOH dissolved in H20 would be present at any one point in time?

As a Chemist you should be able to specify the ratios of each side of the reaction at STP.
I await your reply with the answers to my questions.

Tony,

With the greatest of respect, you just simply do not understand this. I've tried, again and again and again, to explain this. I've explained it in great detail on my website, and I specifically constructed a post on this, where I slowly and painstakingly went through through the chemistry of both the Dr Pepper method and my method. I explained the nature of the equilibrium in fine detail, and explained all the various reactions.

And at one stage I wasn't going to bother even doing this. When I first posted my method a few years ago, I didn't bother answering any of the questions, as I figured I was pretty much wasting my time explaining chemistry to people with no formal qualifications, and many of whom plainly hadn't even done chemistry at high school.

But I changed my mind, and have now gone into great detail attempting to explain this, now with multiple posts, specifically. designed to be understood by the layman. But I'm now starting to wonder whether I've just wasted my time.

There is just no more simpler language I can use to explain this. And this should not really be surprising. This is why we have universities, as some things cannot be understood in five minutes by reading a blog post. Le Chateleir's principle is a 100 level concept, but the mechanism and kinetics of other parts of the process are 200 and 300 level concepts. If you don't understand them, it doesn't mean you're dumb, but simply that you don't have the right educational background.

I once had someone try to explain futures trading to me. I didn't understand a word of it. Concepts that were simple to him were just entirely foreign to me. But one thing I didn't do was to conclude that this guy was wrong simply because I hadn't understood his explanation.

And so if you still don't understand my explanation, then those on this forum have to decide whose advice they want to take - someone with a PhD in Chemistry, and 30y experience in the chemical sector, and who when communicating on this forum always speaks politely and respectfully with people - or someone with no formal qualifications in Chemistry, who takes great personal offence when anyone disagrees with him, and bombards their threads and posts with loud shouty bold text, and has made this forum such an unpleasant place that many people are now frightened to post.

You choose which one has the most credibility.

In any case, I will not be either reading or responding to Tilly's posts. Of all the people I reply to on this forum, he is the one with the least ability to understand simple chemical concepts, and the one with the most childish attitude. It's a complete waste of time to attempt to communicate with him as an adult.

Having said all that, I'll try again. I've explained it again and again and again and again. But I'll try one last time. But this is my last attempt. After this I give up. I'll try to do it in point form.

• I don't know the equilibrium constant, and I don't have to. The reason simply is that it is a dynamic equilibrium. That is, it is constantly reacting in both directions, and will shift according to other reactions occurring. No matter how many times I've attempted to explained this, you just still don't understand how a dynamic equilibrium works.

• Each molecule of KOH reacts with one molecule of MeOH to form one molecule of potassium methoxide and one molecule of water. For every molecule of MeOK you make, you make one molecule of water.
• With the Dr Pepper method, if your concentration of MeOK isn't high enough, there isn't enough catalyst to carry out the reaction. On the other hand, if it's too high, then you have too much water, and the competing saponification reaction makes enough soap to emulsify your batch and it turns to gloop. That's why you have to titrate - to get a successful reaction you have to hit a pretty narrow window of concentration
• Because of this, you can't get a high enough concentration of catalyst to permit the reaction to proceed at room temperature.

This is a very common conundrum in organic synthesis. It's very common to find yourself with competing reactions occurring simultaneously. And so any chemist worth his salt will just simply modify the reaction to favour the desired reaction.

When I looked at this procedure I saw an obvious modification to me. Since I have now had such immense difficulty explaining this to all you non-chemists, I can now see why it wasn't obvious to others.

My modification was simply to remove the water. This simultaneously achieves the following

• it pulls the equilibrium to the right, thus increasing the concentration of catalyst, allowing the reaction to proceed at RT
• It eliminates the competing saponification reaction, so you can shovel in as much KOH as you want
• It results in a deprotonated form of glycerol, resulting in lower viscosity and easier disposal.

Okay that's it. I just can't explain it any more than this and I can't simplify it any more than this. I've explained this over and over and over and over, and if you still don't understand it, enrol in an online chemistry course (if such things exist) or at a university.

Either that or just turn up and watch me make my next batch. You don't have to understand the chemistry, just follow the instructions;.

[Tony From West Oz Vice Chairperson of WARFA Find all posts View Profile Close Tony From West Oz]

Re: The chemistry of the glycerol (and soap) phase

Mark,
I did a quick Google and found (amongst others) http://www.arpnjournals.org/jeas/res..._0816_4863.pdf
This gives the Molar ratios and the reaction coefficients for different reaction temperatures.
Now, this should give us the ability to predict the amount of water for the given ratios of KOH and MeOH.
I believe that Molar ratios are based on the relative mass of the feedstock eg KOH should have K (40), O (16), H (1) = 57, Methanol CH3OH would have C (12) H *4 (4) O (16) = 32
In this case a molar ratio of 0.012 ( KOH:MeOH) would give mass ratios of (32*0.012):57 or 0.0067. Thus for a molar ratio of 0.012 KOH:MeOH you would need to add 6.7g of KOH per Kg of Methanol.

For a reaction temperature of 25° and a molar ratio of 0.012, this would give us:
KOH in solution 1.09%
CH3OH in solution 96.91%
CH3OK in solution 1.64%
H2O in solution 0.81%

Are my figures accurate?

Please comment on this as I have only lower school Chemistry.

[Dr Mark Senior Member Find all posts View Profile Close Dr Mark]

Re: The chemistry of the glycerol (and soap) phase

Hi Mark,
Wrong again, as usual.
You do seem to have a problem understanding what is written.

H2O is only produced when you actually produce MeOK.
As you continually point out, when enough water is produced, the Production of MeOK stops. You may have forgotten, but this is an equilibrium reaction. At the point where the production of MeOK stops because of the presence of too much H2O, the production of H2O also stops. No more MeOK or H2O/ water is produced.
The remaining unreacted KOH is just dissolved in the methanol without producing H2O or MeOK.
A real chemist going by the name Neutral who had over 30 years R&D in the alternate fuel field including actually making "biodiesel" in the 70's {before it was called biodiesel) pointed that out to me years ago.
At the same time he told me that he thought that not much MeOK was actually produced when you mixed KOH with methanol so not much H2O was produced. He thought it was mostly just KOH dissolved in methanol

PS. THIS IS SHOUTING. This is emphasizing an important point.
If you have any further confusion about any of this please do not hesitate to ask.

I give up. I can see now that it was a mistake to attempt to engage with you at all. As has been the case in the past, as you have no formal qualifications in chemistry, and have deluded yourself into thinking that you understand some chemistry just because you have learned a few big words, your posts are an utterly baffling collection of random chemical terms and concepts that make no sense to anyone but you. Consequently I'm just wasting my time, and will just go back to ignoring everything you say.

Please refer to my conversations with Tony - unlike you he is an adult, and is keen to learn from someone that knows a lot more about Chemistry than him.

[Dr Mark Senior Member Find all posts View Profile Close Dr Mark]

Re: The chemistry of the glycerol (and soap) phase

Mark,
I did a quick Google and found (amongst others) http://www.arpnjournals.org/jeas/res..._0816_4863.pdf
This gives the Molar ratios and the reaction coefficients for different reaction temperatures.
Now, this should give us the ability to predict the amount of water for the given ratios of KOH and MeOH.
I believe that Molar ratios are based on the relative mass of the feedstock eg KOH should have K (40), O (16), H (1) = 57, Methanol CH3OH would have C (12) H *4 (4) O (16) = 32
In this case a molar ratio of 0.012 ( KOH:MeOH) would give mass ratios of (32*0.012):57 or 0.0067. Thus for a molar ratio of 0.012 KOH:MeOH you would need to add 6.7g of KOH per Kg of Methanol.

For a reaction temperature of 25° and a molar ratio of 0.012, this would give us:
KOH in solution 1.09%
CH3OH in solution 96.91%
CH3OK in solution 1.64%
H2O in solution 0.81%

Are my figures accurate?

Please comment on this as I have only lower school Chemistry.

Thanks for the paper - I'll have a look at it when I get a chance
T

[tillyfromparadise Senior Member Find all posts View Profile Close tillyfromparadise]

Re: The chemistry of the glycerol (and soap) phase

Hi Mark,

In any case, I will not be either reading or responding to Tilly's posts

Why don't you grow up Tilly? Seriously, why don't you stop being so childish?
I've got two degrees in Chemistry and you don't have any. I have tried and tried and tried to explain Le Chateleir's principle to you, particularly as it pertains to this method, and this sort of childish drivel is the best you can do?

There are people on this thread that want to learn some chemistry. You're not one of them.

We all know the truth. Your fragile ego is threatened by the fact that someone might have developed a superior method than yours. Welcome to the real world son. Get back to me when you either

a. Get a degree in Chemistry

b. Grow up

I think your vow that you "will not be either reading or responding to Tilly's posts" lasted for an even shorter time than my New years commitment to go on a diet.
I forget, please remind me how many degrees you have in chemistry