The chemistry of the glycerol (and soap) phase

[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.

Hi Tony,

At last I've figured out how to post figures. The URL bit still isn't working that I can see, but the thumbnails will do for now.

Right let's go through this step by step.

Firstly, here is the methoxide equilibrium, exactly as I wrote it in my posts



Now lets look at Figure 1:

What information do you get from this?


EDIT: To view the thumbnails - Click on the thumbnail. This will open up the image for viewing. Tony

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

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

What information do I get from the graph? It looks like as the methoxide concentration at 25 degrees centigrade increases so does the water. I don't know the units used in the graph, under magnification they are unclear. I seems to indicate 2% of something being the concentration. Your method of posting information might be improved. As the methoxide concentration increases the water concentration increases and water decomposes the methoxide. The information under magnification is unclear.

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

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

Hi Mark,
Just a reminder, you are supposed to be calculating how far the reaction producing CH3OK (potassium methoxide) goes before it stops due to the presence of the water produced during the reaction. This is without a drying procedure to remove any water such as you claim happens when you add cement to the mixture.

(I know you are dying to tell everyone you have two degrees in chemistry)

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

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

Hi Mark,

Now lets look at Figure 1:
What information do you get from this?

The information I am getting is that you do not know how to answer Tony's Question.
A quick reminder, Tony asked:
"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?"

Surely with all those degrees in chemistry you have, answering something as simple as this would be a formality.

[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.

Mark, Can you please confirm my calculations of the Molar weights of the reactants?
Is the molar weight of KOH (K (40) + O (16) + H (1)) 57?
Is the molar weight of CH3OH (C (12) + H *4 (4) + O (16)) 32?

I was hoping that you would have provided me some guidance in this, as I only have lower school chemistry education.
It appears to me that the way to calculate the molar weight ratios of the reactants is to sum the molecular weights of the elements in the molecules. Thus the relative weights of the reactants relate directly to the formula. If this is correct then the relative mass of the reaction products can easily be calculated, by reference to the graph.
If this is not how to calculate the molar weight ratios, then please educate me how they should be calculated.

thanks in advance,
Tony

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

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

Hi Tony,

The molecular mass for each is as follows:

CH3OH- 32.04186 g/mol
KOH- 56.10564 g/mol
CH3OK- 70.13222 g/mol
H2O- 18.01528 g/mol

Hi Wesley,
"...A number without units is nothing." Error noted and correction made

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

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

Tony from West Oz; The molecular mass is calculated by adding the atomic mass of the elements present in the molecule. Hydrogen 1.00794 oxygen 15.9994 carbon 12.011 potassium 39.0983 the units used are grams per mole. A number without units is nothing. The number must have units attached to it ,a mole is 6.023 times ten to the twenty third power of atoms or molecules if I recall correctly. For practical purposes using 1 gram per mole as the hydrogen element atomic mass works usually.

[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

Tilly and WesleyB. Thank you for the precise numbers. It looks like my use of the atomic mass (10% accuracy) was reasonably accurate for a non-chemist.

So, with the aid of the document I referenced earlier, the amount of water produced when KOH us dissolved in Methanol is fairly small in the scale of things.

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%

This was what I posted previously and I believe that I am in the 'ball park" with the percentage of each of the constituents of the solution.

The majority of the Methanol is still Methanol. the majority of the KOH is still KOH.
A small amount of MEOK has been created which resulted in a very small amount of H2O being produced.

Neutral, we have proven your statement:
"Not much MeOK is actually produced when you mix KOH with methanol, so not much H2O is produced, it is mostly just KOH dissolved in methanol."

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

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

[quote]Mark said:
...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...


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...


Iwill 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.


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

Please refer to my conversations with Tony - unlike you he is an adult,...

Please either alter the instructions for the Dr Pepper method to make it a RT method by removing any mention of heating, or find another thread to fill with your pointless posts.

Respond to this post with a vague, pointless, and childish post in big shouty text.


Unlike me, you have no formal qualifications in chemistry. Given the quality of the advice you give to others (which is mostly wrong), I doubt that you've even completed high school chemistry.[/quote}

Hmm. I wonder what Mark says when he is not speaking politely and respectfully with people

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

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

Would it be possible for Mark and Tilly to respond to this post in a reasonably way, Tony can interject also if he wants.

OK, so I've read most of the important info on here but there are some wee niggley naggley issues?

KOH is usually 90% pure, the rest being mostly water, so does this water reduce/stop the production of potassium methoxide. If so, when using Sodium does this mean more Sodium Methoxide is produced because of it being relatively dry.

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

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

smithy: The equilibrium Dr Mark refers to is calculated mathmatically by multiplying the concentration of the products of a reaction divided by the concentration of the reactants of a reaction. This was used in calculatiion of the strength of weak acids like vinegar or weak bases like ammonia gas dissolved in water. But the equilibrium of methoxide a strong base has methanol as a solvent and water is a product of the reaction. It's a different approach to an equilibrium equation. With weak acids and bases the concentration of water (the solvent) is ignored. The concentration of products (pH) with weak acids and bases is very small. But Dr Mark's proceedure of removing water in the production of methoxide with calcium oxide or cement complicates the equation. The methoxide concentration may be much higher. Maybe Dr Mark can say how to figure the equilibrium constant product in the case since we were given the concentration of the reactants and products at 25 degrees centigrade by Tony. The water present in potassium hydroxide does affect the concentration of methoxide. I vaccum distilled water out of potassium hydroxide solid. I damaged expensive equipment distilling out water from potassium hydroxide solid because it was so hot, over 200 Degrees centigrade.

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

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

What information do I get from the graph? It looks like as the methoxide concentration at 25 degrees centigrade increases so does the water. I don't know the units used in the graph, under magnification they are unclear. I seems to indicate 2% of something being the concentration. Your method of posting information might be improved. As the methoxide concentration increases the water concentration increases and water decomposes the methoxide. The information under magnification is unclear.

Yes - my apologies - the thumbnails aren't much use and despite my best efforts I just cannot work out how to post the images on my desktop onto this site. The method works for other images, but not for the images I'm attempting to post from from my image hosting account. Hopefully the administrator can help me
Y

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

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

Mark, Can you please confirm my calculations of the Molar weights of the reactants?
Is the molar weight of KOH (K (40) + O (16) + H (1)) 57?
Is the molar weight of CH3OH (C (12) + H *4 (4) + O (16)) 32?

I was hoping that you would have provided me some guidance in this, as I only have lower school chemistry education.
It appears to me that the way to calculate the molar weight ratios of the reactants is to sum the molecular weights of the elements in the molecules. Thus the relative weights of the reactants relate directly to the formula. If this is correct then the relative mass of the reaction products can easily be calculated, by reference to the graph.
If this is not how to calculate the molar weight ratios, then please educate me how they should be calculated.

thanks in advance,
Tony

Yes, quite correct. I'll talk you through the processes involved just as soon as I work out how to post the images on my desktop onto this site

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

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

smithy: The equilibrium Dr Mark refers to is calculated mathmatically by multiplying the concentration of the products of a reaction divided by the concentration of the reactants of a reaction. This was used in calculatiion of the strength of weak acids like vinegar or weak bases like ammonia gas dissolved in water. But the equilibrium of methoxide a strong base has methanol as a solvent and water is a product of the reaction. It's a different approach to an equilibrium equation. With weak acids and bases the concentration of water (the solvent) is ignored. The concentration of products (pH) with weak acids and bases is very small. But Dr Mark's proceedure of removing water in the production of methoxide with calcium oxide or cement complicates the equation. The methoxide concentration may be much higher. Maybe Dr Mark can say how to figure the equilibrium constant product in the case since we were given the concentration of the reactants and products at 25 degrees centigrade by Tony. The water present in potassium hydroxide does affect the concentration of methoxide. I vaccum distilled water out of potassium hydroxide solid. I damaged expensive equipment distilling out water from potassium hydroxide solid because it was so hot, over 200 Degrees centigrade.

Quite correct but the whole point of my method is that all water from all sources is removed. So whatever water is present in the KOH is removed by the drying agent. I'll put up some calculations once I can figure out how to post images
Q

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

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

Hi Mark,

Quite correct but the whole point of my method is that all water from all sources is removed.

An inaccurate statement. Even if your method does remove all the water produced when making methoxide, and you have never shown any testing to support this claim, you keep omitting to take into account the water that WILL be present in the oil.

whatever water is present in the KOH is removed by the drying agent.

You say the water is removed, however you have never provided any testing to show this is actually what happens.
You also say the normal method of making biodiesel requires heat and will not proceed at room temperature and we all know this is not true. You did not even perform a simple 15 minute test to verify this most important fact.
More importantly you continue to ignore research papers and experiments performed by people on this forum that shows the standard base method of making biodiesel works fine at room temperature.
You also claimed that the FFA content of the oil does not matter and that is not true either.

When you did your two degrees in chemistry that you are clearly so proud of, did they teach you that once you have a degree in chemistry you know everything and do not need to perform any experimentation in support of your claim and to ignore all valid information that does not support your claim?

I'll put up some calculations once I can figure out how to post images

It would be a whole lot more meaningful if you actually did some testing and posted some test results.