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

  1. #31
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    Re: The chemistry of the glycerol (and soap) phase

    Quote Originally Posted by Tony From West Oz View Post
    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

    Click image for larger version. 

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    Now lets look at Figure 1:

    What information do you get from this?

    Click image for larger version. 

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    EDIT: To view the thumbnails - Click on the thumbnail. This will open up the image for viewing. Tony
    Last edited by Tony From West Oz; 10th May 2019 at 12:25 AM.

  2. #32
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    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.

  3. #33
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    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)
    Last edited by tillyfromparadise; 9th May 2019 at 11:58 AM.

  4. #34
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    Re: The chemistry of the glycerol (and soap) phase

    Hi Mark,

    Quote Originally Posted by Mark View Post
    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.

  5. #35
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    Re: The chemistry of the glycerol (and soap) phase

    Quote Originally Posted by Tony From West Oz View Post
    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
    Last edited by Tony From West Oz; 13th May 2019 at 12:34 AM.

  6. #36
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    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





    Last edited by tillyfromparadise; 13th May 2019 at 09:51 PM.

  7. #37
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    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.

  8. #38
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    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."
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