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I think the calculations are off... 7g NaOH = (23/40)*7 (g Na) + (17/40)*7 (g OH) 4.025 g Na 2.975 g OH 17.052 ml 25% NaOCH3/HOCH3 Solution at 25% (by weight) solution. 0.945 density. 17.052 ml * .945 g/ml = 16.114 g If 25% is NaOCH3 (by weight)... you get 4.029 g NaOCH3 Oops, there's where you missed on your calculation. Above, you calculated grams Na in NaOH solution. Here you calculated grams NaOCH3 in solution. So, if you work the other way. 4.025g Na * 54/23 g NaOCH3/g Na = 9.45 g NaOCH3 9.45 * 4 = 37.8 g (25% by weight) solution. 37.8 g solution / .945 g/ml = 40 ml So, if you wish to have 250 ml solution. You would add 210 ml of Methanol + 40 ml of your 25% solution. ----------- Now, if 40ml (25% solution) is equivalent to 7 gm NaOH You can make your conversion factor. 40/7 ml/g = 5.714 ml (25% NaOCH3) / g (NaOH) So... 5 g NaOH is equivalent to 28.57 ml of the solution. 10 g of NaOH is equivalent to 57.14 ml of the solution. ------------ Another interesting (theoretical) calculation. If you take 7 gm NaOH per 250 ml Methanol, or (28 gm NaOH / 1000 ml Methanol) Theoretically, you are adding: 28 g * 17/40 g OH/g NaOH = 11.9 g OH 11.9 g OH * 18/17 g H2O / g OH = 12.6 g H20 So, in effect, for every liter of Methanol, you are adding 12.6 ml of water. Or... adding 1.26 percent water if you use Sodium Hydroxide. If you are using anhydrous Sodium Methylate, then that 1.26 percent water never gets added. Of course, all that also depends on the purity of your reagents. This message has been edited. Last edited by: keelec, |
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Thank you, Double D,
I've learned something today. Brian 1996 K2500 4x4 6.5TD |
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In titration base numbers, Sodium Hydroxide is 5grams per liter. Potassium Hydroxide is now 7.5g.
If it titrate tests the same way as NaOH and KOH, All one would really need to know is the established base number. This is math I cant stay focused on very well. But I do see how the percentages and weights are being used. I'll just keep reading. Ready to try one. Brian 1996 K2500 4x4 6.5TD |
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member 2009 Sponsor |
What I had is:
1g NaOH = .575g Na per gram NaOH .945*.25= .236g Na per ml NaOCH3 It looks like your calculations are based on the NaOCH3 consisting of 25% Sodium methylate and 75% Methanol? I think  I was figuring the NaOCH3 consisting of 25% Sodium and 75% Methanol. Correct me if I'm wrong, Lord knows I have been before. -Dave- |
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Dave,
My Moles and Normal Solutions, and etc... well, all of that stuff is kind of rusty. Obviously, if you have a salt, you can not have a cation without an anion. So, Pure Sodium Methylate is a salt, which apparently has a crystalline form, Na+ -OCH3 Theoretically, you could mix it with anything, although it would be kind of pointless, and dangerous to mix it with water since it is a stronger base than NaOH. So, if I had pure Sodium Methylate salt, and was told to make a 25% solution, I would, for example take 25 grams of Sodium Methylate powder, and add it to 75 grams Methanol (94.7 ml since the density is .7918). -------------------------------- I suppose that is why many solutions are calculated in Moles, even though the calculations are more complex. Because, then you would know a 1 Molar Solution of Sodium Hydroxide would be equivalent to a 1 Molar solution of Sodium Methylate, and both would be neutralized by an equivalent amount of 1 molar HCL, or Acetic Acid, or whatever acid you wished to use. Oh, some of that Chemistry is coming back. I think 1 Normal Calcium Carbonate is actually a 1/2 Molar solution. ----------------------------------- Anyway, in this case, the solutions are being reported in percentages (presumably weight/weight percentages), so unless proven incorrect using a titration, I'd calculate using the entire weight of the salt. ----- Clifford ----- This message has been edited. Last edited by: keelec, |
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member 2009 Sponsor |
Wouldn't 75 grams of methanol occupy a volume of 94.7 ml at a "density" of 0.7918 (75/0.7918 = 94.7)? Also, when you say "density is .7918" aren't you actually referring to specific gravity? |
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member 2009 Sponsor |
OK. I tried to refigure using 25% soultion of CH3ONa into CH3OH being 75% (by weight) methanol and 25% sodium methylate (by weight). So I have a SG of .945g per ml of the combined soultion.
C = 12 H3 = 3 NA = 23 O = 16 = 54 23(Na) divided by 54 (total molecular weight of Sodium Methylate) = 42.6% (Na 25% (sodium methylate) of .945g = .236g 42.6% of .236=.101(rounded up) So each 1ml of 25% sodium methylate in methanol should contain .101g Na (rounded up from .100536) NaOH: Na = 23 O = 16 H = 1 = 40 23(Na) divided by 40 (Sodium Hydroxide molecular weight) = 57.5% So 57.5% of NaOH is Sodium. So 1 gram NaOH should contain .575g Sodium. So 5g(NaOH)*.575g = 2.875g Na. So 2.875 divided by .101 = 28.47 (rounded up) So wouldn't you then need 28.47ml of 25% Sodium methylate in methanol solution to = 5g NaOH and 56.93ml NaOCH3 to = 10g NaOH. Maybe I need a Chemistry for Dummies book. |
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member 2009 Sponsor |
I think you are right on. Here is another way to look at it. 1.35 gms of pure NaOCH3 provides the same number of moles of Na as 1 gm of NaOH (54/40 = 1.35). So 1 gm of NaOH is equivalent ot 1.35 gms of NaOCH3. 1.35 gms of NaOCH3 is present in 5.4 gms of 25% solution of Sodium methylate (1.35/0.25 = 5.4) Volumetrically, 5.68 ml of 25% NaOCH3 solution contains 5.4 grams of NaOCH3 (5.4/0.95) So 1 gm of NaOH is equivalent to 5.68 ml of 25% NaOCH3 solution. You calculated 5 gm of NaOH to be equal to 28.47 ml of 25% NaOCH3 solution. I agree. 5 gm NaOH x 5.68 ml/gm NaOH equivalent = 28.4 ml. I don't know about you, but to me 28.4 ml is close enough to 28.47 ml, especially when you see I rounded the specific gravity to 0.95 (from Dupont's MSDS for 25% sodium methylate). |
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I use Sodium Methylate all the time now and this is how I work out the quantities. If your oil titrates at 2 then you need 2grm/Litre for that plus a base of 5 grm/Litre. So you need 7grm/Litre of NaOH. As your Sodium Methylate is 25% NaOH, you will need 4 times as much, ie 28 grm/litre That's the easy bit! If you are using, say, 20% of methanol to UCO, you will need 200mL of methanol in total and as the SG of methanol is approximately 0.79 that weighs out at 158grm of methanol. But your Sodium Methylate is 75% methanol, so your 28grm of it contains 28 x 0.75= 21grms of methanol. So the methanol you need to add to get your 20% is 158 - 21 which is 137grms. Hope this helps. |
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Producer,
Thanks for the correction on Density. I forgot to divide,  and have corrected my post.I have always assumed STP, in which the Density of Water is 1 g/cc = 1 g/ml, and the density and specific gravity would be the same. I might have to consider the implications of non STP calculations, especially with respect to the density of liquids. It looks like I also made another typo... which I've also corrected. 57.14/2 is 28.57 (ml equivalent to 5 mg NaOH) rather than 27.57... So, our results are within a half a percent which I presume is a rounding error somewhere, or forgetting a Hydrogen. Note, of course, the atomic weights that I used are also rounded to the nearest whole number which makes calculations much easier. Anyway, I'd probably try rounding off to the equivalences of: 25 ml (25%) <--> 5 g 50 ml (25%) <--> 10 g Just to make everything easy, and considering that the Sodium Methoxide made with Sodium Hydroxide contains about 1% excess water as calculated above... I guess the difference between 50 and 57 ml is about 14%... so perhaps that is too much simplification. |
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Zebedee,
I understand, and thanks for that. I see a common answer going on, thanks Producer, Keelec and Double D. I'll do a liter on Saturday morning. Brian 1996 K2500 4x4 6.5TD |
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member 2009 Sponsor |
Thanks Producer, I was running out fingers and toes in my house to count with. I even thought I might have to stop drinking  |
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member 2009 Sponsor |
It sounds like you titrate using NaOH. Do you account for impurities when making your titration solution? Since 25% Sodium methylate in methanol solution is 25% NaOCH3 as opposed to NaOH you may be a little light on the catalyst. Since you're using a purer catalyst though I'm curiuos. Have you ever performed a 3/27 test on your fuel after using the quantities you posted? -Dave- |
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Dave,
You're right, I titrate using 1grm/Litre NaOH. I have 2 titration solutions, one is made up using de-ionised water from the car shop and the other is made with condensate from a dehumidifier. I was worried about impurities in the dehumidifier water but not anymore as I get identical titration results whichever solution I use. Two titrations is the minimum that I do, I've not found it necessary to do a third one. I also take a separate sample of oil for each of the titrations. Only on the odd occasion do I seem to be short of catalyst, so it is near enough, considering all the other variables. One other thing though, I always use the 2-base method with a 80/20 split as my oil is usually over used, If it is particularly bad I give it a glycerol wash and then titrate again before I do the sums. Processing the first stage takes about an hour before I see the glycerol starting to fall out of a small sample. At that stage I let the batch stand for 1/2 - 3/4 hour then drain off the glycerol. Then I add the remaining 20% sodium methylate and process until I get a 27/3 test pass. Should the brew 'stick' for any reason I add another 20% sodium methylate and process further. A7% pre-wash follows a 27/3 test pass with some of the glycerol from the first stage added to reduce the risk of an emulsion forming. Mix for 1/2 Hr, settle and drain the glycerol. Pump into a settling tank and bubble overnight. Allow to settle for a few hours then pass through eco2pure and a 1micron filter. Job done. |
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member 2009 Sponsor |
Zebedee,
Thanks for the reply. Sounds like you found a system that works for you. |
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Member |
Hi, I tried this catylist this weekend. I did it using a digital scale in grams, just as Zebebee explained. The process was excellent, so far. The fuel passed the 27/3 test, total solubility(sp?) at room temp. The biodiesel is still pretty dark and a bit hazy. I havent decided on how to wash/dry it. I hate to put water to it now. I have most of the liter bubbling on 130*F now. The glycerin seems thinner than NaOH Im used to. Could be psycological. I didnt do an NaOH control batch. I'm saving it for a while to see what it does in time. Above the glycerin is only biodiesel! There are no little islands of soap/whatever. Just 2 definitive layers. So far, I like it. Brian 1996 K2500 4x4 6.5TD |
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member 2009 Sponsor |
Probably not psychological. Probably real. I have processed many batches with NaOH and many batches with NaOCH3. The glycerol is always thinner when using NaOCH3, assuming other parameters are basically equal. The NaOCH3 is a "purer" catalyst and produces less soap than NaOH. Soap is the "bad actor" in glycerol: less soap, thinner glycerol. |
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Acknowledged. Makes perfect sense. Here is a picture of what you all helped me thru.  My system is all set to start its maiden voyage. Still trying things till I get a break in the weather. I have some KOH to try yet. Thank you all. Brian 1996 K2500 4x4 6.5TD |
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Update;
Hi, I passed the 27/3 to the best of my knowledge @ 28g NaOCH3 weighed on a digital scale. I sent it thru the QTA machine at work and it said it was under-reacted. 'Forgot the numbers but it was close. I may have misinterpreted the formula as 28g NaOCH3 = 7g NaOH. Must have meant (=5g NaOH) This is a single stage conversion in a soda bottle, which Ive done many of since last summer with 100% NaOH. Keelec's post saying; "It looks like I also made another typo... which I've also corrected. 57.14/2 is 28.57 (ml equivalent to 5 mg NaOH) rather than 27.57... Mad Anyway, I'd probably try rounding off to the equivalences of: 25 ml (25%) <--> 5 g 50 ml (25%) <--> 10 g", is hitting me. I feel that I'm shy a bit on NaOCH3, thus heavy on Methanol-<-no problem, there. 40'ish liquid grams NaOCH3 seems more like the "recipe" for my oil. No? Keelec also says; "So, if you wish to have 250 ml solution. You would add 210 ml of Methanol + 40 ml of your 25% solution." Perhaps I will try this mix. 'Seems a mason jar of this stuff goes a long way. As I've said, I'm a mechanical thinker, not a chemist nor mathematician. Thanks to you guys and the internet (and the great guys at my new job) I can learn, even at 40 years old. A lot of what has been said in thread is Greek to me, but so was basic titration at first. Thank you. Brian 1996 K2500 4x4 6.5TD |
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