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How much water?
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Kind of a chemistry question here: How much water can I expect from esterifying oil that is 5% ffa? (5% is considered low quality, right?)

My guess is one mol of water is produced for every mol of methanol consumed. Let's say you had 1000ml of wvo that was 5% or 50ml FFA. I've noticed that generally you can expect to use ~ 25% methanol to drive transesterification, but of that 25% only about 3/5ths is used.

Therefore is it safe to assume i would consume about 7.5ml or 6g or .2 mol of MeOH to esterify, and get about .2 mol or 3.6 g of H20?

Is that a reasonable guess?
 
Registered: March 16, 2008Reply With QuoteReport This Post
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5% FFA would be 5% by weight.
Assuming a liter of your oil weighs 920g.
920x.05= 46g/280= .164mol (the 280 represents a possible fatty acid molecular weight).
Assuming you esterified all the FFA (not really possible) you could create .164mol H2O per liter of oil.
.164/18g = 2.95g H2O. (18g represents the MW of H2O).

CH3OH+R-COOH -> R-COOCH3 + H2O
One mole of FFA combines with one mole methanol and produces one mole of water.

.164 x 32.05 = 5.26g/.792= 6.64ml methanol consumed.


It looks like your math is correct.
I think our difference comes from yours being rounded up.
I drag my math out to three significant digits then round up at the end. It makes a difference on small quantities.
5% FFA is not really that high.

That equates to a NaOH titration of ~6.5ml
 
Location: central virginia | Registered: March 13, 2008Reply With QuoteReport This Post
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Thanks! That was a more cogent and succinct answer than I could have hoped for. I have a few followups if you (or anyone) could find the time:

First: If 5% FFA (by weight) is not a poor quality WVO then what is? 15%?

Secondly: I'm also assuming 1% water (by weight) is a good estimate for content of emulsified water in WVO, but maybe 2% or even 5% is better (for poor quality WVO).

For both of these questions I'm just wondering what the best / mean / worst case scenario is, based on the forum's experience.

Finally, how do you derive the titration amount of NaOH to be 6.5g? Is it based on the amount of 'H' in the catalyst available to 'loan' a proton? Would it take twice was many mol of HCL to catalyze the essterification of FA as of H2SO4, and three times as many as mol as anhydrous H3PO4?

In other words, if we assume that 1 L of 5% FFA oil contains .164 mol of FFA, then could we derive:

  • that we need .164 mol of NaOH to saponify the FFA (.164 mol * 40 (g/mol) = 6.5g); or
  • .164 mol of HCL to esterify it (.164 mol * 36 (g/mol) = 5.9g); or
  • .164/2 mol of H2SO4 (.164/2 mol * 98 (g/mol) = 8.0g), or
  • .164/3 mol of (anhydrous) H3PO4 (.164/3 mol * 98 (g/mol) = 5.3g)

?
I think I might have just answered my last question.

Thanks again, and thanks in advance! Wink
 
Registered: March 16, 2008Reply With QuoteReport This Post
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Furious,

quote:
that we need .164 mol of NaOH to saponify (.164 mol * 40 (g/mol) = 6.5g);


This is correct.

quote:

.164 mol of HCL to esterify (.164 mol * 36 (g/mol) = 5.9g); or

.164/2 mol of H2SO4 (.164/2 mol * 98 (g/mol) = 8.0g), or

.164/3 mol of (anhydrous) H3PO4 (.164/3 mol * 98 (g/mol) = 5.3g)



Almost.
The calculations you have based on the Hydrogen equavelence is accuracte for acid/base neutralization reactions but not necessarily for use as a catalyst.

Most of us use way less than 8g H2SO4 to catalyse an esterification reaction of 5% FFA.

The problem with HCL is it doesn't come in a pure enough form. I think the purist % you can get is about 40% with the other 60% being water.

H3PO4 is considered a weak acid in that it doesn't completely ionize, it only donates one (or slightly more) hydrogen. Combined with it not coming in over 85% purity creates a problem.

We use Sulfuric acid mainly because it is inexpensive, easily aquired, and very effective.
 
Location: central virginia | Registered: March 13, 2008Reply With QuoteReport This Post
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quote:
Originally posted by Double D:

quote:

.164 mol of HCL to esterify (.164 mol * 36 (g/mol) = 5.9g); or

.164/2 mol of H2SO4 (.164/2 mol * 98 (g/mol) = 8.0g), or

.164/3 mol of (anhydrous) H3PO4 (.164/3 mol * 98 (g/mol) = 5.3g)



Almost.
The calculations you have based on the Hydrogen equavelence is accuracte for acid/base neutralization reactions but not necessarily for use as a catalyst.

...

H3PO4 is considered a weak acid in that it doesn't completely ionize, it only donates one (or slightly more) hydrogen. Combined with it not coming in over 85% purity creates a problem.


Let's assume that the H3PO4 is in solution with MeOH, not water. Can I assume that on average 1.2 (or something) 'H' is ionized per molecule and use the equations above? What property of an acid tells you how much 'H' is ionized?

Thanks again
 
Registered: March 16, 2008Reply With QuoteReport This Post
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quote:
Originally posted by FuriousGeorge:


Let's assume that the H3PO4 is in solution with MeOH, not water. Can I assume that on average 1.2 (or something) 'H' is ionized per molecule and use the equations above? What property of an acid tells you how much 'H' is ionized?

Thanks again


I don't know what can be assumed as far as how much H is dissocated.
You could create a solution using sulfuric and a solution of Phosphoric and compare the PH of the two, increase the molar concentration of the phoshporic until it reaches the PH of the sulfuric solution.

The Dissocation ~Constant or Kpa is what tells you roughly how much (H) an acid will dissocate.
At least that's how I understand it. Smile
 
Location: central virginia | Registered: March 13, 2008Reply With QuoteReport This Post



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Thanks for the reply. I'll see if I can figure it out. At this point I'm just curious as to how this whole chemistry thing works Wink
 
Registered: March 16, 2008Reply With QuoteReport This Post
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Your welcome, that's what we do.
Me too. Wink
 
Location: central virginia | Registered: March 13, 2008Reply With QuoteReport This Post
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