Keith,
I was hoping to hear from someone with greater expertise in the area,
but will pass along my thoughts.
I'm not sure that the calculations provided from Biofuels Systems are
accurate for the following reasons.
1. The % FFA calculations were based on the neutralization of Oleic Acid
(molecular mass = 283).
The idea, as I see it, is that calculations can be made to determine
the mass of substances consumed in reactions, if we know the mass of one of
the consumed reactants and know the ratio(s) of the involved reactants.
Ex. If we know the mass of NaOH needed to neutralize a known volume of a
specific acid, and we know that there exists a 1-to-1 ratio for the
reaction, we can calculate the mass of the acid that is neutralized.
-However many moles of NaOH were consumed will equal the moles of acid
neutralized.
-The mass of the acid neutralized = moles neutralized X mol. mass of
the acid
Problem:
Used vegetable oils have a mix of free fatty acids, with different molecular
masses, and hence different masses per mole. To do an accurate calculation
of % FFA in veg oil, I think we would need to know the specific FFAs
present, in what concentrations, and the molecular mass of each. After all,
We wouldn't use a titration solution with an unknown mix of NaOH and KOH to
do a quantitative analysis.
2. To determine %, one divides the total mass by the mass of the
individual component and then multiply by 100
Ex. 1 g of NaOH + enough distilled water to equal 1 L
The total mass of the solution = 1000g
1g NaOH divided by 1000g = .001 X 100 = 0.1%
The Biofuel Systems calculations state that there is 0.7063 g of FFA in
100ml of oil. It then divides mass by volume (????). It assumes that 100ml
of the oil has a mass of 100g. This would be true if it was water instead of
oil. The veg oils I am familiar with float on water indicating a lower
density ....... I think along the order of 0.9g/ml
At 0.9g/ml, the 100ml sample of veg oil would have a mass of 90g
0.7063g divided by 90g X 100 = 0.785%
vs The 0.71% calculated in the example provided by Biofuel systems.
If the oil being tested is very high in oleic acid, and we use the
density of the oil when calculating %, I suspect we could get a "ballpark"
figure for % FFAs in a sample of veg oil. We might just as well use 0.75 -
0.80 as a coefficient (constant) and multiply it times the NaOH titration
result rather than going through the laborious calculations described.
Although I've heard it referred to, I don't know why one would want to
know %FFA in veg oil.
Best to You,
Tom
----- Original Message -----
From: "Keith Addison" <[EMAIL PROTECTED]>
To: <biofuel@sustainablelists.org>
Sent: Monday, March 09, 2009 7:01 AM
Subject: [Biofuel] %FFA of vegetable oil
Hi all
I've seen several conflicting ratios for converting homebrewer
titration results of # ml 0.1% NaOH solution to % FFA.
This below is from Biofuel Systems in the UK
<http://www.biofuelsystems.com/>. Can anyone confirm whether it's
correct or not?
----
How to determine percentage free fatty acid (%FFA) of vegetable oil /
used cooking oil
Add 10ml of oil to 100ml of isopropyl alchohol (propan-2-ol)
Mix thoroughly until oil dissolves. Add a few drops of Universal
Indicator solution (UI)
Measure how much 0.025M sodium hydroxide solution (1g NaOH / 1 litre
water) is required to neutralise the oil solution - ie. raise pH to 8
/ turn UI blue/green
While continually stirring this mixture, add the NaOH solution drop
by drop until the mixture turns and remains green / blue. Note the
number of millilitres (ml) of NaOH solution required to do this.
In stoichiometric terms, 1 mole NaOH will neutralise 1 mole of oleic acid
(OA)
Firstly, determine how many moles of NaOH have been usedS
Moles = molarity (mol/l) x volume (l) = [ molarity x volume (ml) ÷ 1000 ]
Example
If 10ml of NaOH solution was used
Moles of NaOH = (molarity of solution x volume in ml) ÷ 1000
= (0.025 x 10) ÷ 1000
= 0.00025 moles of NaOH
Therefore, the equivalent to 0.00025 moles of OA have been neutralised
mass = moles x molecular weight (molecular weight of OA is 282.52 Daltons)
mass of OA in 10ml sample = 0.00025 x 282.52
= 0.07063
= 0.7063g per 100ml of oil
....rounding to 2 decimal places, %FFA = 0.71
Put simply...
%FFA = number ml 0.025M NaOH solution used x 0.07063
-----
Since most homebrewers use 1 ml of oil in 10 ml of isopropanol, that would
be:
%FFA = number ml 0.025M NaOH solution used x 0.7063
I guess blue/green Universal Indicator solution is the same as
magenta phenolphthalein, no?
Thanks!
Best
Keith
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