I'm having a brain freeze here.
Soy ffa molecular weight is about 279. (What's canola?)
Methanol is 32.
Therefore 20:1 is (32*20)/279 or 640g of methanol for every 279g of ffa.
640g methanol is the right answer for a batch containing 279 g FFA if you want 20 moles of methanol per mole of FFA.
Some FFA molecular weights: coconut, 213; tallow, 273; cottonseed, 274; soy, 279; sunflower, 280; canola, 281.
i read before, to produce BD, we need methanol about 20-25% of oil.
my question. if molar ratio is 6:1
i use spent bleaching earth residual oil (from palm oil) with average molar mass = 270.
so i need:
methanol ; 6 x 32.04 = 192.24g
oil ; 1 x 270 = 270g
to produse my BD. right? but it is not 25% amount of oil.
any explaination? 10Q
I think you should specify that you're using the mean molecular weight (or some other approximation) for each feedstock- as no oil or fat has a noncomplex fatty acid profile.
The molecular weight of palm oil is going to be closer to 900.
In that case your 6:1 molar ration would equate to roughly 21% methanol (192.24/900=0.2136). To clarify, you are talking about transesterification, where the methanol ratio is a factor of triglyceride weight. This post was referring to acid esterification, where methanol ration is a factor of free fatty acid weight.
Hope that helps.
Newport Biodiesel, LLC
I am trying to produce biodiesel from tung oil (mol. wt. 888g/mol) using a mixture of methanol and ethanol. The methanol/ethanol ratio to oil is 20:1. Would like to know if the formulation below is correct/in-line with 20:1 molar ratio:
20(32+46)/888 = 156/88.8
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