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.

Right?

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.

Thanks neutral

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

Neutral, Dropout-

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.

gjoe-

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.

-Nick