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Urgent help on yeilds in fermentation (1 Viewer)

beardedwoman

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We were to conduct a fermentation experiment on glucose to produce carbon dioxide & ethanol.

They want us to calculate the theoretical mass of carbon dioxide produced based on a 14% yield. Does anyone know the formula for this, and can you provide an example if you're bothered or something...

thanks in advance.
 

mitochondria

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Right, first of all, you need to get your equation right and balanced. Given that you get CO2 and C2H5OH from C6H12O6, you write:

C6H12O6 ----> C2H5OH + CO2

Balancing this gives:

C6H12O6 ----> 2C2H5OH + 2CO2


Keep this in mind - almost all yield calculations involve converting things into moles first ;):

MWglucose = 180.16 g.mol-1


So, for a given mass of glucose mglucose, the amount in moles, mglucose, is:

nglucose = (mglucose/180.16-1) mol


The theoretical amount of CO2 you will get out of this is:

nCO2 = 2 x nglucose (cf balanced equation)


And the corresponding amount of CO2 produced, in grams is simply:

mCO2 = nCO2 x 44.01 g.mol-1


Since you only get 14% yield, it means that the final amount of CO2 is 14% of what you would expect, so you simply multiply your final yield by 0.14:

mCO2 x 0.14 = mCO2 Expected


This can equally well be done before you calculated the mass, that is:

mCO2 Expected = (0.14 x nCO2) x 44.01 g.mol-1



Hope that helps! :)
 
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beardedwoman

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Yeah, we did this to find the theoretical mass of carbon dioxide produced, but my friend and I don't understand where the 14% yield is, or how to get it.
 

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