# Urgent help on yeilds in fermentation (1 Viewer)

#### beardedwoman

##### Member
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...

#### mitochondria

##### *Rawr*!
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! Last edited:

#### beardedwoman

##### Member
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.

#### mitochondria

##### *Rawr*!
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.
Apologies! That was silly of me... It's been edited in now. #### beardedwoman

##### Member
Ah, okay. Thank you!