Urgent help on yeilds in fermentation (1 Viewer)

[beardedwoman]

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]

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 :

MW_glucose = 180.16 g.mol^-1


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

n_glucose = (m_glucose/180.16^-1) mol


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

n_CO2 = 2 x n_glucose (cf balanced equation)


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

m_CO2 = n_CO2 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:

m_CO2 x 0.14 = m_{CO2 Expected}


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

m_{CO2 Expected} = (0.14 x n_CO2) x 44.01 g.mol^-1



Hope that helps!

 

[beardedwoman]

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]

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]

Ah, okay. Thank you!