Question

Aerobic biomass has a yield coefficient value of 0.4 for glucose (molecular weight = 180 g/mole) substrate. The bacteria is represented as \( {C}_5{H}_7{O}_2{N} \) (molecular weight = 113 g/mole). Assume that no endogenous metabolism occurs. The percentage of carbon going into CO2 from 1 mole/L glucose is ________ % (rounded off to two decimal places).

Hint:

In stoichiometric calculations for microbial growth, always account for the yield coefficient and perform a mass balance to find the distribution of carbon between biomass and CO2.

Answer 1

Step 1: Write the balanced equation for microbial growth.
The general stoichiometric equation for microbial growth is: \[ \text{Glucose} + \text{Oxygen} \rightarrow \text{Biomass} + \text{CO}_2 \] The yield coefficient \( Y_X/S \) represents the amount of biomass produced per unit of substrate consumed. In this case, the yield coefficient for glucose is given as \( 0.4 \) g of biomass per g of glucose consumed.
Step 2: Calculate the amount of carbon in the biomass.
The bacterial formula is \( \text{C}_5\text{H}_7\text{O}_2\text{N} \). The molecular weight of biomass \( \text{C}_5\text{H}_7\text{O}_2\text{N} \) is 113 g/mole. The amount of carbon in one mole of bacteria is 5 moles of carbon (since \( \text{C}_5 \)).
Step 3: Calculate the total carbon in glucose.
The molecular weight of glucose is 180 g/mole. The amount of carbon in one mole of glucose is 6 moles of carbon (since \( \text{C}_6 \) in glucose).
Step 4: Set up the mass balance.
We assume that 1 mole of glucose (180 g) is consumed. The biomass produced will have a yield of \( 0.4 \) g of biomass per 1 g of glucose. Therefore, the amount of biomass produced is: \[ 0.4 \times 180 = 72 \, \text{g biomass} \]
The biomass consists of 5 moles of carbon per mole of biomass. So, the amount of carbon in the biomass is: \[ \frac{72}{113} \times 5 \times 12 \, \text{g of carbon} \]
Step 5: Calculate the carbon going into CO2.
The remaining carbon will go into CO2. The carbon balance is as follows: \[ \text{Carbon from glucose} = \text{Carbon in biomass} + \text{Carbon in CO}_2 \]
The total carbon in glucose is: \[ \frac{1}{180} \times 6 \times 12 \, \text{g of carbon} \]
Now calculate the percentage of carbon going into CO2.
Step 6: Conclusion.
The percentage of carbon going into CO2 from 1 mole/L glucose is approximately: \[ \boxed{47.30\%} \]