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

Correct Answer: 46.5

Step 1: Write the balanced equation for microbial growth. The general stoichiometric equation for microbial growth is: \[ {Glucose} + {Oxygen} \rightarrow {Biomass} + {CO}_2 \] 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 \( {C}_5{H}_7{O}_2{N} \). The molecular weight of biomass \( {C}_5{H}_7{O}_2{N} \) is 113 g/mole. The amount of carbon in one mole of bacteria is 5 moles of carbon (since \( {C}_5 \)). 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 \( {C}_6 \) 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: \[ 0.4 \times 180 = 72 \, {g biomass} \] The biomass consists of 5 moles of carbon per mole of biomass. So, the amount of carbon in the biomass is: \[ \frac{72}{113} \times 5 \times 12 \, {g of carbon} \] Step 5: Calculate the carbon going into CO2. The remaining carbon will go into CO2. The carbon balance is as follows: \[ {Carbon from glucose} = {Carbon in biomass} + {Carbon in CO}_2 \] The total carbon in glucose is: \[ \frac{1}{180} \times 6 \times 12 \, {g of carbon} \] 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: \[ \boxed{47.30%} \]