Answer:
0.37 g
Explanation:
The molecular weight for Glycerol = 92
Number of Carbon atoms in glycerol (x) [tex]C_3H_8O_3[/tex] = 3
Molecular weight of the Biomass ( Klebsiella aerogenes )
= [tex]CH_{1.73}O_{0.43}N_{0.24}[/tex]
= [tex]\frac{23.97}{0.92}[/tex]
= 26.1
From the molecular weight of the Biomass, we can deduce the Degree of reduction for the substrate(glycerol denoted as [tex]\delta _g[/tex]) as follows:
= (4×1)+(1×1.73)-(2×0.43)-(3×0.24)
= 4.15
Given that the yield of the Biomass = 0.40 g
However;
C = [tex]Yield of Biomass *\frac{Molecular weight of substrate}{Molecular weight of the Biomass}[/tex]
C = [tex]0.40*\frac{92}{26.1}[/tex]
C = 1.41 g
Now , the oxygen requirement can be calculated as:
= [tex]\frac{1}{4}*(n*S - C * \delta _{g})[/tex]
= [tex]\frac{1}{4}(3*4.7-1.41*4.15)[/tex]
= 2.1 g/mol
Hence, we can say that the needed oxygen = 2.1 g/mol of the substrate consumed.
Now converting it to mass terms; we have:
= [tex]2.1*\frac{number of mole of oxygen}{molecular weight of glycerol}[/tex]
= [tex]2.1 * \frac{16}{92}[/tex]
= 0.3652 g
≅ 0.37 g
∴ The oxygen requirement for this culture in mass terms = 0.37 g
