Question

Estimate the mass of carbon that is found as cell mass or transformed into CO2 for a bacterial culture that grows using either one mole of octadecane (C18H36, Y = 1.49) or mole of pentachlorophenol (C6HOCl5, Y = 0.05).

Answer 1

OCTADECANE

1 mole of octadecane C18H36 = 254.5g/mole

Y (cell yield) = cell mass produced / substrate consumed

therefore, cell mass produced = substrate mass x cell yield

= 254.5 x 1.49 = 379.205g

cell mass is estimated as C5H7NO2, so molecular weight = 113g/mol

therefore mol. cell mass = 379.205/ 113 = 3.3557mol cell mass

Now in terms of carbon,

cell mass= 3.335 (mol cell mass) x 5mol (C/mol. cell mass) x 12g (mol C)

Substrate= 1 mol (substrate) x 18 mol (C/ mol sub) x 12 (mol C)

therefore % of substrate found in cell mass = (3.35 x 5x 12/ 1x 18 x 12 ) x 100

= 93%

therefore the remaining 7% is the carbon released CO2

PENTACHLOROPHENOL

1 mole of pentachlorophenol (C6HOC15 ) = 266.35

Cell mass produced = 266.35 x o.o5 = 13.317g

Molecular cell mass = 13.317/ 113 = 0.117 mol cell mass

In term of carbon,

% of substrate carbon found in cell mass = (0.117 x 5 x 12/ 1 x 6 x 12) x 100 = 9.75% = 10%

therefore the remaining 90% is the carbon releases as CO2.