A microorganism is capable of using glucose, methanol, and hexadecane, singly. I

Question

A microorganism is capable of using glucose, methanol, and hexadecane, singly. Its average cell composition (by weight) was analyzed to be 47% carbon, 6.5% hydrogen, 31% oxygen, 10% nitrogen, and the rest is ash. During active metabolism, the microorganism converts the substrate, oxygen, and ammonia into biomass, carbon dioxide, and water. A batch culture was carried out to estimate the cell yield based on substrate and oxygen (kg cells/kg substrate or oxygen). Air was supplied continuously and the exhaust gas was vented from the top of the bioreactor. A mass spectrometer was available for gas composition analysis. The water vapor was removed from both inlet and outlet gases before analysis. The inlet gas was ambient air (21% oxygen and 79% nitrogen). The composition of the off-gas on a volumetric basis is listed below. What are the estimated cell yield coefficients based on ammonia and oxygen for the consumption of glucose?

Substrate: glucose, %Nitrogen: 78.8, %Carbon Dioxide: 10.2, %Oxygen: 11.0

Explanation / Answer

The growth of biomass with time is given by ; Substrate +cell gives product + more cell

Total biomass yeild = mass of biomass produced per unit / moles of subsrate consumed .

carbon =12 .hydrogen is 1.66,nitrogen 0.18,oxygen is 16 molecular weight . so the total molecular weight of the compound is 24.18 g, glucose molecular weight is 180g, number of moles is one mole

Y glucose = 24.18 *2.32 / 180 *1 mol = 0.31g of cell yeild upon consumption of oxygen

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