Solve the following problem. Use conversion factors provided in your textbook, a

Question

Solve the following problem. Use conversion factors provided in your textbook, as needed.

The following chemical formula is often used for representation of a bacterial cell, C60H87O23N12P1

(a) Determine the mass (mg) of each element in 3000 mg of bacterial cells.

Mass of C is  mg
Mass of H is  mg
Mass of N is  mg
Mass of N is  mg
Mass of P is  mg


(b) Suppose 25 mg/L of ammonia nitrogen and 5 mg/L of orthophosphate as phosphorus are available for growing bacteria. If other nutrients are in abundance, which is the limiting nutrientnitrogen or phosphorus?
---Select--- Nitrogen Phosphorus

(c) What mass of bacterial cells could be produced in terms of milligrams of bacterial cells per liter of water based on the limiting nutrient in Part b?
mg/L

(d) Suppose the nitrogen source was cut to 15 mg/L of ammonia nitrogen. How much bacterial cell mass (mg/L) could be produced?
mg/L

(e) Suppose the phosphorus source was cut to 4 mg/L of orthophosphate as phosphorus. How much bacterial cell mass (mg/L) could be produced?
mg/L

Explanation / Answer

masC60H87O23N12P1

MW C = 12, MW H = 1, MW of O = 16, MW of N = 14, MW of P = 31

MW = (C*60) + (87*H) + (23*O) + (12*N) + (1*P)

MW = (12*60) + (87*1) + (23*16) + (12*14) + (1*31)

MW = 1374 g/mol

moels in 3000 mg = 3 g

mol = mass/M W= 3/1374 = 0.002183 mol

Mass of "i" = MW of i * No of i * mol of Species * 1000 mg/g

Mass of C is  mg --> 12*60*0.002183 *1000 = 1571.76
Mass of H is  mg   --> 1*87*0.002183 *1000 = 189.9
Mass of O is  mg   --> 16*23*0.002183 *1000 = 803.344
Mass of N is  mg   --> 14*12*0.002183 *1000 = 366.744
Mass of P is  mg   --> 31*1*0.002183 *1000 = 67.673

(b) Suppose 25 mg/L of ammonia nitrogen and 5 mg/L of orthophosphate as phosphorus are available for growing bacteria. If other nutrients are in abundance, which is the limiting nutrientnitrogen or phosphorus?
---Select--- Nitrogen Phosphorus

from

Mass of N is  mg   --> 14*12*0.002183 *1000 = 366.744
Mass of P is  mg   --> 31*1*0.002183 *1000 = 67.673

N/P = 366.744/67.673 = 5.42

mol of N = 25/17 = 1.47

mol of P = 5/95 = 0.0526

1.47/0.0526 = 27.9, meaning that there are more N than P

P is limiting...

(c) What mass of bacterial cells could be produced in terms of milligrams of bacterial cells per liter of water based on the limiting nutrient in Part b?

0.0526 mol of P --> 0.0526 mol of C60H87O23N12P1

mass = mol*MW = 0.0526 *1374 = 72.27 mg

(d) Suppose the nitrogen source was cut to 15 mg/L of ammonia nitrogen. How much bacterial cell mass (mg/L) could be produced?

mol of N = 15/17 = 0.8823

mol of P = 5/95 = 0.0526

0.8823/0.0526 = 16.77

still we have more N

mass = mol*MW = 0.0526 *1374 = 72.27 mg

(e) Suppose the phosphorus source was cut to 4 mg/L of orthophosphate as phosphorus. How much bacterial cell mass (mg/L) could be produced?

mol of N = 4/17 = 0.2352

mol of P = 5/95 = 0.0526

0.2352/0.0526 = 4.47

now..

N/P = 366.744/67.673 = 5.42

P is more, therefore, N limits

12 mol of N = 1 mol of bacteria

0.2352 mol = 0.0196 mol of

mass = mol*MW = 0.0196*1374 = 26.93 mg

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