Question

It is desired to cultivate 150 mg (dry weight) microbial cells with an empirical formula of \(C_5H_7O_2N\). If the phosphorus requirement for the cells is 20% of the nitrogen requirement (on a weight/weight basis), the minimum amount of phosphorous required to be added to the cultivation medium is mg. (rounded off to one decimal place)

Hint:

Find the N mass fraction in \(C_5H_7O_2N\) (molar mass 113), apply it to 150 mg of cells to get the N requirement, then take 20% of that for phosphorus.

Answer 1

Correct Answer: 3.7

Step 1: Understanding the Question.
The empirical formula \(C_5H_7O_2N\) is the standard representative formula used for microbial biomass. It tells us how much nitrogen is locked up in a given mass of dry cells. Once we know the nitrogen mass, the phosphorus requirement follows directly from the 20% ratio given in the question.

Step 2: Find the molar mass of the biomass formula.
\[ M = 5(12) + 7(1) + 2(16) + 1(14) = 60 + 7 + 32 + 14 = 113 \ \text{g mol}^{-1} \]

Step 3: Find the mass fraction of nitrogen in the biomass.
Nitrogen contributes 14 out of the 113 mass units in the formula, so
\[ \text{fraction N} = \frac{14}{113} = 0.1239 \]

Step 4: Find the nitrogen needed for 150 mg of cells.
\[ \text{N required} = 150 \times 0.1239 = 18.58 \ \text{mg} \]

Step 5: Apply the 20% phosphorus-to-nitrogen ratio.
\[ \text{P required} = 0.20 \times 18.58 = 3.72 \ \text{mg} \]

Final Answer:
Rounded to one decimal place, the phosphorus that must be supplied is
\[ \boxed{3.7 \ \text{mg}} \]