Question

Your boss has instructed you to make up a medium for growing Gram-negative procaryotes. She wants you to be able to grow 100 mg/l of volatile suspended solids. You looked up some papers on the microorganisms and found that the empirical chemical formula for them is C5H7O2NP0.1 are minimum concentrations of NH4+-N and PO4-P that you must add to the medium?

Answer 1

The empirical chemical formula = C₅H₇O₂NP_0.1

Molar mass of C₅H₇O₂NP = 148 g/mol

Molar mass of nitrogen = 14 g/mol

So, percent of nitrogen in that compound = (molar mass of nitrogen/molar mass of compound) x 100 = (14 g/mol/148 g/mol) x 100 = 9.46 %

Molar mass of phosphorous = 30 g/mol

So, percent of phosphorous in that compound = 0.1 x (molar mass of phosphorous/molar mass of compound) x 100 = (30/148) x 100 = 2.027 %

source of nitrogen is ammonium ion and source of phosphorous - phosphate

ammonium ion cane obtained from ammonium hydroxide and phosphate from potassium hydrogen phosphate

molar mass of NH₄OH = 35 g/mol

percent of nitrogen in ammonium hydroxide = 14 g/mol/35 g/mol= 46.67 %

molar mass of K₂HPO₄ = 174 g/mol

percent of phosphorous in K₂HPO₄ = 30 g/mol/174 g/mol) x100 = 17.24 %

100 mg/l of volatile suspended solids i.e 0.01 % of C₅H₇O₂NP is present i.e 1.48 g of C₅H₇O₂NP is required

Hence 9.46 % of nitrogen in 1.48 g == 0.133g of nitrogen

2.027 % of phosphorous in 1.48 g = 0.0299 g of phosphorous

So, concentration of nitrogen = mass/molar mass/vol in lt = 0.133 g/14 g/mol/1lt = 0.0095 M

Concentration of phosphorus = mass/molar mass/vol in lt = 0.0299 g/30 g/mol/1 lt = 0.00099 M