Question

A 0.608 g sample of fertilizer contained nitrogen as ammonium sulfate. It was analyzed for nitrogen by heating with sodium hydroxide. The ammonium was collected in 46.3 mL of 0.213 M HCL with which it reacted. The solution was titrated with excess HCL requiring 44.3 Ml of 0.128 M NaOH. What is the percentage of nitrogen in the fertilizer?

Answer 1

no of mol of HCl taken = M*V

                       = 0.213*46.3

                       = 9.862 mmol

no of mol of HCl unreacted = no of mol of NaoH consumed = 0.128*44.3

                                                        = 5.67 mmol

no of mol of HCl reacted = 9.862-5.67 = 4.2 mmol

1 mol HCl = 1 mol NH3

no of mol of NH3 formed = 4.2 mmol

1 mol NH3 = 1 mol N

no of mol of N in sample = 4.2 mmol

amount of N in sample = 4.2*10^-3*14 = 0.0588 g

% of N in sample = 0.0588/0.608*100

                 = 9.67%