Back Titration Prac SAC Question

[Christiano]

Okay so I have 1.234g of lawn fertilizer containing NH4+. It's mixed and boiled with excess NaOH. The resulting excess NaOH was titrated with known concentration of HCl. I get to this part of the question:

d) Amount of NH4+ ions in the 1.234g fertilizer
step 1
n(NH4+) = 1 mol NH4+/1 mol NaOH x 0.00091 mol NaOH = 0.00091 mol NH4+

step 2 Calculate the amount of fertiliser, NH4+, in the 250.0 mL flask, remembering that only 20.00 mL was removed from the 250.0 mL flask for the titration. This is the same amount of NH4+ as is present in 1.234 g of fertiliser.

0.00091 mil x 250/20 (dilution factor) = 0.0114 mol NH4+

then e) The percentage by mass of nitrogen in the fertilizer, assuming all the nitrogen is present as ammonium ions.

n(N)   = n(NH4+)
m(N)   = 0.01144 mol x 14.01g/mol
        = 0.1603 g
then:
% N   =  0.1603g/1.234g x 100%
   = 13.0% (three significant figures)

My question is, if e) was modified to find the percentage by mass of the ammonia, would it be correct in calculating the mass of ammonia by doing 0.01144 mol x 18g/mol then dividing the resulting mass by 1.234g to get the percentage of ammonium ions?

[nacho]

yes it would, as 18g/mol is the molar mass of ammonia

[Mao]

My question is, if e) was modified to find the percentage by mass of the ammonia, would it be correct in calculating the mass of ammonia by doing 0.01144 mol x 18g/mol then dividing the resulting mass by 1.234g to get the percentage of ammonium ions?

If I understand this correctly, then I think there is a trick to this.

we know that n(N) = n(NH4+) = 0.01144 mol

Then,

%N = 0.01144 mol x 14g/mol / 1.234g x 100% (nitrogen)
%NH4+ = 0.01144 mol x 18g/mol / 1.234g x 100% (ammonium)

If the question asks about ammonia, then %NH3 = 0, since the initial sample contained NH4+, not NH3 (assuming that NH4+ is the only source of N)

[Christiano]

That's it