Solve Back Titration Problem to Find Ammonium Sulfate Mass in Fertiliser

[Identity]

Please Help! Back titration!

Find the mass of ammonium sulfate in lawn fertiliser.

The experiment in a nutshell:

A sample of 1.301g lawn fertiliser is dissolved in de-ionised water and diluted to 250mL. A 20mL aliquot is taken and an excess of 0.1M $$NaOH$$ is added. The remaining $$NaOH$$ is titrated with 0.09933M $$HCl$$.

The average titre was 16.335mL.


I get 0.3098g as my mass for ammonium sulfate. But for it to be concordant with my gravimetric analysis, that value should be between 0.600g-0.700g!
Can someone please tell me what's gone wrong or what to do to ge the right answer? Thanks!

My working:

Original n(NaOH) = $$0.1 \times 0.02$$ = 0.002 mol

$$HCl + NaOH -> NaCl + H_2O$$

$$n(HCl) = 0.09933 \times 0.016355$$ = 0.00162454 mol

$$n(NaOH)_{unreacted with ammonium sulfate}$$ = 0.00162454 mol

$$n(NaOH)_{reacted with ammonium sulfate}$$ = 0.002 - 0.00162454 = 0.000375458mol

The ionic reaction between ammonium sulfate and NaOH is as follows:

$$NH_4^{+} + OH^{-} -> NH_3 + H_2O$$

$$n(NH_4^{+}) = n(OH^{-}) = 0.000375458 mol$$

$$[NH_4^{+}]_{20.00mL aliquot} = \frac{0.000375458}{0.02} = 0.01877 M$$

$$[NH_4^{+}]_{20.00mL aliquot} = [NH_4^{+}]_{250.00mL volumetric flask} = 0.01877 M$$

$$n(NH_4^{+})_{250mL volumetric flask} = 0.01877 \times 0.25 = 0.00469 mol$$

$$n(NH_4^{+}) = 2 \times n((NH_4)_2SO_4)$$

so $$n((NH_4)_2SO_4) = \frac{0.00469}{2} = 0.002345$$ mol

$$m((NH_4)_2SO_4) = 0.002345 \times 132.1 = 0.3098g$$

 

[chemisttree]

How much 0.1 M NaOH did you add?

 

[symbolipoint]

I have the same question as post #2. You apparently also confused the volume of aloquat with the unstated volume of NaOH.