A 0.608 g sample of fertilizer contained nitrogen as ammonium sulfate, (NH4)2SO4

Question

A 0.608 g sample of fertilizer contained nitrogen as ammonium sulfate, (NH4)2SO4.  It was analyzed for nitrogen by heating with sodium hydroxide.  Please show work thanks!

(NH4)2SO4 (s)  +  2NaOH(aq)  ------------>  Na2SO4 (aq) + 2H2O (l)  +  2NH3 (g)

The ammonia was collected in 46.3 mL of 0.213 M HCl with which it reacted.

NH3 (g)  +  HCl (aq)  ----------------->  NH4Cl (aq)

The solution was titrated with for excess HCl requiring 44.3 mL of 0.128 M NaOH.

NaOH (aq)  +  HCl (aq) ------------------->  NaCl (aq)  +  H2O (l)

What is the percentage of nitrogen in the fertilizer?

Explanation / Answer

Start with the back titration of NaOH with the unreacted HCl.
Using 44.3 mL of 0.128 M NaOH, find the number of moles of NaOH (= V in litres x M). I will call this x moles NaOH.
From the last equation, x moles NaOH react with x moles HCl.
So now you know how much HCl was left over after treatment with NH3.
This is x moles.
Then find the number of moles of HCl added at the start (46.3 mL of 0.213 M HCl).
Subtract 'x' moles from this and this will tell you how many moles of HCl (I will call this 'y' moles) reacted with NH3.
From the equation NH3(g)+HCl(aq)->NH4Cl(aq), y moles HCl reacted with y moles NH3.
Find out what y moles of NH3 weigh (1 mole weighs about 17g). I will call this z grams of NH3.
You then need to find out the mass of nitrogen (w grams) in z grams of NH3 (this is about 14 g in 17 g NH3 so how many grams N in z gram NH3)
You now know 0.608 g of fertiliser contained w grams of Nitrogen
You can find the %.

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