What is the percentage of ammonia in the sample?

[tag16]

Homework Statement

Ammoniacal nitrogen can be determined by treatment of the sample with chloroplatinic acid; the product is slightly soluble ammonium chloroplatinate: H2PtCl6 + 2NH4 --> (NH4)2PtCl6 + 2H

The precipitate decomposes on ignition, yielding metallic platinum and gaseous products:
(NH4)2PtCl6 --> Pt(s) + 2Cl2(g) + 2NH3(g) + 2HCl(g)

Calculate the percentage of ammonia in a sample if 0.2115g gave rise to 0.4693g of platinum.

Homework Equations

The Attempt at a Solution

0.4693(1/(14+1+1+1))= 0.0276
0.0276(2)(159)= 8.778
8.778/0.2115= 41.5%

did I do this right? The back of my chemistry book says that the answer should be 38.74% NH3, so if this is correct how would you arrive at this answer?

 

[Lok]

Did you use NH4 or NH3 as ammonia is NH3. Might be a mistake there as for Molecular mass 17 the result is the one in the book.

 

[Borek]

0.4693(1/(14+1+1+1))= 0.0276
0.0276(2)(159)= 8.778
8.778/0.2115= 41.5%

Please elaborate, I have no idea what you did. You started dividing mass of platinum by the molar mass of ammonia, next you multiplied the result by 2 (stoichiometric coeffcient) and by molar mass of platinum (with digits reversed to make it harder to guess what you did). From what I understand at this stage your result is in rather unexpected units:

\frac {(g_{Pt})^2 (mol_{NH_3})^2} {g_{NH_3} (mol_{Pt})^2}

I doubt that's what you wanted to do, from the last operation seems like you think at this stage you have mass of ammonia multiplied by 100.

Book answer is correct.

 

[Lok]

Did you use NH4 or NH3 as ammonia is NH3. Might be a mistake there as for Molecular mass 17 the result is the one in the book.

Btw. With 18 I get the same wrong result as you. The things you wrote here make sense but weird that you don't get the result.