So I am trying to find the amount of copper in a 10g brass sample. I first dissolved the brass in 6M nitric acid and then put it in a 250ml volumetric flask to make a copper solution. A 60ml aliquot was then taken of the solution and it was mixed with excessof 6M NaOH. This was then heated to form CuO + H20. It was filtered and left to dry. It was found that there was 3.05g of CuO.To find the amount of Copper in the original brass sample(of 10g) I did the following:Find percentage of Cu in CuO. Ar(Cu)/Mr(CuO) = 63.5/79.7 = 79.8%Mass of Cu in CuO precipitate: .798x3.05=2.43gMass of Cu in 250ml Solution: 2.43x(250/60)= 10.125gIs this the correct calculation procedure?There is obviously some error made in doing this experiment. What are some factors that could have led to this impossible percentage of copper in the brass sample when doing the experiment? (e.g spectator ions in the preciptate)
dissolving brass (Cu + Zn) in HNO3 will produce a Cu 2+ and Zn 2+ solution. you cannot ignore the Zn 2+. everything that occurs to the Cu 2+ will occur to the Zn 2+. it will form Zn(OH)2 and ZnO with heat, etc. how can you be sure that you have a 'pure' Cu 2+ solution? 10 g alloy in 6M HNO3 --> Cu 2+ + NO3- + Zn 2+ as i said before, brass is made of copper and zinc. HNO3 reacts just as well with copper as it does with zinc. so, the zinc and copper ions will be in solution together. there is no way to separate the zinc from the copper with NaOH because both metals will form an insoluble hydroxide. you would be better off separating out the Cu with the addition of NaCl or KCll. CuCl2 would form and precipitate while ZnCl2 will remain aqueous. using NaOH does nothing to separate out the 2 metal ions.
