Finding the percent yield of cobalt in an analysis of a coordination compound

The purpose of the lab is to analysis this coordination compound Co(x)NH3(y)Cl(z) and give the percent yield of colbalt. I started out with 8.007 g of CoCl2*6H20 and 4.007 g of NH4Cl.  
The experiment required the following procedures:
Mixed the 4.007g of NH4Cl with 25mL 15 M NH3.  Then added the 8.007g of CoCL2*6H20.  Then added 6 mL of 30% H2O2. Then added 25mL 12 M HCl.  Heated the mixture @ 80 degrees C for about 30 minutes.  Filtered the mixture through a Buchner funnel.  Put solid into beacker and added 30mL of 12 M Nh3 and 30 mL of H20 and heated for a couple of minutes to dissolve solid.  Poured in 60 mL of 12 M HCl.  Heated solution for 30 min. Filtered solution and then poured 10 mL 95% ethanol on crytals in funnel and and 10 mL of acetone.  
The final weight of compound was 4.717g.  
We are asked to find formula Co(x)NH3(y)Cl(z) and give the percent yield of cobalt.