Ghada Aljaber Thank you for your question. This is relatively simple to answer with a grasp of chemistry but does need some care with the calculations.

So, you want to end with 3wt% Co in NiFe2O4 – that is, 0.03 g Co plus 0.97 g NiFe2O4 for a gram of final material. OK, so how do we get there? Consider NiFe2O4 as an equimolar ‘mix’ of NiO plus Fe2O3. Let’s consider getting to a gram of the doped material as then if we want to make more then we simply have to multiply the amounts of the starting precursors by the appropriate factor.

The MW of the precursors and the decomposition route are key:

• Fe(NO3)3.9H2O MW 403.999 g/mol gives rise to Fe2O3 (MW 159.6882 g/mol) on decomposition. We note that we’d need 2 moles of the nonahydrate to give rise to 1 mole of the oxide

• Ni(NO3)2.6H2O MW 290.79 g/mol gives rise to NiO (MW 74.6928 g/mol) on decomposition

• Co(NO3)2·6H2O MW 291.03 g/mol gives rise to CoO (MW 74.93 g/mol) on decomposition. You’re specifying Co (atomic mass 58.93) not CoO

• NiFe2O4 MW 234.381 g/mol. In 1 g of this material, we’d have 159.6882/234.381 g of Fe2O3 and 74.6928/234.381 g of NiO. For 0.97 g of NiFe2O4 we’d need 0.97*(159.6882/234.381) g of Fe2O3 and 0.97*(74.6928/234.381) g of NiO

Let’s start with the dopant (Co) first. To get 0.03 g of cobalt would require 0.03*(74.93/58.93) g of CoO which would be produced from a multiplier of (291.03/74.93) g of the hexanitrate. Thus, we’d need 0.03*(74.93/58.93)*(291.03/74.93) g of the precursor hexanitrate to lead to 0.03 g of Co

Now, the iron nonahydrate. (2 X 403.999) g of this gives rise to 159.6882 g of Fe2O3, so we need (2 X 403.999)/159.6882 g of the nonahydrate to give rise to 1 g Fe2O3. We actually require 0.97*(159.6882/234.381) g of Fe2O3 equivalent in the final product. So our starting nonahydrate would be 0.97*(159.6882/234.381)*[(2*403.999)/159.6882] g of this precursor

By a similar argument, for the NiO equivalent, then to get 0.97*(74.6928/234.381) g of NiO requires a multiplier of (290.79/74.6928) of the hexahydrate. That is (290.79/74.6928)*0.97*(74.6928/234.381) of the nickel nitrate hexahydrate for a gram of the final NiFe2O4.

You need to check my math and that I haven’t made a typo (or several typos) and also that I have the molecular masses of the compounds correct. I’ll leave you to multiply out the final values and check that they’ll make 1 g of final material. For larger amounts you can multiply by the appropriate factor, as I said earlier.