posted by Sarah on .
A piece of nickel foil, 0.550 mm thick, 1.25 cm width and 1.00 cm long, is allowed to react with fluorine, F2 to give a nickel fluoride. Density of nickel is 8.908 g/cm3. How many moles of nickel foil were used? If you isolated 1.261 g of nickel fluoride, what is its formula?
This is what I have come up with so far...
V = 0.055 cm x 1.25 x 1.00= 0.0688 cm^3
Mass = 0.0688 x 8.908 g/cm^3 = 0.612 g
moles Ni = 0.612 / 58.69 g/mol= 0.0104
Is that right? At all?
It's ok for the first part; i.e., mols nickel. The question has another part to it.
For that part I thought that the formula would be NiF2.
I think the author of the problem expected you to calculate the formula.
You have mols Ni = 0.0104.
You know g nickel fluoride = 1.261 g.
less g nickel = 0.6124
= 0.6486 g fluorine.
Determine mols fluorine then find the ratio. I get a funny number when I do it. like 1:3.28. If I thought the accuracy was very good that would lead me to think the formula is Ni3F5 which isn't likely. If I thought the accuracy was suspect, I would round the 3.28 to 3 and call it NiF3. There IS a NiF3 compound although it isn't common.
To find this part are you using F, or F2. F2 was in the original question.
It's true F2 was in the original problem but only as a means of telling you what reacted. F2 is simply a synonym for fluorine.
To determine the formula, we determine the mols Ni and mols F (not F2), determine the ratio of the two and that gives us the empirical formula.