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?
chemistry - DrBob222, Wednesday, November 12, 2008 at 9:26pm
It's ok for the first part; i.e., mols nickel. The question has another part to it.
chemistry - Jordan, Wednesday, November 12, 2008 at 10:33pm
For that part I thought that the formula would be NiF2.
chemistry - DrBob222, Wednesday, November 12, 2008 at 10:54pm
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.
chemistry - Sarah, Thursday, November 13, 2008 at 12:16am
To find this part are you using F, or F2. F2 was in the original question.
chemistry - DrBob222, Thursday, November 13, 2008 at 10:32am
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.