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Posted by on Wednesday, November 12, 2008 at 9:12pm.

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?

THANK YOU!

  • chemistry - , 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 - , Wednesday, November 12, 2008 at 10:33pm

    For that part I thought that the formula would be NiF2.

  • chemistry - , 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 - , Thursday, November 13, 2008 at 12:16am

    To find this part are you using F, or F2. F2 was in the original question.

  • chemistry - , 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.

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