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!
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.
Your calculations are almost correct, but there is a small mistake in the calculation of the volume (V) of the nickel foil.
To find the volume of a rectangular solid, you need to multiply its length, width, and thickness. In this case, the thickness is given as 0.550 mm, which needs to be converted to centimeters before the calculation.
1 mm = 0.1 cm
So, the thickness in centimeters is 0.550 mm x (0.1 cm/1 mm) = 0.055 cm
Now, we can calculate the volume (V) using the corrected thickness:
V = 0.055 cm x 1.25 cm x 1.00 cm = 0.06875 cm^3
The rest of your calculations are correct:
Mass = 0.06875 cm^3 x 8.908 g/cm^3 = 0.612 g
moles Ni = 0.612 g / 58.69 g/mol = 0.010416 moles
So, the number of moles of nickel foil used is approximately 0.0104 moles.
To determine the formula of the nickel fluoride, we need to work with the mass of nickel fluoride produced.
Given: Mass of nickel fluoride (NiF2) = 1.261 g
To find the molar mass of nickel fluoride, we need to sum the atomic masses of its constituent elements. Nickel (Ni) has a molar mass of approximately 58.69 g/mol, and fluorine (F) has a molar mass of approximately 19.00 g/mol.
Molar mass of NiF2 = (58.69 g/mol) + 2(19.00 g/mol) = 96.69 g/mol
Now, we can calculate the number of moles of nickel fluoride:
moles NiF2 = 1.261 g / 96.69 g/mol ≈ 0.013 moles
Since the moles of nickel fluoride are not an exact whole number (0.013), we need to divide it by the smallest value to get the simplest ratio:
0.013 / 0.013 ≈ 1
0.013 / 0.013 ≈ 1
This indicates that the formula of the nickel fluoride is NiF2.