What is the empirical formula of 2.241 grams of nickel that are heated in air to produce 3.158 grams of nickel oxide.

Answer 1

I'll add my 2 cents to the answer provided by Robert S. We do arrive at the same answer so you now have 2 ways of looking at the problem (they are essentially the same, really).

Ni(s) + O₂(g) ==> NiO(s)

mass of NiO = 3.158 g

mass Ni = 2.241 g

mass O = 3.158 g - 2.241 g = 0.917 g

mols Ni used = 2.241 g Ni x 1 mol Ni / 58.69 g = 0.03817 mols Ni

mols O = 0.917 g O x 1 mol O / 16 g = 0.05731 mols O

Divide both by 0.03817 to try to get whole numbers

Ni = 0.03817 / 0.03817 = 1.0

O = 0.05731 / 0.03817 = 1.50

Still not whole numbers, so we can multiply by 2 to get whole numbers:

Ni = 1.0 x 2 = 2.0

O = 1.5 x 2 = 3. 0

Empirical formula = Ni₂O₃

This is nickel peroxide or nickel (III) oxide

Answer 2

Hello, Ashtin,

I got a result which I feel is suspicious, but I don't see anything incorrect in the derivation, so I'll add my approach and you can judge whether it makes sense. We can assume that the increase in mass, 0.917 g, is due only to oxygen. So we have the masses of both the nickel and oxygen. Convert those masses into moles of each. It is the ratio of the moles that should yield the empirical formula of the nickel oxide. If we divide both values by the number of moles of Ni, we get "1" for Ni and 1.50 for O. That would mean NiO_1.5, which should be changed to Ni₂O₃. I don't know if this structure exists, but the data suggests it has this empirical formula.

grams g/mole moles Ratio to Ni

Ni 2.241 58.69 0.0381 1

O 0.917 15.99 0.0573 1.5