2.241g sample of nickel reacts w oxygen to form 2.85g of the metal oxide. Calculate the empirical formula of the oxide?

Convert 2.241 g Ni to mols.

Determine mass oxygen fron mass metal oxide - mass metal = mass oxygen.
Convert g oxygen to mols oxygen ATOMS.
Deterime ratio of atoms. Those will be the subscripts. Post your work if you get stuck.

n= 2.241/14.00=.1600

.1600/.0382=4.189

o=.611/16.00= .0382
.0382/.0382=0

I cant seem to get a whole # for the nickel

Nickel is Ni. You are using N = 14.00

Not the same thing.

i looked up the wronge symbol. the answer i got was NiO

NiO is correct.

To calculate the empirical formula of the oxide, we need to determine the ratio of atoms present in the compound.

Step 1: Find the moles of nickel and oxygen in the given sample.
- Moles of nickel = mass of nickel / molar mass of nickel
- Moles of oxygen = mass of oxygen / molar mass of oxygen

The molar mass of nickel (Ni) is 58.6934 g/mol, and the molar mass of oxygen (O) is 15.999 g/mol.

Moles of nickel = 2.241 g / 58.6934 g/mol = 0.03814 mol
Moles of oxygen = 2.85 g / 15.999 g/mol = 0.17814 mol

Step 2: Divide the number of moles of each element by the smallest number of moles.
- This gives us the simplest whole number ratio between the elements.

Smallest number of moles = 0.03814 mol

Moles of nickel / Smallest number of moles = 0.03814 mol / 0.03814 mol = 1
Moles of oxygen / Smallest number of moles = 0.17814 mol / 0.03814 mol ≈ 4.67

Step 3: Round off the ratio to the nearest whole number.
- Multiply all the subscripts by the same factor to get the simplest whole number ratio.

Rounded ratio:
Nickel: 1
Oxygen: 5

Step 4: Write the empirical formula using the whole number ratio.
- The empirical formula of the oxide is NiO5, which can be simplified to NiO.

Therefore, the empirical formula of the metal oxide is NiO.