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.