When 6.25grams of pure iron are allowed to react with oxygen, a black oxide forms. If the products weighs 8.15g, what is empirical formula of the oxide?
Fe + O2 ==> FexOy
6.25g + ? = 8.15g
Therefore, the amount of oxygen added must be 8.15-6.25 = 1.90.
%Fe = (6.25/8.15)*100 = about 77% but you can (and must) do that more accurately.
%O = (1.90/8.15)*100 = about 233%. Do all of these plus all that follows more accurately.
Take 100 g sample which gives you
approx 77 g Fe
approx 23 g O
Convert to mols.
approx 77/55.85 = approx 1.4
approx 23/16 = approx 1.4
So the ratio is 1:1 and the formula si FeO.
Remember all of my numbers above are estimates.
Convert to mols.
Fe1.4O1.4 = FeO
To find the empirical formula of the oxide, we need to determine the ratio of the elements in the compound. In this case, we have iron reacting with oxygen.
1. First, calculate the mass of oxygen in the compound:
Mass of oxygen = Mass of products - Mass of iron
Mass of oxygen = 8.15 g - 6.25 g
Mass of oxygen = 1.9 g
2. Next, convert the masses of iron and oxygen to moles using their molar masses:
Molar mass of iron (Fe) = 55.85 g/mol
Molar mass of oxygen (O) = 16.00 g/mol
Moles of iron = Mass of iron / Molar mass of iron
Moles of iron = 6.25 g / 55.85 g/mol
Moles of oxygen = Mass of oxygen / Molar mass of oxygen
Moles of oxygen = 1.9 g / 16.00 g/mol
3. Divide the number of moles of each element by the smallest number of moles to obtain the mole ratio:
Moles of iron = 0.1115 mol (rounded to four decimal places)
Moles of oxygen = 0.1188 mol (rounded to four decimal places)
Divide both moles by 0.1115 mol:
0.1115 mol / 0.1115 mol = 1
0.1188 mol / 0.1115 mol ≈ 1.065
The mole ratio is approximately 1:1.
4. Finally, write the empirical formula using the mole ratio:
The empirical formula of the oxide is FeO, indicating that the ratio of atoms in the compound is 1 iron atom to 1 oxygen atom.
To determine the empirical formula of the oxide, we need to find the ratio of the elements present in the compound.
First, we need to calculate the number of moles of iron (Fe) and oxygen (O) in the reaction. We can use the molar mass of iron (Fe) and oxygen (O) to convert the given masses to moles.
The molar mass of iron (Fe) is approximately 55.845 g/mol, and the molar mass of oxygen (O) is approximately 16.00 g/mol.
Moles of iron (Fe) = mass of iron (Fe) / molar mass of iron (Fe)
Moles of iron (Fe) = 6.25 g / 55.845 g/mol
Moles of oxygen (O) = mass of oxygen (O) - mass of iron (Fe) / molar mass of oxygen (O)
Moles of oxygen (O) = (8.15 g - 6.25 g) / 16.00 g/mol
Next, we need to find the smallest whole number ratio between the moles of iron (Fe) and oxygen (O) to determine the empirical formula.
Divide both the moles of iron (Fe) and oxygen (O) by the smallest value obtained.
Let's calculate the moles of iron (Fe) and oxygen (O) using the given masses.
Moles of iron (Fe) = 6.25 g / 55.845 g/mol
≈ 0.1117 mol
Moles of oxygen (O) = (8.15 g - 6.25 g) / 16.00 g/mol
≈ 0.1194 mol
The smallest whole-number ratio is obtained by dividing both moles by 0.1117 mol (which is approximately 0.1117 mol ÷ 0.1117 mol = 1) to get:
Fe:O ≈ 1:1
Therefore, the empirical formula of the oxide is FeO.