determine the empirical formula of an oxide of iron which has 69.9 iron and 30.1 dioxygen by mass
Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% dioxygen by mass
Answer:
To determine the empirical formula of the iron oxide compound with the given percentages of iron and oxygen by mass, we need to follow the steps below:
Step 1: Convert Mass Percent to Moles
First, we need to convert the mass percentage of each element into moles.
The molar mass of iron (Fe) is 55.85 g/mol.
The molar mass of oxygen (O) is 16.00 g/mol.
For iron:
69.9 grams of Fe x (1 mol Fe / 55.85 g) ≈ 1.25 moles of Fe
For oxygen:
30.1 grams of O x (1 mol O / 16.00 g) ≈ 1.88 moles of O
Step 2: Determine the Mole Ratios
Next, we determine the mole ratios by dividing the moles of each element by the smallest number of moles.
Fe: 1.25 moles / 1.25 moles = 1
O: 1.88 moles / 1.25 moles ≈ 1.5
Step 3: Write the Empirical Formula
The empirical formula is written using these mole ratios.
The empirical formula for the oxide of iron is FeO1.5. However, since the formula should be written with whole numbers, we need to multiply all subscripts by 2 to get rid of the fractional subscripts. Therefore, the empirical formula becomes Fe2O3.
So, the empirical formula of the oxide of iron is Fe2O3.