A compound contains 69.9% iron and 30.1% oxygen. What is its empirical formula?
Fe2oe
Fe2o3
To find the empirical formula of the compound, we need to determine the simplest whole number ratio between the elements present.
First, we need to convert the percentages to grams. Let's assume we have 100 grams of the compound.
Iron: 69.9% of 100 grams = 69.9 grams
Oxygen: 30.1% of 100 grams = 30.1 grams
Next, we need to convert the grams to moles using the respective atomic masses of iron (Fe) and oxygen (O). The atomic mass of iron is 55.84 g/mol, and the atomic mass of oxygen is 16.00 g/mol.
Iron (Fe): 69.9 grams / 55.84 g/mol = 1.25 moles
Oxygen (O): 30.1 grams / 16.00 g/mol = 1.88 moles
Now, we need to find the simplest whole number ratio between the moles of each element. We can divide both numbers by the smaller number (1.25) to simplify it.
Iron (Fe): 1.25 moles / 1.25 = 1 mole
Oxygen (O): 1.88 moles / 1.25 ≈ 1.5 moles
Since we can't have a fraction in an empirical formula, we need to multiply both elements by 2 to get a whole number ratio.
Iron (Fe): 1 mole * 2 = 2 moles
Oxygen (O): 1.5 moles * 2 = 3 moles
Therefore, the empirical formula of the compound is Fe2O3.
69.9 g Fe /55.845 g Fe = 1.252 mol Fe
30.1 g O /15.9994 g O = 1.881 mol O
1.252 mol Fe/1.252 = 1 mol Fe
1.881 mol O/1.252 = 1.502 mol O
FeO2?
Take a 100 g sample which gives you
69.9 g Fe
30.1 g O
mols Fe = grams/atomic mass = ?
mols O = grams/atomic mass = ?
Now determine the ratio of one element to the other with the smallest being no less than 1.00. The easy way to do that is to divide both numbers by the smallest number, then determine the whole number ratio from that.