Question

a compound contains 69.94 percent iron and 30.06 percent oxygen. what is its molecular formula if the molar mass of the compound is 199.55 grams per mole?

Answer 1

The compound's empirical formula is calculated as Fe2O3 by finding the mole ratio of iron to oxygen and converting it to the smallest whole numbers. The molecular formula is also Fe2O3 because the molar mass of the compound is close to that of the empirical formula, indicating a one-to-one ratio.

Step-by-step explanation:

To determine the molecular formula of a compound with 69.94 percent iron (Fe) and 30.06 percent oxygen (O), given that its molar mass is 199.55 grams per mole, we need to calculate the empirical formula first and then the molecular formula. The mass percentages provide a direct way to find the empirical formula, which expresses the smallest whole number ratio of moles of each element in the compound.

To calculate the moles of Fe and O in 100 g of the compound, we divide the mass of each element by its atomic mass:

• Iron: 69.94 g Fe / 55.85 g/mol Fe = 1.252 moles of Fe
• Oxygen: 30.06 g O / 16.00 g/mol O = 1.878 moles of O

Divide each by the smallest number of moles to get the mole ratio:

• Fe/O ratio = 1.252 / 1.252 = 1.000
• Fe/O ratio = 1.878 / 1.252 = 1.500

The ratio is 1.000 mol of iron to 1.500 mol of oxygen (Fe1O1.5). We multiply the ratio by two to obtain whole numbers:

• 2 (Fe1O1.5) = Fe2O3

The empirical formula is Fe2O3. Since the empirical formula represents Fe2O3 with a molar mass of 159.69 g/mol, we can determine the molecular formula by comparing the empirical formula mass to the given molar mass (199.55 g/mol) of the compound. Dividing the molar mass of the compound by the molar mass of the empirical formula gives us:

Molecular formula = (199.55 g/mol) / (159.69 g/mol) = 1.25

Since we cannot have a fraction of an atom in a molecular formula, we round to the nearest whole number, which gives us 1. Thus, the molecular formula is the same as the empirical formula in this case, Fe2O3.

Answer 2

The percentages are given in weight. You need to transform them into moles.

Take a basis of 100 grams of compound. Then there are 69.94 grams of iron and 30.06 grams of oxygen.

Use the atomic masses of iron and oxygen to obtain each number of moles.

Molar mass of iron: 55.85 g/mol
# moles of iron = 69.94 g / 55.85 g/mol = 1.252

Molar mass of oxygen: 16 g/mol
# moles of oxygen = 30.06 g / 16 g/mol =1.87875

Proportions

1.87875 moles of Oxigen/ 1.252moles of iron = 1.50 /1

3moles of Oxygen: 1 mol of Iron

Empirical formula: FeO3

Molar mass of the empirical formula: 55.85 g/mol + 3(16g/mol) = 103.85 g/mol

Molar mass of the compound / molar mass of the empirical formula = 199.55 g/mol / 103.85 g/mol = 1.92

Round to 2.

Then the molecular formula is the empirical formula times 2.

Fe2O6