Question

A 115.0-g sample of iron (Fe) is burned in excess oxygen. The result is a pure solid with a mass of 164.5 grams. What is the empirical formula of the solid?

Answer 1

The empirical formula is Fe2O3

Step-by-step explanation:

Step 1: Data given

Mass of the iron sample = 115.0 grams

Mass of the pure solid = 164.5 grams

Molar mass of Fe = 55.845 g/mol

Molar mass of O = 16 g/mol

Step 2: Calculate mass of oxygen

mass of oxygen = mass of solid - mass of iron

Mass of oxygen = 164.5 grams - 115.0 grams

MAss of oxygen = 49.5 grams

Step 3: Calculate moles of oxygen

Moles oxygen = mass oxygen / molar mass oxygen

Moles oxygen = 49.5 grams / 16.0 g/mol

Moles oxygen = 3.09 moles

Step 4: Calculate moles of Fe

Moles Fe = mass Fe / molar mass Fe

Moles Fe = 115.0 grams / 55.845 g/mol

Moles Fe = 2.06 moles

Step 5: Calculate mol ratio

We divide by the smallest amount of moles

O: 3.09/2.06 = 1.5

Fe: 2.06 / 2.06= 1

This means for 1 mol Fe we have 1.5 mol O OR for each 2 moles Fe we have 3 moles O.

The empirical formula is Fe2O3

Answer 2

Fe₂O₃

Step-by-step explanation:

When Fe is burned it can form iron (II) oxide or iron (III) oxide. The reactions are:

a) Fe + 1/2 O₂ → FeO

b) 4 Fe + 3 O₂ → 2 Fe₂O₃

In the reaction "a", the mass ratio of FeO to Fe is 71.84 g FeO:55.85 g Fe =

1.29 g FeO: 1 g Fe

In the reaction "b", the mass ratio of Fe₂O₃ to Fe is 319.3 g Fe₂O₃: 223.4g Fe =

1.43 g Fe₂O₃ : 1 g Fe

According to the experimental data, the mass ratio of the product to Fe is 165.5g compound: 115.0 g Fe =

1.43 g compound: 1 g Fe

Since this ratio is equal to that of reaction "b", the empirical formula of the solid is Fe₂O₃.