How many grams of iron are in a 54.39g sample of iron(III) oxide?

Answer 1

Just to add to the answer provided by Russ F, and to show an easier way to solve this problem, we can use % composition.

atomic mass Fe = 55.85 g/mol

molar mass Fe2O3 = 159.69 g/mol

There are 2 Fe in Fe2O3 so the total mass of Fe = 2 x 55.85 = 111.7 g Fe

% Fe = 111.7 g / 159.69 g = 69.95%

Mass of Fe in 54.39 g of Fe2O3 = 0.6995 x 54.39 g = 38.04 g Fe --> ANSWER

Answer 2

Grams Fe in 54.39g Iron (III) Oxide?

Step 1: Determine Molecular Formula

The Roman numeral indicates the charge (oxidation state) of the metal (Fe).

The oxidation state of oxygen can be determined from its Group number on the periodic table (6A). This tells you that oxygen needs 2 more electrons to achieve an octet. When this occurs, oxygen attains a -2 charge.

Use the criss-cross method to arrive at the formula.

Fe³⁺ O^2- → Fe₂O₃

2 mol Fe + 3 mol O → Fe₂O₃

Step 2: Convert grams Fe₂O₃ to moles Fe₂O₃

2 Fe: 56g/mol x 2 = 112g/mol

3 O: 16g/mol x 3 = 48g/mol

Fe₂O₃ = 160g/mol

54.39g Fe₂O₃ [1 mol Fe2O3/160g Fe₂O₃] = 0.340 mol Fe₂O₃

We see that there are 2 mol Fe and 3 mol O to create 1 mole Fe₂O₃. 

Step 3: Use the mole ratio to determine mole Fe

0.340 mol Fe₂O₃ [2 mol Fe/1 mol Fe₂O₃] = 0.680 mol Fe

Step 4: Convert mole Fe to grams Fe

0.680 mol Fe [56g Fe/1 mol Fe] = 38.1g Fe