   # 3.117 For the oxides of iron, FeO, Fe 2 O 3 , and Fe 3 O 4 , describe how you would determine which has the greatest percentage by mass of oxygen. Would you need to look up any information to solve this problem? ### Chemistry for Engineering Students

4th Edition
Lawrence S. Brown + 1 other
Publisher: Cengage Learning
ISBN: 9781337398909 ### Chemistry for Engineering Students

4th Edition
Lawrence S. Brown + 1 other
Publisher: Cengage Learning
ISBN: 9781337398909
Chapter 3, Problem 3.117PAE
Textbook Problem
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## 3.117 For the oxides of iron, FeO, Fe2O3, and Fe3O4, describe how you would determine which has the greatest percentage by mass of oxygen. Would you need to look up any information to solve this problem?

Interpretation Introduction

Interpretation: The greatest percentage by mass of oxygen in oxides of iron, FeO, Fe2O3 and Fe3O4 should be determined.

Concept Introduction:

• Mass Percentage is one of the ways to represent the concentration of the solute in solution.
• Mass Percentage is the mass of element divide by the mass of the compound. Then the result is multiplied by 100.

### Explanation of Solution

Formula used:

Mass percentage = (mass of element in 1 mole of the compound/mass of 1 mole of compound) ×100

Calculation:

To calculate mass percentage of oxygen in the given oxides of iron we must know the values for molar mass of iron and oxygen.

We know that

Molar mass of iron, Fe=55.845 g/mol

Molar mass of oxygen, O=16.00 g/mol

Now we will calculate the molar mass of each iron oxide and mass of oxygen in 1 mole of that oxide.

1. FeO

Molar mass of FeO = molar mass of Fe + molar mass of O

=55.845+16.00=71.845 g/mol

Now, molar mass implies mass of one mole of oxide and that is 1 mole of FeO contains 16.00 g of oxygen.

Thus, Mass Percentage of O in FeO = (1671.845)×100=22.27%

2. :Fe2O3

Molar mass of Fe2O3 = (2 × molar mass of Fe ) + (3 × molar mass of O )

=(2×55.845)+(3×16.00)=111.69+48.0=159.69 g/mol

Now molar mass implies mass of 1 mole of oxide and that is one mole of Fe2O3 contains= ( 3×16

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