Let's complete your problem:
1. Mass of crucible: 27.50 g
2. Mass of crucible + Fe: 28.62 g
3. Mass of Fe: Mass of crucible + Fe - Mass of crucible
Mass of Fe = 28.62 g - 27.50 g
= 1.12 g
Now, let's calculate the remaining values:
4. Mass of crucible + iron oxide: 29.10 g
5. Mass of iron oxide: Mass of crucible + iron oxide - Mass of crucible
Mass of iron oxide = 29.10 g - 27.50 g
= 1.60 g
To determine the empirical formula of iron oxide, we need to compare the moles of iron (Fe) and oxygen (O). To calculate the moles, we need the molar masses of Fe and O.
The molar mass of Fe is 55.845 g/mol.
The molar mass of O is 16.00 g/mol.
6. Moles of Fe: Moles of Fe = Mass of Fe / Molar mass of Fe
Moles of Fe = 1.12 g / 55.845 g/mol
≈ 0.020 moles
7. Mass of oxygen: Mass of oxygen = Mass of iron oxide - Mass of Fe
Mass of oxygen = 1.60 g - 1.12 g
= 0.48 g
8. Moles of oxygen: Moles of oxygen = Mass of oxygen / Molar mass of O
Moles of oxygen = 0.48 g / 16.00 g/mol
≈ 0.030 moles
Now, we can determine the empirical formula by finding the simplest ratio between the moles of Fe and O.
Since the moles of Fe are approximately 0.020 and the moles of oxygen are approximately 0.030, we can simplify these numbers by dividing them both by 0.020, giving us a ratio of 1:1. This means the empirical formula of iron oxide is FeO.
Therefore, the empirical formula of iron oxide is FeO.