Determine the empirical formula of an oxide of iron, which has 69.9% iron and 30.1% dioxygen by mass

General
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determine the empirical formula of an oxide of iron, which has 69.9% iron and 30.1% dioxygen by mass.

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Determine the empirical formula of an oxide of iron, which has 69.9% iron and 30.1% dioxygen by mass

Answer:

To determine the empirical formula of the oxide of iron, given that it consists of 69.9% iron and 30.1% oxygen by mass, we need to convert these percentages into moles and find the mole ratio between iron and oxygen to determine the empirical formula.

  1. Calculate the Moles of Iron and Oxygen:

Given:

  • Mass percent of iron = 69.9%
  • Mass percent of oxygen = 30.1%

Let’s assume we have 100 g of the compound.

  • Mass of iron = 69.9 g
  • Mass of oxygen = 30.1 g

Now, we need to convert these masses into moles by dividing by their respective molar masses from the periodic table.

  • The molar mass of iron (Fe) is approximately 55.85 g/mol.
  • The molar mass of oxygen (O) is approximately 16 g/mol.

Moles of iron = \frac{69.9 \text{ g}}{55.85 \text{ g/mol}} \approx 1.25 \text{ mol}

Moles of oxygen = \frac{30.1 \text{ g}}{16 \text{ g/mol}} \approx 1.88 \text{ mol}

  1. Determine the mole ratio:

Next, we need to find the simplest ratio between the moles of iron and oxygen.

Dividing both by the smallest value (1.25):
Moles of iron: \frac{1.25}{1.25} = 1
Moles of oxygen: \frac{1.88}{1.25} \approx 1.5

The empirical formula of the iron oxide is FeO_{1.5}.

However, we cannot have a non-integer value in a chemical formula. Hence, we need to multiply all the subscripts by a number that will result in whole numbers.

Multiplying by 2 to get rid of the fraction, the empirical formula of the iron oxide is Fe_2O_3 .

Therefore, the empirical formula of the oxide of iron, based on the given percentages, is iron (III) oxide (Fe2O3).