Question

In a reaction involving iron, Fe, and oxygen, O. it was determined that 4.166 grams of iron reacted with 1.803 grams of oxygen. From this information, determine the empirical formula of the compound that resulted.a. FEO2b. FeO3c. Fe2Od. Fe2O3

Answer 1

Answer: The empirical formula for the given compound is $Fe_2O_3$

Explanation:

We are given:

Mass of Fe = 4.166 g

Mass of O = 1.803 g

To formulate the empirical formula, we need to follow some steps:

• Step 1: Converting the given masses into moles.

Moles of Iron =$\frac{\text{Given mass of Iron}}{\text{Molar mass of Iron}}=\frac{4.166g}{55.85g/mole}=0.0746moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.803g}{16g/mole}=0.113moles$

• Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0746 moles.

For Iron = $\frac{0.0746}{0.0746}=1$

For Oxygen = $\frac{0.113}{0.0746}=1.5$

Converting the mole ratio into whole number by multiplying with '2'

Mole ratio of Fe = (1 × 2) = 2

Mole ratio of O = (1.5 × 2) = 3

• Step 3: Taking the mole ratio as their subscripts.

The ratio of Fe : O = 2 : 3

Hence, the empirical formula for the given compound is $Fe_2O_3$