Answer: To determine the empirical formula of aluminum fluoride, we need to find the ratio of the number of atoms of each element present in the compound. We are given the masses of aluminum (Al) and fluorine (F) in the reactants and product.
The balanced chemical equation for the reaction between aluminum and fluorine to form aluminum fluoride is:
2 Al + 3 F2 → 2 AlF3
According to the stoichiometry of the balanced equation, 2 moles of aluminum react with 3 moles of fluorine gas to produce 2 moles of aluminum fluoride.
Now, let's calculate the number of moles of aluminum and fluorine present in the reactants and product:
Moles of Aluminum (Al):
Given mass of aluminum (m_Al) = 3.15 g
Molar mass of aluminum (M_Al) = 26.98 g/mol
Moles of Al = m_Al / M_Al
Moles of Al = 3.15 g / 26.98 g/mol ≈ 0.117 mol
Moles of Fluorine (F):
Given mass of aluminum fluoride formed (m_AlF3) = 9.81 g
Molar mass of aluminum fluoride (M_AlF3) = 83.98 g/mol
Moles of AlF3 = m_AlF3 / M_AlF3
Moles of AlF3 = 9.81 g / 83.98 g/mol ≈ 0.117 mol
Now, let's find the simplest whole-number ratio between moles of aluminum and moles of fluorine:
Moles of Al : Moles of F = 0.117 mol : 0.117 mol
The ratio is 1:1. This means that there is an equal number of moles of aluminum and fluorine in the reactants and product.
Therefore, the empirical formula of aluminum fluoride is AlF3.