Aluminum reacts with oxygen to yield aluminum oxide. If 5.0 g of Al reacts with 4.45 g of O2, what is the empirical formula of aluminum oxide?

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction. It allows us to calculate the amounts of substances consumed and produced in a reaction based on balanced chemical equations. Understanding stoichiometry is essential for determining the empirical formula, as it helps in converting masses of reactants to moles and subsequently to the mole ratio of the elements.

The empirical formula of a compound represents the simplest whole-number ratio of the elements present in that compound. It is derived from the mole ratio of the elements obtained from stoichiometric calculations. For aluminum oxide, determining the empirical formula involves finding the ratio of aluminum to oxygen after calculating the moles of each element based on the given masses.

Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It is crucial for converting between grams and moles in stoichiometric calculations. For this question, knowing the molar masses of aluminum (approximately 27 g/mol) and oxygen (approximately 32 g/mol) allows for the accurate conversion of the given masses of Al and O2 into moles, which is necessary for determining the empirical formula.