A1.32-g sample of aluminum reacted with 1.174 g oxygen to form aluminum oxide. determine the empirical formula of the aluminum oxide.

Al₂O₃

Explanation:

Given 1.132 g Al and 1.174 O,

Number of moles of Al = mass of Al * 1 mole of Al/ molar mass of Al

Molar mass of Al = 26.98 g/mol

∴ number of moles of Al = 1.32 * 1/26.98 = 0.04892 moles

Number of moles of O = mass of O * 1 mole of O/ molar mass of O

Molar mass of O = 16g/mol

Number of moles of O = 1.174* 1/16 = 0.073375 moles

Aluminium / Oxygen = 0.04892/0.073375 = 0.6667 = 2/3

So, 2 moles of Al with 3 moles of O will be required