Mass of crucible and iridium oxide: 39.73 g The answer is 2 times the above empirical formula, so the molecular formula is C 2H 4O 2Įxample #3: A compound made of two elements, iridium (Ir) and oxygen (O), was produced in a lab by heating iridium while exposed to air. The molecular weight of this molecule is 30 g/mole, so you divide 60/30 to find how many times you must multiple your empirical formula. Solution: 12.0 g carbon is about 1 mole of carbon 2.0 g of H is about 2 moles and 16.0 g O is about one mole. What is the molecular formula of the compound? (no calculator allowed!) This is a 1:1 molar ratio between Cu and O.Įxample #2: On analysis, a compound with molar mass 60 g/mol was found to contain 12.0 g of carbon, 2.0 g of hydrogen and 16.0 g of oxygen. Determine the empirical formula of the oxide.Ĭu -> 2.50 g / 63.546 g/mol = 0.03934 mol ChemTeam: Calculate empirical formula when given mass data Calculate empirical formula when given mass dataĬalculate empirical formula when given percent composition dataĭetermine identity of an element from a binary formula and a percent compositionĭetermine identity of an element from a binary formula and mass dataĮxample #1: A sample of copper metal weighing 2.50 g is heated to form an oxide of copper.
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