Copper+Lab - 1. Data (a) mass of copper powder used (g):...

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1. Data (a) mass of copper powder used (g): 25.000 g (b) mass of copper powder and beaker (g): 72.420g (c) mass of copper oxide and beaker (g): 78.714g 2. Calculate the following for the copper: (a) The number of moles of copper powder, given a molecular weight of copper equal to 63.55 g/mole. (1 pt) {Convert the value from 1(a) to moles} moles Cu = 25.000 g x 1 mol = 0.3934 mol 63.55 g 3. Calculate the following for the oxygen: (a) The mass of the beaker (g) {1(b) - 1(a)} = 47.420 g (b) The final mass of copper oxide (g) {1(c) - 3(a)} = 31.300g (c) The gain in mass in the beaker which is equal to the added mass of oxygen (g) {3(b) - 1(a)} = 6.3000 g (d) The number of moles of oxygen in the copper oxide, given a molecular weight of oxygen (O) equal to 16.00 g/mole. {convert 3(c) to moles} moles O = 6.300 g x 1 mol = 0.3938 mol 16.00 g 4. Calculate the molar ratio of copper to oxygen from: (moles copper powder) / (moles of oxygen in the copper oxide) = 0.3934 mol Cu / 0.3938 mol O = 0.9990 5. According to this molar ratio, what is the empirical formula of copper oxide? Since the mole ratio is effectively 1:1, the empirical formula is Cu
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This note was uploaded on 02/06/2011 for the course CHEM 115L taught by Professor Nakkiew,pichaya during the Spring '10 term at Grand Canyon.

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Copper+Lab - 1. Data (a) mass of copper powder used (g):...

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