Experiment Five Pre-laboratory Empirical Formula of an Oxide PDF

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Nhi Chung General Chemistry I - Chem1411, HCC Alief Campus October 16, 2017

Pre-lab n. # Five Before using a loose magnesium ribbon, grab a clean, dry crucible and cover and weigh them on a balance or digital scale until they reach to 0.01 g. After that, place a loose magnesium ribbon that is about twenty to thirdly centimeters in a crucible, then weigh and report its mass on the lab report sheet. More importantly, place the covered crucible in a clay triangle on an iron ring, adjust the height of the ring stand, and heat it about two minutes from the Bunsen burner flame. Once two minutes has passed, allow the crucible to cool down until it has reached to its normal or room temperature; thus, a reaction occurs with an oxygen. Most importantly, the crucible must be heated again on the high setting about three minutes; therefore, its constant mass is completed and it must be recorded on the lab report sheet.

Pre lab questions: 1. When doing the experiment, it is very important to use and cover the lid on a crucible, so that the magnesium would not expose to the air. Furthermore, by using the lid for the crucible, the metal would not burn and cause temporary blindness. 2. Mole is based on Avogadro's number, which is meant to be a small number, whereas molar mass has one mole of substance in grams, which is normally

equals to the atomic mass units. Molar masses of magnesium are 24.30 g/mol and atomic oxygen is 16 g/mol. 3. A) Mg(s) + O(l) —> MgO B) Mg(s) + N2(g) —> MgN2 4. A) Mole of Al = (mass of Al) / (molar mass of Al) = (0.0200 g)/ (26.98 g/mol) = 7.412 mol Mole of Cl = (mass of Cl) / (molar mass of Cl) = (0.0600 g)/ (35.45 g/mol) = 0.0017 mol Al (ratio) = (7.412) / (7.412) = 1 Cl (ratio) = (0.0017) / (7.412) = 2.3 Moles of Al atoms = 1.0 x 2 = 2 Moles of Cl atoms = 2.3 x 2 = 4.6 Al2Cl4 B) Mole of Ba = (mass of Ba) / (molar mass of Ba) = (0.0800 g)/ (137.33 g/mol) = 5.82 mol Mole of S = (mass of S) / (molar mass of S) = (0.0800 g)/ (32.06 g/mol) = 0.0025 mol Mole of O = (mass of O) / (molar mass of O) = (0.320 g)/ (16.0 g/ mol) = 0.02 Ba (ratio) = (5.82) / (5.82) = 1 S (ratio) = (0.0025) / (5.82) = 4.3 O (ratio) = (0.02) / (0.0025) = 8 Moles of Ba atoms = 1.0 x 2 = 2 Moles of S atoms = 4.3 x 2 = 8.6

Moles of O atoms = 8 x 2 = 16 Ba2S8O16 5. Mass of O2 = 0.606 g – 0.424 g = 0.182 g Mole of Fe = (mass of Fe) / (molar mass of Fe) = (0.424 g)/ (55.84 g/mol) = 0.0076 mol Mole of O2 = (mass of O2) / (molar mass of O2) = (0.182 g)/ (16.00 g/mol) = 0.011 mol Fe (ratio) = (0.0076) / (0.0076) = 1 O2 (ratio) = (0.011) / (0.0076) = 1.45 Moles of Fe atoms = 1.0 x 2 = 2 Moles of O2 atoms = 1.45 x 2 = 2.9 Fe2O3...