CHEM
Empirical Formula Determination Formal Lab Report.docx

While burning the powder lit on fire the flame was

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the wire coil. While burning, the powder lit on fire. The flame was blue and produced a faint smog. After burning, the wire turned into tiny, brittle, black and red pieces. Data: Part 1- Mass of crucible and lid 28.481 g Mass of crucible, lid, and Mg ribbon 28.7192 g Mass of Mg ribbon 0.238 g Combined mass after heating 28.904 g Combined mass after first repeat heating 28.905 g Mass of oxygen present in sample 0.1858 g Percentage of Mg by weight of oxide 56.2% Moles of Mg 0.0098 Moles of O .01161 Empirical Formula of Magnesium Oxide MgO Part 2- Mass of crucible and lid 29.6475 g Mass of crucible, lid and Cu 29.9615 g Mass of Cu 0.314 g Combined mass after heating 30.0753 g Combined mass after first repeat heating 30.0863 g Mass of sulfur present in sample 0.5075 g Moles of S 0.0158 Moles of Cu 0.005 Empirical formula of copper sulfide CuS 3 Formula for determining empirical formula:
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0.238 gMg 24.31 molecular weightMg = 0.0098 moles Mg . Round to successive whole number: 1 for subscript. 15.9994 molecular weightO ¿ = 0.01161 moles O. 0.1858 gO ¿ Round to successive whole number:1 for subscript. Therefore, the empirical formula for Magnesium Oxide is: MgO. In addition, based off of class data, the average percentage of Mg in MgO is 59.95%. The standard deviation for the percentages found was 5.03 and the percent error is 6.33%. Part 3- Mass of 1 square foot of aluminum foil 3.809 grams Cost of one 75 square foot roll of aluminum foil $6.07 Cost of 1 square foot of aluminum foil $0.08 Finding the cost of 1 square foot of aluminum foil: 6.07 75 = 0.08 Determining cost of one atom of aluminum foil: $ 0.08 1 sq.ft × 1 sq .ft . 3.809 g × 27 gAl 1 mol × 1 mol 6.02 x 10 23 = 9.42 x 10 25 Using this equation, it is found that one atom of aluminum foil costs: $ 9.42 × 10 25 Discussion:
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  • Spring '13
  • ABC
  • Magnesium, Cheryl Coolidge

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