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Lab 9

Lab 9. Empirical Formula Determination. 19.33 g. 19.57 g. 19.57 g -. 19.33 g. = 0.24 g Mg. 1 mol Mg. 0.24 g Mg x. 0.0099 mol Mg. = 0.00987 mol Mg =. 24.305 g Mg. 19.33 g. 19.57 g. 19.71 g. 19.71 g -. 19.33 g. = 0.38 g Mg x O y. Answer to #1. Answer to #3.

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Lab 9

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  1. Lab 9 Empirical Formula Determination

  2. 19.33 g 19.57 g 19.57 g - 19.33 g = 0.24 g Mg 1 mol Mg 0.24 g Mg x 0.0099 mol Mg = 0.00987 mol Mg = 24.305 g Mg

  3. 19.33 g 19.57 g 19.71 g 19.71 g- 19.33 g = 0.38 g MgxOy Answer to #1 Answer to #3 0.38 g MgxOy – 0.24 g Mg = 0.14 g O Two methods 19.71 g – 19.57 g = 0.14 g O 1 mol O 0.14 g O x 0.0088 mol O = 0.00875 mol O = 15.9994 g O

  4. 19.57 g - 19.33 g = 0.24 g Mg 1 mol Mg 0.24 g Mg x 0.0099 mol Mg = 0.00987 mol Mg = 24.305 g Mg 19.71 g- 19.33 g = 0.38 g MgxOy 0.38 g MgxOy – 0.24 g Mg = 0.14 g O Two methods 19.71 g – 19.57 g = 0.14 g O 1 mol O 0.14 g Mg x 0.0088 mol O = 0.00875 mol O = 15.9994 g O

  5. 19.57 g - 19.33 g = 0.24 g Mg 1 mol Mg 0.24 g Mg x 0.0099 mol Mg = 0.00987 mol Mg = 24.305 g Mg 19.71 g- 19.33 g = 0.38 g MgxOy 0.38 g MgxOy – 0.24 g Mg = 0.14 g O Two methods 19.71 g – 19.57 g = 0.14 g O 1 mol O 0.14 g Mg x 0.0088 mol O = 0.00875 mol O = 15.9994 g O

  6. 1 mol Mg 0.24 g Mg x 0.0099 mol Mg = 0.00987 mol Mg = 24.305 g Mg 1 mol O 0.14 g Mg x 0.0088 mol O = 0.00875 mol O = 15.9994 g O Whole # ratio 0.0099 mol Mg = 1.125 = 1.1 1:1 0.0088 mol O 1 1

  7. Whole # ratio 0.0099 mol Mg = 1.125 = 1.1 1:1 0.0088 mol O 1 1 1:1 ratio means the empirical formula is: Mg1O1 or just MgO + 0.1 1 - 1.1 - 0.1 x 100 = x 100 = 10 % % error = x 100 = 1 1 1

  8. 1 mol Mg 0.24 g Mg x 0.0099 mol Mg = 0.00987 mol Mg = 24.305 g Mg 1 mol O 0.14 g Mg x 0.0088 mol O = 0.00875 mol O = 15.9994 g O Since moles of Mg were higher than moles of O it would seem that the group could have the following errors that would support this error: 1. The sample was not heated long enough. This would result in not all of the Mg completely reacting leading to less oxygen in the MgO combustion product. 2. The sample lost some mass when poked with the spatula. This would show the final mass lower than normal and since oxygen was calculated by subtracting Mg from final mass the O would have been less. 3. The sample lost some mass when smoke escaped. All error examples represent a ratio greater than 1:1.

  9. If moles of Mg is lower than moles of O it would seem that the group could have the following errors: 1. The crucible was still wet when it was weighed. The water would have evaporated affecting the masses recorded later. 2. The magnesium strip may have not been cleaned well or had a coating of oxide already on the strip. This would make the original mass of Mg incorrect due to the added O. All error examples represent a ratio less than 1:1.

  10. See Excel spreadsheet on http://kitzchem.wikispaces.com Day 36 for class results and form your response based on this data.

  11. One molecule of C2H6 contains 2 atoms of C and 6 atoms of H. One mole of C2H6 contains 2 moles of C and 6 moles of H.

  12. To answer this question, think about what else is in the air that could react with magnesium.

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