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Determination of the empirical formula of magnesium oxide exemplar

Determination of the empirical formula of magnesium oxide exemplar. The purpose of this experiment is to experimentally determine the percent composition and empirical formula of magnesium oxide.

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Determination of the empirical formula of magnesium oxide exemplar

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  1. Determination of the empirical formula of magnesium oxide exemplar

  2. The purpose of this experiment is to experimentally determine the percent composition and empirical formula of magnesium oxide. Magnesium will be heated in the presence of air to produce magnesium oxide and magnesium nitride. Water will be added to react with the magnesium nitride, producing more magnesium oxide: (Write the balanced chemical equations for the following magnesium + oxygen  magnesium oxide magnesium + nitrogen  magnesium nitride magnesium nitride + water  magnesium oxide + ammonia (NH3)

  3. Theoretical calculations: The empirical formula of magnesium oxide is MgO. The ratio of magnesium atoms to oxygen atoms is expected to be 1 to 1. The molar mass of MgO = 24.305 + 15.999 = 40.304 g/mole Percent of magnesium in magnesium oxide: Mass Mg/mass MgO = 24.305/40.304 x 100 =60.304% Percent of oxygen in magnesium oxide: Mass O/mass MgO = 15.999/40.304 x 100 = 39.696%

  4. Materials and Procedure Perform the method as written on the instruction sheet

  5. Data collection Masses of the crucible and cover, magnesium and final product. +-0.01g

  6. Observations • Blah….. Blah….

  7. Data Processing: masses of substances Masses of the magnesium, oxygen and final product +-0.02 grams

  8. Data processing empirical formula: Moles of magnesium = mass magnesium molar mass of Mg Moles Mg = 0.40 g/24.305 = 0.016 moles Mg Moles of oxygen = mass oxygen/molar mass of O Moles oxygen = 0.17 g/15.999 = 0.011 moles O Ratio of Mg to O 0.016 moles Mg/0.010 moles 0 = 1.5 Mg to 1.0 O Empirical formula: Mg3O2

  9. Data processing Percent composition: Percent of magnesium in magnesium oxide: Mass Mg/mass MgO = 0.40/0.57 x 100 = 70% Percent of oxygen in magnesium oxide: Mass O/mass MgO = 0.17/0.57 x 100 = 30%

  10. Data processing: percent error Percent error = (experimental – theoretical) x 100 theoretical value Percent error of the percent magnesium: (70– 60.304) x 100 = 60.304 16% error ( too high) Percent error of the percent of oxygen: (39.696 – 30) x 100 = 39.696 24% error (too low)

  11. Data presentation Data presentation: percentages of magnesium and oxygen and empirical formula

  12. Conclusion In this experiment, magnesium ribbon was heated in a crucible in the presence of air to produce magnesium oxide. The theoretical percentages of magnesium and oxygen were expected to be 60.304% and 39.696%. Experimentally, the percent of magnesium was found to be 70%, which is higher than expected with a percent error of 16%. The percent of oxygen was found to be 30% which is higher than the theoretical value with a percent error of 24%. The empirical formula for magnesium oxide is MgO. Experimentally, it was determined to be Mg3O2. This result agrees with the percent composition values as the amount of magnesium was higher than the theoretical value. The experimental formula for magnesium oxide is does not agree what is theoretically expected. Magnesium is an alkaline earth metal and has a charge of +2. Oxide has a charge of -2.

  13. Evaluation Although the magnesium was carefully heated, the experimental percentage of magnesium was too high and the value for oxygen was too low. The following errors occurred. The first is that the magnesium was not heated long enough or hot enough. In the observations… blah….blah… Weaknesses… Limitations.. Improve each…

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