Lab Report 5
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Chemistry
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Apr 3, 2024
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CHEM 2115 Magnesium Oxide Lab Report Experiment
#5 Chem I Lab
Name Alexis Cervenak
Section# 008 Station# B-5 Date 2.25.24
1.
Data and calculations: Attach sample calculations of each type necessary to obtain the results in the Calculations section of the table. These should include the equations used, substituted values, appropriate units, and results. Trial 1 Trial 2 Data
Mass of crucible and lid/g 42.045 42.061 Mass of crucible, lid, and Mg/g 42.265 42.269 Mass of crucible, lid, and product/g
42.403 42.414 Calculations
Mass of Mg/
g
0.220 0.208 Mass of O/
g
0.138 0.145 Chemical amount of Mg/mol 9.05 x 10
-3 8.56 x 10
-3 Chemical amount of O/mol 8.62 x 10
-3 9.06 x 10
-3 Mg/O mole ratio/mol 0.00905/ 0.00862 0.00856/ 0.00906 Average Mg/O mole ratio/
mol
0.00881 / 0.00884 Empirical formula of product Mg
2
O
2 2. Data Analysis: Class data for this experiment is available on Blackboard. Use Excel to calculate each item in the Calculations section of the table above for each trial in the class data file, to calculate the average of the chemical amount of Mg, the chemical amount of O, and the Mg/O mole ratio. Use the average Mg/O mole ratio to find an empirical formula for the magnesium oxide product based on the class average Mg/O mole ratio. Average Values Chemical amount of Mg/mol 1.00 x 10
-2 Chemical amount of O/mol 9.80 x 10
-3 Class average Mg/O mole ratio/mol 0.01007 / 0.00980 Class average empirical formula of product Mg
2
O
2
The most common ionic forms of magnesium and oxygen are Mg
2+
and O
2−
. Based on the formulas of these ions, the predicted Mg/O mole ratio is 1:1 and the predicted empirical formula is MgO. Calculate the relative error of the class average Mg/O mole ratio. Relative error is defined on page IV-3 of the lab manual. Discuss how well the class average result agrees with the prediction. Relative error ranges of −2% to 2%, −5% to 5%, and −10% to 10% signify excellent, very good, and good agreement between the class average and predicted results. My Mg/O mole ratio: 0.00881 / 0.00884 = 0.99661 Class average Mg/O mole ratio: 0.01007 / 0.00980 = 1.0276 *I divided the ratio to make the calculations easier [ (0.99661 –
1.0276) / (1.0276) ] x 100% = -3.016 The relative error is -3.016, since the error is not between −2% to 2%, −5% to 5%, and −10% to 10%
. This does not signify a good agreement between the class average and predicted results. A negative relative error means that my value was higher than the class average. The class results do agree with the prediction because of the increase in values throughout the class data the average is raised, compared to my data. 3. Questions A. Suppose that the magnesium is not polished at the start of the experiment. Would the reported Mg/O mole ratio be too high, too low, or unaffected owing to this procedural error? If the Magnesium was not polished at the start of the experiment the reported Mg/O mole ratio would be too high. Magnesium is polished to remove any oxide that has formed on the surface of the Magnesium. If the Magnesium is not polished the oxide would remain and ultimately raise the amount of Oxygen in the Mg/O mol ratio. B.
Suppose the crucible tips over and some of the product is lost before the mass of the crucible, lid, and product are determined. Would the reported Mg/O mole ratio be too high, too low, or unaffected by this mishap? If the crucible was to tip over and some of the product was lost the reported Mg/O mole ratio would be too low. This would cause the Mg/O mole ratio to be too low because to calculate the mass of O you need to have the full mass of the crucible, lid, and product and mass of Magnesium. If some of the product is lost before mass is determined the mass would not be full and correct.
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