Empirical Formula Lab Report

606 Words3 Pages
Empirical Formula: MgO7 The results shows the masses and the percent composition of the products theoretically and experimentally. The mass of magnesium oxide was calculated, when the mass of crucible, lid, and magnesium oxide was subtracted from the mass of clean, empty crucible and lid. The percent composition theoretically was found, when the mass of magnesium was divided by the mass of magnesium oxide and later was subtracted by 100 percent to get the percent composition of oxygen. Theoretically, the molar mass of magnesium must be divided by the molar mass of magnesium oxide, and also later subtract by 100 percent to find the percent composition of oxygen. Discussions These results indicate…show more content…
Using the masses of Mg and MgO that were calculated from the experiment, it was determined that the percent composition of Mg in MgO is 17 %, and the percent error was 72 percent. The experiment that was conducted by carefully applying heat and oxygen to a strip of magnesium, it was shown that magnesium oxide was created with magnesium and oxygen that has a ratio of 1:7. During the experiment, the calculations confirm that magnesium added to oxygen became Mg. The calculations shown in appendix, show that that the percent of magnesium in MgO was 17 %. This did not approximately match the 60.304% magnesium that should be realistic as along as the formula is MgO. The percent error was 72 percent for magnesium. This was likely caused by an avoidable human error. Sources of Error: One of the major causes for such a high percent error was that MgO gas was lost through the opening in the top of the crucible. This, simply enough, would explain lost product. Another reason, would be simply human error. One could have simply read the measuring scale wrong, or has setup the lab incorrectly. There are other minor sources of error that could have contributed to a high percent
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