Post Lab Number Five Empirical Formula of an Oxide PDF

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Nhi Chung General Chemistry I - Chem 1411, HCC 31 October, 2017

Post Lab n. 5 Empirical Formula of an Oxide INTRODUCTION The purpose of this lab is to understand how empirical and molecular formulas of compounds work when experimenting with a magnesium ribbon (Mg). As for the concept, it is important to recognize and interpret how this chemical formula affect its mass and moles. When calculating the magnesium ribbon's moles, it is very important to use the empirical and molecular formulas: molecular weight of magnesium/ empirical formula weight of magnesium. Furthermore, the data are obtained by dividing the mass of the magnesium ribbon, which would give its ratios.

EXPERIMENTAL This experiment is done by obtaining a clean, dry crucible and cover, then weigh them on a digital balance scaler. Next, obtain and place about a foot-long magnesium ribbon in the crucible and weigh it. After that, place the crucible in a clay triangle with the ring stand and make sure that they are in the appropriate adjustment when using the Bunsen burner flame. More importantly, heat the crucible about ten minutes until its color changes to red, which allows the reaction to flow constantly with oxygen. When this is done, it is important to allow the crucible to cool off for about a minute or more and then use a tong to gently remove it on a paper towel. The crucible should be weigh on a digital balance scaler to see if there are any changes in its mass, which should end up some mass increases for trial one. Furthermore, the magnesium nitride

(Mg3N2) will be converted to oxide by dropping at least ten drops of deionized water in the crucible if necessary. Again, repeat the process of weighing and heating the crucible, but this time it should take about no more than three minutes. Therefore, it would be possible to obtain a constant mass when the crucible is dried. Thus, the experiment is completed, record the mass of the crucible in the lab report sheet, and carefully empty the rest of its remaining contents in the waste container, and clean it thoroughly.

DATA Table 1 1. 2. 3. 4. 5. 6. 7.

Mass of empty crucible + cover Mass of crucible, cover, and magnesium Mass of magnesium Mass of crucible, cover, and magnesium oxide (after heating and cooling) Mass of magnesium oxide Mass of oxygen in magnesium oxide Moles of magnesium (show your calculations)

45.32g 45.57g 0.25g Trial 1 45.69g 0.37g 0.07g 0.0152 mol.

Mg: 0.37g/ 24.30g/ mol. = 0.0152 mol. Mg 8. Moles of oxygen (show your calculations)

0.0043 mol.

O: 0.07g/ 16g/ mol. = 0.0043 mol. O 9. Simplest (empirical) formula of the oxide of magnesium (show your

Mg2O

calculations) Mg: 0.0152 mol./ 0.0152 mol. = 1 1x2=2 O: 0.0043 mol./ 0.0152 mol. = 0.28 0.28 x 2 = 0.56 10. 2Mg + O2 —> 2MgO

RESULTS AND DISCUSSION

The determination of the empirical formula is done by dividing the mass of each element by the corresponding molecular mass. Magnesium is measure directly at the beak of the experiment while oxygen is measure as the difference of the mass of magnesium oxide and magnesium. The data is shown on table one should revealed that the molecular ratio of mass one to one as predicted by the oxygen number of magnesium and oxide. Thus, the result should have a constant mass when calculating and finding the empirical formula of magnesium and oxygen. The purpose of the experiment is to calculate and find the appropriate mass of the magnesium and magnesium oxide after the heating process. Moreover, the mass of the magnesium oxide in the crucible plays a significant role because its mass could affect the ratio and molecular. In other words, it is very important not to neglect how long should the magnesium and magnesium oxide be heated and stopped by judging its color. The outcome is somewhat unexpected because somehow the mass of the magnesium oxide did not raise enough, which is something has to do with its oxygen.

POST LAB QUESTIONS 1. If I add water rapidly to my product in a hot crucible, then it would have reacted greatly and somewhat out of control. 2. A) Mg3N2(s) + 6H2O(l) —> 3Mg(OH)2(s) + 2NH3(g) B) Mg(OH)2(s) —> MgO(s) + H2O(g) 3. The Mg to O ratio would have been affected if I had failed to convert all of the Mg3N2 present to magnesium oxide by having the wrong molecular, atoms, and empirical formulas. 4. moxide = mO + mFe mO = 14.3g - 11.15g = 3.15g O

11.15g x 1 mol. Fe/ 55.84g = 0.2059 mol. Fe 3.15g x 1 mol. O/ 16 g = 0.1968 mol. O Fe: 0.2059 mol./ 0.2059 mol. = 1 1x2=2 O: 0.1968 mol./ 0.2059 mol. = 0.9558 0.9558 x 2 = 1.9116 (Fe1O0.9558)2 = Fe2O2 is the simplest formula of the compound. 5. 1.32g - (0.288g O - 0.974g Ag) = 0.110g C C: 0.110g C x 1 mol. C/ 12.01g C = 0.00915 mol. C O: 0.288g O x 1 mol. O/ 16g O = 0.0180 mol. O Ag: 0.974g Ag x 1 mol. Ag/ 107.86g Ag = 0.00903 mol. Ag C: 0.00915 mol./ 0.00915 mol. = 1 or 1.01 O: 0.0180 mol./ 0.00903 mol. = 1.99 or 2 Ag: 0.00903 mol./ 0.00903 mol. = 1.00 The empirical is AgCO2. (1 x 107.86g/ mol. Ag) + (1 x 12.0107g/ mol. C) + (2 x 16g/ mol. O) = 151.9g and the empirical is Ag2C2O4. 303.8g/ 151.9g = 2 and the molecular formula is Ag2C2O4....