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Finding the Empirical Formula for Magnesium Oxide

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Introduction

Finding the Empirical Formula for Magnesium Oxide The Results In order to work out the ratio for magnesium and oxygen, I will have to calculate the amount of magnesium and oxygen used. From these results, and knowing that the Ar for Magnesium is 24, and the Ar for Oxygen is 16, I can find the number of moles for Magnesium and Oxygen. Investigation 1: Magnesium = 36.08 - 36.04 = 0.04g Oxygen = 36.11 - 36.08 = 0.03g Magnesium Ar = 24 Oxygen Ar = 16 Investigation 2: Magnesium = 36.10 - 36.03 = 0.07g Oxygen = 36.14 - 36.10 = 0.04g Magnesium Ar = 24 Oxygen Ar = 16 To ensure that my final answer was as accurate as possible, I did not round any numbers until the end to get the most accurate final equation. Likewise for test 1, 3, and the average. Investigation 3: Magnesium = 36.10 - 36.03 = 0.07g Oxygen = 36.19 - 36.12 = 0.07g Magnesium Ar = 24 Oxygen Ar = 16 To ensure that my final answer was as accurate as possible, I did not round any decimals off until the end to maintain accuracy throughout the calculation. ...read more.

Middle

Oxygen is in the sixth group of the periodic table, and therefore has 6 electrons in it's outer shell, so needs to gain 2 electrons to achieve a full outer shell, as shown in the following diagram. So when the two of these elements react with each other, magnesium oxide is formed. In this, there are magnesium ions, which have a positive charge x 2, and negative oxide ions, (double negative charge). This shows that for each magnesium atom there is, one oxygen atom will react with it, as only one magnesium atom is needed to react with one oxygen atom for the outer shell of each to become full; two magnesium atoms are not necessary; the diagram clearly shows for each magnesium atom, one oxygen atom will react with it. The two electrons on the outer shell of the magnesium atom are transferred over to the gap of the outer shell of the oxygen atom, leaving both atoms with a full outer shell. Therefore, the formula for Magnesium Oxide is Mg O. ...read more.

Conclusion

To avoid this, the crucible with the magnesium oxide inside should have been reheated and then weighed again after the final weighing, until a constant mass could be found. It would be very difficult to perform this experiment accurately for tests 1, 2, and 3. Doing the experiment again, I would try and keep this part more accurate, but it is very difficult to maintain this level of accuracy. To further my experiment, perhaps I would experiment with different elements, e.g. use another element in group 2 of the periodic table with oxygen, to see if the formula would involve the same number of atoms, e.g. for Beryllium Oxide, I would expect the equation to be BeO, as magnesium and beryllium have a similar atom structure. I could also react magnesium with another element in group 6, such as sulphur. For this, I would expect the formula to be Mg S, because Sulphur has a similar atomic structure to oxygen, both holding 6 electrons in their outer orbital. Joanna Sedgwick 1 ...read more.

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