Finding the Empirical Formula for Magnesium Oxide

Finding the Empirical Formula for Magnesium Oxide

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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.

[IMAGE]

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
--------------------------------

[IMAGE]


Investigation 2:

Magnesium = 36.10 - 36.03

= 0.07g

Oxygen = 36.14 - 36.10

= 0.04g


Magnesium Ar = 24 Oxygen Ar = 16
--------------------------------

[IMAGE]

Formula:


Magnesium 42 : 36 Oxygen [ -- 3
-------------------------------

Magnesium 14 : 12 Oxygen [ -- 2

Magnesium 7: 6 Oxygen

= Mg 0


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
--------------------------------

Moles:

Magnesium. Mass Oxygen. Mass

Ar Ar

= 0.07 = 0.07

24 16

= 0.002916 = 0.004375



[IMAGE]
=======


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.

Average of Investigation 1, 2, and 3

Magnesium mean = (0.04 + 0.07 + 0.07)

3

= 0.06g

Oxygen mean = (0.03 + 0.04 + 0.07)

3

= 0.047 (2 s.f.)


Magnesium Ar = 24 Oxygen Ar = 16

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Related Searches

--------------------------------

Moles:

Magnesium. Mass Oxygen. Mass

Ar Ar

= 0.06 = 0.047

24 16

= 0.0025 = 0.0029375



[IMAGE]
=======


Conclusion

Looking at the formula for the average mass of Magnesium and Oxygen, I
can see that a number of moles of magnesium will react with the same
number of moles of oxygen to form magnesium oxide. This means that 1
atom of magnesium will react with 1 atom of oxygen. This means that
the empirical formula of magnesium oxide is MgO.

Looking at the periodic table, I can see how Magnesium and oxygen
would bond during a reaction.

Atoms bond with each other to be stable, which involves gaining a full
outer shell. Magnesium is a metal, and oxygen is a non-metal (a gas),
so therefore the type of bonding occurring would be ionic bonding. In
this process, electrons are transferred

from the metal atom (magnesium) to the non-metal atom (oxygen). The
atoms "want to" get a full outer shell, and as most atoms don't have a
full outer orbital, electrons had to be transferred.

Magnesium is in the second group of the Periodic Table, meaning that
it would hold two electrons in it's outer orbital, it's structure
pattern is 2-8-2. It "wants to" give away 2 electrons to be able to
have a full outer shell and to gain the structure of a noble gas. The
2 electrons that the magnesium atom "wants to" get rid of are
transferred onto the oxygen atom. 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.

[IMAGE]

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.

Evaluation

The scientific information in the conclusion part of this experiment
supports my average of the results, which say the formula for
magnesium oxide is MgO.

I decided to take the overall formula from my experiment as Mg O,
rather than Mg O from the first experiment, Mg O from the second
experiment, or

Mg O in the final experiment. This is because for two of the
experiments, there was one more oxygen atom to react with magnesium,
and for one of the experiments, there was one more magnesium atom to
react with oxygen, so from this, I found the average, it being that
the number of atoms was equal.

When working out the number of moles, the ratio, and the formula for
each test, I worked as accurately as possible, not rounding off any
decimal places right up until the very end of the calculation, to
achieve the most accurate results possible and to ensure that accuracy
was attained throughout the calculation. The repeat readings show that
for each experiment, the number of magnesium and oxygen atoms reacting
was very close, with only a difference of one atom for each test. My
experiment was also accurate in the sense that when the crucible was
measured, a top pan balance of an accuracy of one hundredth of a gram
was used.

My final result displaying the empirical formula for Magnesium Oxide
(being MgO) seems to be reliable; because scientific evidence talked
about in the conclusion supports my results, saying that for each
magnesium atom, one atom of oxygen will react with it to form
magnesium oxide.

During the experiment, when the lid had to be lifted to allow air into
the crucible (for the oxygen in the air to react with the magnesium in
the crucible), some magnesium oxide that had already reacted may well
have been lost when the lid was lifted. 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.
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