MASS OF MAGNESIUM = 0.324g ± 0.002g
- When that crucible and its cover had cooled, we folded the magnesium ribbon and placed it in the crucible. Before we heated the crucible again, we recorded the mass of the crucible, the lid and the magnesium together.
COMBINED MASS OF THE OBJECTS BEFORE = 33.279g ±0.002g
- Then we heated up the crucible with magnesium. It took very long time until the magnesium began to ignite, but when it did, it reacted very rapidly. While the magnesium was reacting, we lifted the lid several times for approximately 2 seconds. We did this to let inn oxygen into the crucible so that it could react with the magnesium. While doing this, we saw smoke rising from the crucible and thus closed the crucible with the lid to prevent too much lose of the smoke.
- Finally, when the magnesium that finished reacting, we let it cool. When it had cooled, we recorded the mass of crucible with magnesium oxide and its lid.
COMBINED MASS OF THE OBJECTS AFTER = 33.450g ±0.002g
Result
Now I have every mass that I need to find the empirical formula for magnesium oxide, except the mass of oxygen. By subtracting the mass that the object before the reaction from what it had after magnesium was burned, we find the mass of oxygen.
COMBINED MASS OF THE OBJECTS BEFORE - COMBINED MASS OF THE OBJECTS AFTER = MASS OF OXYGEN
(33.450g – 33.279g) ± (0.002g + 0.002g) = 0.171g ±0.004g
A summary of all the mass is listed in the chart below.
By finding the amount of moles of these two atoms, we can determine the empirical formula for magnesium oxide. But first we have to find the percentage of the uncertainty, so that we can find the final uncertainty.
Relative uncertainty in O = 0.004 *100 = 2.339 %
0.171
Relative uncertainty in Mg = 0.002 *100 = 0.617 %
0.324
Then we take the mass of oxygen and magnesium and divide it with its molar mass to find it in moles. We know their molar mass and know that the molar mass has 0% uncertainty.
O: 0.171g ± 2.339% = 0.0106 mol ± 2.339 %
16.00g mol-1 ± 0.0 %
Mg: 0.324g ±0.617% = 0.0133 mol ± 0.617%
24.31g mol-1 ±0.0%
Now that the moles of each element are known, a stoichiometric comparison between the elements can be made to determine the empirical formula. This is achieved by dividing each of the mole quantities by which ever mole quantity is the smallest number of moles. Here the mole of oxygen is the smallest.
0.0106 mol O ± 2.339 % = 1O ±4.678%
0.0106 mol ±2.339%
0.0133 mol Mg ±0.617% = 1.2547 ≈ 1 Mg ± 2.956%
0.0106 mol ± 2.339
The ratio of oxygen atoms to magnesium in magnesium oxide is 1:1
Conclusion
The empirical formula for magnesium oxide is “Mgo”; containing one atom of magnesium and one atom of oxygen. However, we did not got the ratio exact 1:1
in experiments like this, when we ignited the magnesium to react with oxygen, we will not get the same amount of moles for both of them. In the air, heated magnesium reacts not only with oxygen, but also with nitrogen.
Errors
The mole of magnesium and oxygen was not the same. I found out that the mass of oxygen was smaller then it was supposed to be. This error can have occurred by three things:
1. We did not found out the empirical formula for magnesium oxide properly; because magnesium also reacted with nitrogen. It takes three magnesium atoms, while it takes only two nitrogen atoms to form a compound. Nitrogen is also lighter then oxygen; thus our combined mass after the reaction became smaller then if the magnesium had totally reacted with oxygen.
2. When we lifted the lit while magnesium ribbon was reacting, we let out smoke. That was magnesium oxide in vapour.
3. There is a possibility that maybe not all of the magnesium reacted.
Improvements:
If we had carried the experiment a little more, we could have removed the magnesium nitride. By adding some drops of water and then heating it up, we will get this formula:
Mg3N2 + 6H2O → 3Mg(OH)2 + 2NH3
The result will be ammonia gas, which will depart when heated.
