calculating percentage error in experiment- help!?

Question

We did a lab where we burned magnesium (to create magnesium oxide), then added water and heated again in a crucible with the cover ajar to release any Magnesium Nitride created.

The question on the lab report is this: if water had not been added to your initial product, a mixture of MgO and Magnesium Nitride (Mg3N2?) what error in the percentage magnesium determined would have resulted?

I will greatly appreciate any help on this! :)

Answer 1

I find 16.4%
Here we go:
1) As air is a mixture of O2 and N2 two reactions are taking place:
2 Mg + O2 -> 2 MgO
3 Mg + N2 -> Mg3N2
There are 4 parts of N2 for one part of O2, so the second reaction takes place four times as frequently as the first. Effectively we have
14 Mg + O2 + 4 N2 -> 2 MgO + 4 Mg3N2.

2) The molar mass of MgO is 40.3 gram. In MgO the magnesium makes up 24.3/(24.3+16.0) = 60.3 % of the mass.
The molar mass of Mg3N2 is 100.9 gram, the magnesium makes up 3*24.3/(3*24.3 + 2*14.0) = 72.2 % of the mass.

3) Assume you have 1000 gram of reaction product. If it is pure MgO you will find 603 gram of Mg. But in the mixture you will have
two moles of Mg2N3 (201.8 g) for every mole of MgO (40.3 gram).
In 1000 gram of mixture, 833.5 gram of Mg2N3 and 166.5 gram of MgO. This gives 72.2% * 833.5 + 60.3% * 166.5 = 702 gram of Mg.

4) Finally: the error is (702-603)/603 = 16.4%