Magnesium buring in a crucible experiment question?

Question

We did an experiment with magnesium verifying the law of conservation of mass. we weighed the crucible, then the crucible and the magnesium together, then a final time after heating the magnesium. We had to answer questions at the end and this one has me stumped. She asks:

If the formation of magnesium nitride is not taken into account and the gray residue after the initial heating is assumed to be wholly magnesium oxide, the experiment would take much less time. Explain what errors might result, i.e., would the mass of the MgO/ Mg3N2 residue be greater or less than the mass of pure MgO residue? How would this affect the final percentage yield?

my thoughts on this is that the residue would actually be less than the mass of pure MgO residue. the reasoning behind that is the weight of the MgO compared to the weight of the Mg3N2. if i had .5g of MgO residue it would be .0124mol. Compared to if i had 1/2 MgO and Mg3N2 i would have .00867mol of combined. what do you guys think?

Answer 1

Doing the sums, 72g of Mg gives 100g of Mg3N2, a 28% increase in mass.
24g of Mg gives 40g of MgO, a 40% increase in mass. This difference is due to O atoms being heavier than N atoms, of course. But it does help you to answer your question - the residue would be lighter than it should be.
Incidentally, you can tell in practice whether you've got any nitride there by adding water. If you smell ammonia, you did!