Assume that some of the magnesium nitride is not completely converted to

magnesium hydroxide and subsequently to magnesium oxide. Will your empirical
formula be too high in magnesium or too high in oxygen? Please show any
calculations you used to determine your answer.
Where would you even start to answer this?

2Mg + O2 ==> 2MgO

3Mg + N2 ==> Mg3N2
Mg3N2 + 6HOH ==> 3Mg(OH)2 + 2NH3

I would make up some convenient numbers (such as 24.3g Mg) and calculate how much MgO was produced and how much O in the MgO, then calculate the formula from that. Then go through the same calculation but change the O and see how that affects the formula.

How would you calculate how much MgO was produced? Would you just use the first reaction 2Mg+O2 --> 2MgO or would you have to put the amount of Mg through the other reactions with the mole to mole ratios and then find the amount of MgO produced?

To answer this question, we need to understand the concept of empirical formula and the reaction involved. Let's break it down step by step:

1. Understanding empirical formula: The empirical formula represents the simplest whole number ratio of the elements in a compound. It does not provide information about the actual number of atoms present in the compound, but rather the relative proportions of the elements.

2. Reaction involved:
Magnesium nitride (Mg3N2) + Water (H2O) → Magnesium hydroxide (Mg(OH)2) + Ammonia (NH3)

Mg3N2 + 6H2O → 3Mg(OH)2 + 2NH3

Magnesium hydroxide (Mg(OH)2) → Magnesium oxide (MgO) + Water (H2O)

Mg(OH)2 → MgO + H2O

3. Calculation approach:
Let's assume we start with 100 g of magnesium nitride (Mg3N2) and that some of it is not completely converted to magnesium hydroxide and subsequently to magnesium oxide. We'll use these numbers for ease of calculation, but the approach can be applied to any amount given.

a) Determine the moles of magnesium nitride (Mg3N2) present:
- Molar mass of Mg3N2 = (3 * atomic mass of Mg) + (2 * atomic mass of N)
= (3 * 24.31 g/mol) + (2 * 14.01 g/mol)
= 72.93 g/mol + 28.02 g/mol
= 100.95 g/mol

- Moles of Mg3N2 = mass of Mg3N2 / molar mass of Mg3N2
= 100 g / 100.95 g/mol
≈ 0.990 mol

b) Determine the theoretical moles of magnesium hydroxide (Mg(OH)2) produced:
- From the balanced equation, we know that the mole ratio between Mg3N2 and Mg(OH)2 is 1:3, respectively.
- Therefore, the theoretical moles of Mg(OH)2 = (0.990 mol of Mg3N2) * 3
= 2.970 mol

c) Calculate the moles of magnesium oxide (MgO) expected:
- From the balanced equation, we know that the mole ratio between Mg(OH)2 and MgO is 1:1, respectively.
- Therefore, the theoretical moles of MgO = 2.970 mol of Mg(OH)2

d) Compare the theoretical moles of MgO with the actual moles obtained:
- If the actual moles of MgO obtained are less than 2.970 mol, it suggests that some of the magnesium nitride was not completely converted, leading to a higher proportion of magnesium in the empirical formula.
- If the actual moles of MgO obtained are equal to or more than 2.970 mol, it implies that all the magnesium nitride was converted, and the empirical formula will have a correct proportion of magnesium and oxygen.

Now, perform the experiment and determine the empirical formula to check whether it is too high in magnesium or oxygen, based on the actual results obtained.