Question

 

Mg₃N₂ + 3H₂O --> 3MgO + 2NH₃

	Trial 1	Trial 2
Mass of Mg in g ±0.0001 g	0.3059g	0.2846g
Mass of crucible and lid + residue in g ±0.0001 g	28.9329g	29.6265g
Mass of residue in g ±0.0001 g	0.4788g	0.5016g
Mass of used oxygen in g ±0.0001 g	0.1729g	0.2170g
Moles of Mg	0.012586	0.011710
Moles of O	0.010806	0.013562
Normalized moles of Mg	1.1647 mol of Mg	1 mol of Mg
Normalized moles of O	1 mol of O	1.16 mol of O
Empirical formula	MgO	MgO

How do the theoretical and experimental empirical formula can be compare?

How an incomplete conversion of Mg₃N₂ affect the results

From the result, is this a plausible method to determine the formula of metal oxides?

Answer 1

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1)The theoretical formula of MgO is that in which molar ratio of Mg and O is 1:1 but in experimental empirical formula the molar ratio of Mg and O is not exactly 1:1. It differs. For trial 1 the molar ratio of Mg and O is 1.1647:1 and for trial 2 the molar ratio of Mg and O is 1:1.16. For trial 1, the product is slightly magnesium-rich the ratio of Mg: O is greater than the 1-to-1 expected. For trial 2, the product is slightly magnesium-poor product would have a ratio of Mg: O that is less than the 1-to-1 expected.

2) Magnesium reacts vigorously when heated in the presence of air. Although there is a higher percentage of N₂ gas in the atmosphere than O₂ but oxygen is more reactive than N₂. So, more magnesium oxide is formed. The small amount of nitride that forms can be removed with the addition of water, which converts the nitride to magnesium hydroxide and ammonia gas. Heating Mg(OH)₂ causes the loss of water and conversion of the hydroxide to the oxide.

Mg₃N₂ + 3H₂O --> 3MgO + 2NH₃

Mg₃N₂(s) + H₂O (l) ---> Mg(OH)₂ + NH₃(g)

Mg(OH)₂ heat ----->MgO + H₂O(l)

3) Yes, this is a plausible method to determine the formula of metal oxides