Question b5c9b

Dec 20, 2015

Here's what I got.

Explanation:

So, you're doing a lab on determining the empirical formula of magnesium oxide, $\text{MgO}$, by heating a strip of magnesium metal in the presence of air.

As you know, nitrogen gas, ${\text{N}}_{2}$, is far more abundant than oxygen gas, ${\text{O}}_{2}$, in air, which means that some of the magnesium will react with nitrogen gas to form magnesium nitride, ${\text{Mg"_3"N}}_{2}$.

Despite the fact that nitrogen is present in a greater proportion, is is not as reactive as oxygen, so you can expect to see much more magnesium oxide formed upon heating.

Now, some of the magnesium present in the initial sample will now be a part of the magnesium nitride. To convert the magnesium nitride to magnesium oxide, you first need to add water.

Magnesium nitride will react with water to form magnesium hydroxide, "Mg"("OH")_2#, and ammonia, ${\text{NH}}_{3}$

${\text{Mg"_3"N"_text(2(s]) + 6"H"_2"O"_text((l]) -> 3"Mg"("OH")_text(2(s]) + 2"NH}}_{\textrm{3 \left(g\right]}} \uparrow$

Heating the magnesium hydroxide will convert it to magnesium oxide

${\text{Mg"("OH")_text(2(s]) -> "MgO"_text((s]) + "H"_2"O}}_{\textrm{\left(g\right]}}$

You know that not all magnesium nitride was converted back to magnesium oxide.

This means that the crucible will contain magnesium oxide and some of the water you added to convert the magnesium nitride to magnesium hydroxide and ammonia.

The total mass of the oxide will come you to be bigger than what it actually is.

Simply put, the ratio between magnesium and oxygen in the oxide will appear to be smaller than in reality, since you will conclude that for the initial sample of magnesium you have more oxygen in the oxide than you actually have.

Therefore, the ratio between magnesium and oxygen will come out to be too low.