Question

I did a Magnesium Oxide Lab.Magnesium often reacts with nitrogen in the air. If some of the magnesium had reacted with the nitrogen, would it make the % composition of ur sample higher or lower than it should be? Explain (Hint-Try a sample calculation).

Answer 1

Okay, so you've performed the classic magnesium oxide lab. I assume that you have your data, so I'll just walk you through this from a theoretical standpoint.

So, you heat magnesium in the presence of air according to this reaction

`2Mg_((s)) + O_(2(g)) -> 2MgO_((s))`

If you let some smoke escape the crucible, the mass of magnesium oxide you'll measure at the end of the experiment will be lower than what was actually produced,

Since the mass of magnesium is recorded before the reaction, the percent composition of magnesium in the final product will be higher.

Moreover, some of the magnesium can react with the nitrogen present in the air to form magnesium nitride, `Mg_3N_2`.

However, this is usually not a very big problem because you can react this product with water to get magnesium hydroxide and ammonia

`Mg_3N_(2(s)) + 6H_2O_((l)) -> 3Mg(OH)_(2(s)) + 2NH_3(g)`

When you heat magnesium hydroxide, it decomposes into water vapor and magnesium oxide.

`Mg(OH)_(2(s)) -> MgO(s) + H_2O_((g))`

If you perform these aditional reactions, you'll recover whatever magnesium oxide lost when magnesium reacted with nitrogen.

Like your hint says, try a sample calculation with your data. If you performed this experiment correctly, you'll get the percent of magnesium in magnesium oxide to be around 60%, more or less depending on how you did the experiment.

Here's a sample calculation to go by - keep in mind these are not actual values, I'm just using them as an example.

Let's say you have 24.3 g of magnesium which you react with enough oxygen to produce magnesium oxide. According to the first equation, you have a `"1:1"` mole ratio between magnesium and magnesium oxide.

You calculate how many moles of magnesium you have

`"24.3 g" * "1 mole"/"24.3 g" = "1.00 mole Mg"`

This means you'll produce

`"1 mole" * "40.3 g"/"1 mole MgO" = "40.3 g MgO"`

The percent composition of `MgO` will be

`24.3/40.3 * 100 = 60.3%` `Mg`

Assuming you lost some magnesium oxide in the process (in the smoke), you find that you get 35.5 g of `MgO`. In this case, you'll get

`24.3/35.5 * 100 = "68.5%"` `Mg` `->` the percent of magnesium will be higher.