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).

Question

Henry Antrim · Accepted Answer

Alright, so you've completed the traditional magnesium oxide lab. Since I'm assuming you already have your data, I'll just guide you through this theoretically.

In light of this reaction, you heat magnesium in the presence of air.

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

The mass of magnesium oxide you measure at the end of the experiment will be less than what was actually produced if you allow some smoke to escape the crucible.

The final product will have a higher percentage of magnesium because the mass of magnesium was measured before the reaction began.

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

But since you can react this product with water to produce ammonia and magnesium hydroxide, this is typically not a major issue.

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

Magnesium hydroxide breaks down into magnesium oxide and water vapor when heated.

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

You can recover the magnesium oxide that was lost during the magnesium-nitrogen reaction by carrying out these additional reactions.

Try doing a sample calculation with your data, as suggested in your hint. If you carried out this experiment correctly, the percentage of magnesium in magnesium oxide should be approximately 60%, depending on how you carried out the experiment.

Here is an example calculation to follow; please note that these are only examples and not the actual values.

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 figure out how much magnesium you have in moles.

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

This implies that you will generate

#"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.

Serenity Allen · Answer

undefined If some of the magnesium had reacted with the nitrogen in the air, it would result in the formation of magnesium nitride (Mg3N2), which would increase the mass of magnesium in the sample. This would lead to a higher percentage composition of magnesium in the sample than it should be.

For example, let's assume you started with 10 grams of magnesium and some of it reacted with nitrogen to form magnesium nitride. The actual mass of magnesium in the sample would then be less than 10 grams. However, when calculating the percentage composition of magnesium, you would still use the initial mass of magnesium (10 grams) as the denominator, resulting in a higher percentage composition of magnesium in the sample.

So, the presence of magnesium nitride due to the reaction with nitrogen in the air would lead to a higher percentage composition of magnesium in the sample than it should be.