Question

In the lab you measure a clean dry crucible and cover to be 24.36 grams. You obtain a 2cm piece of pure magnesium metal. After sand papering the magnesium you place it in into your crucible and the mass of crucible, cover and magnesium masses out to be 24.66 grams. You observe that the magnesium is gray and shiny. you heat the crucible gently at first then vigorous for 10 minutes until the magnesium ignites. After the crucible cools you add a few drops of distilled water and heat again. The mass of the magnesium compound, crucible, and cover masses out to be 24.85 grams.
What is the empirical formULa of the compound? Show all work and units.
1) MAss of the Mg
2) Mass of the MgO
3) Mass of 0
4) Moles of Mg & Moles of O
5) Show mole to mole MgxOy
6) Empricial Formula of compound

Answer 1

Answer:

1) 0.3g Mg

2)0.5g MgO

3)0.2g O

4)0.01mol Mg & 0.01mol O

5)0.01mol MgO

6) Empirical formula MgO

Explanation:

The mass og Mg is obtained by substracting 24.36g from 24.66g:

24.66 - 24.36 = 0.3g Mg

The ignition of Mg means that it's reacting with oxygen to form an oxide. The increase in the crucible mass after the Mg ignition is due to the addition of oxygen. However, the addition of few drops of water produces a new compound: a hydroxide. According to the oxidation state og Mg (2+), the only magnesium oxide possible is MgO. It happens because the oxidation state of oxygen in oxides is 2-. Which means that just one oxygen atom is required to electrically neutralize one magnesium atom.

We can use a conversion factor to know how much MgO is made from from 0.3 g of Mg:

$0.3g Mg$*$\frac{16gO}{24.3gMg}$= 0.2g O

Thereby the mass of the oxide is 0.2g O + 0.3g Mg = 0.5g MgO

We convert the mass of oxygen and magnesium to the respective amounts in moles by using conversion factors:

$0.2g O$*$\frac{1 mol O}{16g O}$= 0.01mol O

$0.3g Mg$*$\frac{1mol Mg}{24.3g Mg}$= 0.01mol Mg

The moles of MgO can be obtained from:

$0.5g MgO$*$\frac{1mol MgO}{40.3g MgO}$= 0.01mol MgO

To obtain the empirical formula, the amount fo moles of each elements must be divided by the smallest one, in this case, 0.01.

The result for both number of  Mg atoms and O atoms is 1. This can be interpreted to mean that there is a Mg atom for each O atom forming the  formula unit of the compound.

The step when water is added to the compound resulting after heating does not affect the calculations necessary for the magnesium oxide.