# Determining the Empirical Formula of Magnesium Oxide

1511 Words Nov 29th, 2012 7 Pages
Chemistry Lab Report example:

Chemistry Laboratory Report (Magnesium Oxide)

INTRODUCTION:

As we learned before on how to determine the empirical formula of a compound based on the test and also chemical analysis on it. Hence this experiment is mainly goes around with how to determine the empirical formula of Magnesium Oxide following various tight procedures in order to get the knowledge and apply it onto another compounds. We are investigating the empirical formula of Magnesium Oxide in this experiment.

RESEARCH QUESTION:

How empirical formula of Magnesium Oxide is obtained by heating Magnesium in the presence of air?

HYPOTHESIS:

When Magnesium and Oxygen are heated together, they readily undergo a chemical change

Magnesium +
The ribbon is not. The crucible is heated for another 10 minutes. Step 9 is repeated until the ribbon become whitish. Then the crucible is allowed to cool.

10. The crucible and the end product is weighed.

DATA TABLE:

Items
Mass(+/- 0.001)g
Mass of crucible + lid
21.880
Mass of crucible + lid + magnesium
22.190
Mass of crucible + lid + magnesium oxide
22.380
Mass of magnesium oxide
0.500
Mass of magnesium
0.310
Mass of oxygen
0.190

DATA PROCESSING:

Mass of magnesium used

= 22.190g – 21.880g

= 0.310g

Moles of magnesium

=0.310g/24.310 g mol -1

=0.01275 mol

Mass of oxygen that combines with magnesium

= 22.380g – 22.190g

= 0.190g

Moles of oxygen

= 0.190g/16.000g mol -1

= 0.011875 mol

Empirical formula

Magnesium (Mg)
Oxygen (O)
Mass
0.310g
0.190g
Mole
=0.310g/24.310g mol -1
=0.012752 mol
=0.190g/16.000g mol -1
=0.011875 mol
Ratio
= 0.012752mol/0.011875mol
=1.1
=1
=0.011875mol /0.011875mol
= 1
Formula
MgO
MgO
Experimental percent oxygen in Magnesium Oxide

Mass of MgO = 0.190g

=0.190g/0.500g x 100

=38%

Experimental percent magnesium in Magnesium Oxide

Mass of magnesium= 0.310g

=0.310g/0.500g x 100

= 62%

Theoretical percent oxygen in Magnesium Oxide

Formula: 2Mg + O2 → 2MgO

Moles of MgO = 0.500g/(24.310+16.000)g mol -1

= 0.01240387 mol

From the equation:

2 mol of MgO→1 mol of O2

0.01240387 mol→0.006219 mol

Mass = 0.062019g x 32 g mol -1

= 0.1984000g

Percent =