Lab 3
Determination of an Empirical Formula
Diana Bakos
Sabrina Panetta
SCH3U. 02
Wednesday, November 27, 2013
Introduction
Stoichiometry is the study of quantitative relationships among the amounts of reactants used and
amount of products produced. One of the main parts of stoichiometry is that it gives you a balanced
equation which gives you the mole ratios from the coefficients, which also represents relative number of
particles involved in the chemical reaction. For the following experiment used magnesium, this reacted
with oxygen from the air in a crucible, while it was heated over a Bunsen burner. The masses before and
after the reactants were oxidized were measured. With the results, the experimental empirical formula
of magnesium oxide could be determined and compared with the theoretical empirical formula.
Purpose
1. To prepare a binary compound
2. To find the percent composition by mass of the compound
3. To determine the formula of the compound
Materials
Please refer to page 1 of 2 of the lab handout
Procedure
Please refer to page 1 of 2 and 2 of 2 of the lab handout.
Observations/Data
Mass of magnesium, crucible and lid
Mass of crucible and lid
Mass of magnesium
Mass of product crucible and lid
Mass of product
Increase in mass
Mass of element that combined with magnesium
25.88g
25.57g
0.31g
26.03g
0.46g
0.19g
0.15g
Calculations
Calculate the percentage
composition by mass of
the compound
Experimental percentage
composition of Mg and O
Experimental percentage
composition of Mg and O
you calculated in #1
determine the EF of the
compound.
Mg = Mass of magnesium/ Mass of product x 100
Mg = 0.31/0.46 x 100% = 67%
O=Mass of element that combined with magnesium/ Mass of product x 100
O = 0.15/0.46x 100% = 33%
Assume 100 g
Mg = 67gMg x 1 mol/24.32 MgO =2.75mols
O = 33g O x 1mol/15.99gMgO = 2
Mg:O
2.75:2
1.38:1
MgO
Theoretical percentage
composition of Mg and O
you calculated in #1
determine the EF of the
compound.
Mg = 24.32gMg/(24.32+15.99MgO) x 100% = 60%
O = 15.99gMg/(24.32 +15.99MgO) x 100% = 40%
EF
Assume 100g
Mg = 60gMg x 1 mol/24.32 MgO =2.47mols
O = 40g O x 1mol/15.99gMgO = 2.50
Mg:O
2. 47:2.50
1:1
Calculate the % error:
Actual Yield
Theoretical Yield
2Mg + O2 → 2MgO
MgO
Mg = (67-60)/60 x 100% = 7%
O = (33-40) /40 x 100% = 7%
Discussion
1. The observation that suggests that the formation of the compound was a chemical change
rather than a physical one is oxidation of magnesium into magnesium oxide and a new
substance that formed because heat burned the magnesium.
2. As you know, when magnesium is burnt it reacts with oxygen to produce magnesium oxide; thus
the additional mass of the oxygen is being added to the original mass of magnesium to form
magnesium oxide, which composes a compound with an increased mass.
Magnesium + Oxygen → Magnesium Oxide
2Mg + O2 → 2MgO
3. One would want to heat the crucible strongly after the reaction had apparently been completed
just in case there was some leftover residue that didn't react, and by heating it further one is
increasing the likelihood of eliminating all that was left unreacted, thereby getting a more
accurate weight of the yield.
4. The reason the lid must be left slightly ajar is so the combustion reaction can take place by
allowing the oxygen in the air to enter in the crucible and react with the magnesium, but at the
same time to stop to many fumes from escaping because that would reduce the mass of the
magnesium.
5. The results of the experimental composition compared to the theoretical composition only have
a 7% worth of error so
Source of experimental error
While the Mg ribbon was being heated over the Bunsen burner, to check The atmosphere is composed
of 78.08% nitrogen, 20.95% oxygen, 0.93% argon, 0.039% carbon dioxide and about 1% water vapor
depending on the temperature in the room. It’s unlikely that the Mg reacted with the 1% of H2O vapor,
Argon or the CO2 that was in the air, but there is a likely chance. However, some of the nitrogen,
because there it is present is a large quantity of it, could have reacted with the Mg, especially when the
lid was lifted to check if the oxidization was occurring, which could have allowed for some of the MgO
fumes to escape, and nitrogen to react. It’s difficult to indicate, which reaction took place because both
Mg3N2 and MgO produce a powder. In order to have avoided this, water should have been added to
form Mg(OH)2 and NH3 then one would again heat the product and produce MgO.
Conclusions
In conclusion, the empirical formula of Magnesium oxide (MgO) was determined. To find the empirical
formula, first the mass of the O2, which was 0.15g, and mass of the MgO, which was 0.46g, had to be
calculated. Then those digits, along with the original mass of the Mg, which was 0.31g, were used to
calculate the experimental composition. For Mg the experimental composition was 67% and for O2 the
experimental composition was 33%. Compared to the theoretical composition of Mg (60%) and O2
(40%) there was a 7% error for both, which is not accurate because it’s know, by mass, that the
experimental composition should be close to 60%. However, dispite the 7% error the experiment was
still a success as the empirical formula showed to be MgO, when Magnesium reacts with oxygen from
the air. This also proves the law of definite proportions because the chemical compound contained
exactly the same proportion of elements by mass.