Empirical Formula of Magnesium Oxide by Experiment Chemistry Tutorial

Key Concepts

• Empirical formula of a compound gives the lowest whole number ratio of atoms of each element present in the compound.
• Empirical formula of magnesium oxide is determined by reacting magnesium metal with oxygen from the air to produce the magnesium oxide.
• Known quantities:

  (a) mass of magnesium used

  (b) mass of magnesium oxide produced

• Required calculations:

  (i) mass oxygen = mass magnesium oxide - mass magnesium

  (ii) moles magnesium = mass magnesium ÷ molar mass magnesium

  (iii) moles oxygen = mass oxygen ÷ molar mass oxygen

• Lowest whole number ratio of moles of magnesium to moles of oxygen is determined.

  Usually requires both the moles of magnesium and the moles of oxygen to be divided by whichever is the lowest number.

• Empirical formula of magnesium oxide is written:

  ⚛ with the symbol for magnesium (Mg) written before the symbol for oxygen (O)

  ⚛ using the lowest whole number ratio of moles of magnesium (x) to moles of oxygen (y), the subscripts for Mg and O are added to give a formula of the type Mg_xO_y

  Note that a subscript number is NOT included in the formula if it would be a 1

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Theory Behind the Experimental Determination of the Empirical Formula of Magnesium Oxide

Magnesium metal reacts with oxygen from the atmosphere in a combustion reaction to produce grey-white solid magnesium oxide.

magnesium + oxygen gas → magnesium oxide

Since the product, magnesium oxide, contains only magnesium "atoms" and oxygen "atoms"⁽¹⁾, we could write the formula Mg_xO_y in which:

x represents the number of Mg "atoms"

y represents the number of O "atoms"

We cannot see the atoms of each element in the compound because they are much too small, so we can't count them just be looking at 1 "molecule" of magnesium oxide. We need a different way to determine the values of x and y.

There is a relationship between the mass of an element and the number of "atoms" of that element:

6.02 × 10²³ atoms of an element has a mass equal to its atomic weight expressed in grams

The atomic weight of each element is listed in the Periodic Table of the Elements:

atomic weight of magnesium (Mg) = 24.31

atomic weight of oxygen (O) = 16.00

Therefore, if we know the mass of an element we can calculate how many atoms of that element are present:

mass of 6.02 × 10²³ atoms of an element = element's atomic weight expressed in grams (molar mass)

mass of 1 atom of an element in grams = molar mass ÷ 6.02 × 10²³

So, if we know the mass of magnesium and the mass of oxygen making up our sample of magnesium oxide, Mg_xO_y, product, then:

the number of magnesium atoms in a given mass of magnesium atoms can be calculated:

and, the number of oxygen atoms in a given mass of oxygen atoms can be calculated:

In this experiment we will determine the empirical formula for magnesium oxide, that is, we will determine the lowest whole number ratio of x to y:

Note that the quantity equal to an element's mass divided by its molar mass is measured in units of moles:

So, we only need to measure the mass of magnesium and the mass of oxygen present in the magnesium oxide sample in order to determine the ratio of moles of magnesium to moles of oxygen, from which we can determine the empirical formula of the magnesium oxide.

The Law of Mass Conservation tells us that during a chemical reaction mass can neither be created nor destroyed, so the total mass of the system before the chemical reaction must be equal to the total mass of the system after completion of the chemical reaction:

Magnesium metal and magnesium oxide are both solids at room temperature and pressure, so we can easily weigh these in order to determine their mass.
We can not easily weigh the amount of oxygen gas used to combust the magnesium, but we don't have to because we can use the Law of Mass Conservation to calculate how much oxygen is present in the magnesium oxide we produce:

mass oxygen in compound = (mass magnesium oxide) - (mass magnesium used)

The mass of magnesium used and the mass of oxygen atoms we calculate can then be used to determine the ratio of magnesium atoms to oxygen atoms in the compound using the relationship we derived above, that is:

However, the value of x and the value of y will probably be fractions (or decimals) rather than whole numbers:

for example x : y is calculated to be 0.072 : 0.069

In order to force this into a ratio of whole numbers we will divide both x and y by the lowest number (0.069 in this example):

We then write the empirical formula for the magnesium oxide (Mg_xO_y) by replacing the x and y with the values we have calculated, for example:

Note that if either x=1 or y=1 then the subscript 1 is not written in the formula, that is, Mg₁O₂ is actually written as MgO₂, and, Mg₂O₁ would only ever be written as Mg₂O.

Experiment to Determining the Empirical Formula of Magnesium Oxide

In this experiment we will be using a piece of apparatus known as a crucible.
A crucible is ceramic vessel with a lid used to hold substances that are heated to high temperatures.
A crucible is preferable to using a glass test tube in this experiment because:

• the crucible is less likely to break at high temperatures than a glass test tube
• the lid of the crucible allows some air to pass into the crucible while preventing the powdery solid produced from escaping

Setting up the Experiment

1. Preparing the crucible.⁽²⁾

Place a pipe clay triangle in a ring stand. (3) Place a crucible with its lid slightly off-center in the pipe clay triangle. Slide a Bunsen Burner on a blue flame (not a smokey yellow flame) under the crucible so that the hottest part of the flame (the top of the inner blue cone) is directly heating the bottom of the crucible. Heat for 5 minutes to burn off any contaminants that may be present. The bottom of the crucible should glow red-hot for about 20 seconds. This is known as heating to incandescence. Slide the Bunsen Burner, still on a blue flame, out from under the crucible before turning it off. This should ensure that no soot attaches to the bottom of the crucible. Use tongs to re-position the lid so that it covers the open crucible. Cool the clean, empty crucible and lid to room temperature without removing it from the pipe clay triangle.(4) You can use your hands to make a tent shape over the crucible at a little distance from it so that you can feel if any heat is still being radiated off the crucible. DO NOT touch the crucible, if it is hot you will be burnt, even if it is cool you will be adding contaminants from your hands! Remove the crucible from the pipe clay triangle using tongs to prevent contaminants from your hands being transferred to the crucible. If you have to walk with the crucible to the balance, hold the crucible in the tongs and support it with a heat resistant mat under it to take it to the balance for weighing. Use tongs to transfer the crucible to the balance. Weigh the clean, empty crucible and lid and record its mass.

Safety Wear safety goggles (eye protection). Do not place anything near the open flame. (Keep your pens, notebooks, etc away from the flame) Do not breathe in any fumes coming out of the crucible. Assume that anything that is being heated or has been heated is hot. Do not touch these items without first trying to feel for heat being radiated off them without touching them. DO NOT look directly into the crucible while it is being heated! DO NOT place anything hot directly onto the lab bench, always place hot objects on a heat resistant mat. DO NOT place anything hot on a balance!

2. Producing the magnesium oxide.

1. Clean a 35 mm strip of magnesium (about 0.3 g) using emery papery or steel wool so that it is silver and shiny.
2. Wind the magnesium strip around a clean glass stirring rod to obtain a loose coil and place this inside the crucible and place the lid on the crucible.
3. Weigh the crucible, lid and magnesium (remember not to handle the crucible with your hands, use tongs).
   Record this mass.
4. Using tongs, place the crucible on the pipe clay triangle and position the lid so that it is slightly off-center to allow air to enter but prevent the magnesium oxide from escaping.
5. Light the Bunsen Burner, and obtain a blue flame, and use this flame to brush the bottom of the crucible for about 1 minute until the magnesium starts to burn as evidenced by a bright glow within the crucible, then place the Bunsen Burner under the crucible and heat strongly until all the magnesium turns into a grey-white powder (about 10 minutes).⁽⁵⁾
6. Use tongs to re-position the lid so that it covers the crucible.
   This is done to prevent contaminants from the air, especially water, from entering the crucible while it is cooling.
7. While still on a blue flame, slide the Bunsen Burner out from under the crucible before turning it off to prevent soot attaching to the crucible.
8. Allow the crucible to cool to room temperature while still on the pipe clay triangle.
9. Weigh the crucible, lid and contents.
   Record this mass.
10. Use tongs to re-position the crucible back into the pipe clay triangle with the lid slightly off-center as before, and heat strongly for a few more minutes.
    Remove the Bunsen Burner, and allow the crucible to cool back to room temperature, then weigh the crucible and lid again.
    Record this mass.
11. Continue this heating, cooling and weighing process until you achieve a constant mass.

Sample Results

Mass crucible + lid / g		24.62
Mass crucible + lid + magnesium / g		24.89
Mass crucible + lid + magnesium oxide/ g	Trial 1	24.54
	Trial 2	25.06
	Trial 3	25.06

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Calculating the Empirical Formula of Magnesium Oxide Using the Results of the Experiment

1. Calculate the mass of magnesium used in the experiment:

   mass magnesium used	=	(mass of crucible + lid + magnesium)	-	(mass of crucible + lid )
   	=	24.89 g	-	24.62 g
   	=	0.27 g

2. Calculate the mass of magnesium oxide produced:

   Note: we will use the "constant mass" of the crucible + lid + magnesium oxide in the results table and disregard earlier, lighter masses which indicate that the reaction had not yet gone to completion.

   mass magnesium oxide produced	=	(mass of crucible + lid + magnesium oxide)	-	(mass of crucible + lid )
   mass magnesium oxide produced	=	25.06	-	24.62
   	=	0.44 g

3. Calculate the mass of oxygen present in the magnesium oxide:

   mass oxygen in magnesium oxide	=	(mass of magnesium oxide produced)	-	(mass of magnesium used )
   	=	0.44 g	-	0.27 g
   	=	0.17 g

4. Calculate the moles of magnesium present in the magnesium oxide compound:

   moles magnesium	=	mass magnesium (g)	÷	molar mass magnesium (g mol-1)
   	=	0.27	÷	24.31 (from periodic table)
   	=	0.0111 mol

5. Calculate the moles of oxygen present in the magnesium oxide compound:

   moles oxygen	=	mass oxygen (g)	÷	molar mass oxygen (g mol-1)
   	=	0.17	÷	16.00 (from periodic table)
   	=	0.0106 mol

6. Calculate the mole ratio magnesium:oxygen in the magnesium oxide compound:

   	moles magnesium	:	moles oxygen
   moles calculated from experimental data	0.0111	:	0.0106
   divide each number by the smallest number	0.0111 ÷ 0.0106	:	0.0106 ÷ 0.0106
   mole ratio	1.05	:	1.00
   since 1.05 ≈ 1.00 the ratio of lowest whole numbers is	1	:	1

7. Write the empirical formula for the magnesium oxide:
   We have calculated that the ratio of moles (and hence the ratio of number of atoms) of Mg to O in magnesium oxide is 1:1
   We can write this formula as Mg₁O₁, but, because we do not write the subscript 1 in a chemical formula, the empirical formula is simply written as MgO

Sources of Error in the Experimental Determination of the Empirical Formula of Magnesium Oxide

• If the mass of magnesium oxide is too high

  Contamination is the most likely problem. Water may have entered your crucible while it was cooling (especially if your lid was not completely covering your crucible), or soot or other particulate matter may have adhered to the top, sides, or lid of your crucible.

  Consider the impact of a drop of water, say 0.05 g, on your experimental results:

  	"magnesium oxide"	magnesium	oxygen
  Mass / g	0.44 + 0.05 = 0.49	0.27	0.49 - 0.27 = 0.22
  moles / mol		0.27/24.31 = 0.011	0.22/16.00 = 0.014
  mole ratio		0.011/0.011 = 1.0	0.014/0.011 = 1.3 ≈ 4/3
  × 3		1 × 3 = 3	4/3 × 3 = 4
  lowest whole number ratio		3	4
  empirical formula	Mg3O4	calculated mass of O is too high (0.23 > 0.17) so moles O is too high

• If the mass of magnesium oxide is too low.

  Magnesium oxide product is a very fine powder which can easily escape whenever the lid on the crucible is raised.
  Consider, if you lost just 0.05 g of the total 0.44 g of your MgO before the final weighing:

  	magnesium oxide	magnesium	oxygen
  Mass / g	0.44 - 0.05 = 0.39	0.27	0.39 - 0.27 = 0.12
  moles / mol		0.27/24.31 = 0.011	0.12/16.00 = 0.0075
  mole ratio		0.011/0.0075 = 1.5 = 3/2	0.0075/0.0075= 1.0
  × 2		3/2 × 2 = 3	1.0 × 2 = 2
  lowest whole number ratio		3	2
  empirical formula	Mg3O2	calculated mass of O is too low (0.12 < 0.17) so moles O is too low

  The mass of magnesium oxide will also be too low if some of the magnesium metal does not combust.

  Consider an experiment in which 0.07 g of magnesium has not combusted and is still present as the metal in amongst the product magnesium oxide which will all be recorded as the mass of "magnesium oxide":

  	"magnesium oxide"	magnesium	oxygen
  Mass / g	0.40	0.27	0.40 - 0.27 = 0.13
  moles / mol		0.27/24.31 = 0.011	0.13/16.00 = 0.00813
  mole ratio		0.011/0.00813 = 1.35 ≈ 4/3	0.00813/0.00813 = 1.0
  × 3		4/3 × 3 = 4	1 × 3 = 3
  empirical formula	Mg4O3	calculated mass of O is too low (0.13 < 0.17) so moles O is too low

• Competing Reactions⁽⁶⁾

  Air contains not only oxygen gas but also a large proportion of nitrogen gas. Although nitrogen gas is not particularly reactive, magnesium is a very reactive metal, and, at the temperatures achieved during the combustion of magnesium some magnesium will also react with nitrogen gas, that is:

  word equation :	magnesium metal	+	nitrogen gas	→	magnesium nitride
  balanced chemical equation :	3Mg	+	N2	→	Mg3N2

  If about half of the original 0.27 g of magnesium metal, 0.13 g, were to react with nitrogen in this way, then

  word equation :	magnesium metal	+	nitrogen gas	→	magnesium nitride
  balanced chemical equation :	3Mg	+	N2	→	Mg3N2
  moles / mol	0.13/24.31 = 0.0053				1/3 × 0.0053 = 0.0018
  mass / g = moles × molar mass					0.0018 × 100.95 = 0.18

  Our final "magnesium oxide" product would contain 0.18 g of magnesium nitride. We can calculate how much MgO would be included in the product:

  word equation :	magnesium metal	+	oxygen gas	→	magnesium oxide
  balanced chemical equation :	Mg	+	½O2	→	MgO
  mass / g	0.27 - 0.13 = 0.14
  moles / mol	0.14 ÷ 24.31 = 0.0058				0.0058
  mass = moles × molar mass					0.0058 × 40.31 = 0.23

  The apparent mass of "magnesium oxide" product would therefore be 0.18 g + 0.23 g = 0.41 g
  which is less than the 0.44 g we would expect without the magnesium nitride contaminant.
  We could then calculate the mass of O = 0.41 - 0.27 = 0.14
  which is less than the expected 0.17 g of O based on complete combustion of all 0.27 g of Mg
  mole ratio Mg : O would be 0.27/24.31 : 0.14/16 or 0.011 : 0.0088
  dividing through by 0.0088 the ratio Mg : O is 0.011/0.0088 : 0.0088/0.0088 is 1.25 : 1 or ⁵/₄ : 1
  multiplying throughout by 4, the mole ratio of Mg : O is 5 : 4
  empirical formula would be Mg₅O₄