Empirical and Molecular
Formulas
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C ONCEPT
Concept 1. Empirical and Molecular Formulas
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Empirical and Molecular
Formulas
Lesson Objectives
• The student will reduce molecular formulas to empirical formulas.
• Given either masses or percent composition of a compound, the student will determine the empirical formula.
• Given either masses or percent composition of a compound and the molar mass, the student will determine the
molecular formula.
Introduction
The empirical formula is the simplest ratio of atoms in a compound. Formulas for ionic compounds are always
empirical formulas but for covalent compounds, the empirical formula is not always the actual formula for the
molecule. Molecules such as benzene, C6 H6 , would have an empirical formula of CH.
Finding Empirical Formula from Experimental Data
Empirical formulas can be determined from experimental data or from percent composition. Consider the following
example.
Example 11
We find that a 2.50 gram sample of a compound contains 0.900 grams of calcium and 1.60 grams of chlorine. The
compound contains only these two elements. We can calculate the number of moles of calcium atoms and the
number of moles of chlorine atoms in the compound. We can then find the ratio of moles of calcium atoms to moles
of chlorine atoms and from this; we can determine the empirical formula.
Solution
First, we convert the mass of each element into moles.
0.900 g
= 0.0224 mole Ca
40.1 g/mol
1.60 g
moles of Cl atoms =
= 0.0451 mole Cl
35.5 g/mol
moles of Ca =
At this point, we have the correct ratio for the atoms in the compound, Ca0.0224 Cl0.0451 , except that this isn’t an
acceptable formula. We need to find the simplest whole number ratio. To find a simple whole number ratio for these
numbers, we divide each of them by the smallest of them.
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0.0224
= 1.00 Ca
0.0224
0.0451
moles of Cl =
= 2.01 Cl
0.0224
moles of Ca =
Now, we can see the correct empirical formula for this compound is CaCl2 .
It is important to note that when solving for empirical formulas, we are determining the number of atoms of
each element in the compound. Therefore, those substances which occur in nature as diatomic molecules such
as Cl2 , O2 , H2 , N2 , and so on, are dealt with as atoms in this procedure.
Finding Empirical Formula from Percent Composition
When finding the empirical formula from percent composition, the first thing we do is to convert the percentages into
masses. For example, suppose we are given the percent composition of a compound as 40.0% carbon, 6.71% hydrogen, and 53.3% oxygen. Since every sample of this compound regardless of size will have the same composition in
terms of ratio of atoms, we could choose a sample of any size. Suppose we choose a sample size of 100. grams. The
masses of each of the elements in this sample will be 40.0 grams of carbon, 6.71 grams of hydrogen, and 53.3 grams
of oxygen. These masses can then be used to find the empirical formula. You should note that you could use any size
sample. You could choose a sample size of 167.8 grams and take the percentages of this sample to get the masses of
the individual elements. We choose a sample size of 100. grams because it makes the arithmetic simple.
Example 12
Find the empirical formula of a compound whose percent composition is 40.0% carbon, 6.71% hydrogen, and 53.3%
oxygen.
Solution
We choose a sample size of 100. grams and multiply this 100. gram sample by each of the percentages to get masses
for each element. This would yield 40.0 grams of carbon, 6.71 grams of hydrogen, and 53.3 grams of oxygen. The
next step is to convert the mass of each element into moles.
40.0 g
= 3.33 moles C
12.0 g/mol
6.71 g
moles of H =
= 6.64 moles H
1.01 g/mol
53.3 g
moles of O =
= 3.33 mole Ca
16.0 g/mol
moles of C =
Then, we divide all three numbers by the smallest one to get simple whole number ratios:
C=
3.33
=1
3.33
H=
6.64
=2
3.33
O=
3.33
=1
3.33
and finally, we can write the empirical formula, CH2 O.
Sometimes, the technique of dividing each of the moles by the smallest to get a whole number ratio does not yield
whole numbers. Whenever the subscript for any element in the empirical formula is “1,” dividing each of the moles
by the smallest will yield a simple whole number ratio but if none of the elements in the empirical formula has a
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Concept 1. Empirical and Molecular Formulas
subscript of “1,” then this technique will not yield a simple whole number ratio. In those cases, a little more work is
required.
Example 13
Determine the empirical formula for a compound that is 66.0% calcium and 34.0% phosphorus.
Solution
We choose a sample size of 100. grams and multiply the 100. grams by the percentage of each element to get masses.
This yields 66.0 grams of calcium and 34.0 grams of phosphorus. We then divide each of these masses by their molar
mass to convert to moles.
66.0 g
= 1.65 moles Ca
40.1 g/mol
34.0 g
moles of P =
= 1.10 moles P
31.0 g/mol
moles of Ca =
We then divide each of these moles by the smallest.
Ca =
1.65
= 1.50
1.10
P=
1.10
= 1.00
1.10
In this case, dividing each of the numbers by the smallest one does not yield a simple whole number ratio. In such a
case, we must multiply both numbers by some factor that will produce a whole number ratio. If we multiply each of
these by 2, we get a whole number ratio of 3Ca to 2 P. Therefore, the empirical formula is Ca3 P2 .
Finding Molecular Formulas
Empirical formulas show the simplest whole number ratio of the atoms in a compound. Molecular formulas show
the actual number of atoms of each element in a compound. When you find the empirical formula from either
masses of elements or from percent composition, as demonstrated in the previous section, for the compound N2 H4 ,
you will get an empirical formula of NH2 and for C3 H6 , you will get CH2 . If we want to determine the actual
molecular formula, we need one additional piece of information. The molecular formula is always a whole number
multiple of the empirical formula. That is, in order to get the molecular formula for N2 H4 , you must double each
of the subscripts in the empirical formula. Since the molecular formula is a whole number multiple of the empirical
formula, the molecular mass will be the same whole number multiple of the formula mass. The formula mass for
NH2 is 14 g/mol and the molecular mass for N2 H4 is 28 g/mol. When we have the empirical formula and the
molecular mass for a compound, we can divide the formula mass into the molecular mass and find the whole number
that we need to multiply by each of the subscripts in the empirical formula.
Example 14
Suppose we have the same problem as in example 12 except that we are also given the molecular mass of the
compound as 180 grams/mole and we are asked for the molecular formula. In example 12, we determined the
empirical formula to be CH2 O. This empirical formula has a formula mass of 30.0 g/mol. In order to find the
molecular formula for this compound, we divide the formula mass into the molecular mass (180 divided by 30) and
find the multiplier for the empirical formula to be 6. As a result, the molecular formula for this compound will be
C6 H12 O6 .
Example 15
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Find the molecular formula for a compound with percent composition of 85.6% carbon and 14.5% hydrogen. The
molecular mass of the compound is 42.1 g/mol.
Solution
We choose a sample size of 100. g and multiply each element percentage to get masses for the elements in this
sample.
This yields 85.6 g of C and 14.5 g of H.
Dividing each of these by their atomic mass yields 7.13 moles of C and 14.4 moles of H.
Dividing each of these by the smallest yields a whole number ratio of 1 carbon to 2 hydrogen.
Thus, the empirical formula will be CH2 .
The formula mass of CH2 is 14 g/mol.
Dividing 14 g/mol into the molecular mass of 42.1 g/mol yields a multiplier of 3.
The molecular formula will be C3 H6 .
Lesson Summary
• The empirical formula of a compound indicates the simplest whole number ratio of atoms present in the
compound.
• The empirical formula of a compound can be calculate from the masses of the elements in the compound or
from the percent composition.
• The molecular formula of a compound is some whole number multiple of the empirical formula.
Review Questions
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
4
What is the empirical formula for C8 H18 ?
What is the empirical formula for C6 H6 ?
What is the empirical formula for WO2 ?
A compound has the empirical formula C2 H8 N and a molar mass of 46 grams/mol. What is the molecular
formula of this compound?
A compound has the empirical formula C2 H4 NO. If its molar mass is 116.1 grams/mole, what is the molecular
formula of the compound?
A sample of pure indium chloride with a mass of 0.5000 grams is found to contain 0.2404 grams of chlorine.
What is the empirical formula of this compound?
Determine the empirical formula of a compound that contains 63.0 grams of rubidium and 5.90 grams of
oxygen.
Determine the empirical formula of a compound that contains 58.0%Rb, 9.50%N, and 32.5%O.
Determine the empirical formula of a compound that contains 33.3%Ca, 40.0%O, and 26.7%S.
Find the molecular formula of a compound with percent composition 26.7%P, 12.1%N, and 61.2%Cl and
with a molecular mass of 695 g/mol.
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Concept 1. Empirical and Molecular Formulas
Further Reading / Supplemental Links
• Zumdahl, Steven S. and Zumdahl, Susan A., Chemistry, Fifth Edition, Chapter 3: Stoichiometry, Houghton
Mifflin Company, New York, 2000.
• Website with lessons, worksheets, and quizzes on various high school chemistry topics.
• Lesson 5-5 is on Empirical Formulas.
• http://www.fordhamprep.org/gcurran/sho/sho/lessons/lesson55.htm
• http://misterguch.brinkster.net/molarmass.html
• http://www.ausetute.com.au/percentc.html
• http://www.chem.tamu.edu/class/majors/tutorialnotefiles/empirical.htm
Vocabulary
empirical formula The formula giving the simplest ratio between the atoms of the elements present in a compound.
molecular formula A formula indicating the actual number of each kind of atom contained in a molecule.
Labs and Demonstrations for
Teacher’s Pages for Empirical Formula of Magnesium Oxide
Investigation and Experimentation Objectives
In this activity, the student will collect and interpret data, use mathematics to determine a solution, and communicate
the result.
Lab Notes
It is not crucial that the magnesium ribbon be exactly 35 cm, but it should be clean. A good way to clean the ribbon
is to dip it in 0.1 MHCl for a couple of seconds, then rinse it in distilled water, and dry it in alcohol or acetone.
The more finely divided the ribbon is, the faster it will react. 0.5 cm to 1.0 cm pieces seems to work best. The
crucibles often react with the magnesium during this process. This can cause greenish-black discoloration to the
crucibles. This does not really affect chemical behavior for later reactions, as long as they are cleaned. However,
the crucibles will often crack during this procedure. If a crucible cracks, discard it. Stress to the students that the
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crucibles get extremely hot. A hot crucible can cause a very serious burn. Show the students how to handle a crucible
properly by using crucible tongs. A common source of error in this experiment is not to react the nitride. Make sure
the students do this portion of the lab – it is not washing.
Answers to Pre-Lab Questions
1. Assume you have 100 grams of the compound. This then changes the percent composition to grams of each
element.
Find the number of moles of each element present. (Divide the grams of each element by its molar mass.)
Divide each of these answers by the smallest answer. This will give the empirical formula.
2. The molecular weight of the compound is needed.
3. Yes. The charge of magnesium increases from 0 to +2, and the charge of the oxygen is reduced from 0 to −2.
4. 3.93 g for MgO, 3.28 g for Mg3 N2
Lab - Empirical Formula of Magnesium Oxide
Background Information
In this lab, magnesium metal (an element) is oxidized by oxygen gas to magnesium oxide (a compound). Magnesium
reacts vigorously when heated in the presence of air. The Mg − O2 reaction is energetic enough to allow some Mg
to react with gaseous N2 . Although there is a higher percentage of N2 gas in the atmosphere than O2 , O2 is more
reactive, and the magnesium oxide forms in a greater amount than the nitride. The small amount of nitride that forms
can be removed with the addition of water, which converts the nitride to magnesium hydroxide and ammonia gas.
Heating the product again causes the loss of water and conversion of the hydroxide to the oxide.
The unbalanced equations are:
Mg(s) + N2(g) + O2(g) → MgO(s) + Mg3 N2(s)
MgO(s) + Mg3 N2(s) + H2 O(L) → MgO(s) + Mg(OH)2(s) + NH3(g)
MgO(s) + Mg(OH)2(s) → MgO(s) + H2 O(g)
Pre-Lab Questions
a. If the mass percent of each element in a compound is known, what steps are taken to determine the compound’s
empirical formula?
b. If the empirical formula of a compound is known, what additional information is required to determine the
molecular formula of the compound?
c. Is the reaction of magnesium metal and oxygen gas an oxidation-reduction reaction? If so, what is the change
in oxidation number of each type of atom?
d. What is the theoretical yield in grams of MgO if 2.37 g Mg metal reacts with excess O2 ? What is the theoretical
yield of Mg3 N2 if the same amount of Mg reacts with excess N2 ?
Purpose
To determine the empirical formula of magnesium oxide, and to reinforce the concepts of the law of mass conservation and the law of multiple proportions.
Apparatus and Materials
• Safety goggles
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•
•
•
•
•
•
•
•
Concept 1. Empirical and Molecular Formulas
Magnesium ribbon, Mg
Balance (to 0.01 g or better)
Ring stand
Bunsen burner
Ring support with clay triangle
Crucible with lid
Crucible Tongs
Heat resistant tile or pad
Safety Issues
The crucible and all of the apparatus gets very hot. The ammonia emitted during the secondary reaction is irritating.
Open flames can be dangerous. Do not place a hot crucible on an electronic balance. It can damage the electronics.
In addition, a hot crucible causes the air above it to become buoyant. If placed on a balance, the buoyant air will
cause a mass reading, which is less than the actual mass.
Procedure
1. Heat the empty crucible and lid for about 3 minutes to remove water, oils, or other contaminants and to make sure
there are no cracks. The bottom of the crucible should glow red-hot for about 20 seconds. Remove the flame and
cool the crucible with lid.
2. Record the mass of crucible and lid once it has cooled. Handle the crucible with tongs.
3. Obtain about 0.3 g (35 cm) magnesium ribbon (do not handle the ribbon with your hands). Cut the magnesium
into 0.5 − 1.0 cm pieces with scissors.
4. Record the mass of the magnesium ribbon, lid and crucible.
5. Place the crucible securely on the clay triangle. Set the lid slightly off-center on the crucible to allow air to enter
but to prevent the magnesium oxide from escaping.
6. Place the Bunsen burner under the crucible, light it, and heat the bottom of the crucible with a gentle flame for
about 1 minute; then, place the burner under the crucible and heat strongly.
7. Heat until all the magnesium turns into gray-white powder (around 10 minutes).
8. Stop heating and allow the crucible, lid and contents to cool.
9. Add about 1 ml (approx. 10 drops) of distilled water directly to the solid powder. Carefully waft some of the gas
that is generated toward your nose, but be very careful. Record any odor.
10. Heat the crucible and contents, with the lid slightly ajar, gently for about 2 minutes and then strongly for about
another 3 to 5 minutes.
11. Allow the crucible to cool and then record the mass of the crucible, lid and contents.
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12. Follow instructions for oxide disposal given by your teacher. Clean all equipment thoroughly.
Data
Mass of crucible and lid = _______ g
Mass of the crucible, crucible lid, and the magnesium = _______ g
Mass of the crucible, crucible lid, and magnesium oxide = _______ g
Post-Lab Questions
1. Determine the mass of magnesium ribbon used in the experiment by subtracting the mass of the crucible and lid
from the mass of the crucible, lid, and magnesium.
Mass of magnesium = _______ g
mass
2. Determine the number of moles of magnesium used. Remember: atomic
weight = number of moles. The atomic
weight of magnesium is 24.3 g/mol.
Number of moles of magnesium = _______ mole
3. Determine the mass of magnesium oxide that was formed by subtracting the mass of the mass of the crucible and
lid from the mass of the crucible, lid, and magnesium oxide.
Mass of magnesium oxide formed = _______ g
4. Determine the mass of oxygen that combined with the magnesium.
Mass of oxygen = mass of magnesium oxide - mass of magnesium
Mass of oxygen that combined with the magnesium = _______ g
5. Determine the number of moles of oxygen atoms that were used. This is elemental oxygen so use 16.0 g/mol for
the atomic weight.
Number of moles of oxygen atoms that were used = _______ mole
6. Calculate the ratio between moles of magnesium atoms used and moles of oxygen atoms used. Remember, this
is simple division. Divide the number of moles of magnesium by the number of moles of oxygen. Round your
answer to the nearest whole number, as we do not use part of an atom. This represents the moles (and also atoms)
of magnesium. The moles (and also atoms) of oxygen, are represented by 1, because it was on the bottom of the
division.
Moles of Magnesium : Moles of Oxygen
_______:_______
Teacher’s Pages for Water of Hydration Lab
Investigation and Experimentation Objectives
In this activity, the student will collect and interpret data, use mathematics to determine a solution, and communicate
the result.
Lab Notes
This lab is fairly straightforward and simple in execution. The biggest hazard is with the crucibles. They become
very, very hot – enough to weld your skin on to it. Make sure to handle hot crucibles with crucible tongs, and never
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Concept 1. Empirical and Molecular Formulas
to weigh a hot crucible. Teach your students how to gauge the temperature of a piece of glassware by having them
approach the hot item with the back of their hand. If they can bring their hand to within a centimeter of the piece of
glassware and it is hot, it is too hot to handle! If it is too hot for the back of your hand, it is certainly too hot for the
front. Instruct the students on the proper procedure for using a desiccator.
Water of Hydration Lab
Background Information
When ionic crystals form, they often incorporate water molecules within their structure. This water within the
crystal is called the water of hydration, and the compounds themselves are called hydrates. This water of hydration
can often be removed by simply heating the hydrate, because the water molecules are only weakly attracted to the
ions present. When this is done, the resulting leftover substance is called the anhydrous form of the crystal.
The amount of water present is often in a whole-number stoichiometric amount relative to the anhydrous form. Examples include barium chloride-2-hydrate, BaCl2 · 2H2 O, and cobalt nitrate-6-hydrate, Co(NO3 )2 · 6H2 O. Hydration
numbers are most often integers, but in calcium compounds they are often fractional. Portland cement is an example
of an important hydrate. When water is introduced to the anhydrous form, the water incorporates into the structure,
and the cement hardens. Since the removal of water requires an input of heat, it should not be surprising that adding
water to a hydrate gives off heat, and it gives off as much heat as was put into the system to remove the water in
the first place. This is a problem for civil engineers who pour large amounts of cement: the heat given off by the
hardening cement can be so great as to break the cement that has already hardened, due to heat stress. Steps must be
taken to remove this heat. The Hoover Dam is such a large piece of concrete (cement + aggregate) that the dam is
still cooling – and the last of the cement was poured in 1935.
Purpose
To determine the hydration number and empirical formula of copper(II) sulfate hydrate.
Apparatus and Materials
•
•
•
•
•
•
•
•
•
•
Ring Stand and Ring
Crucible
Clay Triangle
Bunsen burner
Wash bottle
Matches
Electronic Balance
Copper(II)sulfate hydrate (approximately 3 g per lab group)
desiccator
watch glass
Safety Issues
Always handle crucibles with crucible tongs. Never place a hot crucible on a balance. It can damage the electronics
and give a measurement, which is less than the actual mass.
Procedure
1. Clean a porcelain crucible with soap and water. Rinse and dry the crucible by placing the crucible and cover on a
clay triangle over a laboratory burner and heating until red-hot.
2. Carefully remove the crucible and cover with crucible tongs and let it cool. Handle the crucible and cover with
tongs for the remainder of the experiment.
3. Measure the mass of the empty crucible and cover to the nearest 0.01 gram.
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4. Add about 3 g of CuSO4 hydrate crystals to the crucible, replace the cover, and measure the mass to the nearest
0.01 g.
5. Begin heating slowly. Increase the heat until you have heated the crucible strongly for about 10 minutes.
6. Remove the crucible from the triangle support, let it cool in a desiccator, and measure the mass.
7. Reheat with a hot flame for a few minutes, cool, and measure the mass again. If the mass is different from that
recorded in Step 6, continue to heat and measure until the masses agree.
8. Remove the button of anhydrous copper sulfate by tipping it into a watch glass. Add a few drops of water to the
anhydrous copper sulfate, and record your observations below.
TABLE 1.1: Data
Number
1.
2.
3.
4.
5.
6.
7.
8.
Object
Mass of Crucible + Cover
Mass of Crucible + Cover + Copper
Sulfate Hydrate
Mass of Crucible + Cover + Anhydrous Copper Sulfate
Mass of Water (2 − 3)
Mass of Anhydrous Copper Sulfate
(3 − 1)
Moles of Anhydrous Copper Sulfate
Moles of water driven off
Ratio of moles of �water
� to moles of
anhydrous CuSO4 76
Mass (grams)
_______________ grams
_______________ grams
_______________ grams
_______________ grams
_______________ grams
_______________ moles
_______________ moles
9. Describe the behavior of anhydrous copper sulfate when water is added.
Post-Lab Questions
a. Compare the number of moles of anhydrous CuSO4 to the number of moles of water in the hydrate. Use the
ratio of these two values to predict a formula for the hydrated CuSO4 .
b. Why is it necessary to let the crucible cool before measuring mass? Why should the mass of the crucible be
measured immediately after the crucible cools, and not later?
c. In this experiment, you cooled your crucible in a desiccator. What is a desiccator? How does a desiccator
work?
d. How would your experimental results be affected if you did not use a desiccator when cooling the crucible
and contents?
e. How can you account for the behavior of the anhydrous form of the copper sulfate when the water was added?
What do you think the new substance is?
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