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General Chemistry I
CHEM-1030
Laboratory Experiment No. 3
The Empirical Formula of a Compound
(revised 05/21/2015)
Introduction
An initial look at mass relationships in chemistry reveals little order or sense. Mass ratios of elements
in a compound, while constant, do not immediately tell anything about a compound’s chemical
formula. For instance, water always contains the same proportions of hydrogen (11.11% by mass) and
oxygen (88.89% by mass) but these figures do not tell how the formula H2O is obtained.
A chemical formula is a usually a whole number ratio showing the relative numbers of moles of each
element present. A chemical formula can be determined from the mass of each element present in a
sample of compound by taking into account their relative atomic masses. The unit of chemical
quantity is not mass but the gram-molecular weight, or mole, which is the formula mass of a substance
expressed in grams. As a result, a mole of any substance can be weighed out. Moreover, a mole of
any substance contains the same number (Avogadro’s number, 6.022 x 1023) of formula units. For
example, a mole of U weighs 238.029 g and contains Avogadro's number of uranium atoms. One mole
of H2O weighs 18.015 g and contains Avogadro's number of H2O molecules. One mole of NaC1
weighs 58.443 g and contains Avogadro's number of NaC1 formula units, that is, NaCl ion pairs.
In many cases, it is experimentally straightforward to find the mass of each element in a sample of
compound. Dividing the mass of each element by the mass of one mole of the element (its atomic
mass) gives the moles of each element present. The resulting molar ratio, when simplified to small
whole numbers, gives the empirical formula of the compound. The molecular formula, which gives
the number of each kind of atom in a molecule of compound, may be identical to or a multiple of the
empirical formula. It is necessary to know the molecular weight of a compound to obtain the
molecular formula from the empirical formula. For instance, hydrogen peroxide has the empirical
formula HO. (This formula only means the hydrogen peroxide molecule contains equal numbers of
hydrogen and oxygen atoms.) Its molecular weight, however, is 34, not 17 Daltons. The 34-Dalton
formula mass corresponds to the molecular formula H2O2.
Example 1: To find the empirical formula of a sulfide of copper, 1.956 g of copper wire is heated
with a large excess of elemental sulfur. An excess of sulfur ensures that all the copper reacts.
Unreacted sulfur burns off to leave behind copper sulfide weighing 2.477 g. The calculations are as
follows:
Mass of Cu wire:
Mass of copper sulfide formed:
Mass of combined sulfur:
1.956 g
2.447 g
0.491 g
The molar amount of each element is found by multiplying its mass by the conversion factor derived
from the equality: one mol of any element = the atomic mass of the element expressed in grams.
1mol Cu
= 3.078 x10 -2 mol Cu
63.546g Cu
1molS
0.491gS x
= 1.53x10 -2 molS
32.066gS
1.956g Cu x
The molar amounts of each element depend on the amount of copper starting material. More important
is the molar ratio, which remains the same no matter how much starting material you have. Find the
smallest whole number molar ratio by dividing both molar amounts by the smaller of the two values.
3.078 x 10-2 mol Cu
= 2.01mol Cu
1.53 x 10-2
1.53 x 10-2 molS
= 1.00 molS
1.53 x 10-2
The molar ratio of the two elements is close to 2:1, so the empirical formula of the compound must be
Cu2S. This is an ionic compound; there are no molecules present so the question of a molecular
formula does not arise.
Example 2: Commercial analytical laboratories often provide the percent composition of a compound
sent in for analysis. (Example 2 does not apply directly to this experiment report.) Percent
composition values require an extra computational step to find the empirical formula. For instance, the
composition of an organic compound is reported to be 53.38% carbon, 11.18% hydrogen and 35.53%
oxygen by mass. For simplicity, assume that you have exactly 100 g of the compound so you can
work with the mass quantities 53.38 g C, 11.18 g H and 35.53 g O. Next, find the number of moles of
each element in the hypothetical 100-gram sample.
53.38g C x
1mol C
= 4.444 mol C
12.011g C
11.18g H x
1mol H
= 11.09 mol H
1.00794 g H
35.53g O x
1mol O
= 2.221mol O
15.9994 g O
As before, convert the molar amounts to ratios by dividing each molar amount by the smallest number
of moles present, in this case 2.221 (mol of oxygen).
4.444 mol C
= 2.001 mol C  2 mol C
2.221
11.09 mol H
= 4.993 mol H  5 mol H
2.221
2.221mol O
= 1.000 mol O  1 mol O
2.221
The empirical formula of this molecular compound is C2H5O, which has a 45.012 g/mol formula mass.
A separate experiment gives a compound molecular mass of 90.12 g/mol. Therefore, the molecular
formula must be twice the empirical formula or C4H10O2. Verify this for yourself as an exercise.
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At times, computations like the previous ones give decimal fraction molar ratios that need to be
converted to whole numbers by multiplying all the ratios by the same small whole number such as 2, 3
etc. Thus an atomic or molar ratio of 1:1.33:1 becomes 3:4:3. Similarly, NO1.5 becomes N2O3.
Percent Composition
Since percent composition is a common way of reporting analytical results, it is important to know
how to calculate percent compositions from analytical data. The mass fraction of each element present
is the mass of an element in the sample divided by the total sample mass. Each fraction multiplied by
100% gives the percent of that element in the compound.
For the 2.447 g sample of Cu2S in Example 1,
1.956 g Cu
x 100 % = 79.93% Cu by mass
2.447 g of compound
0.491g S
x 100 % = 20.1%Sby mass
2.447 g of compound
Experimental
You will do two determinations of the empirical formula of an oxide of tin, that is, a compound of tin
and oxygen alone. Since the final tin oxide product will be weighed in a porcelain crucible, it is best to
begin by weighing the tin metal in the same crucible. You will determine the mass of a sample of tin
and then treat the tin with concentrated nitric acid to obtain a wet tin oxide/tin nitrate mixture. You
will heat the mixture to drive off water and nitrogen oxides to leave behind only tin oxide. (Subtract
the mass of the empty crucible each time to get the mass of the tin and the mass of the tin oxide.) The
difference between the original mass of tin and the mass of the resulting tin oxide is the calculated
mass of oxygen in the compound.
Obtain a clean porcelain crucible. Support the crucible in a clay triangle and heat it to redness with a
strong, blue burner flame for about 4 min. Heating removes absorbed water from the crucible surface
and burns off any volatile chemicals from previous experiments. When an object is heated, it usually
loses some mass due to the release of absorbed water. (Since you will heat your crucible to convert tin
nitrate to tin oxide, you need to remove this absorbed water prior to starting the experiment.
Otherwise, the loss of water from the crucible surface will lead to an apparent smaller amount of
oxygen gained by the tin sample.) Use clean metal tongs to transfer the crucible to a wire gauze square
at your workstation and let the crucible cool to room temperature. Weigh the cooled crucible to 0.001
g on a hanging pan balance. Note: It is not necessary to zero in the balance but you must use the same
balance for all your measurements. Any mass error will be the same for each weighing and will cancel
out each time you subtract the mass of the empty crucible. However, you should check your balance
for consistency by moving the balance masses back to zero and reweighing the empty crucible three
separate times to make sure you get close to the same mass each time. Show your instructor your first
weighings before you proceed further. Do not move or adjust your balance after you begin using it.
Place pieces of tin foil totaling between 1.1 to 1.3 g in the heated and cooled crucible, loosely folded
so they half-fill the crucible. Determine the total mass of the tin and crucible three times on the same
balance you used before. In the fume hood, pour a 10-15 mL supply of concentrated nitric acid (conc.
HNO3) into a small beaker. In the hood, use a medicine dropper to transfer nitric acid from the beaker
to the tin in the crucible until all evolution of brown NO2 gas ceases, always holding the dropper
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upright to keep nitric acid away from the rubber bulb. (Do not put your eyedropper into the laboratory
acid supply bottle.) Be sure to wet all the tin with nitric acid but do not add a large excess of HNO3
solution. Be extremely careful of concentrated nitric acid. It can cause serious eye injuries and severe
skin burns. If any nitric acid gets on your skin, rinse it off immediately with water and notify your
instructor.
When a ring stand setup is available in the fume hood, use tongs to transfer your crucible to the clay
triangle on the iron ring. Be sure the ring is positioned about 5 cm above the top of the Bunsen burner
you will be using. Light and adjust the burner so you have a cool, blue flame with some yellow color
at its top. To begin driving off excess water and nitric acid, hold the burner by its base and carefully
wave the flame along the sides, not the bottom, of the crucible. Exercise caution and patience while
heating the wet product, because rapid heating will cause spattering. Observe the crucible closely for
signs of boiling and listen for a sizzling sound that may signal imminent spattering. (If a significant
amount of tin oxide spatters out, remove the crucible from the ring stand, let it cool, dispose of the
material as heavy metal waste and start the experiment over.)
When the tin oxide appears to be dry, heat it more strongly with longer applications of the flame to
drive off any remaining water. When you are sure no moisture remains in the tin oxide, hold a strong,
blue flame under the crucible. Finally, place the burner with a hot, blue flame under the crucible and
heat the crucible to redness. Note the time when evolution of NO2 gas ceases completely and heat the
crucible to redness for an additional 4 min. The tin oxide will acquire an uneven yellow color. With
tongs, remove the crucible from the hood, place it on a wire gauze square at your bench and let it cool
to room temperature before weighing it on the same balance you used before. Do not place the hot
crucible on the balance pan. Do not move or adjust your balance at any time during the experiment.
After you weigh the cooled crucible containing tin oxide, scrape the tin oxide out of the crucible and
dispose of it as heavy metal waste. Wash the crucible with soapy water and a test tube brush. Do a
second trial either concurrent with or after the first trial. Do a third trial if one of your determinations
was not done properly. Present only your two best trials in your report. Do not average your results;
report each trial separately.
Safety
Be especially careful of concentrated nitric acid. One drop in your eye can cause blindness. If you get
any nitric acid on your skin, rinse it off immediately to avoid a serious burn and inform your instructor.
If you spill a significant amount of nitric acid anywhere in the laboratory, do not clean it up with paper
towels or cloth. Ask your instructor for help. Concentrated HNO3 is a powerful oxidizing agent that
can ignite cellulosic materials. Chemical splash goggles and a waterproof apron must be worn at all
times during this and all chemistry experiments, from the very beginning to the very end of the time
you spend in the laboratory.
Disposal
Dispose of leftover nitric acid and the tin oxide product in the heavy metal hazardous waste container.
Cleanup
Remove the rubber bulb from your glass dropper and rinse the parts separately with tap water before
putting them away. At the conclusion of the lab period, wipe down all work surfaces with a damp
sponge. If any concentrated nitric acid contacts your sponge, rinse the sponge thoroughly with tap
water before you put it away.
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General Chemistry I
CHEM-1030
Laboratory Experiment No. 3
The Empirical Formula of a Compound
Alternate Experimental Procedure
Rationale
The tin oxide preparation produces toxic nitrogen dioxide gas and requires the use of a fume hood. If
enough laboratory hood space is not available, your instructor will assign an alternate procedure, the
determination of the magnesium oxide formula to some students to avoid long waits for hood space.
The magnesium oxide preparation described below can be safely done on an open laboratory bench.
Experimental
There are two principal differences between this magnesium oxide procedure and the one for tin oxide.
First, molten magnesium metal corrodes porcelain so it is preferable to use old crucibles to avoid
damaging clean new ones. To protect the crucibles from damage, you will put a layer of sand inside them.
Second, it is necessary to cover burning magnesium during most of the procedure to avoid losing
magnesium oxide, which, unlike tin oxide, may escape as a dense, white smoke. Covers limit the rate of
the reaction and capture most of the airborne magnesium oxide that would otherwise escape the system.
Wash and rinse an old crucible and a crucible cover. Choose a cover of a size that will span the entire
crucible opening and that you can easily adjust to control the amount of air entering the crucible. Place an
approximately 0.5 cm layer of dry sand in the bottom of the crucible to offer some protection from the
corrosive hot magnesium. Note: The presence of sand in the crucible will not affect your weighing
results. Sand is extremely resistant to heat and to chemical changes. Once you dry the sand at the start of
the experiment, its mass will not change and it will not react with the burning magnesium. Consider the
sand as part of the container in which you weigh your magnesium and magnesium oxide. As long as the
container mass does not change, the mass of a chemical you weigh in it will always equal the total system
mass minus the container mass.
Support the crucible containing sand with a cover in a clay triangle mounted on a ring stand and heat the
crucible to redness with a strong blue Bunsen burner flame for about 4 min. After the crucible, sand and
cover cool to room temperature, handle them with clean tongs and weigh them together to 0.001 g. Check
your balance for consistency as described on page 3 by weighing the crucible, cover and sand three or
more times until you get three close readings. If your balance does not read consistently, tell your
instructor. It is not necessary to zero in your balance. Do not move your balance, adjust the zero knob or
switch to another balance once you perform your initial weighing.
Place 0.8 to 0.9 g of magnesium turnings on the sand in the bottom of the cooled crucible and measure the
total mass of crucible, sand, magnesium and cover. Put the covered crucible back on the clay triangle and
heat the assembly strongly for ten minutes with the cover slightly ajar. The magnesium will glow as it
reacts with air and forms a mixture of magnesium oxide and magnesium nitride. Do not remove the cover
completely because this will admit too much air and accelerate the rate of the reaction. This will allow
some combustion products to escape as visible white smoke, reducing the amount of oxide in the crucible.
After ten minutes, stop heating and remove the cover to admit air. The magnesium may flare up slightly
in response to incoming oxygen. With the cover off, reheat the crucible until the magnesium stops
glowing. Then heat the crucible with the cover ajar to redness for 4 min more to complete the reaction.
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From now on, handle the crucible carefully in case it has cracked from the corrosive effects of hot
magnesium. A crack will not affect your results as long as the crucible does not break apart and spill its
contents.
Let the crucible cool to room temperature. Add enough distilled water to wet the oxide/nitride mixture
and sand completely. This will convert the magnesium nitride side product to magnesium hydroxide and
ammonia (Equation 1). Waft some of the escaping vapors toward you to smell the ammonia. Carefully
heat the crucible with the cover ajar to evaporate all the water without spattering the product. Then heat
the crucible to redness with the cover ajar for 4 minutes to drive off residual water vapor and convert any
magnesium hydroxide present to magnesium oxide (Equation 2). Let the crucible cool to room
temperature and determine the total mass of crucible, cover, sand, and magnesium oxide.
Mg3N2(s)+ 6 H2O(l)  3 Mg(OH)2(s) + 2 NH3(g)
Mg(OH)2(s)  MgO(s) + H2O(g)
(Equation 1)
(Equation 2)
After you weigh the cooled crucible, scrape out the sand and magnesium oxide and dispose of them as
heavy metal waste. Neither component in the crucible meets the criteria of hazardous waste, but is
preferable to dispose of them this way, rather than as ordinary trash. Do a second trial either
concurrent or after the first trial. Your instructor will examine your data to see if you did the experiment
correctly. Do a third trial if one the instructor feels one of your determinations did not go well. Report
only your two best trials separately. Do not average your results.
Record your data on the page provided for tin oxide. Wherever the word tin appears on the data page and
elsewhere, change it to magnesium. In your report, convert the mass of magnesium to moles of
magnesium and the mass of oxygen in the compound (mass of magnesium oxide minus mass of
magnesium) to moles of oxygen. The Mg/O molar ratio, expressed as small whole numbers, will give the
empirical formula of the magnesium compound.
Safety
Chemical splash goggles and a waterproof apron must be worn at all times during this and all
chemistry experiments, from the very beginning to the very end of the time you spend in the
laboratory.
Disposal
Dispose of the sand and magnesium oxide product in the heavy metal hazardous waste container as
explained above.
Cleanup
Use a test tube brush and soapy water to wash the crucibles and rinse them with tap water. Keep one
crucible in your laboratory locker and return all others to the laboratory supply area. At the conclusion
of the lab period, wipe down all work surfaces with a damp sponge.
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General Chemistry I
CHEM-1030
Laboratory Experiment No. 3
The Empirical Formula of a Compound
Data Page
Trial Number
1
2
Mass of Empty Crucible
Mass of Empty Crucible
Mass of Empty Crucible
Mass of Empty Crucible
(average)
Mass of Crucible and Tin
Mass of Crucible and Tin
Mass of Crucible and Tin
Mass of Crucible and Tin
(average)
Mass of Tin
(By Subtraction)
Mass Crucible & Tin Oxide
Mass Crucible & Tin Oxide
Mass Crucible & Tin Oxide
Mass Crucible & Tin Oxide
(average)
Mass of Tin Oxide
(By Subtraction)
Mass of Oxygen
(By Subtraction)
7
3
4
8
General Chemistry I
CHEM-1030
Laboratory Experiment No. 3
The Empirical Formula of a Compound
Report
For full credit, clear, concise setups are required for all calculations. Express all answers with proper
units, the proper number of significant figures and, where appropriate, in proper scientific notation.
1. Calculate the empirical formula of tin oxide using the data from your two best trials.
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(12 points)
2. Calculate the percent composition of the tin oxide from the masses of the elements in your two
best trials.
(4 points)
3. If some tin oxide were lost by spattering, would it appear that your oxide product contained more
or less oxygen? Explain the reasoning for your answer.
(2 points)
4. For one of your trials, use the experimental masses of tin and oxygen to calculate the number of tin
atoms and the number of oxygen atoms in your tin oxide sample.
(2 points)
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1)
General Chemistry I
CHEM-1030
Laboratory Experiment No. 3
The Empirical Formula of a Compound
Prestudy
What is the difference between an empirical and a molecular formula?
(1 point)
2)
What is the proper empirical formula for the compound ClO3.5?
(1 point)
3)
A compound has the empirical formula CH2O and a molecular weight of 150.0 Daltons. What
is the molecular formula?
(1 point)
4)
Calculate the mass of 1.00 mole of F2.
(1 point)
5)
How many moles of cobalt are contained in 4.780 g of Co?
(1 point)
6)
What is the empirical formula of a compound that consists of 38.35% Fe and 61.66% O? Show
your calculation steps.
(3 points)
7)
What is the percent composition of a compound if a 3.25 mg sample contains 3.00 mg of
carbon and 0.25 mg of hydrogen? Show your setup.
(2 points)
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