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Experiment 8
Determination of Empirical Chemical Formulae
OBJECTIVES:
1. To determine the formula for the compound between magnesium and oxygen by determining the
weight of oxygen (from air) that reacts with a weighed amount of magnesium ribbon.
2. To determine the formula (that is, the number of molecules of water in a mole) for the hydrate of
magnesium sulfate (epsom salt).
Dalton’s atomic theory was consistent with the experimental laws of definite composition and multiple
proportions which gave it great credibility. However, it did not indicate what the weights of the postulated
atoms were nor what the formulae for the compounds they formed were. Consider the following:
1. Weight percent composition of compounds (which elements and how much of each)
2. Weights of atoms (relative or absolute)
3. Formulae of compounds
If one knows any two of these three, the other can be calculated. The problem facing chemists in the period
from 1800 to 1855 was that only #1 was known from experimental data. There were opinions about the other
two but no conclusive experimental evidence. Cannizzaro (in 1857) first worked out an interpretation of
existing data that gave both the formulae of compounds and the atomic weights of atoms simultaneously. He
used Avogadro’s hypothesis (first given in 1811, no less) about equal volumes of gases containing equal
numbers of molecules at the same temperature and pressure. But that’s a story for another day.
In this experiment the atomic weights will be assumed to be known and the percent composition (or
equivalent data) will be determined in the laboratory. Using these, the formula for the compound will be
calculated.
A symbol is assigned to each element but when a “symbol” for a compound is desired, it is more useful and
convenient to symbolize the compound by indicating which elements are in the compound and how much of
each is present. The symbols of the elements are used to indicate which elements are in the compound; how
much of each one is present is given in terms of the relative number of atoms; this representation is called the
formula for the compound.
Thus the formula Fe2O3 indicates that the compound contains the elements iron and oxygen and that there are
2 iron atoms for 3 oxygen atoms. The formula SO2 indicates a compound composed of sulfur and oxygen
with 2 atoms of oxygen for one atom of sulfur. The use of symbols and formula have evolved over the
decades and present meanings are given here.
Avogadro’s number (6.02 x 1023, an almost incomprehensibly large number) of atoms or molecules is called a
mole of the element (if atoms) or of the compound (if molecules). If one takes an amount of the compound
Fe2O3 that contains 2 x (6.02 x 1023) atoms of iron, then there will be 3 x (6.02 x 1023) atoms of oxygen.
These amounts are 2 moles of iron and 3 moles of oxygen and reacted together they are one mole of Fe2O3.
The subscripts on the symbols in a formula give the relative number (ratio) of atoms that exist in the
compound and also give the relative number of moles of each element that exists in the compound. The
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formula Fe2O3 says therefore that 2 moles of iron will require 3 moles of oxygen to react to give one mole of
the compound Fe2O3, assuming that neither oxygen nor iron will be left over. One can discuss the concept
of limiting reagent starting here, but that will be done in a later experiment.
Two moles of iron is 2 molar masses of iron which is 2(55.85) =111.7 grams of iron. Three moles of oxygen
is 3 molar masses of oxygen which is 3(15.9994) = 47.998 grams of oxygen. One mole of Fe2O3 is 159.7
grams (111.7 + 47.998). Thus iron and oxygen will react to form Fe2O3 in the ratio of 111.7 g Fe to 47.998 g
O or 2.327 grams of Fe for 1.00 gram of oxygen.. All this information is present in the formula if one has a
table of atomic weights and knows how to read the information in the formula.
EXAMPLE OF THE CALCULATION OF THE EMPIRICAL OR SIMPLEST FORMULA OF A METAL
OXIDE FROM EXPERIMENTAL DATA
It is found that 2.325 grams of metallic yttrium (molar mass 88.905) combines with oxygen to produce 2.953
grams of yttrium oxide. Calculate the formula for this compound.
total weight of Y and O (the oxide)
weight of Y in the oxide
weight of oxygen reacted
=
=
2.953 g
2.325 g
0.628 g
In the sample of 2.953 grams of yttrium oxide there are then
(3-1)
moles of Y =
2.325 g
= 0.02615 moles
88.905 g/mole
(3-2)
moles of O = 0.628 g
= 0.03925 mole
15.9994 g/mole
There are 0.02615 moles of Y and 0.03925 moles of O in the 2.953 grams of yttrium oxide. The formula for
yttrium oxide could be written
(3-3)
Y0.02615O0.03925
This formula gives the relative ratios of moles. However, it is much more useful if the ratio (the subscripts
on the symbols) is given in integers rather than decimals. The number moles of oxygen present if there is
one mole of yttrium present can be calculated from
(3-4)
0.02615 mole Y
0.03925 mole O
=
1 mole Y
x mole O
Solution of this equation gives x = 1.5 moles O per mole of Y or 3 moles of oxygen for 2 moles of Y. This
leads to the formula Y2O3 for yttrium oxide.
Note that solution of eq (3-4) for the moles of oxygen per mole of yttrium gives the same result as the
following:
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Mole ratios
(3-5)
Y
0.02615 mole Y
0.02615 mole Y
(3-6)
O
0.03925 mole O
0.02615 mol Y
=
=
1.000 mole Y/mole Y
1.501 mole O/mole Y
Integer mole ratios
2
3
The usual way this problem is worked is using eqs (3-5) and (3-6). One divides the moles of each
component by the number of moles of the component with the lesser or least number of moles. The resulting
numbers give the number of moles of each element present for one mole of the element with the least number
of moles in the compound.
The formula Y2O3 obtained above is called the empirical (or simplest) formula for yttrium oxide. 2 and 3 are
the smallest integers that give the correct ratio of atoms (or moles) in the compound. 4 and 6 are another set
that give the correct ratio but they are larger than 2 and 3. The formula with the smallest ratio is the
empirical formula. Empirical formulae obtained as above do not allow the actual calculation of the number
of atoms of each element in a molecule of the compound. The formula that shows the actual number of
atoms is called the molecular formula.
The empirical formula and molecular formula are the same thing for ionic compounds since molecules do not
exist in ionic compounds. In covalent compounds the empirical formula and the molecular formula may be
different. Thus the molecular formula for hydrogen peroxide is H2O2 and the empirical formula is HO. The
number of atoms of each element in a molecule of the molecular formula is an integral multiple of the
number of that element in the empirical formula.
HYDRATES
The formulae CuSO4.5H2O, CoCl2.6H2O, and NiSO4.6H2O are typical of the formulae for many metal
compounds. These compounds are called hydrates because of the molecules of water involved in their
formula. Some (sometimes all) of the water molecules in these compounds are chemically bonded to the
metal atoms by coordinate covalent bonds.
Some of the water molecules are simply part of the crystal structure of the solid compound. In CuSO4.5H2O,
four of the waters are attached to the Cu(II) ion and the other one is part of the solid crystal structure. In fact,
it is the water molecules (attached to the Cu(II) ion) that makes the copper(II) sulfate blue. Removing the
water causes the compound to change from blue CuSO4.5H2O to white CuSO4. The number of water
molecules per molecule of metal compound is more often an even number (2, 4, or 6) than an odd number (1,
3, 5, 7).
The water molecules in hydrates can often be removed by heating but not always. If CrCl3.6H2O is heated,
HCl is evolved instead of water and one has Cr2O3 left instead of CrCl3. The bond between the chromium
and the oxygen is very strong relative to that between chromium and chlorine.
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EXAMPLE OF THE DETERMINATION OF THE NUMBER OF WATER MOLECULES IN A
HYDRATE
An empty crucible is weighed, a sample of BaCl2.nH2O is added and the crucible is weighed again. After
heating (to constant weight - see below for discussion of heating to constant weight) to remove all the water
molecules, the crucible is weighed again. Typical results are given below:
(3-7) weight of evaporating dish + BaCl2.nH2O.......................................... 35.913 grams
weight of evaporating dish + BaCl2 (after heating and cooling)
1st weighing 35.895 grams
2nd weighing 35.882 grams
(3-8)
3rd weighing 35.880 grams
(3-9) weight of empty evaporating dish......................................................35.681 grams
From these data one can calculate
(3-10)
weight of sample of BaCl2.nH2O (3-7 minus 3-9)
= 35.913 - 35.681 = 0.232 grams
(3-11)
weight of BaCl2 (3-8 minus 3-9)
= 35.880 - 35.681 = 0.199 grams
(3-12)
weight of water lost (3-7 minus 3-8)
(Or 3-10 minus 3-11)
= 35.913 - 35.880 = 0.033 grams
These data can be used to calculate the empirical formula for the hydrate of barium chloride:
Mole ratio
(a) BaCl2
0.199 g
208.24 g/mole
= 9.556 x 10 -4 moles
9.556 x 10-4
9.556 x 10-4
= 1.00 mole BaCl2
per mole BaCl2
1.832 x 10-3
9.556 x 10-4
= 1.92 moles H2O per
mole BaCl2
(3-13)
(b) H2O
0.033 g
18.015 g/mole
= 1.832 x 10-3 moles
Allowing for experimental errors and uncertainties, and 1.92 is close to 2, one takes the value of n = 2 and
the formula for the hydrate is BaCl2.2H2O. The compound is barium chloride dihydrate.
From the above data the percent water in BaCl2.2H2O is
(3-14) % water =
.
Gms water
(100) =
Total grams BaCl2.2H2O
(0.033)(100) = 14.22 % water
(0.232)
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HEATING TO CONSTANT WEIGHT
In the above example about BaCl2, the crucible containing the hydrate is heated to remove the water
molecules. If one heats for a time and then weighs the crucible, it is not certain that all the water has been
driven off. The sample is then heated again, cooled and weighed. If the weight is the same or close, then all
the water has been removed. If significant weight is lost during the second heating then a third heating is
done. This is continued until the weight reaches a constant value at which time it is assumed that all the
water has been removed. This is called heating to constant weight. One must heat at least two times but
more than two may be necessary.
When a weighing is obtained that is the same as the previous weighing or different from it by 0.001 to 0.002
gram (for this course), then this can be taken as sufficient and further heating is not needed. This is true
beginning with the second weighing.
ALWAYS ALLOW AN OBJECT TO COOL BEFORE WEIGHING IT ON THE BALANCE. IF IT IS
COOL TO THE TOUCH, THEN IT CAN BE WEIGHED. HOT OBJECTS HEAT THE AIR AROUND
THEM WHICH CHANGES THE BUOYANCY OF THE OBJECT (ARCHIMEDES PRINCIPLE) AND
COULD AFFECT THE MASS VALUE OBTAINED.
PRECAUTIONS SUCH AS THESE ARE TAKEN NOT BECAUSE IT IS CERTAIN THAT THE
RESULTS WILL BE SIGNIFICANTLY AFFECTED IF THEY ARE NOT TAKEN BUT BECAUSE ONE
IS CONCERNED THEY MIGHT BE.
APPROXIMATELY ACCURATELY
In the example above with yttrium it is not necessary that the weight of yttrium be exactly 2.325 grams.
Certainly any weight near this (in this case not even too close) would have worked as well since the amount
of oxygen reacted would have been adjusted by nature. The calculations would have lead to the same result
no matter what the weight (within wide limits) of yttrium used. However, the weight of yttrium that is
used must be known accurately in order to get good results. One can say to use approximately accurately
2.5 grams of yttrium for the experiment. This means that the weight of yttrium can be in the neighborhood of
2.5 grams but does not have to be exactly 2.5 grams; the 2.325 grams used above is approximately 2.5 grams
but is accurately known. Whatever the mass used it must be known accurately. It is also true that the weight
of BaCl2.nH2O above (eq 3-10) does not need to be exactly 0.232 gram.
The above is commonly found when doing experiments in the laboratory. The phrase approximately
accurately will be used to describe the amount of weight to use in these contexts.
REACTION OF Mg WITH AIR
When Mg is heated in air, the major reaction that occurs is with the oxygen in the air to form MgO.
However, air is 80% N2 and at the high temperatures of reaction, some Mg reacts with N2.
(3-15) Mg + O2 ---------> MgO
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(3-16) Mg + N2 ---------> Mg3N2
(Magnesium nitride)
If one is going to weigh the result and assume that it is all MgO, there will be an error introduced. This can
be corrected by allowing the crucible to cool somewhat and adding water.
(3-17)
Mg3N2
+
H2O --------> Mg(OH)2 + NH3
(ammonia or nitrogen trihydride)
Heating the resulting magnesium hydroxide converts it into magnesium oxide and drives off the ammonia.
(3-18)
Mg(OH)2 ----------> MgO + H2O
(heat)
At this point all the solid in the crucible is MgO.
The reaction of magnesium with the nitrogen above is an example of a side reaction. A side reaction is one
that occurs along with the desired reaction. When gasoline burns in oxygen in an automobile engine the
temperature becomes very high (1000 to 1500 oC) and the nitrogen and oxygen of air react in a side reaction
to form NO and NO2. These two compounds (referred to collectively as NOx) are instrumental in the
formation of atmospheric smog as well as possible respiratory distress for people.
The magnesium oxide experiment described here is similar to the yttrium oxide example used above.
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LABORATORY PROCEDURES
YOUR INSTRUCTOR WILL CHOOSE PART I OR PART II (OR POSSIBLY BOTH).
SAFETY WARNINGS:
HOT OBJECTS –
High temperatures from hot plates and laboratory burners can cause severe burns. Use proper care when
handling hot objects and when near hot plates or burners. Only handle hot glassware with tongs or
approved devices.
Always allow an object to cool before weighing it on the balance. If it is cool to the touch, then it can be
weighed. You should test to see if objects are hot without touching them. Bring the back of your hand
near the evaporating dish, and feel for any heat. If you feel heat, it is still too hot to handle.
I.
DETERMINATION OF THE FORMULA OF MAGNESIUM OXIDE
NOTE: It is most efficient if two samples are run simultaneously. One crucible can be heated while
waiting for the other to cool. Your instructor will direct you if the experiment is to be performed in this
fashion.
1. Obtain a piece of magnesium ribbon weighing approximately 0.5 gram. This is a piece about 60 to
65 cm long. Clean the ribbon with steel wool or equivalent until it is shiny.
2. Obtain a crucible and cover; remove any material from the bottom. Then heat on the burner for a
couple of minutes. Allow to cool. Tare the balance. Place the crucible (with cover) on the balance and
record the mass to 3 decimal places.
3. Cut the magnesium ribbon into pieces about 3/4" to 1" in length. Bend each of them into a V shape
and place them in the crucible. The V shape prevents the pieces from stacking with large areas of their
surfaces touching; the stacking should be done so that this is true.
4. Weigh the crucible and cover with the magnesium in the crucible.
5. Put the crucible on a clay triangle on a ring attached to a ringstand (or use a tripod). Heat the
crucible with the burner with the cover on but slightly ajar. At a certain temperature the magnesium
pieces will begin to glow (they are burning). At this time smoke (which is MgO in the air) may appear.
Heat the crucible and open and close the cover so as to keep the magnesium glowing but minimizing
the escape of the MgO smoke. The burner may be removed and replaced as well as opening and closing
the cover in order to maintain control. Heat until the magnesium stops glowing. Now, heat for several
minutes with the burner close enough to the crucible to keep the bottom of the crucible at red heat. (A
hot flame is important. The tip of the inner blue cone should touch the bottom of the crucible.)
6. Allow the crucible to cool sitting on the ring or tripod. Break up the large particles with a stirring rod
avoiding loss of the solid.
7. Add about 10 drops of distilled water using a medicine dropper or a disposable pipet. Spread the
drops of water over the whole of the solid in the crucible. It may be possible to smell ammonia coming
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from the crucible at this time.
8. Cover the crucible and reheat (carefully at first until the water has reacted or evaporated) for several
minutes with the bottom of the crucible at red heat.
Heating to a constant weight.
9. Remove the crucible and cover from the tripod and sit it on a clay triangle on the bench to cool.
Then weigh it and record the weight. Don’t handle the crucible while cooling.
10. Place the crucible and cover on the tripod and heat very hot again for 2 or 3 minutes. Continue to
heat, cool and weigh until a constant weight is obtained. (When a mass is obtained that is the same as
the previous mass or different from it by 0.001 to 0.002 gram, then this can be taken as sufficient and
further heating is not needed. This is true beginning with the second weighing).
11. Repeat the experiment and using data from two runs, calculate the formula for magnesium oxide.
II.
DETERMINATION OF THE FORMULA OF EPSOM SALT (MgSO4.nH2O)
NOTE: It is most efficient if two samples are run simultaneously. One crucible can be heated while
waiting for the other to cool. Your instructor will direct you if the experiment is to be performed in this
fashion.
1. Obtain a clean evaporating dish and heat on the burner for a few minutes. Allow to cool. Zero the
balance and place the clean dry evaporating dish on the balance. Record the weight of the dish.
2. Place approximately accurately 1.2 grams of epsom salt into the dish and record the weight of the
dish and compound.
3. Place the dish on a ring or tripod, cover with a clean ribbed watch glass, and heat. Heat gently. As
the water is driven out of the solid particles some of it will condense on the watch glass. The crystals of
epsom salt will probably split and spatter as the water escapes, jumping out of the dish (if no cover is
used) or they may stick to the wet watch glass, both of which causes an error. Keep increasing the
temperature slowly, however, while minimizing the amount of spattering loss.
4. When the water appears to have been removed (the watch glass is dry), heat with a hot flame (inner
blue cone just below the bottom of the evaporating dish) for 2 or 3 minutes.
5. Remove the watch glass cover. If there are any crystals of epsom salt on the underside of the watch
glass brush them off into the evaporating dish with a glass rod and lean the watch glass against the base
of the ring stand to cool. Remove the evaporating dish from the tripod and sit it on a clay triangle on
the bench to cool. Then weigh it and record the weight. Don’t handle the evaporating dish while
cooling.
6. Place the evaporating dish (the watch glass cover is not needed after the first heating) on the tripod
and heat very hot again for 2 or 3 minutes. Continue to heat, cool and weigh until a constant weight is
obtained.
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7. Using a new sample of epsom salt, repeat the experiment (steps 2 to 6) at least once more.
8. Using the data obtained, calculate the number of water molecules per “molecule” of MgSO4 in epsom
salts for each run separately. Average the results for two runs.
REPORT
Give all data and calculations and any other things in a form as requested by your instructor.
QUESTIONS AND PROBLEMS
1. For a 5.0 gram sample of each of the following, what weight of water will be lost if the sample is heated to
constant weight: CoCl3.9H2O; CuSO4.5H2O; ZnSO4.4H2O; BaCl2.2H2O.
2. Heating the empty evaporating dish (step II-1 above) before using it removes moisture and other material
that might add an error to the experimental results. Suppose the dish was not heated and it was weighed
initially containing contaminants that were lost during later heating of the hydrated compound. How would
this affect the results?
3. In steps 1 and 2 of part II above one could determine the weight of MgSO4.7H2O used by tarring the
balance with the evaporating dish on it, then adding some epsom salt until approximately accurately 1.2
grams have been added as read from the digital readout of the balance. While this will accurately determine
the weight of sample taken, why is this not a good procedure in this case?