


Common Experiments
Chromatography
The purpose of chromatography is to separate out parts
of a solution—to isolate substances. You might have used paper chromatography
in your chemistry lab. In paper chromatography, a small drop of
the substance to be separated is placed on one end of the chromatography
paper. A pencil is used to mark the spot where the substance was
placed, and then the tip of the paper is placed into a container
with solvent. As the solvent travels up the paper, the substance
separates into its various components. Whatever component is most
like the solvent travels the greatest distance. At the end of the
experiment, measurements are taken of how far each component traveled.
The distance that the solvent traveled and the distance that the
solutes (the components) traveled are usually measured in centimeters.
A ratio, called the R_{f} value,
is then calculated for each component. This information can be used
to identify various parts of the mixture.
The formula for the calculation is
Example
Data:
Distance solvent traveled:  0.0 cm 
Distance red dye traveled:  7.0 cm 
Distance blue dye traveled:  4.0 cm 
Calculate the R_{f} for
the red dye and the blue dye.
Explanation
Just plug your numbers into the equation:
R_{f} for
red dye:
R_{f} for
blue dye:
Density of Liquids and Solids
Density is defined as a pure substance’s
mass over its volume. Density is a property of matter that is often
used to identify an unknown substance since pure substances have
known densities. The units of density are usually grams divided
by milliliters or cubic centimeters:
Density of a solid: Typically the solid sample
is massed on the balance first. The mass is recorded in grams. If
the solid is a regularly shaped object, the length, width, and height may
be measured with a metric ruler. These three measurements are then
multiplied together to obtain the volume in cubic centimeters (cm^{3})
or some similar unit. If the solid is irregular, the volume can
be obtained by water displacement. A known amount of water is recorded,
the object is immersed, and the final volume of water is recorded.
The difference in volumes will give the volume of the object. Density
can then be calculated by dividing the mass by the volume.
Density of a liquid: The density of a liquid
is obtained in much the same way as above. To obtain the mass of
the liquid, the mass of a container must first be measured, the
liquid poured in, and the total mass recorded. The difference in
mass is the mass of the liquid. It is often convenient to measure
the liquid in a graduated cylinder. Now try a density problem.
Example
Data (for an irregular solid):
Mass of the solid:  5.00 g 
Initial volume of water:  30.0 mL 
Final volume of water:  32.5 mL 
Find the density of the unknown solid.
Explanation
Volume of solid: (final  initial volumes) = 32.5  30.0
= 2.5 mL
Density of solid:
Titration
A titration (also called volumetric analysis)
is a laboratory procedure that usually involves either an acid and
base neutralization reaction or a redox reaction. In a titration,
two reagents are mixed, one with a known concentration and known
volume (or a solid with a known mass) and one with an unknown concentration.
The purpose of a titration is to find the concentration of the unknown
solution. There must be some way to indicate when the two reagents
have reacted essentially completely, and at the end of the titration
the unknown solution’s concentration can be calculated since the
volume of the solution required to complete the reaction has been
accurately measured.
The titrant is the solution of known concentration
and is usually placed in the burette. The burette must be rinsed
with the solution to be placed in it before filling.
The solution from the burette is added to a flask that
contains either a measured volume of a solution or a weighed quantity
of solid that has been dissolved. An indicator that changes color
at or near the equivalence point is usually added to the solution
to be analyzed before titration. The solution of known concentration
is then added to the flask from the burette until the color changes.
The equivalence point is the point in the reaction
where enough titrant has been added to completely neutralize the
solution being analyzed. The end point is the point
during the titration where the indicator changes color. It is important
to choose an indicator that has an end point that is at the same
pH as your expected equivalence point. The burette has graduations
that are used to read the volume of titrant that’s added to the
flask.
The data required for titrations include the mass of the
dry substance to be analyzed or an accurately measured
volume of the substance to be analyzed, the initial volume
and final volume of titrant required to reach the
end point, and the molarity of the titrant. At the equivalence point,
the moles of the titrant will be equal to the moles of the substance analyzed.
To obtain the moles of the unknown substance, multiply the molarity
of the titrant by the volume (in liters) of the titrant. Once moles
are known, just divide moles by volume and you have the molarity
of the unknown substance
If the substance to be analyzed is a solid, you will be
trying to calculate the molecular weight of the unknown solid. Remember
that molecular weight is grams per mole. The mass in grams will
be known from the beginning of the experiment, when the solid sample was
massed. You can find the moles of the unknown substance by multiplying
the molarity of the titrant by the volume (in liters) of the titrant.
Divide grams by moles to get molecular weight.
If you are doing a titration of a strong base with a strong
acid, the equivalence point occurs at a pH of 7.00. The dilution
formula can be used to calculate the moles of acid, which will equal
the moles of base at the equivalence point:
M_{1}V_{1} = M_{2}V_{2}
or moles of acid = moles of base. Don’t
try to use this formula if either the acid or base is weak!
Example
Data:
Volume of unknown acid sample:  10.0 mL 
Initial volume of titrant (base):  0.0 mL 
Final volume of titrant (base):  20.0 mL 
Molarity of titrant (base) (must be given):  1.0 M 
Find the molarity of the unknown acid solution.
Explanation
Don’t forget to change mL to L.
0.020 L (titrant)1.0 mole/liter (titrant molarity)
= 0.020 mole of base titrant
At the equivalence point: moles acid = moles base = 0.020
mole of acid
Calorimetry
Calorimetry is used to determine the amount of heat released
or absorbed during a chemical reaction. In the lab we can experiment
with finding the energy of a particular system by using a coffeecup
calorimeter. The coffeecup calorimeter (shown below) can
be used to determine the heat of a reaction at constant (atmospheric)
pressure or to calculate the specific heat of a metal. The coffeecup
calorimeter is a double plastic foam cup with a lid; the lid has
a hole in it where the thermometer pokes through.
The data to be collected include the volumes of the solutions
to be mixed, the initial temperatures of each solution, and the
highest temperature obtained after mixing. Accurate results depend
on measuring precisely and starting with a dry calorimeter. The
total volume recorded must be changed into grams (use the density
and multiply densityvolume = grams).
The change in temperature must be calculated by subtracting the
final and initial temperatures. To find the heat of reaction, multiply
the specific heat capacity, the mass, and the change in temperature: q = mC_{p}DT.
Example
Data (for a specific heat of a
metal):
Mass of the solid metal:  24.00 g 
Initial temperature of the metal:  100.0ºC 
Mass of water in plastic foam cup:  100.00 g 
Initial temperature of the water:  25.0ºC 
Highest temperature of the water:  30.0ºC 
Find the specific heat of the metal.
q = mC_{p}DT
(C_{p} = 4.18
J/g ºC)
Explanation
Temperature change for the metal = 100.0  30.0 = 70.0ºC
Temperature change for the water = 30.0  25.0 = 5.0ºC
We can assume that the heat lost by the metal should equal
the heat gained by the water.
Calculate the heat gained by the water:
Heat gained by water = (sp. heat water)(mass
of water)(DT water)
= (4.18 J/g ºC)(100.0 g)(5.0ºC)
= 2090 joules gained
Then find the specific heat of metal:
(Small values for metals are very typical!)
Stoichiometry
Many experiments require the use of stoichiometry to
find an unknown. One typical experiment is the neutralization of
an acid with a base to produce a salt and water. If a known volume
and concentration of an acid and a base are reacted, the amount
of salt produced can be predicted. Typical data would require the
accurate recording of molarities and volumes of the acid and the
base. The mass of the reacting vessel must be recorded, and then the
two solutions are mixed. The mixed solution is then evaporated over
a low flame (to avoid spattering and loss of mass) to dryness. The
vessel is allowed to cool, and the final weight is then recorded.
The vessel should be heated to a constant mass in this type of experiment.
This requires heating, cooling, and weighing until two consecutive
measurements are within an acceptable range of each other. The mass
of the solid salt obtained is calculated by subtracting the mass
of the vessel from the total mass of the vessel with the salt. This
mass is known as the actual (or experimental) yield. To find the
theoretical yield (what should have been produced), a balanced chemical
equation must be written. The moles of each substance reacted must
be determined (multiply molarity by volume) and also the moles of
the limiting reagent (using coefficients). From the moles of limiting reagent,
multiply by the mole ratio and then convert to grams using the formula
weight of the salt from the periodic table. This gives the theoretical
yield of salt. It is typical to calculate the percent yield. The
closer to 100%, the better your results were!
