Common Experiments
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 Rf 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 Rf for the red dye and the blue dye.
Explanation
Just plug your numbers into the equation:
Rf for red dye:
Rf 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 (cm3) 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:
M1V1 = M2V2
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 coffee-cup calorimeter. The coffee-cup 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 coffee-cup 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 = mCpDT.
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 = mCpDT  (Cp = 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!
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