Materials Included In Kit

Additional Materials Required

Prelab Preparation

Activity A. Phase Change—Cooling Curve
Lauric acid samples

• Obtain about 6 g of lauric acid in a weighing dish and transfer the solid to a clean, dry test tube. Prepare two samples, one for each workstation.

Activity D. Properties of Metals—Crystal Structure and Heat Treatment
Connectors, 5-cm

• Take one 40-cm floral wire and cut it into 5-cm lengths using a wire cutter or heavy-duty scissors. Repeat this for six more wires to produce 54 5-cm connectors.

Safety Precautions

Ethyl alcohol, acetone, hexane, heptane and isopropyl alcohol are flammable liquids and a dangerous fire risk. The addition of denaturants makes ethyl alcohol poisonous. Avoid contact of all liquids with heat, flames or other sources of ignition. Acetone and heptane are slightly toxic by ingestion or inhalation. Do not allow chemicals to come into contact with eyes and skin. Perform this experiment in a well-ventilated lab only. Exercise care when working with hot metals. Wear chemical splash goggles whenever working with chemicals, heat or glassware in the laboratory. Please review current Safety Data Sheets for additional safety, handling and disposal information. Remind students to wash hands with soap and water before leaving the lab.

Disposal

Please consult your current Flinn Scientific Catalog/Reference Manual for general guidelines and specific procedures, and review all federal, state and local regulations that may apply, before proceeding. Lauric acid may be disposed of according to Flinn Suggested Disposal Method #24a. Alternatively, the lauric acid samples may be recycled from class to class and also from year to year. To recycle the samples from one lab period to the next, simply collect the test tubes at the end of the period and place them in a central hot water bath—the test tubes will be ready to use in Part A, step 6. To recycle the lauric acid samples for future use, stopper the test tubes and store them in a labeled container. Isopropyl alcohol and ethyl alcohol solutions may be rinsed down the drain with plenty of water according to Flinn Suggested Disposal Method #26b. Hexanes, heptanes, and acetone may be disposed of according to Flinn Suggested Disposal Method #18a.

Lab Hints

• For best results, set up two stations for each activity throughout the lab. This will allow eight groups of students to rotate through four activity stations in a 50-minute lab period, if needed. A double lab period (two 50-minute class periods) will allow time both for a review of basic properties of solids and liquids before lab and for a collaborative class discussion after lab.
• The activities may be completed in any order. Also, since each activity is a self-contained unit, the experiment may be set up with as many or as few of the activities as the teacher desires. Students should need only 8–10 minutes per station—keep the pace fairly brisk to avoid dawdling. Questions in the Observations and Analysis section may be answered during downtime between stations.
• Prelab preparation is an essential component of lab safety, and it is also critical for success in the lab. (Standing in front of the lab station is not a good time for students to be reading the activity for the first time.) Having students complete the written prelab assignment for this lab will help teachers ensure that students are prepared for and can work safely in the lab.
• In Activity A, the experiment may also be carried out using solid, low-melting organic “unknowns” for the students to identify. Suitable unknowns that may be used in addition to lauric acid include cetyl alcohol (1-hexadecanol, C₁₆H₃₃OH, mp 54–56 °C), stearic acid (octadecanoic acid, C₁₇H₃₅CO₂H, mp 67–69 °C), and BHT (butylated hydroxytoluene, 2,6-di-tbutyl-4-methylphenol, C₁₅H₂₃OH, mp 69–71 °C). Phenyl salicylate (salol, HOC₆H₄CO₂C₆H₅, mp 44–46 °C) has a melting point close to that of lauric acid. Acetamide, a traditional favorite for melting point and heat of fusion experiments, is classified as a possible carcinogen and is not recommended.
• In Activity A, the experiment may be extended to determine the heat of fusion of lauric acid. See “Hot Wax” in the Demonstrations section of the Flinn ChemTopic™ Solids and Liquids, Volume 11, for a sample calorimetry procedure and results. The heat of fusion procedure gives good results but requires a larger sample size (about 10 g of lauric acid), which increases the time needed to complete the experiment. This extension may be more suitable for an honors or advanced class. It is not recommended as a general procedure for all classes.
• In Activity B, water is not attracted to polyethylene (there is no adhesion between the drop and the polymer). Each molecule in the water drop is attracted to the other water molecules in the drop. This causes the water to pull itself into a shape with the smallest amount of surface area, a bead (sphere). All the water molecules on the surface of the bead are “holding” each other together or creating surface tension. However, the water is attracted to the glass because of the ions encapsulated in the glass. The ion–dipole attraction makes the water droplet spread on the surface and also creates an interesting attraction between the two glass plates when they are compressed.
• In Activity B, cohesive forces are the electrostatic forces that hold molecules together. In water, these are primarily hydrogen bonding and dipole–dipole attraction. Adhesive forces are electrostatic forces between molecules of different substances. In the interaction of a glass tube with water, the adhesive forces between the polar water molecules and the polar Si—O bonds at the surface of the glass are greater than the cohesive forces between the water molecules. The water is pulled up the sides of the tubing until the weight of the water column just balances the total adhesive forces between the glass and water molecules (see Figure 4 in the Procedure). The smaller the diameter of the tubing, the greater the height of the column.
• In Activity C, two groups can combine if only two temperature probes are available. Temperature measurements may alternately be made using bare temperature probes that have been dipped into the liquid. The liquid will evaporate very quickly, however, so the temperature will bottom out after about 30 seconds and will then start to increase.
• In Activity C, many different solvent pairs may be tested in this experiment. We tried to balance availability and relative toxicity of different solvents against the desire for good pair-wise comparisons (e.g., effect of molar mass in alkanes, polar versus nonpolar, presence or absence of hydrogen bonding). Mindful of the inhalation hazard for both students and the teacher in this “evaporation lab,” we decided to limit the total number of solvents to no more than four (two trials). Hexane was selected rather than pentane for the alkane series because pentane has a very low flash point and is narcotic in high concentrations. Isopropyl alcohol is recommended rather than n-propyl alcohol (1-propanol) because it is more readily available in high school chemistry labs and because it is less toxic by inhalation (the TLV is 983 mg/m³ for isopropyl alcohol compared to 492 mg/m³ for n-propyl alcohol).
• In Activity D, on average, a paper clip will break after it has been straightened out and rebent ten times. The annealed paper clips are easier to bend and more difficult to break—the treated clips can be bent back and forth about 18 times before breaking. After the paper clips are hardened, they are extremely hard to bend and break easily—on average, the paper clips break on the first try! After tempering, the paper clips are hard but “springy”—they do not break.

Teacher Tips

• “Phase-change wallboard” has been developed as a passive thermal storage construction material to improve energy efficiency in heating and cooling buildings. Phase-change wallboard contains paraffin wax embedded in gypsum. Paraffin is used as a “phase change material” (PCM); it undergoes reversible phase changes when heated or cooled, absorbing or releasing large amounts of heat and maintaining a constant temperature in the process. Consider how a PCM might be used in buildings in a mild-climate area. During the day, when the outside temperature increases, the solid PCM melts and absorbs (stores) the excess heat energy. This cools the building and reduces the need for external cooling (air conditioning). The reverse occurs at night—when the outside temperature decreases, the liquid PCM solidifies and releases its “stored” heat energy.
• Evaporative cooling is one of the oldest and most energy efficient methods of cooling a home. A typical evaporative cooler uses only about one-fourth as much electricity as an air conditioner during peak demand. It also does not require ozone-depleting gases, and is thus an environmentally friendly alternative to air conditioning. An evaporative cooler basically consists of a large fan that draws outside air through a water-soaked pad—evaporation of water cools the air and increases the moisture content in the air. The resulting temperature decrease that takes place depends on the temperature and the relative humidity of the incoming air. Outside air at 100 °F and a relative humidity of 30%, for example, will be cooled to about 82 °F. Direct evaporative coolers are a viable alternative to air conditioners in hot and dry climates such as the desert Southwest. They are not useful in hot and humid areas such as the Midwest or the South Atlantic region.
• Metals that crystallize in the BCC crystal structure include vanadium, chromium, manganese, iron, and all of the alkali metals. Metals that crystallize in the FCC crystal structure include aluminum, lead, copper, silver, and gold. The FCC structure is an example of “closest packing” of solids—identical atoms are packed as closely as possible into a given space. If one assumes that the atoms behave as small spheres, the atoms occupy 74% of the volume of the FCC crystal structure and have a coordination number of 12. This is the maximum coordination number and maximum density possible for atoms in a solid lattice composed of small “spheres.”

Answers to Prelab Questions

In Activity A, the temperature and the energy changes that take place when a liquid freezes will be investigated by observing the cooling curve for lauric acid.

In Activity B, the intermolecular forces that act between molecules will be compared for two compounds, water and ethyl alcohol.

In Activity C, the effect of intermolecular attractive forces on evaporation rate will be studied by comparing pair-wise the evaporation rates of four liquids.

In Activity D, the effect of heating and cooling on the structure of steel will be investigated by observing the changes heating and cooling make to the properties of paper clips.

Sample Data

Activity A

Activity B Activity C Activity D

Answers to Questions

Activity A

1. Prepare a graph of temperature on the y-axis versus time on the x-axis. Draw a smooth (continuous) curve through the plotted points for the series of data.
   {12754_Answers_Figure_5}
2. Label the following regions (A–C) on the cooling curve: A, only liquid is present; B, liquid and solid are present together; C, only solid is present.
3. What happens to the temperature of a substance while it is freezing? Estimate the freezing point of lauric acid from the cooling curve.

   The temperature of a pure substance should remain constant at the freezing point or melting point as long as both phases are present. The freezing point of lauric acid is estimated from the “flat” region of the cooling curve, 44.0 °C.

4. Circle the correct choices: Freezing is an (exothermic/endothermic) process—the liquid (absorbs/releases) heat from or to its surroundings. At the freezing point, the average (kinetic energy/potential energy) of the molecules (increases/decreases) and the liquid solidifies.

   Note: Question 4 is difficult for students to grasp. Energy is released during freezing but the temperature does not change. Average kinetic energy is dependent on temperature so it does not change. The potential energy must decrease and this is due to the formation of intermolecular forces holding the solid together. Molecules in the liquid state have a higher potential energy than molecules in the solid state.

5. (Optional) Lauric acid is a fatty acid (a component of fats and oils). What factor might explain the regular increase in the melting point of the following fatty acids as the number of carbon atoms increases?

   Lauric acid, CH₃(CH₂)₁₀CO₂H, mp 43.2 °C
   Myristic acid, CH₃(CH₂)₁₂CO₂H, mp 54.0 °C
   Palmitic acid, CH₃(CH₂)₁₄CO₂H, mp 61.8 °C
   Stearic acid, CH₃(CH₂)₁₆CO₂H, mp 68.8 °C
   In general, the greater the attractive forces between the molecules in a molecular solid, the higher its melting point. This implies that the strength of the attractive forces increases as the number of carbon atoms increases. Note: The properties of the fatty acids are dominated by the long nonpolar hydrocarbon “tail.” The principal attractive forces acting between nonpolar molecules are London dispersion forces. The strength of London dispersion forces increases as the size of the molecules increases.

Activity B

1. Which liquid was least attracted to each of the surfaces? How could you tell that this was true?

   Water was least attracted to the plastic slide as was ethyl alcohol. The least attracted liquid to a particular surface formed a bead. The more attracted spread out on the surface.

2. In step 3, which of the drops was the flattest and widest? What does this mean about the attraction of the molecule to the surface?

   The water drop on the glass slide was the flattest. The attraction of water molecules to glass formed the strongest attraction.

3. Can you determine whether the polyethylene is made of polar or nonpolar molecules? Explain.

   If polyethylene were polar, water—being more polar than ethyl alcohol—would spread out flatter across the surface than ethyl alcohol.

4. When you compressed the slides in step 8, which slide seemed the most difficult to separate?

   The glass slides with water between them.

5. The glass capillary tube contains encapsulated ions at the surface; based on this fact, which molecules show the greatest ion–dipole attraction?

   Since the water rose higher in the tube than did ethyl alcohol, water would have the greater ion–dipole attraction.

6. In step 16 when the drops run down the dry erase board, which liquid seemed to spread out on the surface?

   Ethyl alcohol

7. Which liquid took longer to evaporate? What does this imply about the attraction of the molecules to each other?

   Water took the longest time to evaporate. The attraction between water molecules is greater than that between ethyl alcohol molecules.

8. Which liquid has weaker intermolecular attractions and which has the stronger intermolecular attractions?

   Based on the data, water has stronger intermolecular attractions than ethyl alcohol.

Activity C

1. Describe in words a typical temperature versus time graph for the evaporation of a liquid—be as specific as possible. Explain the graph in terms of the cooling effect of evaporation.

   The initial temperature of the liquid is constant at or near room temperature for the first 10 seconds or so, corresponding to the first 3–4 data points, when the temperature probe is still in the liquid (see step 13 in the Procedure). There is then a sharp decrease in the temperature over the next 30 seconds as the liquid begins to evaporate. The temperature will usually continue to decrease over the entire time period (180 sec), although the rate at which the temperature decreases begins to slow down or level off after about 100 seconds. The minimum temperature is generally observed after about 150 seconds. The temperature decrease occurs because a liquid cools as it evaporates and absorbs heat from the surroundings. The rate of evaporation of a liquid decreases at lower temperatures. The rate at which the temperature decreases thus reflects the rate of evaporation—as the liquid cools, it evaporates more slowly.

2. Summarize the results of this experiment in Data and Results Table C. Subtract the initial temperature from the minimum temperature to determine the temperature change due to evaporation. What is the relationship between the temperature change due to evaporation and the rate of evaporation of a liquid? Explain.

   See the Sample Data. The observed temperature change due to evaporation correlates with the ease of evaporation of a liquid and its rate of evaporation. Liquids that evaporate very fast show large temperature changes (lower final temperatures).

3. Compare the results obtained for hexane and heptane in trial A. Which compound evaporated more quickly? How are these compounds similar? How are they different? Which compound has stronger attractive forces? Explain.

   Hexane reached a lower minimum temperature than heptane. This means that hexane evaporated more quickly than heptane. Hexane and heptane have very similar structures—they are both nonpolar hydrocarbons consisting of C—C and C—H bonds. Heptane has one more —CH₂— group in its “chain” and thus has stronger attractive forces than hexane. (Dispersion forces are stronger for “bigger” molecules.)

4. Compare the results obtained for acetone and isopropyl alcohol in trial B. Which compound evaporated more quickly? How are these compounds similar? How are they different? Which compound has stronger attractive forces? Explain.

   Acetone reached a lower minimum temperature than isopropyl alcohol. This means that acetone evaporated more quickly than isopropyl alcohol. Acetone and isopropyl alcohol have similar molar masses and both are polar compounds. Isopropyl alcohol, however, has an –OH group in its structure and is thus capable of forming hydrogen bonds with neighboring molecules. Isopropyl alcohol has stronger attractive forces than acetone.

5. (a) Is there a pattern between the molar mass of a compound and the temperature change observed due to evaporation? (b) Why would it not be fair to conclude that “dispersion forces are stronger than hydrogen bonding” by comparing the results for hexane and isopropyl alcohol in this experiment?
   1. In general, compounds with larger molar masses exhibited smaller temperature changes due to evaporation (they evaporated more slowly). This was true in the first series being compared—hexane and heptane in Trial A.
   2. Hexane has a larger molar mass than isopropyl alcohol (86 versus 60 g/mole), but it evaporated much more quickly, that is, it has weaker attractive forces.
6. Rank the four liquids tested from most volatile to least volatile based on their observed temperature changes due to evaporation.

   From most volatile to least volatile:
   Acetone > hexane > isopropyl alcohol > heptane

7. (Optional) Look up the boiling points of the four compounds tested in this experiment. Is there a relationship between the rate of evaporation of a liquid and its boiling point?

   The rate of evaporation increases as the boiling point decreases. Thus, compounds that have lower boiling points exhibited larger temperature changes due to evaporation. Note: The data are tabulated below. There is a relationship between the temperature change due to evaporation and the boiling point, but it is not linear.

   {12754_Answers_Table_6}

Activity D

1. Examine the BCC model: Why is this arrangement of atoms called a body-centered cubic structure? How many “nearest neighbors” surround the central atom of the structure?

   One atom occupies the center of a cube formed by the eight atoms of layers one and three. Eight atoms surround the central atom.

2. Examine the FCC model: Why is this arrangement of atoms called a face-centered cubic structure?

   One atom occupies the center of each face of the six faces of the cube formed by the outer four atoms of layers one and three.

3. Repeat step 17 to construct another “middle” layer of four balls and add this to the top of the structure as a fourth layer. Count the number of balls that surround the center atom in the third layer. How many “nearest neighbors” surround this atom?

   Four from each of layers two, three and four surround the center ball in layer three. Each atom in the FCC crystal structure is surrounded by 12 “nearest neighbors”—this is the coordination number for a metal in a FCC structure.

References

These experiments have been adapted from Flinn ChemTopic™ Labs, Volume 11, Solids and Liquids; Cesa, I. Ed., Flinn Scientific: Batavia, IL, 2006.

Introduction

The properties of solids and liquids provide a mirror for us to “see inside” the world of atoms and molecules—to understand the motion of molecules and to compare the interactions among different types of molecules. These properties offer convincing evidence for the kinetic-molecular theory, the most important model for explaining the physical properties of matter. Use this set of four mini-lab activities to explain and predict the properties of solids and liquids.

Concepts

• Phase changes
• Dipole–dipole interactions
• Polar vs. non-polar compounds
• Melting point
• Kinetic-molecular theory
• Properties of metals
• Hydrogen bonding
• Surface tension
• Crystal structure
• Evaporation
• Capillary action

Background

Activity A. Phase Change—Cooling Curve
The temperature changes and energy changes that occur when a liquid freezes can best be understood by imagining what solids and liquids look like at the level of molecules or ions. Solids and liquids differ in how ordered or rigid their structures are and in the range of motion that the molecules or ions are allowed. Molecules in a crystalline solid are packed together in an ordered three-dimensional pattern, called the crystal lattice, where they are held in place by attractive forces between the molecules. The motion of molecules in the solid state is limited to vibrations (stretching and rocking motions)—the molecules are not free to move away from their fixed positions. The forces between molecules in the liquid state are less well understood. Molecules in the liquid state are free to move and are not locked in position. Attractive forces between molecules, however, tend to keep the molecules close together, so that their motion is perhaps best described as coordinated rather than independent.

A solid and its corresponding liquid are in equilibrium at the melting point, the temperature at which a crystalline solid becomes a liquid. The melting point of a pure substance is a characteristic physical property that can be used to identify a substance and to determine its purity. When a solid is heated, the temperature of the solid will increase until it reaches the melting point.

Temperature is related to the average kinetic energy of the molecules—as the temperature increases, the average kinetic energy increases and the molecules begin to vibrate more rapidly. At the melting point, the vibrations become so rapid that the molecules begin to “break loose” from their fixed positions and melting occurs. The temperature of the solid–liquid mixture will remain constant at the melting point until all the solid has melted. Although the temperature remains constant at the melting point, heat must be added to break the attractive forces between molecules. In general, the more orderly the packing arrangement of molecules in the solid state and the stronger the attractive forces between molecules, the higher the melting point will be and the more heat that will be needed to melt the solid. The amount of heat energy required to melt a solid at its melting point is called the heat of fusion. The reverse process occurs when a liquid freezes. When a liquid freezes, energy in the form of heat is released to the surroundings. The same amount of heat required to melt a solid will be released by the liquid when it freezes or solidifies.

Activity B. Intermolecular Forces
Intermolecular forces include dipole–dipole attractions, hydrogen bonding, dipole-induced dipole attraction and London dispersion forces. All of these types of forces are electrostatic in nature. Electrostatic forces arise when the molecules contain or are capable of creating areas of charge separation. For the two compounds to be studied in this experiment, ethyl alcohol and water, all four types of intermolecular forces may exist between molecules. However, dipole–dipole interactions and hydrogen bonding play the most important roles in determining the overall properties of the compounds. Dipole–dipole interactions occur only in polar compounds. The greater the polarity of the molecules, the larger the force of attraction between those molecules.

Many compounds that contain an O—H or N—H bond exhibit a specialized form of dipole–dipole attraction called hydrogen bonding. Hydrogen bonding occurs in molecules where hydrogen is bonded to a highly electronegative atom (X). The difference in electronegativity between H and X creates a large charge separation in the bond.

δ–  δ+
X—H

If the molecule also contains a highly electronegative atom with a lone pair of electrons, this lone pair is strongly attracted to the now partial positive charge on the hydrogen atom in a neighboring molecule. This is hydrogen bonding.

δ– δ+ δ–
X—H- - -:Y—

Both water and ethyl alcohol contain —O—H bonds capable of forming strong hydrogen bonds. Because they possess strong intermolecular attractive forces, water and ethyl alcohol have higher melting and boiling points than similar-sized nonpolar molecules. For different compounds to form solutions, the intermolecular forces between the molecules must be similar to allow for the separation and mixing of the two substances. Without this similarity, the substances will remain separated. Hence the phrase, “like dissolves like.” The interactions of noncrystalline solids and liquids are also a function of the molecular forces that occur at the surface of the solid and liquid. The attractive forces between molecules of the different substances are called adhesive forces while those between the molecules of the same substance are called cohesive forces.

These forces come into play in surface tension and capillary action. At the surface of a liquid, the only forces on the molecules are inward. Unless the molecules touch a solid surface, the liquid will contract on itself and form a spherical drop. Surface tension is a measure of the force needed to break through the surface of the drop and spread the substance out as a film. The greater the forces of attraction between the molecules of a liquid, the greater the surface tension.

Activity C. Evaporation
Vaporization is the process by which a substance changes from a liquid to a gas or vapor. Vaporization that occurs gradually from the surface of a liquid is called evaporation. Evaporation is an endothermic process—energy is required for molecules to leave the liquid phase and enter the gas phase. The most common way to provide energy for the vaporization of a liquid is by heating it. Water evaporating from the Earth’s oceans, for example, absorbs heat energy from the sun and helps to moderate the temperature around large bodies of water. When the heat energy for vaporization comes from the immediate surroundings rather than from external heating, the temperature of the surroundings will decrease when a liquid evaporates. This is the origin of the cooling effect of evaporation. Water evaporating from the surface of the skin by perspiration, for example, cools the body.

Evaporation and the cooling effect of evaporation may be explained using the kinetic molecular theory. According to this model, molecules in the liquid state are in constant motion, and interactions among neighboring molecules influence the motion of the molecules and the properties of the liquid. The temperature of a substance is proportional to the average kinetic energy, and thus the average speed, of the molecules. Evaporation occurs when fast-moving molecules near the surface of a liquid have enough energy to break free of their interactions with neighboring molecules and “escape” into the gas phase. Molecular energy, and thus the temperature, of the remaining molecules decreases—a liquid cools as it evaporates. This phenomenon is known as evaporative cooling. The rate of evaporation of a liquid increases at higher temperatures, because more molecules have enough energy to break free of the liquid’s surface.

The rate of evaporation of a liquid depends on the nature of the liquid and the type of attractive forces between molecules. Strong intermolecular attractions hold the molecules in a liquid more tightly. Liquids with weak intermolecular attractive forces have low heats of vaporization and are volatile—they evaporate easily. Liquids with strong intermolecular attractive forces evaporate more slowly, because a greater amount of energy is needed to overcome the attractive forces between the molecules.

Nonpolar compounds generally have very weak attractive forces, called London dispersion forces, between molecules. The strength of London dispersion forces increases in a regular manner as the size of the molecules increases. Dipole interactions occur when polar molecules are attracted to one another. Because dipole interactions are stronger than dispersion forces, polar compounds generally have higher heats of vaporization and evaporate more slowly than nonpolar compounds (assuming that the molecules have similar molar masses). Hydrogen bonding represents a special case of dipole interactions, in which F–H, O–H and N–H groups in molecules associate very strongly with electronegative atoms in adjacent molecules. Hydrogen bonds are the strongest type of intermolecular attractive forces. Hydrogen bonding in water, for example, leads to a high degree of association among water molecules in the liquid and solid state. As a result, water is a very unusual liquid in many ways. It has an unusually high heat of vaporization and a very high boiling point compared to other compounds that are about the same size or have similar structures. Evaporation of water acts as a “heat sink” for energy from the sun. A significant portion of the Sun’s energy that reaches Earth is spent evaporating water from the oceans, lakes, and rivers rather than warming the Earth.

Activity D. Properties of Metals—Crystal Structure and Heat Treatment
Paper clips are made of steel—iron that has been alloyed with about 1% carbon to improve its hardness and toughness. Heat treatment affects the crystal structure of the metal. At room temperature, steel crystallizes in a BCC structure called alpha-ferrite. This BCC structure does not dissolve carbon and is soft and ductile. Heating the BCC form transforms it into a FCC crystal structure that dissolves carbon and is very hard. Sudden cooling of the high-temperature FCC structure by quenching it in water (hardening) causes the dissolved carbon atoms to become trapped in the BCC lattice. The resulting stress and distortions in the crystal structure make the metal extremely hard but also very brittle. This form of iron is called martensite. Slow cooling of the high-temperature FCC structure (annealing) allows the iron to crystallize in the stable BCC form and the carbon to precipitate out in the form of large particles that cause minimal disruption or dislocation of the crystal structure. The result is a soft, nonbrittle, very workable form of the metal. Gentle reheating of the hardened form followed by slow cooling (tempering) allows the trapped carbon to precipitate and removes many of the internal stresses in the distorted martensite crystal structure. This reduces the extreme hardness of the metal but also eliminates the brittleness. The tempered metal is very strong yet still “workable.”

Experiment Overview

The purpose of this activity stations lab is to investigate phase transitions, identify the accompanying energy changes, and recognize the underlying influence of attractive forces between molecules. Each activity focuses on different properties and is a self-contained unit.

1. Phase Change—Cooling Curve

   When freezing weather is predicted, Florida’s orange growers spray their trees with water to prevent the fruit from freezing. As the water freezes, it releases heat to the surroundings and protects the fruit from damage. The temperature of the freezing water mixture will remain at the freezing point as long as both ice and water are present. Investigate the temperature changes and the energy changes that take place when a liquid freezes.

2. Intermolecular Forces

   The forces that act between molecules, called intermolecular forces, play a significant role in many aspects of chemistry, from boiling point trends and the solubility of gases, liquids, and solids to the structure of DNA and proteins. A series of experiments will be performed to investigate the effect of intermolecular forces on the properties of compounds.

3. Evaporation

   It’s a hot and sunny summer day, and you step out of the pool, cool and refreshed. Soon, however, your teeth start chattering and your lips turn blue. Water evaporating from your skin draws heat from your body, leaving you feeling cold. The cooling effect of evaporation is nature’s most important way of cooling not only our bodies but also the Earth! How cool is evaporation?

4. Properties of Metals—Crystal Structure and Heat Treatment

   Heat treatment of metals is used to increase their hardness and their “workability”—their ability to be bent and shaped. Annealing, hardening, and tempering are examples of changes that occur in the properties of metals as they are heated and cooled. Models of metal crystal structure can help us visualize the changes that take place during the heat treatment of a metal.

Materials

Prelab Questions

Read the Background material and Procedure for each activity A–D. Write a brief, one- to two-sentence description of each experiment. Example: In Activity A, the temperature and the energy changes that take place when a liquid freezes will be investigated by observing the cooling curve for lauric acid.

Safety Precautions

Read the entire Procedure before beginning each experiment. Work carefully to avoid scalding skin with hot water. Exercise care when working with hot metals. Acetone, ethyl alcohol, hexane, heptane and isopropyl alcohol are flammable liquids and dangerous fire risks. Avoid contact of all liquids with heat, flames or other sources of ignition. The addition of denaturants makes ethyl alcohol poisonous. Acetone and heptane are slightly toxic by ingestion or inhalation. Perform this experiment in a well-ventilated lab only. Avoid contact of all chemicals with eyes and skin. Wear chemical splash goggles, chemical-resistant gloves and a chemical-resistant apron. Wash hands thoroughly with soap and water before leaving the laboratory.

Procedure

Activity A. Phase Change—Cooling Curve

1. Fill a 400-mL beaker two-thirds full with hot tap water. Heat the water to approximately 80 °C on a hot plate. Proceed to steps 2–5 as the water is heating. Note: After the lauric acid test tube has been removed from the hot water bath (step 5), lower the setting on the hot plate and reduce the temperature of the bath to about 60 °C.
2. Obtain a test tube containing the lauric acid sample.
3. Add about 100 mL of cold tap water (15–20 °C) to a Styrofoam cup and nest the cup containing the tap water inside a second Styrofoam cup. Place the nested Styrofoam cups in a 250-mL beaker and set the beaker on the base of the support stand.
4. Using a clamp or test tube holder, place the test tube containing lauric acid into the hot water bath (step 1) at 80 °C.
5. Insert a digital thermometer into the lauric acid. When the temperature is about 75 °C, using a test tube clamp, remove the test tube from the hot water bath and clamp the test tube to the support stand. Lower the temperature setting on the hot plate as described in step 1.
6. Measure and record the precise temperature of the melted lauric acid in the test tube and then immediately lower the test tube into the cold water bath in the Styrofoam cup. Start timing.
7. Carefully stir the lauric acid with the digital thermometer and measure the temperature every 30 seconds for 8 minutes, or until the temperature is about 30 °C (whichever comes first). Record all time and temperature readings in Data Table A. Note: Continue stirring the sample until it is no longer possible to do so (when the lauric acid has solidified).

Activity B. Intermolecular Forces

1. Obtain two clean slides, one glass and one polyethylene.
2. Using a micro-tip pipet, add a drop of water to each slide.
3. Using a millimeter ruler, measure the drop width and height on both slides. Record these values in the Data Table B.
4. Thoroughly dry the slides and repeat steps 2 and 3 using a clean micro-tip pipet and ethyl alcohol.
5. Thoroughly dry the glass slide and polyethylene slide. Obtain a second glass slide and another polyethylene slide.
6. Wipe off the surfaces of the four slides. Place a drop of water on one glass slide and one drop on the polyethylene slide.
7. Place the second glass slide over the first one. Place the second polyethylene slide over the first one.
8. Compress the water between the slides. Try separating the slides. Record the relative force needed to separate the slides in Data Table B. Separate and dry the four slides.
9. Obtain a disposable Petri dish. Fill the top with 10 mL of water and the bottom with 10 mL of ethyl alcohol.
10. Obtain two capillary tubes.
11. Place one capillary tube in the Petri dish with water. Use the ruler to measure, in millimeters, the height of water in the capillary tube. Record this value in Data Table B.
12. Repeat step 11 using a new capillary tube for the Petri dish bottom containing ethyl alcohol.
13. Use a dry erase marker to place a mark half-way up a dry erase board.
14. Fill the micro-tip pipets, one with water and one with ethyl alcohol.
15. Place a drop of each liquid at the mark. Note the path or pattern that each liquid takes to reach the bottom. Record your observations in the data table.
16. Use the micro-tip pipets to place a drop of each liquid on a glass slide. Determine the time, in seconds, for each drop to evaporate. Record this time in the data table.
17. Dispose of the ethyl alcohol as directed by the instructor.

Activity C. Evaporation

1. Cut four small pieces of filter paper or cotton gauze, approximately 3 cm² each.
2. Label four test tubes 1–4 and place them in a test tube rack. Obtain 2–3 mL of the appropriate solvent in each test tube, according to the following scheme. Stopper the test tubes with corks or rubber stoppers until needed.
   {12754_Procedure_Table_1}
3. Obtain two temperature probes. Wrap one filter paper square around each temperature probe and secure the filter paper with a small rubber band.
4. Plug the temperature probes (sensors) into CH1 and CH2 on the interface system (LabPro) and connect the LabPro to the computer or calculator.
5. Open the data collection software (LoggerPro) on the computer. Select Setup and Sensors from the main screen and choose the appropriate temperature probe for both CH1 and CH2.
6. Select Experiment and Data Collection from the main screen and choose Mode—Time Based.
7. Enter the following choices in the dialog box to set the conditions for the experiment: Length – 180 sec; Sampling speed – 2 seconds/sample.
8. Choose Save from the File menu and save the experiment file. Make sure Over-sampling is not checked.
9. Select Experiment and Remote from the main screen. Check the OK box for remote setup. (The yellow light on the LabPro will come on at this point, indicating that the LabPro is ready.)
10. Disconnect the LabPro from the computer.
11. Place the CH1 temperature probe into test tube 1 (hexane) and the CH2 temperature probe into test tube 2 (heptane). The liquid level in each test tube should be above the filter paper to ensure that the paper is thoroughly soaked with liquid. Allow the temperature probes to soak in the liquid for about 30 seconds. (This is Trial A.)
12. Press the Start/Stop button on the LabPro.
13. The yellow light will go off and a green light will blink as each data point is collected. Collect 3–4 data points, then remove the temperature probes from the test tubes and carefully extend the probes over the test tube rack as shown in Figure 1. Avoid jostling the temperature probes (air drafts may change the rate at which the liquids evaporate).
    {12754_Procedure_Figure_1}
14. The yellow light on the LabPro will briefly flash when the sampling period is over and data collection is done.
15. Open the experiment file on the computer and select Remote followed by Retrieve Data. Reconnect the LabPro to the computer and hit OK.
16. The data will be displayed on the computer screen in two formats—a table of Time versus Temperature and a graph of Temperature versus Time. Save the experiment file and print the table and graph, if possible.
17. Remove the filter paper squares from the temperature probes and carefully dry the probes with a paper towel.
18. Repeat steps 3–17 using acetone (test tube 3) and isopropyl alcohol (test tube 4) in CH1 and CH2, respectively. (This is Trial B.) Be sure and use a new filter paper square on each probe.

Activity D. Properties of Metals—Crystal Structure and Heat Treatment

Part A. Heat Treatment of a Metal

1. Unwind a paper clip so that the curved sections are opposite to each other. Determine the average number of times the small curved end of a paper clip can be bent back and forth before it will break (see Figure 2). Record the results in the Untreated row of the Data Table D.
   {12754_Procedure_Figure_2}
2. Repeat step 1 with three more untreated paper clips.
3. Hold the large end of an unfolded paper clip with crucible tongs. Heat the bend of the small end in a burner flame until the metal is red hot. Place the paper clip on a heat-resistant surface and allow the pin to cool to room temperature. Repeat to obtain four samples for testing.
4. Test the properties of the metal: Count the number of times the small ends of the treated paper clips can be bent back and forth before breaking as described in step 1.
5. Step 3 represents the annealing process. Annealing is the process of strong heating followed by slow cooling. Annealing softens a metal and makes it less brittle. Record the results in Data Table D for the annealed paper clips.
6. Hold the large end of an unfolded paper clip with crucible tongs. Heat the bend of the small end in a burner flame until the metal is red hot. Immediately drop the paper clip into a beaker of cold water. Repeat with three more unfolded paper clips.
7. Remove the paper clips from the water and dry them. Test the properties of the metal: Count the number of times the treated paper clips can be bent back and forth before breaking.
8. Steps 6 and 7 represent the hardening process. Hardening is the process of strong heating followed by “quenching” or rapid cooling. Hardening makes a metal very rigid and brittle. Record the results for the hardened paper clips.
9. Heat the small end of an unfolded paper clip until it is red hot as in step 6, and then drop it into cold water to cool it quickly. Dry the paper clip and gently reheat the paper clip by holding it above a burner flame until it acquires a blue oxide coating. Place the paper clip on a heat-resistant surface and allow to it cool to room temperature. Repeat for the three other paper clips.
10. Test the properties of the metal: Count the number of times the treated paper clips can be bent back and forth before breaking.
11. Step 10 represents the tempering process. Tempering is the process of strong heating and rapid cooling followed by gentle reheating and slow cooling. Tempering reduces the extreme hardness of the metal but increases its “toughness.” The tempered metal is nonbrittle. Record the results for the tempered paper clips in Data Table D.

Part B. Models of BCC and FCC Crystal Structure

12. Using 5-cm floral wire stems, attach four Styrofoam balls together to form a square, as shown in Figure 3a. Note: Leave a short, 0.5-cm space (stem) between the balls to separate them. This space is necessary for packing the balls into subsequent layers.
13. Place a single ball so that it occupies the “hole” created in the middle of the first square of balls (Figure 3b).
14. Repeat step 12 to form another layer of four balls arranged in a square. Place this layer on top of the middle ball so that the balls are aligned directly over the balls in the lowest layer, as shown in Figure 3c. This is a body-centered cubic (BCC) crystal structure.
    {12754_Procedure_Figure_3_Body-centered cubic crystal structure}
15. (Optional) If desired, attach the ball in the middle layer diagonally to two balls in opposite corners of the lower and upper layers to secure the BCC model.
16. To prepare a model of a face-centered cubic (FCC) crystal structure, attach four Styrofoam balls at 90º angles to a center ball, as shown in Figure 4a. It is not necessary to leave any space or stem showing between the balls. This will be the lower layer of atoms.
17. To make the middle layer of atoms, attach four balls together to form a square (Figure 4b).
18. Repeat step 16 to form another layer of five atoms. This will be the upper layer of atoms in the FCC crystal structure. Arrange the lower, middle, and upper layers together as shown in Figure 3c.
    {12754_Procedure_Figure_4_Face-centered cubic crystal structure}
19. Attach one ball in each of the lower and upper layers to one ball in the middle layer to secure the FCC model.

Student Worksheet PDF

12754_Student1.pdf