Determination of The Molecular Weight by Freezing Point Depression

The objective of this lab is to determine the molecular weight of an unknown solute by utilizing the freezing point depression method. Students will use tools and equipment for scientific analysis, interpret numerical and graphical data, demonstrate safe practices in the chemical laboratory, demonstrate an understanding of chemical reaction and quantities in chemical reactions, demonstrate an understanding of molecules, compounds, and chemical composition, demonstrate the proper use of exponential notation and significant figures, demonstrate an understanding of the composition of matter and energy, record their investigation results, use scientific reasoning to evaluate physical and natural phenomena, and identify unifying themes in the scientific field of study.

The discussion provides an overview of vapor pressure and its relationship to the volatility of liquids. When a solute is added to a solvent, it lowers the vapor pressure of the solvent. This property, known as freezing point depression, is a colligative property that depends on the number of solute particles in a solution. The freezing point depression can be quantitatively related to the concentration of the solution using the equation ΔT = Kc, where ΔT is the freezing point depression, K is the freezing point constant, and c is the concentration of the solution.

The concentration of the solute in the solution can be calculated by dividing the mass of the solute by its molecular weight. By rearranging the equation, the molecular weight of the solute can be directly calculated: M.W. of solute = K(mass solute)(1000)/(ΔT)(mass solvent).

The procedure involves preparing a hot water bath in a 600 ml beaker. P-dichlorobenzene is weighed and transferred to a test tube, which is then clamped in the water bath. The test tube is heated until the solid has melted completely. A thermometer and wire stirrer are inserted into the test tube, and the temperature is recorded at one-minute intervals. The freezing point of p-dichlorobenzene is determined when the first crystals appear, and the temperature levels off and remains constant until all the liquid has solidified.

An unknown solute is obtained in a vial, and its weight is measured. Half of the unknown is added to the test tube containing the solvent p-dichlorobenzene, and the weight is measured again. This creates solution 1. The test tube is placed in the hot water bath, and the temperature is recorded at one-minute intervals. The freezing point of the solution is determined. The remaining contents of the vial are added to the test tube, and the vial is weighed. The procedure is repeated to obtain the freezing point of the new mixture, which is solution 2.

The calculations involve subtracting the freezing point of the two mixtures of different concentrations from the freezing point of pure p-dichlorobenzene. These values are plotted on a graph with the total grams of solute in the mixture on the x-axis and the corresponding freezing point depression values on the y-axis. A straight line is drawn through these points and the origin. The molecular weight of the unknown solute can be determined using the equation M.W. of solute = K(mass solute)(1000)/(ΔT)(mass solvent). The freezing point constant (K) for p-dichlorobenzene is provided as 7.10 kg ∙ °C/mole.

Throughout the lab, students should handle the materials with care, accurately measure weights and temperatures, record data, and construct graphs. The focus is on understanding the relationship between freezing point depression and the concentration of a solution, using this relationship to calculate the molecular weight of an unknown solute, and interpreting the results through graphical analysis.

Objective

·         Use the tools and equipment necessary for basic scientific analysis and research

·         Interpret the numerical and graphical presentation of scientific data

·         Demonstrate safe practices in the Chemical Laboratory

·         Demonstrate an understanding of Chemical Reaction and Quantities in Chemical Reactions

·         Demonstrate an understanding of Molecules, Compounds and Chemical Composition

·         Demonstrate the proper use of Exponential Notation and Significant Figures

·         Demonstrate an understanding of the composition of matter and energy

·         Record the results of investigation through writing

·         Use scientific reasoning to evaluate physical and natural phenomena

·         Identify the unifying themes of the scientific field of study

Materials

·         600 ml beaker

·         p-dichlorobenzene

·         Test Tubes

·         Thermometer

·         Wire Stirrer

·         Unknown Solute 1

·         Unknown Solute 2

·         Wide Mouth Bottle

Discussion

The vapor pressure of a liquid indicates the amount of gas form of the liquid present above its surface. When a liquid has a high vapor pressure, it means there is a significant quantity of vapor above the liquid. Consequently, such a liquid evaporates easily and is referred to as volatile. On the other hand, liquids with low vapor pressures have only a small amount of vapor above their surface, evaporate slowly, and are considered nonvolatile.

When a solute is added to a solvent, the vapor pressure of the solvent decreases. This decrease in vapor pressure is related to colligative properties observed in solutions, including freezing point depression, boiling point elevation, and osmotic pressure. This experiment focuses on freezing point depression. Colligative properties like this depend solely on the number of solute particles in a solution and not on their chemical properties. The more solute added to a fixed amount of solvent, the greater the decrease in freezing point. For instance, dissolving one mole of sucrose in 1000 g of water will decrease the freezing point twice as much as dissolving half a mole of sucrose in 1000 g of water. In general, adding one mole of solute particles (6.0 x 10^23) to 1000 g of solvent will cause the same decrease in freezing point. This decrease has a specific numerical value, which is a constant unique to each solvent. By altering the number of particles in a solution, the concentration changes. This relationship can be expressed mathematically as:

(1)                                             ∆T = KC

Where:

∆T = freezing point depression of the solvent

K = freezing point constant
(temperature decrease in the solvent with one mole of solute per 1000 g of solvent)

C = concentration of the solution

∆T represents the difference between the freezing temperature of the pure solvent and the freezing temperature of the solvent with the solute added.

The concentration of the solute in the solution is determined by the number of moles of solute. The number of moles can be calculated by dividing the mass of the solute by its molecular weight. Thus, the concentration of a solution is related to the molecular weight of the solute:

(2)         C = (((moles solute) (1000)) / (Mass solvent)) = (((Mass solute) (1000)) / ((Mass solvent) (M.W. of solute))                         

By substituting Equation 1 into Equation 2, we obtain a more practical equation for calculating the molecular weight of the solute:

(3)        ∆ T = (K(Grams of solute) (1000)) / ((Molecular weight) (Grams of solvent))   

Equation (3) can be rearranged to directly calculate the molecular weight of a nonvolatile solute:

(4)     M.W. of solute = (K(mass solute) (1000)) / ((∆ T) (mass solvent))                                      

To illustrate, let's consider the following calculation:

When 34.2 grams of sugar are dissolved in 1000 grams of water, the freezing point of water decreases from 0°C to -0.186°C. The freezing point constant for water is 1.86. The molecular weight of the sugar is:

 molecular weight = (1.86 (34.2 g sugar) 1000) / ([0° C - (-0.186)] (1000 g H2O)) = 342g / mole

The instructor will provide a discussion on the experimental setup and the characteristics of the cooling curve, with particular emphasis on preventing supercooling. Figures 16 and 17 depict the setup and cooling curve, respectively.

Procedure

1.       Prepare a hot bath by filling a 600 ml beaker approximately two-thirds full with water. Measure out 25.00 g of p-dichlorobenzene and transfer it to a large test tube. Securely clamp the test tube in the water bath as illustrated in Figure 16, and initiate heating. Once most of the solid has melted, introduce a thermometer and a wire stirrer into the test tube. As soon as all the p-dichlorobenzene has melted, swiftly remove the test tube from the bath, wipe its exterior with a towel, and place it inside a wide mouth bottle. While continuously stirring, take temperature readings at one-minute intervals. The appearance of the first crystals indicates that the freezing point of p-dichlorobenzene has been reached. At this point, the temperature should stabilize and remain constant until the entire liquid solidifies. Construct a graph displaying temperature versus time and determine the freezing point of dichlorobenzene.

2.       Obtain an unknown solute and place it in a vial. Weigh the vial along with its contents. Transfer approximately half of the unknown solute into a tube containing the solvent dichlorobenzene, and then reweigh the vial. This will be referred to as solution 1. Position the test tube in a hot bath and allow all the mixtures to melt completely. Thoroughly stir the mixture to ensure it is uniformly mixed. Once again, transfer the test tube to a wide mouth bottle and record the temperature every minute while continuously stirring. Determine the freezing point of the solution. Empty the remaining contents of the vial into the test tube and weigh the vial. Repeat the aforementioned procedure to obtain the freezing point of the new mixture, which will be referred to as solution 2.

Calculations

Calculate the difference between the freezing point of two mixtures with varying concentrations and the freezing point of pure dichlorobenzene. Then, create a graph on graph paper with the y-axis representing the values of the freezing point difference and the x-axis representing the total grams of solute in each mixture. Draw a straight line passing through these two points and the origin. Determine the molecular weight of the unknown substance using equation (4). Include all graphs and the report sheet in your submission. The constant "k" for p-dichlorobenzene is 7.10 kg ∙ °C/mole.

Report Sheet for MOLECULAR WEIGHT BY FPEEZING POINT DEPRESSION

Name_______________________________________                          Date___________________________

Lab / Section_________________________________

Review Questions

1.       What is the purpose of determining the molecular weight using freezing point depression?

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2.       Describe the relationship between vapor pressure and the volatility of liquids.

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3.       Explain the concept of colligative properties and give examples.

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4.       How does adding a solute to a solvent affect the vapor pressure of the solvent?

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5.       What equation is used to calculate the freezing point depression of a solution?

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6.       How does the concentration of a solute affect the freezing point depression?

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7.       What is the significance of the freezing point constant in the calculation of molecular weight?

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8.       Describe the procedure for preparing the hot water bath in the lab.

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9.       How is the freezing point of p-dichlorobenzene determined in the lab?

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10.    Explain the purpose of weighing the vial containing the unknown solute.

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11.    What is the relationship between the total grams of solute in the mixture and the freezing point depression?

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12.    How are the freezing point depression values used to determine the molecular weight of the unknown solute?

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13.    What precautions should be taken to ensure accurate temperature measurements in the lab?

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14.    Why is it important to stir the mixture thoroughly before determining the freezing point?

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15.    What is the significance of the cooling curve in the lab?

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16.    How does the concept of supercooling relate to the freezing point depression method?

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17.    Explain the difference between a volatile and nonvolatile liquid.

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18.    How does the freezing point depression method demonstrate the colligative properties of solutions?

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19.    Discuss the limitations and potential sources of error in the lab procedure.

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20.    What can be inferred about the molecular weight of a solute based on the freezing point depression observed in the experiment?

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