POSTLAB Experiment No.3

ChE321L Experiment No. 3DETERMINATION OF THE MOLECULAR WEIGHT OF A NON-VOLATILE SOLID BY

THE CRYOSCOPIC METHOD

Prepared by: GROUP No. 3

LUNK, Michael AngeloYUSON, Joana Marie N.

University of Santo TomasFaculty of Engineering

Chemical Engineering Department

I. INTRODUCTION

The objective of this experiment is to determine the molecular weight of an unknown solute using the cryoscopic method or freezing point depression method.

The general definition of freezing point depression is the effect of lowering the freezing point of a substance due to an increased amount of solute added to the solvent. This principle can be explained in three primary equations.

These are:

ΔTf = Tpure solvent - Tsolution

Kf = ΔTf MWsolute msolvent / msolute

MWsolute = Kf msolute/ΔTf msolvent

Equation 1

Equation 2

Equation 2 can also be rearranged to Equation 3

Equation 3

The Cryoscopic constant, Kf (of Glacial acetic acid) was determined. We first determined the freezing point of the pure glacial acetic acid, and then the freezing point of the solutions containing measured masses of glacial acetic acid and benzoic acid. From these experimental data, we have calculated the Kf of Glacial Acetic Acid.

We have prepared a solution of known masses of an unknown substance dissolved in the Glacial acetic acid and and determined the freezing point of the solution. From these data, we have calculated the molar mass (MW) of the unknown substance.

After the experiment, it was realized that the H+ ions within the solutes used have played a big role in the freezing point depression of the solution.

2. TRANSFERRED in the hard glass test tube

II. PROCEDUREA. Determination of Cryoscopic Constant of the Solvent

15mL Glacial Acetic Acid

Crushed Ice

All contents SOLIDIFIED

All contents LIQUEFIED

1. PLACED in the beaker

5. TRANSFERRED into an empty beaker6. RECORDED temperature reading (15 seconds interval)

3. DIPPED in the ice-water mixture4. RECORDED temperature reading (15 seconds interval)

2. TRANSFERRED in the hard glass test tube15mL Glacial Acetic Acid

Crushed Ice1. PLACED in the beaker

3. ADDED with 1-2 grams Benzoic Acid4. STIRRED

Benzoic Acid is Completely

Dissolved

5. DIPPED in the ice-water mixture6. RECORDED temperature reading (15 seconds interval) All contents SOLIDIFIED



*This procedure was done twice using varied masses of Benzoic Acid (between 1.0g -1.5g) and a new volume of Glacial Acetic Acid

5. DIPPED in the ice-water mixture6. RECORDED temperature reading (15 seconds interval)

B. Determination of the Molecular Weight of the Unknown Solute

2. TRANSFER in the hard glass test tube15mL Glacial Acetic Acid

Crushed Ice1. PLACED in the beaker

3. ADDED with 1-2 grams unknown solute4. STIRRED

UNKNOWN SOLUTE

All contents SOLIDIFIED



*This procedure was done twice using varied masses of unknown solute (between 1.0g -1.5g) and a new volume of Glacial Acetic Acid.

III. DATA AND RESULTS

Table 1. Temperature readings to determine the freezing point of the pure Acetic Acid (Tsolvent) and Acetic-Benzoic Acid solutions (Tsolution).

Mixture t/mm:ss T/°C Observation15mL Pure Glacial

Acetic Acid01:30 15 First Crystals appeared14:45 15.5 Pure solid

(01:) 00:45 17 Pure liquid15mL Pure Glacial

Acetic Acid+1.4100 g Benzoic Acid

02:00 10 First Crystals appeared

18:30 12.5 Pure solid42:30 17 Pure liquid

10mL Pure Glacial Acetic Acid+1.4916 g

Benzoic Acid

01:45 9 First Crystals appeared11:30 11 Pure solid22:45 12 Pure liquid

A. DETERMINATION OF THE CRYOSCOPIC CONSTANT OF THE SOLVENT

* Freezing Point

Figure 1. Cooling curve for pure Glacial Acetic Acid.

Time (minutes)

Tem

pera

ture

(°C)

Pure liquidPure solid

Liquid - solidFirst crystals appears

Figure 2. Cooling curve for the solution of Benzoic Acid in Glacial Acetic Acid.

Time (minutes)

Tem

pera

ture

(°C)

Pure liquid

Pure solid


Liquid - solid

Table 2. Temperature readings to determine the freezing point of the Glacial Acetic Acid- Unknown substance solutions (Tsolution).

Mixture t/mm:ss T/°C Observation15mL Pure Glacial

Acetic Acid+1.2768 g Unknown solute

02:30 12.5 First Crystals appeared

04:15 13 Pure solid37:15 18 Pure liquid

10mL Pure Glacial Acetic Acid+1.3745 g

Unknown solute

01:15 11.5 First Crystals appeared12:30 11.5 Pure solid42:15 19.5 Pure liquid

B. DETERMINATION OF THE MOLECULAR WEIGHT OF THE UNKNOWN SOLUTE

* Freezing Point

Figure 3. Cooling curve for the solution of Unknown solute in Glacial Acetic Acid.

Pure liquid

Pure solid


Liquid - solid

COMPUTATIONS*The following formulae were used to obtain the required

valuesin Procedure A:

Where: ΔTf is the lowering of the freezing point in °C. Tpure solvent is the freezing point of pure solvent in °C. Tsolution is the freezing point of the solution in °C.

ΔTf = Tpure solvent - Tsolution

m= ρV m is the mass in g. ρ is the density in g/mL. V is the volume in mL.

Where:

Where: Kf is the cryoscopic constant in °C kg/ mole.ΔTf is the lowering of the freezing point in °C. MWsolute is the molecular weight of solute in °C. msolvent is the mass of solvent in kg. msolute is the mass of solute in g.

Kf = ΔTf MWsolute msolvent / msolute

*The following formulae were used to obtain the required valuesin Procedure B:

m= ρV ΔTf = Tpure solvent - Tsolution

MWsolute = Kf msolute/ΔTf msolvent

A. DETERMINATION OF THE CRYOSCOPIC CONSTANT (Kf) OF THE GLACIAL ACETIC ACID

• MWsolute : MW C6H5COOH = 122 g/mole

• Tpure solvent : T CH3COOH = 15.5 °C (from Table 1)

• ρsolvent : ρ CH3COOH = 1.049 g/mL (from Atkin’s Data Section on page 990)

TRIAL 1• V CH3COOH = 15 mL;

since m= ρV, then

• m CH3COOH = 0.015735 kg

• m C6H5COOH = 1.4100 g• Tsolution = 12.5 °C (from Table 2)

ΔTf = Tpure solvent – Tsolution = 15.5 °C - 12.5 °C

ΔTf = 3 °C

Kf = ΔTf MWsolute msolvent

msolute

= 3 °C (122 g C6H5COOH /mole C6H5COOH) (0.015735 kg CH3COOH) 1.4100 g C6H5COOH

• ΔTf = 3 °C• MWsolute = 122 g/mole C6H5COOH

• msolvent = 0.015735 kg CH3COOH

• msolute = 1.4100 g C6H5COOH

Kf (Trial 1) = 4 °C kg/mole

A. DETERMINATION OF THE CRYOSCOPIC CONSTANT (Kf) OF THE GLACIAL ACETIC ACID

• MWsolute : MW C6H5COOH = 122 g/mole

• Tpure solvent : T CH3COOH = 15.5 °C (from Table 1)



since m= ρV, then

• m CH3COOH = 0.01049 kg

• m C6H5COOH = 1.4916 g• Tsolution = 11 °C (from Table 3)

ΔTf = Tpure solvent – Tsolution = 15.5 °C - 11 °C

ΔTf = 4.5 °C

Kf = ΔTf MWsolute msolvent

msolute

= 4.5 °C (122 g C6H5COOH /mole C6H5COOH) (0.01049 kg CH3COOH) 1.4916 g C6H5COOH

• ΔTf = 4.5 °C• MWsolute = 122 g/mole C6H5COOH

• msolvent = 0.01049 kg CH3COOH

• msolute = 1.4916 g C6H5COOH

Kf (Trial 2) = 3.86 °C kg/mole

Kf (Trial 2) = 3.86 °C kg/moleKf (Trial 1) = 4 °C kg/mole

Kf (Average)= 3.93 °C kg/mole

+

2

Average Kf :

=

=


• Kf = 3.89 °C kg/mole• Tsolvent : T CH3COOH = 15.5 °C (from Table 1)



since m= ρV, then

• m CH3COOH = 0.015735 kg

• m unknown = 1.2768 g• Tsolution = 13 °C (from Table 4)

ΔTf = Tsolvent – Tsolution = 15.5 °C – 13 °C

ΔTf = 2.5°C

MWsolute= Kf msolute

ΔTf msolvent

= 3.93 °C kg CH3COOH /mole unknown (1.2768 g unknown) 2.5 °C (0.015735 kg CH3COOH)

• ΔTf = 2.5 °C • Kf = 3.89 °C kg/mole• msolvent = 0.015735 kg CH3COOH

• msolute = 1.2768 g unknown

Mwunknown solute (Trial 1) = 127.56 g/mole


• Kf = 3.89 °C kg/mole• Tsolvent : T CH3COOH = 15.5 °C (from Table 1)



since m= ρV, then

• m CH3COOH = 0.01049 kg

• m unknown = 1.3745 g• Tsolution = 11.5 °C (from Table 4)

ΔTf = Tpure solvent – Tsolution = 15.5 °C – 11.5 °C

ΔTf = 4°C

MWsolute= Kf msolute

ΔTf msolvent

= 3.93 °C kg CH3COOH /mole unknown (1.3745 g unknown) 4 °C (0.01049 kg CH3COOH)

• ΔTf = 4 °C • Kf = 3.93 °C kg/mole• msolvent = 0.01049 kg CH3COOH

• msolute = 1.3745 g unknown

Mwunknown solute (Trial 2) = 128.74 g/mole

MW (Trial 2) = 128.74 g/moleMW (Trial 1) = 127.56 g/mole

MW Unknown solute (Average)= 128.15 g/mole

+

2

Average Molecular Weight :

=

=

Average Molecular Weight :

Average Kf:

PERCENT ERROR = | Experimental – Theoretical | Theoretical

% error = |3.93 °C kg/mole – 3.9 °C kg/mole| 3.9 °C kg/mole

X 100

% error = 0.77 %

% error = |128.15 g/mole – 128.1632 g/mole| 128.1632 g/mole

X 100

% error = 0.01 %

IV. PERCENT ERROR

V. ANSWERS TO QUESTIONS

1. From the plot of temperature vs. time, the freezing point can be determined by observing the lowest point on the curve. It is the indication when freezing is nearly to occur.

2. Based on the results, the freezing point of pure acetic acid is relatively higher compared to the solutions’ freezing point after a certain amount of solute is dissolved in the acetic acid. As the solvent crystallizes, the solute concentration increases, resulting in further lowering of freezing temperature.

3. The calculated molecular weight of the unknown solute was not affected by the amount of acetic acid used. Instead, it is the freezing point that was affected by the amount of the acetic acid. This can be proven by the equation:

Where MWsolute and Kf are constants.

ΔTf = Kf msolute

MWsolutemsolvent

So if the amount of glacial acetic acid used was more than 15mL, then the freezing temperature would have become higher and lower if it was less than 15mL.

4. If one is to guess what the solid sample is without looking at the result of the experiment, the first logical clue that must be considered is its smell since all the solidified solutions in this experiments looks the same.

5. Supercooling is a phenomenon where in a liquid cools below its freezing point before crystallization occurs. This phenomenon has actually occurred in all parts of this experiment as explained in the graphs.

In the addition of the naphthalene to the Glacial acetic acid solution. The H + ions within the naphthalene cause the freezing point to lower because the ions act to disrupt the bonds between the particles.

VI. CONCLUSION AND RECOMMENDATION

The percent error was calculated to be relatively low (around 0.01%). Errors that contributed to this could include impurities in the Glacial acetic acid-naphthalene mixture and imprecise readings of temperature and masses of substances.

REFERENCES

*Atkin’s Physical Chemistry 8th Edition*Physical Principles 2 Laboratory Manual*Leider’s Physical Chemistry 3rd Edition*http://www.ncbi.nlm.nih.gov/pmc/articles/PMC1253340/pdf/biochemj01101-0190.pdf*http://www.chemistry.ccsu.edu/glagovich/teaching/31

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Definition of Terms

CRYOSCOPY - A phase-equilibrium technique to determine molecular weight and other properties of a solute by dissolving it in a liquid solvent and then ascertaining the solvent's freezing point.

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POSTLAB Experiment No.3