POSTLAB Experiment No.3

University of Santo Tomas Faculty of Engineering Chemical Engineering Department

ChE321L Experiment No. 3 DETERMINATION OF THE MOLECULAR WEIGHT OF A NON-VOLATILE SOLID BY  THE CRYOSCOPIC METHOD Prepared by: GROUP No. 3 LUNK , Michael Angelo  YUSON, Joana Marie N.

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.

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 This principle can be solvent. explained in three primary equations.

 These are: Equation 1

Equation 2

ΔTf  = Tpures esolvent

- Tsolution

K f  = ΔTf  MWsolute msolvent

/

msolute

Equation 2 can also be rearranged to Equation 3 Equation 3

MWsolute = K f msolute /ΔTf 

msolvent

  The Cryoscopic constant, K f  (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 K f 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.

II. PROCEDURE A. Determination of Cryoscopic Constant of the Solvent

Crushed Ice 1. PLACED in the beaker

15mL Glacial Acetic Acid 2. TRANSFERRED in the hard glass test tube 3. DIPPED in the ice-water mixture 4. RECORDED temperature reading (15 seconds interval) All contents SOLIDIFIED 5. TRANSFERRED into an empty beaker 6. RECORDED temperature reading (15 seconds interval) All contents LIQUEFIED

Crushed Ice 1. PLACED in the 15mL Glacial Acetic Acid beaker 2. TRANSFERRED in the hard glass test tube 3. ADDED with 1-2 grams Benzoic Acid 4. STIRRED Benzoic Acid is Completel y 5. DIPPED in the ice-water Dissolved mixture

6. RECORDED temperature reading 7. TRANSFERRED into an empty beaker (15 seconds interval) 8. RECORDED temperature reading (15 seconds interval) All contents LIQUEFIED *This procedure was done twice using varied masses of Benzoic Acid (between 1.0g -1.5g) and a new volume of Glacial Acetic Acid

All contents SOLIDIFIED

B. Determination of the Molecular Weight of the Unknown Solute

Crushed Ice 1. PLACED in the beaker

15mL Glacial Acetic Acid 2. TRANSFER in the hard glass test tube 3. ADDED with 1-2 grams unknown solute 4. STIRRED UNKNO WN SOLUTE 5. DIPPED in the ice-water mixture 6. RECORDED temperature readin (15 seconds interval) All contents SOLIDIFIED *This procedure was 7. TRANSFERRED into an empty beaker done twice using varied 8. RECORDED temperature reading masses of unknown (15 seconds interval) solute (between 1.0g All contents LIQUEFIED

-1.5g) and a new volume of Glacial Acetic Acid.

III. DATA AND RESULTS

A. DETERMINATION OF THE CRYOSCOPIC CONSTANT OF THE SOLVENT  Table 1. Temperature readings to determine the freezing point he pure Acetic Acid (T solvent ) and Acetic-Benzoic Acid solutions (Tsolution

Mixture 15mL Pure Glacial Acetic Acid

t/mm:s T/°C s 01:30 15 14:45 (01:)

15mL Pure Glacial Acetic Acid+1.4100 g Benzoic Acid

00:45 02:00 18:30 42:30

15.5 17 10 12.5 17

Observation First Crystals appeared Pure solid Pure liquid First Crystals appeared Pure solid Pure liquid

)

C

Pure liquid Pure solid

°( er ut ra e p m e T

Liquid - solid First crystals appears

 Time (minutes)

Figure 1. Cooling curve for pure Glacial Acetic Acid.

Pure liquid )

C

°( er ut ra e p m e T

Liquid - solid

Pure solid

Liquid - solid First crystals appears

 Time (minutes)

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

B. DETERMINATION OF THE MOLECULAR WEIGHT OF THE UNKNOWN SOLUTE Table 2. Temperature readings to determine the freezing poi of the Glacial Acetic Acid- Unknown substance solutions (Tsolution

Mixture

t/mm:s T/°C s 02:30 12.5

15mL Pure Glacial Acetic Acid+1.2768 g 04:15 Unknown solute 37:15 10mL Pure * Freezing Point Glacial Acetic Acid+1.3745 g

13 18

01:15

11.5

12:30

11.5

Observation First Crystals appeared Pure solid Pure liquid First Crystals appeared Pure solid

Pure liquid Liquid - solid

Pure solid

First crystals appears Liquid - solid

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

COMPUTATIONS *The following formulae were used to obtain the required values m= Where:in Procedure A:

ρV

m is the mass in g. ρ is the density in g/mL. V is the volume in - Tsolution ΔTf  = Tpuresolvent mL.  



Where:

ΔTf  is the lowering of the freezing point in °C. is the freezing point of pure Tpuresolvent solvent in °C 



K f  = ΔTf  MWsolute msolvent Where:

msolute

/

K f  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*The °C. following formulae were used to obtain the required values in kg. of solvent msolvent is the mass in Procedure B: of solute g.- T msolute is the mass m= ΔTf  = Tpuresolvenin t solution 





 

ρV

MWsolute = K f msolute /ΔTf 

msolvent

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

• MWsolute : MW C 6H5COOH = 122 g/mole • Tpuresolvent : T CH 3COOH = 15.5 °C (from Table 1) • ρsolvent : ρ CH 3COOH = 1.049 g/mL (from Atkin’s Data Section on page 990)

TRIAL 1 • V CH 3COOH = 15 mL; since

• m CH 3COOH

• m C 6H5COOH = 1.4100 g • Tsolution = 12.5 °C (from Table 2)

m= ρV, then = 0.015735 kg

ΔTf  = Tpuresolvent

 – Tsolution = 15.5 °C - 12.5 °C

ΔT

3 °C

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

f 

= 0.015735 kg CH 3COOH = 1.4100 g C6H5COOH

= ΔTf  MWsolute msolvent

msolute 3 °C (122 g

C6H5COOH

K f(Trial 1)

/mole C6H5COOH ) (0.015735 kg 1.4100 g C6H5COOH

= 4 °C kg/mole

CH 3COOH

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

• MWsolute : MW C 6H5COOH = 122 g/mole • Tpuresolvent : T CH 3COOH = 15.5 °C (from Table 1) • ρsolvent : ρ CH 3COOH = 1.049 g/mL (from Atkin’s Data Section on page 990)

TRIAL 2 • V CH 3COOH = 10 mL;

• m C 6H5COOH = 1.4916 g • Tsolution = 11 °C (from Table 3) since m= ρV, then • m CH 3COOH = 0.01049 kg

ΔTf  = Tpuresolvent

 – Tsolution = 15.5 °C - 11

ΔT

°C = 4.5 °C

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

f 

= 0.01049 kg CH 3COOH = 1.4916 g C6H5COOH

= ΔTf  MWsolute msolvent

msolute 4.5 °C (122 g

C6H5COOH

K f(Trial 2)

/mole C6H5COOH ) (0.01049 kg 1.4916 g C6H5COOH

= 3.86 °C kg/mole

CH 3COOH

Average K f  :

=

K f(Trial 1) = 4 °C kg/mole

K f=(Average)

K f(Trial 2) +

= 3.86 °C kg/mole

2

= 3.93 °C kg/mole

B. DETERMINATION OF THE MOLECULAR WEIGHT OF THE UNKNOWN SOLUTE

• K f  = 3.89 °C kg/mole • Tsolvent : T CH 3COOH = 15.5 °C (from Table 1) • ρsolvent : ρ CH 3COOH = 1.049 g/mL (from Atkin’s Data Section on page 990)

TRIAL 1 • V CH 3COOH = 15 mL; since

• m CH 3COOH

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

m= ρV, then = 0.015735 kg

ΔTf  = Tsolvent  – Tsolution = 15.5 °C – 13 °C ΔTf  = 2.5°C

• ΔTf  = 2.5 °C • K f  = 3.89 °C kg/mole • msolvent • msolute

= 0.015735 kg CH 3COOH = 1.2768 g unknown

MWsolute = K f msolute

ΔTf  msolvent =

3.93 °C kg

Mwunknownsolute

/mole unknown (1.2768 g unknown ) 2.5 °C (0.015735 kg CH 3COOH )

CH 3COOH

(Trial1)

= 127.56 g/mole

B. DETERMINATION OF THE MOLECULAR WEIGHT OF THE UNKNOWN SOLUTE

• K f  = 3.89 °C kg/mole • Tsolvent : T CH 3COOH = 15.5 °C (from Table 1) • ρsolvent : ρ CH 3COOH = 1.049 g/mL (from Atkin’s Data Section on page 990)

TRIAL 2 • V CH 3COOH = 10 mL; since

• m CH 3COOH

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

m= ρV, then = 0.01049 kg

ΔTf  = Tpuresolvent

 – Tsolution = 15.5 °C – 11.5 °C ΔTf  = 4°C

• ΔTf  = 4 °C • K f  = 3.93 °C kg/mole • msolvent • msolute

= 0.01049 kg CH 3COOH = 1.3745 g unknown

MWsolute = K f msolute

ΔTf  msolvent =

3.93 °C kg

Mwunknownsolute

/mole unknown (1.3745 g unknown ) 4 °C (0.01049 kg CH 3COOH )

CH 3COOH

(Trial2)

= 128.74 g/mole

Average Molecular Weight :

=

+ (Trial2) MW (Trial1) = 127.56 MW g/mole

= 128.74 g/mol

2

= Unknownsolute MW

(Average)

= 128.15 g/mole

IV. PERCENT ERROR PERCENT ERROR = | Experimental – Theoretical | Theoretical

Average K f :

% error = |3.93 °C kg/mole – 3.9 °C

kg/mole|

X 100

3.9 °C kg/mole % error = 0.77 %

Average Molecular Weight : % error = |128.15 g/mole – 128.1632 g/mole| X 100 128.1632 g/mole % error = 0.01 %

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. ThisK  can be proven ΔTf  = f msolute by the equation: MWsolute msolvent

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.

VI. CONCLUSION AND RECOMMENDATION

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.

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/PMC125 3340/pdf/biochemj01101-0190.pdf  *http://www.chemistry.ccsu.edu/glagovich/teachin g/31

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