molar mass determination using freezing point depression

 

To determine the molar mass of the unknown solute by freezing point depression.

To determine the freezing point of pure Lauric Acid/ Unknown Solute Solution.

To measure thefreezing point of the pure solvent.

Introduction

The main aim of this experiment is to determine the molar mass of an unknown using freezing point depression of a known solvent and from experimental data. The unknown is a nonelectrolyte and nonvolatile substance. Dissolving non-volatile substance in a solvent lowers the freezing point of a solvent. This affects the freezing and boiling temperatures of the solution by lowering the freezing point and increases the boiling point in relation to the pure solvent.The unknown substance properties depend on the number of solute particles and not the properties of the solute particles, these properties are called colligative properties.

Molality, m, is the concentration of the solute expressed by the number of moles of a solute per kilograms of a solvent.

m=moles of the solute/kilograms of the solvent.

The difference between the freezing point of the pure solvent and the solution is used to determine the freezing point depression.

Freezing point depression=∆T=T⁰_{F pure solvent} -T⁰_{F solution}

The relationship between the freezing point of the solution and the pure solvent for non-dissociating substances is given by the equation below.

∆T_F =K_F m

Taking accounts for the number of ions that dissociate in a solution, the equation below is used to relate the freezing point of the solution and the freezing point of the pure solvent.

∆T_F =i K_F m

Where i=Van’t Hoff i-Factor.

Van’t Hoff i-Factor is the total number of the ions produced in solution per unit salt.

To obtain the objectives of this experiment, various tests were run. The molar mass of an unknown substance is obtained using the freezing point depression of the solvent and the solution.The freezing point depression is found in the graph produced after running the tests.  The chemical used in this experiment is Lauric Acid solute solution and Cyclohexane. After the freezing points of solution and solvent are obtained, the molar mass of unknown solute is found by using the formula relating the freezing point depression of solvent and solute given in an equation above.

Procedure

Part I: Determining the Freezing Point of Pure Lauric Acid

 

1. A large test tube was weighed and approximately eight grams of lauric acid was added. The test tube and its content were then re-weighed.
2. The water bath was prepared by filling a 400 ml beaker with tap water and the beaker was supported on a ring clamp furnished with wire gauze. The beaker containing water was heated using a Bunsen burner at around 70-80 °C, making sure that the water levels remain above 200 ml in the beaker.
3. A second beaker was then filled with room temperature tap water on the bench.
4. The test tube containing the lauric acid was then clamped to the ring stand and immersed in the hot water. The addition of 2-3 minutes waited after all the lauric acid has been liquefied.
5. The test tube was lowered into the room temperature bath, a stir bath and thermistor were added making sure that the thermistors are not touching or becoming too close to the stir bar. The thermistor program ran for about 15 minutes until the liquid and the stir bar stops moving. The data obtained was saved in a flash drive.
6. Without pulling the probe out of the solidified sample, a hot water bath to re-melt the acid and the probe was removed gently after the sample has been liquefied. The produce was then repeated for the second time.
7. Once finished the sample was re-melted and the probe was removed. Any excess sample fromthe probe was wiped with a tissue. Spills in the hood and balance area were cleaned up. The data obtained were then saved in a flash drive.

Part III: Measuring the Freezing Point for Lauric Acid/Unknown Solute Solution

1. The contents of the previous steps in the labeled waste container under the hood were discarded. Approximately 1 gram of an unknown solute was weighed to the nearest o.oo1 g.
2. The amount in procedure 1 above was added to a known amount of lauric acid. Once again, approximately 8 grams of the solvent was used having been weighed to the nearest 0.001 g. The solute and solvent were mixed well using a mortar and pestle.
3. Steps 2-7 in Part I were repeated to determine the freezing point of the solution. The data was then saved in a flash drive.
4. The solution was disposed of appropriatelymaking sure the balance area, reagent area and hoods

were cleaned.

Results

 

Figure 1:  This figure shows a graph of determining the Freezing Point of Pure Lauric Acid. The results were obtained during the second trial of the test in part I.

 

 

Figure 2:  This figure shows a graph of determining the Freezing Point of Pure Lauric Acid. The results were obtained during the third trial of the test in part I.

The freezing temperature for the part I is the point where the slope changes abruptly in the graph.

 

 

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Figure 3:  This figure shows a graph ofmeasuring the Freezing Point for Lauric Acid/Unknown Solute Solution. The results were obtained during the second trial of the test in part III.

 

 

Figure 4:  This figure shows a graph ofmeasuring the Freezing Point for Lauric Acid/Unknown Solute Solution. The results were obtained during the third trial of the test in part III.

The final temperature in part III is found by drawing the line which extrapolates the behavior just after freezing began back through the supercooling “dip” to the curve just before freezing began. The intersection of the extrapolated line with this curve is taken as the final freezing temperature.

Table 1: Part I results

Test trials	Trial 1	Trial 2	Trial
Temperature(C0)	35	68	69
	25	42	40
Average (X̅)	X̅= 30	X̅=55	X̅=54.5
STD (+̅S)	+̅S=5	+̅ S=13	+̅ S=14.5

Table 2: Part III results

Test trials	Trial 2	Trial 3
Freezing points(C0)	82.1	83.2
Mass of unknown (kg)	1.096	1.076
Mass of lauric acid (kg)	8.0416	8.076
Molal concentration (moles/kg)	0.2273	0.0231
Molar mass of unknown (g/moles)	107	105

Calculations

Table 1: Part I

X̅ of temperature =35+25/2=30

STD (+̅S) of temperature=√v

V= (X̅-X)²/n

30-35=-5²=25

v=25+25/2=25

+̅S =√25=5

Table 2:Part III

molal concentration of unknown= moles of lauric acid / kg of unknown.

Moles of lauric acid= molar mass /mass

The molar mass of lauric acid=200.3178

Mass of lauric acid =8.0416*1000=8041.6g

Moles of lauric acid=200.3178/8041.6=0.02491 moles

molal concentration of unknown=0.02491 moles/1.096=0.02273 moles/kg

∆T_F =i K_F m

Where:

∆T_F = 83.2

i =Van’t Hoff i-Factor=0.2

K_F=3.90⁰c

M=Molar mass of unknown

83.2=2*3.90*m

M=83.2/0.2*3.9=106.65g/moles

X̅ of the molar mass of unknown =106.65+105.25/2=105.95g/moles

STD (+̅S) of the molar mass of unknown=√v

V= (X̅-X)²/n

106.65-105.25=0.6979²=0.4855

v=4855+0.4855/2=0.4855

+̅S of the molar mass of unknown=√0.4855=0.698

The %error for your Molar mass, assuming your unknown is benzoic acid (C7H6O2= {(Molar mass of benzoic acid (C7H6O2)-Average Molar mass of unknown)/Average Molar mass of unknown} *100

The molar mass of benzoic acid (C7H6O2)=122.1213 g/mol

The average Molar mass of unknown=106g/moles

%Error= [{122.1213 g/mol-105.95g/moles}/105.95 g/moles] *100%=15.2%

Discussion

From the results above, the data in the table shows the freezing temperature of the unknown substance. The average freezing temperature for trail 1, 2 and 3 was carried as shown in table 1 above. The results for trial 1average freezing temperature is constituent with the results obtained from the graphs. Whereby the freezing temperature from the point where the slope changes abruptly in the graph is around 30 for both graphs.

From table 2, the freezing point from trial 1 and 2 are consistent. The molar concentration calculated from trail 1 results=0.2273 and from trail 2=0.0231.

The relationship between the molar concentration and the freezing point depression is that the freezing point depression is the multiplication of molar mass and Can’t Hoff i-Factor.  Molar concentration is the number of moles in a solute divided by the mass of a solvent. This shows that the unknown substance properties depend on the number of solute particles and not the properties of the solute particles, which shows colligative properties. Molar mass is given by mass per mole.

Also, the molar mass of an unknown substance is calculated in table 2 above. The average molar mass of the unknown is equal to 105.95 g/moles. The expected molar mass of the unknown is 122.1213 g/mol.  The difference between the expected and the calculated value of the molar mass of the unknown is due to some error while carrying the experiments.

The percentage error of molar mass of unknown is 15.2 percentage which is too high and not recommended hence the results in the future can be improved by eliminating sources of errors as much as possible. Possible sources of errors include wrong measures of reagents used, improper reading and recording of the data, wrong interpretation of graphs and incorrect calibration of instruments used.

Conclusion

The freezing point is this experiment is used to determine the molar mass of the unknown substance. This takes place as a result of a decrease in the freezing point and increases in the boiling point in relation to the pure solvent. The main objective of this experiment, which was to determine the molar mass of unknown substance was achieved. The molar mass calculated is approximately 85 percent right and 15 percent wrong. The accuracy and certainty of these results can be improved in the future by eliminating the sources of errors during the experiment. The freezing point depression aspectcan also be applied in daily life for example in raising the boiling point of the water that cools cars. Antifreeze decreases the freezing point and increases the boiling point of thecoolant water in the car radiator. This is significant in cooling the car since as soon as the water in the car radiator boilsit does not work anymore as a coolant.

Post lab questions

Question one

A solution is prepared by dissolving 4.9 g sucrose (C12H22O11) in 175 g water. Calculate the

the freezing point of this solution.

∆T_F =K_F m

Kf=1.86°

M=moles of solute/mass of solvent

Moles of solute= mass of solute/molar mass of solute

Mass of solute=4.9 g

Molar mass of sucrose (C12H22O11) =342.2965 g/mol

M=4.9 /342.2965=0.014315 moles

Mass of solvent 175 g

 

∆T_{F =}1.76*0.014315 moles/175 g=1.439*10^{-4 0}c

Question 2

Is the freezing point of 0.01 mKF (aq)higher or lower than that of 0.01 M glucose (aq)? Explain

The freezing point depression depends on the molal concentration of particles in solution

KF dissociates in solution

KF→ K+ + F-

1mol KF produces 2 mol particles (ions)

Therefore a 0.01m solution of KF is 0.02m in particles

Glucose does not dissociate. Therefore a 0.01m solution of glucose is 0.01m in particles.

Therefore a 0.01m KF solution has a lower freezing point than a 0.01m solution of glucose

Question 3

∆T_F =K_F m

Kf=9.1 ⁰c/m

M=moles of solute/mass of solvent

Moles of solute= mass of solute/molar mass of solute

Mass of solute=10 g

The molar mass of t-butanol =74.123 g·mol⁻¹.

M=10/74.123 g·mol⁻¹. =0.1349 moles

∆T_{F =}0.1349/mass of solvent

∆T_F =25.5-24.59=0.91

0.91=0.1349/mass of solvent

Mass of solvent=0.1349/0.91=0.148kg=148g

Question 4

∆T_F =K_F m

0.40=4.70 M

M=0.240/4.70

M=0.05106

M=Moles of solute/mass of solvent

0.05106=moles/15

Moles of solute=15*0.05106=0.7659 moles

Molar mass=0.350*1000/0.7659=456.9 g/moles

Question 5

At this temperature the salting of icy roads with calcium chloride will not be effective in melting the ice since the temperature of the roadway is lower than 15 degrees F hence the salt won’t have any effect on the ice.

Question 6

cyclohexane would give a more accurate determination by freezingpoint depression of the molar mass of a substance that is soluble in either solvent than naphthalene because cyclohexane has a higher freezing point constant than naphthalene. The higher the freezing point constant the higher the freezing depression point hence the higher the accuracy.

Question 7

Part a

∆T_F =K_F m

1.32+ -0.4=5.192 M

M=1.32/5.192=0.254(+ or -)0.4

M=moles of solute/mass of solvent

Moles of solute=m*mass of solvent

Moles of solute =0.254(+ or -)0.04*15.60 (+or-)0.01

0.294*15.61=4.56934

0.214*15.59=3.3363

Moles of solutes ranges from +4.56934 to +3.3363

Molar mass=mass/moles

Mass=1.22 ± 0.01=1.23 to1.21 g

Molar mass=1.23/4.56934=269.185g/moles

=1.21/3.3363=0.3626=362.6772g/moles

Molar mass ranges from 362.6772g/moles to 269.185g/moles

Part b

Absolute error= measured value -actual value

Absolute error=362.6772g/moles-303.35 g/mol=+59.33272

And 269.185g/moles-303.35 g/mol=-34.165

Part c

Yes, the chemist could unequivocally state that the substance is cocaine since theMolar mass calculated ranges from 362.6772g/moles to 269.185g/moles where the molar mass of cocaine as well as codeine fall in between.

Part d

The chemist can increase the precision by paying close attention to detail such as using equipment properly, increasing your sample size, ensure that the equipment is properly calibrated, functioning, clean and ready to use.

 

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