Determining the Molecular Mass from the Freezing Point

Objective The non-polar solvent 2, 6-di-tert-utyl-4-methylphenol has a freezing point of approximately 70 degrees C. A quantity of para- dichlorobenzene will be dissolved in the solvent and the effect on the freezing point determined. The freezing point depression constant will be calculated for the solvent. The experiment will be repeated with two unknown solutions and the molecular weight of the unknowns will be determined from the freezing point depression.

Materials 1 X 25g para-Dichlorobenzene 2 X 25g 2,6-di-tert-butly-4-methylphenol 1 X 25g Stearic Acid 1 X 25g Naphthalene Capillary tubes Beakers Hot plates Balance Mortar and Pestle Small Plastic Spoon/Scraper Weigh boats Thermometer Rubber gloves Safety goggles

Procedure 1.) First use the analytic balance and measure the mass of the empty weigh boat and set the balance so the balance reads zero grams with the weigh boat still on the balance. 2.) Then take the solvent and mass out the needed amount,.5g. You need four amounts of this solvent for finding the M.P./ F.P. of the solvent the known solution and the two unknown solutions.

Procedure Continued… 3.) Afterwards mass out the need amounts of the different solutes,.1g of each, and then put the three amounts into three different beakers. Then add the.5g of solvent to each beaker, crushing/ mixing the solvent and solute into a fine powder. Then melt the solutions using the hotplate. 4.) Take a beaker filled with water and put it on the hotplate. Using the beaker full of warm water melt the sample of the loan solvent and find the M.P./ F.P. using the data studio hardware.

Procedure Continued… 5.) Afterwards, take a test tube and fill it with about 1-2cm of the known solution. Then using the warm water melt the sample and find its M.P./ F.P. using the data studio hardware. Repeat this step for the two unknown solutions. Record the found melting points.

Procedure Continued… 6.) Then calculate the change in temperature by subtracting the solutions’ melting points from the solvent’s melting point in three individual equations. Find the molality of the known solution. Then take the calculated change in temperature of the known solution and divide that number by the molality of the known solution to find the constant M.P/ F.P.

Procedure Continued… 7.) Calculate the number of moles for the unknown solutions by using the freezing point depression equation for the two unknown solutions. 8.) Then calculate the molar mass of the two unknowns by taking the amount of the unknowns solute and dividing it by the moles found in the previous calculation. 9.) Find the percent error after solving the theoretical molecular mass of the two unknown solutions. 10.) Clean up lab.

Data & Observations Solvent melting point - Solution 1 melting point(known) = C C= 2.63 C/m Solvent melting point - Solution 2 melting point(unknown1) = C C=12.32 C/m Solvent melting point - Solution 2 melting point(unknown2) = C C=15.71 C/m (.101 g solute * 1 mole solute) / 147 g solute= mole solute.00069moles C 6 H 4 Cl kgC15H24O =.711 m 2.63 /1.37 m= 1.92 C/m (K FP ) T FP = K FP *m*i (12.32 * ) / (1.92*1)=.0032 mole CH 3 (CH 2 ) 16 COOH (.102gCH 3 (CH 2 ) 16 COOH/.0032 mole CH 3 (CH 2 ) 16 COOH= M (15.71* )/(1.92*1)=.0041mole CH 3 (CH 2 ) 16 COOH.105gCH 3 (CH 2 ) 16 COOH/.0041mole CH 3 (CH 2 ) 16 COOH= M

Data and Observations Continued Actual Solvent melting point - Solution 2 melting point(unknown1) = C C=3.40C/m Actual Solvent melting point - Solution 2 melting point(unknown2) = C C=-7.26C/m T FP = K FP *m*i (3.40* )/(1.92*1)= mole CH 3 (CH 2 ) 16 COOH.102gCH 3 (CH 2 ) 16 COOH/.00089mole CH 3 (CH 2 ) 16 COOH= M (7.26* )/(1.92*1)=. 0019mole CH 3 (CH 2 ) 16 COOH.105gCH 3 (CH 2 ) 16 COOH/.0019mole CH 3 (CH 2 ) 16 COOH= M

Percent Error [ ] X 100% = 72% error [ ] X 100% = 54% error

Conclusion I believe the true purpose behind this lab was to show how the K FP or constant freezing point is an important part in determining how to find the unknown molecular mass of unknown solutions. However, because of errors, it made it harder determining the actual molecular mass of the two unidentified solutions. One of these errors was that the solution was supersaturated by the large quantity of solute added to the amount of solvent. With the solution becoming supersaturated the melting points lowered greatly and that caused errors all over when making the calculations for the molecular mass. Another source of error may have been taking the readings for the temperatures of the solutions to early causing errors with the calculating the change of temperature and the constant freezing point which makes it near impossible to find the correct molecular mass.