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 paradichlorobenzene 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
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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
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Solvent melting point - Solution 1 melting point(known) = 68.54 C- 65.91 C= 2.63 C/m
Solvent melting point - Solution 2 melting point(unknown1) = 68.54 C-56.22 C=12.32 C/m
Solvent melting point - Solution 2 melting point(unknown2) = 68.54 C-52.83 C=15.71 C/m
(.101 g solute * 1 mole solute) / 147 g solute= .00069 mole solute
.00069moles C6H4Cl2
.000505kgC15H24O = .711 m
2.63 /1.37 m= 1.92 C/m (KFP)
TFP = KFP *m*i
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(12.32 * .000505) / (1.92*1)= .0032 mole CH3(CH2)16COOH
(.102gCH3(CH2)16COOH/.0032 mole CH3(CH2)16COOH= 31.88 M
(15.71*.000505)/(1.92*1)= .0041mole CH3(CH2)16COOH
.105gCH3(CH2)16COOH/.0041mole CH3(CH2)16COOH= 25.61 M
Data and Observations Continued
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Actual Solvent melting point - Solution 2 melting point(unknown1) = 73.00 C-69.60 C=3.40C/m
Actual Solvent melting point - Solution 2 melting point(unknown2) = 73.00 C-80.26 C=-7.26C/m
TFP = KFP *m*i
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(3.40*.000505)/(1.92*1)= .00089 mole CH3(CH2)16COOH
.102gCH3(CH2)16COOH/.00089mole CH3(CH2)16COOH= 114.61M
(7.26*.000505)/(1.92*1)= . 0019mole CH3(CH2)16COOH
.105gCH3(CH2)16COOH/.0019mole CH3(CH2)16COOH= 55.26 M
Percent Error
• [114.61- 31.88]
114.61
X 100% = 72% error
• [55.26-25.61]
55.26
X 100% = 54% error
Conclusion
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I believe the true purpose behind this lab was to show how the KFP 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.