# Molar Mass by freezing point depression? Help?

I’m trying to determine the molar mass of an unknown substance by using freezing point!

BHT is my solvent. I use cetyl alcohol to calculate freezing point depression constant

My data from the experiment

Freezing point of pure BHT: 67.0 degrees Celsius

Freezing point of BHT+cetyl alcohol: 59.7 degrees Celsius

Freezing point of BHT+unknown: 63.9 degrees Celsius

Mass of solution #1 BHT + cetyl alcohol

BHT: .5055g

cetyl alcohol: .1094g

Mass of solution #1 BHT + unknown

BHT: .5019

unknown: .9960

My calculations… (the part I’m having trouble with)

I calculated change in freezing point for #1 solution —>7.3 degrees Celsius

change in f.p. for soln #2—> 3.1 degrees Celsius

I calculated molality of cetyl alcohol. m=g solute/kg solvent x molar mass solute

which came out to be .8927 mol solute/ kg solvent

Then i used the equation change in freezing point(delta t) = freezing point depression constant(kfp) x molality(m) to find freezing point depression constant(kfp)

So i did 7.3 = kfp(.8927)

and got 8.2 as my freezing point depression constant

then i used the equation molar mass solute=kfp x g solute / kg solvent x delta t (to find molar mass of the unknown)

which is molar mass= 8.2 x .9960 g unknown / .0005019 kg BHT x 3.1 degrees celsius and got

5249.214 g/mol as my molar mass! did i do something wrong? because the true molar mass of the unknown is 284.49g / mol which is stearic acid… and my answer is…. very off…

• Trevor H
9 days ago

Let us start by calculating the freezing point depression , Kf for BHT using data from solution #1.

Solution is:

BHT: .5055g

cetyl alcohol: .1094g

You need to calculate the molality of the cetyl alcohol solution:

Mass of cetyl alcohol in 1.0kg of BHT = 0.1094/0.5055*1000 = 216.419g in 1.0kg BHT

Molar mass cetyl alcohol = 242.44 g mol−1

Molality of solution = 216.419/242.44 = 0.8927m solution

This decreased the freezing point by 7.3°C

Kf for BHT = 7.3/0.8927 = 8.177°C/m.kg

The freezing point depression for solution #2 was 3.1°C

Therefore the solution was 3.1/8.177 = 0.379 molal

Mass unknown in 1.0kg BHT = 0.9960/0.5019*1000 = 1,984.45g in 1.0kg

Molar mass of unknown = 1,984.45/0.379 = 5,236g/mol

We both get the same answer.

What is the problem: I am concerned about the very high amount of unknown you used in solution #2. 0.996g in 0.5055g solvent. I do not know much about the solubility of stearic acid in BHT, but I wonder if you have produced a true solution or some form of saturated solution or an emulsion. Under these conditions you cannot use Kf to determine the molar mass of the unknown. Was there a special reason that you used such a large amount? I would prefer you to have made a test solution using a similar quantity as you used for the cetyl alcohol. You will see that the molar mass of cetyl alcohol and stearic acid are very similar – so solutions of similar concentration would have produced a better result , I think. You have used about 9 times the mass concentration of the unknown compared to the cetyl alcohol. I think that this is where your problem is.

Recheck you figures for possible mistake in recording the masses or mistake in recording freezing points.

• Rafaellle
6 days ago

RE:

Molar Mass by freezing point depression? Help?

I'm trying to determine the molar mass of an unknown substance by using freezing point!

BHT is my solvent. I use cetyl alcohol to calculate freezing point depression constant

My data from the experiment

Freezing point of pure BHT: 67.0 degrees Celsius

Freezing point of BHT+cetyl…

• Kellie
5 days ago

mass of the solvent cyclohexane = 20.2 mL x .779 g . mL ^-1 = 15.736 g delta t = m x Kf in this case Kf for cyclohexane = 20.2 °C/(mol kg-1 2.5 oC = m mol Kg ^-1 x 20.2 °C/(mol kg-1 m = 2.5 / 20.2 = 0.1238 mol kg ^-1 we have .2436 g dissolved in 15.736 g of solvent = .2436 g x 1000 g / 15.736 g per Kg = 15.48 g in 1 Kg of solvent .1238 moles = 15.48 kg mass of 1 mole = 15.48 g / .1238 moles = 125.1 g / mole molar mass