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...

Please help! Thank you.

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