Freezing Point Depression Lab
Ben Quinones
AP Chem
Mr. Perry
25 January 2013
Molar Mass by Freezing Point Depression
Conclusion: My Lab unsuccessfully found the molar mass of the unknown given to us in the lab. We were tasked with finding the molar mass of an unknown compound using the colligative property of Freezing Point Depression. The way we did this was first finding the temperature at which ice crystals form for just regular BHT. Luckily thanks to our TA we were given temperature probes so we could pinpoint exact temperatures. We found that the temperature at which crystallization of BHT was 68.9 Celsius in our lab. After finding temperature of BHT we were tasked with adding the solute cetyl acid to our BHT mixture. We then heated and melted the two substances
than found the freezing point of the two substances. We found the temperature at which crystals formed of the two to be 63.8 degrees Celsius.
Now the main reason we mixed these two substances was to find the freezing point of BHT. Now to do this we need to take the grams of cetyl acid and find moles of solute and divide it by kilograms of our solvent. This gave the molality of our solution which turned out to be .515 mol/kg. After finding molality we could find kfp or the freezing point. To do this you would use the formula ∆Tfp = kfpm, so we take the change in temperature which is 5.1 degrees Celsius equal to kfp times .515 mol/kg all which turns out to be 9.90 oc/molal.
After we find this crucial piece of information we can add the unknown to our BHT melt the two down and find the ∆T of it which comes out to be 3.3. With all this information given we can finally determine the molar mass of our unknown. To do this we would use the equation MM (unknown solute) = kfp x g(solute)/ kg(solvent) x ∆Tfp now when we plugged all our information into said formula and crunched the numbers we found our molar mass to be 386 g/mol. We found out that the correct molar mass was 284 g/mol and the unknown was stearic acid.
Data Table: | Trial 1 | Trial 2 | Mass of BHT, g | 7.999 g | | Mass of weighing paper, g | .415 g | | Mass of weighing paper plus cetyl alcohol, g | 1.416 g | | Mass of cetyl alcohol | .998 g | | Mass of BHT | | 7.999 g | Mass of Unknown | | 1.027 g |
Temperature in oC Time, in Seconds | Pure BHT | BHT + Cetyl Alcohol | BHT + Unknown | 160 sec | 68.9 | | | 195 sec | | 63.8 | | 145 sec | | | 65.6 |
| BHT | BHT + Cetyl Alcohol | BHT + Unknown | Freezing Point | 68.9 oC | 63.8 oC | 65.6 oC | ∆Tfp | - | 5.1 oC | 3.3 oC | kfp | 9.91 oC | - | - | Molar Mass | - | - | 386 g/mol |
Error Analysis: Im not quite sure what went wrong in this lab perhaps some of the solute was spilled after weighing and didn’t make it to the mixing of solute and solvent. Perhaps simply calculator errors but for now it is unknown. The percent error of this lab was 35%.
AP Chem
Mr. Perry
25 January 2013
Molar Mass by Freezing Point Depression
Conclusion: My Lab unsuccessfully found the molar mass of the unknown given to us in the lab. We were tasked with finding the molar mass of an unknown compound using the colligative property of Freezing Point Depression. The way we did this was first finding the temperature at which ice crystals form for just regular BHT. Luckily thanks to our TA we were given temperature probes so we could pinpoint exact temperatures. We found that the temperature at which crystallization of BHT was 68.9 Celsius in our lab. After finding temperature of BHT we were tasked with adding the solute cetyl acid to our BHT mixture. We then heated and melted the two substances
than found the freezing point of the two substances. We found the temperature at which crystals formed of the two to be 63.8 degrees Celsius.
Now the main reason we mixed these two substances was to find the freezing point of BHT. Now to do this we need to take the grams of cetyl acid and find moles of solute and divide it by kilograms of our solvent. This gave the molality of our solution which turned out to be .515 mol/kg. After finding molality we could find kfp or the freezing point. To do this you would use the formula ∆Tfp = kfpm, so we take the change in temperature which is 5.1 degrees Celsius equal to kfp times .515 mol/kg all which turns out to be 9.90 oc/molal.
After we find this crucial piece of information we can add the unknown to our BHT melt the two down and find the ∆T of it which comes out to be 3.3. With all this information given we can finally determine the molar mass of our unknown. To do this we would use the equation MM (unknown solute) = kfp x g(solute)/ kg(solvent) x ∆Tfp now when we plugged all our information into said formula and crunched the numbers we found our molar mass to be 386 g/mol. We found out that the correct molar mass was 284 g/mol and the unknown was stearic acid.
Data Table: | Trial 1 | Trial 2 | Mass of BHT, g | 7.999 g | | Mass of weighing paper, g | .415 g | | Mass of weighing paper plus cetyl alcohol, g | 1.416 g | | Mass of cetyl alcohol | .998 g | | Mass of BHT | | 7.999 g | Mass of Unknown | | 1.027 g |
Temperature in oC Time, in Seconds | Pure BHT | BHT + Cetyl Alcohol | BHT + Unknown | 160 sec | 68.9 | | | 195 sec | | 63.8 | | 145 sec | | | 65.6 |
| BHT | BHT + Cetyl Alcohol | BHT + Unknown | Freezing Point | 68.9 oC | 63.8 oC | 65.6 oC | ∆Tfp | - | 5.1 oC | 3.3 oC | kfp | 9.91 oC | - | - | Molar Mass | - | - | 386 g/mol |
Error Analysis: Im not quite sure what went wrong in this lab perhaps some of the solute was spilled after weighing and didn’t make it to the mixing of solute and solvent. Perhaps simply calculator errors but for now it is unknown. The percent error of this lab was 35%.