The 1st image is my data table from lab. The 2nd image is post lab table where I have to perform calculations. The 3 empty boxes in both Trial 3 and Trial 4 I am not sure. I did do the calculations (the one in brackets but Im not 100% if its right) so if you could do the calculations in the bracket to double check my answers. The 3rd image is the 1st calculation step to determine the molar mass of Mixture 1 (Trial 3) and Mixture 2 (Trial 4) which goes along with 2nd image. *Note grams of Mixture 2 is about 2 grams (so for my case its 2.003g because 1.001g +1.002 g equals 2.003g. Please write neat and show units. Thank you (: The rest of the images is my work.

After number 1 of calculation, I have to figure out 2 and 3. 2nd image is Table 2

DATA TABLE: A Fron lab A Post Lal Calculations: 1. Separately calculate the molar mass of your unknown acid for maxtures 1 and 2 be careful - the grams of unknown acid for mixture 2 is approximately 2g because it is the total mass of the unknown acid added to make moxture 1 and mixture 2): molar mass =molesunknownacidgramsunknownacid Midtere 1 mass of uniman fatty acid added: 1.002s From Trialxs 3 9.000 grans of SA added Molalify =kssolventnsoule 1.0025 crossmultiply 9gM=M9Sk1.002s1000ks=111.33M/ 9gM1.0025=M9SR1.002S210001cs=11133M/ks 1000kg The Freezing point befeemixing Tial sl. 2oftrals68.91C+68.41C The freezing point after mixing Trial \&2 : 6+27Trin143T=T1T2=68.66C6cr660C6al27C=739C Freezins point depression T AT=1cfmkp=45C1cgg/mat7.39C=4.5C(M111.33)M=7.39Cx4.5111.33mM=67.79M Mixture $2 : mess of unlowown fatty acid adled. 1.001S From Tial 4 Trials 3 Trial|s Molaity=kssolventnsoduleM2.0035crossnulthly10001cS9g=M9s2.003s10001S= Freezing paint befre maing Thal|li. 68.91C+68.41C2137.32C=68.66C Freezing print ifter mixing thid $2 : 40.482CTral 40.98i= 27.68CT T=kfm If=4.5Ckg0l 2). 68C=4.5Clm222.5J lf=4.5Ckg/mol 27.68C=4.5Clm222.55 M=27.68C4.5c/m222.55M 2. Calculate the average molar mass. 3. Using the data from Table 2, determine the identity of your unknown acid and calculate a percent error in molar mass. Table 2. Possible identities of unknown samples DATA TABLE: A Fron lab A Post Lal Calculations: 1. Separately calculate the molar mass of your unknown acid for maxtures 1 and 2 be careful - the grams of unknown acid for mixture 2 is approximately 2g because it is the total mass of the unknown acid added to make moxture 1 and mixture 2): molar mass =molesunknownacidgramsunknownacid Midtere 1 mass of uniman fatty acid added: 1.002s From Trialxs 3 9.000 grans of SA added Molalify =kssolventnsoule 1.0025 crossmultiply 9gM=M9Sk1.002s1000ks=111.33M/ 9gM1.0025=M9SR1.002S210001cs=11133M/ks 1000kg The Freezing point befeemixing Tial sl. 2oftrals68.91C+68.41C The freezing point after mixing Trial \&2 : 6+27Trin143T=T1T2=68.66C6cr660C6al27C=739C Freezins point depression T AT=1cfmkp=45C1cgg/mat7.39C=4.5C(M111.33)M=7.39Cx4.5111.33mM=67.79M Mixture $2 : mess of unlowown fatty acid adled. 1.001S From Tial 4 Trials 3 Trial|s Molaity=kssolventnsoduleM2.0035crossnulthly10001cS9g=M9s2.003s10001S= Freezing paint befre maing Thal|li. 68.91C+68.41C2137.32C=68.66C Freezing print ifter mixing thid $2 : 40.482CTral 40.98i= 27.68CT T=kfm If=4.5Ckg0l 2). 68C=4.5Clm222.5J lf=4.5Ckg/mol 27.68C=4.5Clm222.55 M=27.68C4.5c/m222.55M 2. Calculate the average molar mass. 3. Using the data from Table 2, determine the identity of your unknown acid and calculate a percent error in molar mass. Table 2. Possible identities of unknown samples