PLEASE ONLY DO PART 2 Part 1: Preparation of Benzoic Acid Amounts of reagents KM

Question

PLEASE ONLY DO PART 2

Part 1: Preparation of Benzoic Acid

Amounts of reagents

KMnO4    = 9.14g

Benzyl alcohol    = 4ml

3M H2?SO4 ? = 150ml

mass of crude benzoic acid crystals formed = 2.757g

A) Calculate the limiting reagent, proving which of the three reagents is limiting (write the chemical equation here as well)

B) Calculate the % yield of the benzoic acid.

C) Create a table showing the theoretical, actual and % yield of the benzoic acid prepared.

Part 2: Analysis of benzoic acid

Table 1: Melting Point Data

Sample    Melting point, Trial 1    Melting point, Trial 2

Benzoic Acid Standard    122.8    122.5

Prepared Benoic Acid    121.3    121.5

Literature Value    122?oC

Table 2: Benzoic Acid Titration data using 0.1036M NaOH

Trial 1 Trial 2    Trial 3

Mass of benzoic acid 0.5375g    0.5727g 0.4451g

Final burette reading    38.2ml    39.02ml 30.02ml

Initial burette reading 2.56ml 0.20ml 1.04ml

Volume NaOH used 35.64ml    58.82ml    28.98ml

A) Create a table showing the melting points of the benzoic acid samples, include a literature value with reference

B) What reasons could there be that the literature value and experimental value differ?

C) Calculate the % purity of each benzoic acid sample which was titrated. And create a table showing your results.

Explanation / Answer

part 2) A)

B) The different degrees of sample purity is the reason behind the difference in melting point literature value and experimental value.Impurities may be unreacted reactants that may lead to difference of m.pt

C) Benzoic Acid Titration data using 0.1036M NaOH

                                      Trial 1 Trial 2    Trial 3

Mass of benzoic acid      0.5375g    0.5727g 0.4451g

Volume NaOH used         35.64ml    38.82ml    28.98ml

trial 1) mol of NaOH used =(0.1036mol/L)*0.03564L=0.00369 mol

NaOH+C6H5COOH ---->C6H5COONa +H2O

mol of benzoic acid reacted with NaOH=mol of NaOH used=0.00369mol

mass of benzoic acid (actual mass)=0.00369mol*(122.12g/mol)=0.451 g

% purity =(actual benzoic acid/impure benzoic acid)*100=(0.451g/0.5375g)*100=83.9%

Similarly in trial 2)

mol of NaOH used =(0.1036mol/L)*0.03882L=0.00402 mol

mol of benzoic acid reacted with NaOH=mol of NaOH used=0.00402 mol

mass of benzoic acid (actual mass)=0.00402 mol*(122.12g/mol)=0.491 g

% purity =(actual benzoic acid/impure benzoic acid)*100=(0.491g/0.5727g)*100=85.7%

trial 3)

mol of NaOH used =(0.1036mol/L)*0.02898L=0.003 mol

mol of benzoic acid reacted with NaOH=mol of NaOH used=0.003 mol

mass of benzoic acid (actual mass)=0.003 mol*(122.12g/mol)=0.367 g

% purity =(actual benzoic acid/impure benzoic acid)*100=(0.367g/0.4451g)*100=82.4%

trial 1 trial 2 Melting point of reference sample(std) of benzoic acid 122.8 deg C 122.5 deg C Melting point of prepared sample of benzoic acid 121.3 deg C 121.5 deg C Literature value ( m.pt of benzoic acid) 122 deg C

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