Part 1: Preparation of Benzoic Acid Amounts of reagents KMnO 4 = 9.14g Benzyl al

Question

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 1: Preparation of Benzoic Acid from Benzyl alcohol on oxidation with acidified KMnO4 reagent-

The balanced equation for oxidation rxn is :

3 C6H5CH2OH +4KMnO4 ----->3 C6H5COOH +4 MnO2 + H2O +4KOH

calculation of mol of reagents:

mol KMnO4 = mass/molar mas=9.14g/158.034g/mol=0.0578 mol

Benzyl alcohol    = 4ml

density=1.044g/ml

So, mass of benzyl alcohol=1.044g/ml *4ml=4.176g

mol of benzyl alcohol=4.176g/108.14g/mol=0.0386 mol

mol of 3M H2?SO4 = 150ml*3mol/L=0.150L*3mol/L=0.45mol

A) mol of KMnO4 required to react with given mol of benzyl alcohol=3/4*mol of benzyl alcohol=3/4*0.0386mol=0.0289 mol <<< actual KMnO4 added

Thus, benzyl alcohol is lesser in amount and thuslimiting reactant

B) % yield of benzoic acid=(mass of benzoic acid/theoretical yield of benzoic acid)*100

theoretical yield of benzoic acid=mol of benzyl alcohol

So ,mass of benzoic yield=0.0386 mol*molar mass of benzoic acid=0.0386 mol*(122.12g/mol)=4.714 g

% yield of benzoic acid=(2.757g/4.714 g)*100=58.5%

C)

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