In this particular experiment performed, 3 different metal-organic frameworks we

Question

In this particular experiment performed, 3 different metal-organic frameworks were synthesised in which Cu(II) centres are bridged by pyrazine ligands. Each metal-organic frameworks (MOFs) were synthesised in water (Method 1) and in ethanol (Method 2). Write the complete chemical reactions and calculate the theoretical yield and percent yield for each method (Method 1 and Method 2) for each MOFs (MOF-A1, MOF-A2, MOF-B1, MOF-B2, MOF-C1 and MOF-C2). The procedures followed in the experiment are provided below along with the actual yield obtained in the experiment. If possible, figure out what the value for 'n' should be for each MOF. Note: MOF-A1 refers to the MOF synthesised in water and MOF-A2 refers to MOF synthesised in ethanol. This is the same naming method for the other two MOFs synthesised.

1) MOF A - [Cu(pyz)(NO3)2]n

Method 1 (MOF-A1) - from water

Actual Yield = 0.2318 g

Theoretical Yield = _________ g

Percentage Yield = _________ %

Method 2 (MOF-A2) - from ethanol

Actual Yield = 0.9101 g

Theoretical Yield = _________ g

Percentage Yield = _________ %

Figure 1 (above): Experimental Method for MOF A - [Cu(pyz)(NO3)2]n

2) MOF B - [Cu(pyz)(NO3)2]n

Method 1 (MOF-B1) - from water

Actual Yield = 0.3576 g

Theoretical Yield = _________ g

Percentage Yield = _________ %

Method 2 (MOF-B2) - from ethanol

Actual Yield = 1.244 g

Theoretical Yield = _________ g

Percentage Yield = _________ %

Figure 2 (above): Experimental Method for MOF B - [Cu(pyz)(NO3)2]n

3) MOF C - [Cu(pyz)(NO3)2]n

Method 1 (MOF-C1) - from water

Actual Yield = 0.2948 g

Theoretical Yield = _________ g

Percentage Yield = _________ %

Method 2 (MOF-C2) - from ethanol

Actual Yield = 1.2377 g

Theoretical Yield = _________ g

Percentage Yield = _________ %

Figure 3 (above): MOF C - [Cu(pyz)(NO3)2]n

MOF A [Cu(pyz)(NO3)2ln Method 1 From water solvent Dissolve 1.04 g of Cu(NO3)2.3H20 and 0.34 g of pyrazine separately in 3.5 mL of water each. Note that the solvent volume is crucial and must be measured carefully. Combine the solutions in a 50 mL conical flask and mix well. Cover the opening of the flask with Parafilm and allow the solution to stand undisturbed for 1 week. Crystals should form within a few days. Remove some crystals from the mother liquor and dry them by blotting briefly on a piece of filter paper. Note their shape and colour and report this in the experimental section of your report. Record the IR spectrum of these crystals immediately and conduct thermal decomposition as soon as possible (see below). Collect the remaining crystals by vacuum filtration, wash with a small amount of ice-cold water and allow to air dry for 5 minutes. Note the shape and colour of the crystals and

Explanation / Answer

1) MOF A - [Cu(pyz)(NO3)2]n

in all the cases n=1

the only thing which is changed is moles of pyrazine in the comlex formed 1,2,3 respectively.

note-in present case Cu(NO3)2·3H2O  is limiting reagent and

applying mole concept-

no of moles/mmoles of Cu(NO3)2·3H2O = no of moles/mmoles of [Cu(pyz)(NO3)2]n

so ,mmoles of Cu(NO3)2·3H2O =1.04/241.599 =4.30 mmole (where 241.599 g/mole is molar mass of  Cu(NO3)2·3H2O )

accordingly theoretical yield of [Cu(pyz)(NO3)2]n =1.1507 g

Method 1 (MOF-A1) - from water

Actual Yield = 0.2318 g

Theoretical Yield = 1.1507g

Percentage Yield = 18.945%

Method 2 (MOF-A2) - from ethanol

Actual Yield = 0.9101 g

Theoretical Yield = 1.1507g

Percentage Yield = 79.0909 %

MOF B - [Cu(pyz)2(NO3)2]n

Method 1 (MOF-B1) - from water

note-in present case Cu(NO3)2·3H2O  is limiting reagent and

applying mole concept-

no of moles/mmoles of Cu(NO3)2·3H2O = no of moles/mmoles of [Cu(pyz)2(NO3)2]n

so ,mmoles of Cu(NO3)2·3H2O =1.04/347.689 =4.30 mmole (where 241.599 g/mole is molar mass of  Cu(NO3)2·3H2O )

accordingly theoretical yield of [Cu(pyz)2(NO3)2]n =1.4950g

Actual Yield = 0.3576 g

Theoretical Yield = 1.4950g

Percentage Yield = 23.9197%

Method 2 (MOF-B2) - from ethanol

Actual Yield = 1.244 g

Theoretical Yield = 1.4950g

Percentage Yield =83.2107 %

MOF C - [Cu(pyz)3(NO3)2]n

Method 1 (MOF-C1) - from water

note-in present case Cu(NO3)2·3H2O  is limiting reagent and

applying mole concept-

no of moles/mmoles of Cu(NO3)2·3H2O = no of moles/mmoles of [Cu(pyz)3(NO3)2]n

so ,mmoles of Cu(NO3)2·3H2O =1.04/347.689 =4.30 mmole (where 241.599 g/mole is molar mass of  Cu(NO3)2·3H2O )

accordingly theoretical yield of [Cu(pyz)3(NO3)2]n =1.8394g

Actual Yield = 0.2948 g

Theoretical Yield = 1.8394 g

Percentage Yield = 16.0269%

Method 2 (MOF-C2) - from ethanol

Actual Yield = 1.2377 g

Theoretical Yield =  1.8394 g

Percentage Yield = 67.2882%

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