Calculate percent yield for each reaction: Reaction 1: Preparing the Nitration S

Question

Calculate percent yield for each reaction:

Reaction 1: Preparing the Nitration Solutions----- .6ml of H2SO4 was added to .5ml of HNO3-----Nitrating Acetanilitde---.5g of acetanilitde was dissolved in 1.0ml of H2SO4. The nitrating solution was added to this and the reaction took place for 30 minutes. After this time 10ml of water was placed in a beaker and the solution was added to this water. Crystals formed and my final mass was .243g Please calculate percent yield

Reaction 2: Preparing the Nitration Solutions----- .6ml of H2SO4 was added to .5ml of HNO3-----Nitrating Methyl Benzoate---.5ml of methyl benzoate was dissolved in 1.0ml of H2SO4. The nitrating solution was added to this and the reaction took place for 30 minutes. After this time 10ml of water was placed in a beaker and the solution was added to this water. Crystals formed and my final mass was ..184g Please calculate percent yield  

Explanation / Answer

Reaction 1 :

moles acetanilide = 0.5 g/135.17 g/mol = 0.0037 mol

Theoretical moles nitrated product formed = 0.0037 mol

Theoretical yield of product = 0.0037 mol x 180.163 g/mol = 0.67 g

percent yield = (0.243/0.67) x 100 = 36.46%

Reaction 2 :

moles methyl benzoate = 0.5 ml x 1.08 g/mol/136.15 g/mol = 0.004 mol

Theoretical moles of nirated product = 0.004 mol

Theoretical yield of nitrated product = 0.004 mol x 181.147 g/mol = 0.720 g

percent yield = (0.184/0.720) x 100 = 25.55%

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