11.A student was given a sample of crude salicylic acid to recrystallize. The in

Question

11.A student was given a sample of crude salicylic acid to recrystallize. The initial mass of the the crude acid was 161 mg. The mass after recrystallization was 121 mg. Calculate the percent recovery from recrystallization.

2.A student measures out exactly 0.109 g of salicylic acid and runs the experiment as dictated in the lab manual. They obtain 0.119 g of aspirin. What is the percent yield for their reaction

3.If 91.0 g of oxalic acid and 79.0 g of methanol react to form 89.0 g of dimethyl oxalate, what is the percent yield?(Molar mass of oxalic acid=90.04g/mol,MM of methanol=32.04,MM of dimethyl oxalate=118.1,MM of water=18.02)

Explanation / Answer

11.A student was given a sample of crude salicylic acid to recrystallize. The initial mass of the the crude acid was 161 mg. The mass after recrystallization was 121 mg. Calculate the percent recovery from recrystallization.

Solution :-Formula to calculate the percent yield is as follows

% yield =( actual yield / theoretical yield )*100%

                                  = (121 mg / 161 mg)*100 %

                                  = 75.15 %

So the percent yield of the recrystallization is 75.15 %

2.A student measures out exactly 0.109 g of salicylic acid and runs the experiment as dictated in the lab manual. They obtain 0.119 g of aspirin. What is the percent yield for their reaction

Solution :- mole ratio of the salicylic acid and aspirin is 1 : 1

Molar mass of the salicylic acid = 138.121 g per mol

Molar mass of aspirine is 180.157 g per mol

Using the mole ratio lets calculate the theoretical yield of the aspirine

0.109 g salicylic acid * 180.157 g aspirin / 138.121 g salicylic acid = 0.142 g aspirin

Now lets calculate the percent yield

% yield =( actual yield / theoretical yield )*100%

                                  = (0.119 g / 0.142g)*100 %

                                  = 83.80 %

3.If 91.0 g of oxalic acid and 79.0 g of methanol react to form 89.0 g of dimethyl oxalate, what is the percent yield?(Molar mass of oxalic acid=90.04g/mol,MM of methanol=32.04,MM of dimethyl oxalate=118.1,MM of water=18.02)

Solution :-

Oxalic acid + 2 mole methanol ----- > dimethyl oxalate + 2 mol water

Mole ratio is 1 : 2

So lets calculate the mass of the dimethyl oxalate that can be formed form each reactant

91.0 g oxalic acid * 118.1 g dimethyl oxalte / 90.04 g oxalic acid = 119.3 g dimethyl oxalate

79.0 g methanol * 118.1 g dimethyl oxalate / (32.04 g methanol * 2) = 145.6 g dimethyl oxalate

Since oxalic acid gives less amount of the dimethyl oxalate therefore theoretical yield is 119.3 g dimethyl oxalate.

Actual yield is 89.0 g dimethyl oxalate

Now lets calculate the percent yield

% yield = (actual yield / theoretical yield )*100%

              = ( 89.0 g / 119.3 g) *100 %

              = 74.6 %

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