7. 1-propanol 0.090 L (60 g/mole, d = 0.81 g/mL) and 0.075 L acetic anhydride (6

Question

7. 1-propanol 0.090 L (60 g/mole, d = 0.81 g/mL) and 0.075 L acetic anhydride (60 g/mole, 1.05 g/mL) are mixed in esterification reaction and catalyst used is 1.8 mL of sulfuric acid (d = 1.84g/mL). The crude distillate is washed with two 25-ml. portions of water, then with 20-ml. portions of 10% sodium carbonate until the ester layer is neutral to litmus; the product is finally dried over anhydrous sodium sulfate. After removal of the drying agent, the ester is fractionally distilled through an efficient fractionating column. The pure ester distills between 102°C - 103°C and distillate weighed 20 g.

Explanation / Answer

amount of propanol = 0.09 L

density = 0.81 g/ml

hence 0.09 L or 90 ml will have -----> 90X 0.81

= 72.9 g

or moles = 72.9 / 60 = 1.215

amount of acetic anhydride = 0.075 L

density = 1.05 g/ml

hence , 75 ml will have ---> 75 X 1.05

78.75 g

or moles of it = 1.3125

sulphuric acid is catalyst ...but still amount of it used = 1.8 ml

or density = 1.84g/ml

hence 1.8 ml will have -----> 1.8 X 1.84 = 3.312 g

or moles of acid = 3.3312 / 98

= 0.0337 moles

further limiting reacgent = propanol as it's moles aare less than tha of acetic anhydride..

CH3CH2CH2OH + CH3COOH ------> CH3COOCH2CH2CH3

1 mol of propanol gives 1 mol of ester

hence

1.215 moles of propanol should give 1.215 moles of ester...

hence amount of ester produced = 1.215 X 102

= 123.9 g this is the theoretical yield

practical yield = 20 g

hence % yield =(theoretical yield / practical yield) X 100

=(20/123.9)

= 16.14 %

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