A mixture of 40 mL of methyl butyrate (MW+102.1 g/mol, d= 0.89 g/mL) and 10 mL o

Question

A mixture of 40 mL of methyl butyrate (MW+102.1 g/mol, d= 0.89 g/mL) and 10 mL of butyric acid (MW=88g/mol, d=0.96 g/mL) is distilled. Calculate the mole percent of each component and then answer these questions:

-What is the initial boiling point of this mixture?

-What is the composition of the vapor in equilibrium with this liquid?

-What is the composition of the initial condensate from simple distillation?

-Instead of a simple distillation, you set up a fractional distillation. Assuming three theoretical plates, what is the composition of the first fraction collected?

Explanation / Answer

methyl butyrate moles = 40*0.89/102.1 = 0.3486 mole

butyric acid moles = 10*0.96/88 = 0.109 mole

mole percent of methyl butyrate = 0.3486*100/0.3486+0.109 = 76.18 %

mole percent of butyric acid = 0.109*100/0.3486+0.109 = 23.82 %

The total pressure is P=Pa + Pb

methyl butyrate Pa = (0.3486*40/0.3486+0.109)    (since vapour pressure of methyl btyrate= 40 mm Hg)

                            = 30.47 mm Hg

butyric acid Pb = (0.109*43/0.3486+0.109)    (since vapour pressure of butyric acid = 43 mm Hg)

                            = 9.53 mm Hg

P=Pa + Pb = 30.47 + 9.53 = 39.99 mm Hg

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