Question

Diethyl ether (C4H10O) has a vapor pressure of 379.6 torr at 20.0 °C. If diethyl ether is place in a 10.0 L container at atmospheric pressure and sealed once equilibrium has been established, what will the mass percent of diethyl ether in the gas-phase be (assume air has an average molecular mass of 28.97 g/mol and diethyl ether liquid has a negligible volume).

Answer 1

Solution :-

Vapor pressure of diethyl ether = 379.6 torr* 1 atm / 760 torr = 0.50 atm

Volume = 10.0 L

Molar mass of diethyl ether = 74.12 g /mol

Molar mass of air = 28.97 g/mol

Temperature = 20 C +273 = 293 K

Lets first calculate the total moles of gas and diethyl ether present in the flask

PV= nRT

PV/RT=n

1 atm * 10.0 L / 0.08206 L atm per mol K * 293 K = n

0.416 mol = n

Now lets calculate the moles of diethyl ether in the flask using the vapor pressure diethyl ether

PV= nRT

PV/RT=n

0.5 atm * 10.0 L / 0.08206 L atm per mol K * 293 K = n

0.2077 mol = n

Now lets calculate the moles of air

Moles of air = total moles – moles of diethyl ether

                      = 0.416 mol – 0.2077 mol

                    =0.2083 mol air

Now using the moles and molar mass of each we can find their masses

Mass= moles x molar mass

Mass of diethyl ether = 0.2077 mol * 74.12 g per mol = 15.4 g

Mass of air = 0.2083 mol * 28.97 g per mol = 6.03 g

Now lets calculate the mass percent of diethyl ether

% diethyl ether= [mass of diethyl ether / (mass of air + mass of diethyl ether)] * 100 %

                           = [15.4 g /(6.03 g + 15.4 g)]*100%

                           = 71.9 %

Therefore the mass percent of the diethyl ether is 71.9 %