Calculate the partial pressure of diethylether.

Question

A sample of 5.50 mL of diethylether (C2H5OC2H5; density= 0.7134 g/mL) is introduced into a 6.50 -L vessel that already contains a mixture of N2 and O2, whose partial pressures are Pn2= 0.751 atm and Po2= 0.209 atm. The temperature is held at 35.0 degrees C and the diethylether totally evaporates.

Answer 1

molar mass C2H5OC2H5 = 94.28 g/mole

mass C2H5OC2H5 = 5.50 ml x 0.7134 g/ml = 3.9237 g

moles C2H5OC2H5 = 3.9237 g x 1 mole/94.28 g = 0.04162 moles

PV = nRT where P=?; n=0.04162 moles; V=6.50L; R=0.0821 Latm/Kmol; T=35ºC +273 = 308K

P = nRT/V = (0.04162)(0.0821)(308)/6.50

P = 0.162 atm (to 3 significant figures (this is the partial pressure of diethyl ether).

I used the incorrect molar mass for diethyl ether. It is not 94.28 but should be 74.12 g/mole. The correct answer is... molar mass C2H5OC2H5 = 94.28 g/mole mass C2H5OC2H5 = 5.50 ml x 0.7134 g/ml = 3.9237 g moles C2H5OC2H5 = 3.9237 g x 1 mole/74.12 g = 0.05294 moles PV = nRT where P=?; n=0.05294 moles; V=6.50L; R=0.0821 Latm/Kmol; T=35ºC +273 = 308K P = nRT/V = (0.05294)(0.0821)(308)/6.50 P = 0.206 atm (to 3 significant figures (this is the partial pressure of diethyl ether).