What is the mass of a 22.41*L volume of ethane that has a pressure of 780*mm*Hg?

1 Answer
Apr 26, 2017

We get about 16*g, i.e. a molar quantity.

Explanation:

1*atm pressure will support a column of mercury that is 760*mm*Hg high. Measurement of pressure in mm*Hg is thus useful (because a mercury column is easy to set up!), because it can measure a pressure of about an atmosphere, or very LOW pressures, such as when you do a vacuum distillation, and evacuate the system out to 0.01*mm*Hg.

If you attempt to measure pressures much HIGHER than 1 *atm with a mercury column, YOU WILL GET metallic mercury all over the laboratory, and this is a major clean-up job. Nevertheless, mercury barometers sometimes read a pressure of slightly greater than 760*mm.

So let's convert the pressure to "atmospheres".

"Pressure"=(780*mm*Hg)/(760*mm*Hg*atm^-1)=1.026*atm.

The Ideal Gas Equation tells us that n=(PV)/(RT)

i.e. "Mass"/"Molar mass"=(PV)/(RT)

OR "Mass"="Molar Mass"xx(PV)/(RT)

And now we fill in the "noombers":

"Mass"=(16.33*g*cancel(mol^-1)xx1.026*cancel(atm)xx22.410*cancelL)/(0.0821*cancel(L*atm*K^-1*mol^-1)xx273.13*cancelK)

=??*g

We get an answer in "grams" so we must be doing something right.

We could have also used the fact that under standard conditions of 273*K, and 1*atm, ONE mol of Ideal Gas will occupy 22.4*L. And we know the molar mass of methane.