Ok, for 3) you need the volume of butane you measured, and the pressure of that butane. The pressure of the butane is atmospheric minus water vapor pressure (there is water vapor in there).
You get the water vapor pressure from a water vapor pressure table, knowing the temperature.
Then, you have volume, pressure, and number of moles. Now correct it to stp
PV=nRT and (letting small letters meaning your conditions, capital letters at stp)
pv=nRt
dividing the first equation by the second
PV/pv=T/t
and you are solving for V
V=pvT/Pt
density at stp
V/massbutane
I did this experiment where I filled a graduated cylinder with water. The I put a stopper on it and put it under water and removed the stopper. The with a modified lighter (that wouldn't release sparks) I added butane gas. I got the following results:
mass of lighter = 15.18 g
final mass of lighter = 15.00 g
volume of butane = 80.0 mL
temperature = 25.0 degrees celsius
atmosperic pressure = 101.4 Kpa
1) What would the mass of butane be?
0.18 grams
2) Number of mols of butane?
0.003170846 mol.
3)Molar mass of butane?
56.8 g/mol
3) Calulate the volume of the number of moles found in questin 2 of butane at STP then use that value to find density. I don't know how to do this at all. ???
3 answers
Isn't the density the mass of butane/V
?
?
yes, you are correct, I was in too much of a hurry. Correct that.