[laxplayer]

The problem states:
An aerosol spray can with a volume of 250 mL contains 1.90 g of propane gas (C3H8) as a propellant.
If the can is at 18°C, what is the pressure in the can?

I found the answer to be 412 atm by using the ideal gas law (PV=nRT). I found there to be 4.308 moles, the temperature in kelvin to be 291.15, and the volume to be .25 L. I multiplied (4.308) (.0821) (291.15) to get 411.9. I then divided this by .25.

It keeps saying my answer is wrong, and I have no idea why.

[billnotgatez]

are you sure you got the moles correctly

[laxplayer]

I think so, heres what I did:
(190g)(1 mol/44.094g) and I got 4.309

[billnotgatez]

from wiki

The standard scientific unit for dealing with atoms in macroscopic quantities is the mole (mol), which is defined arbitrarily as the amount of a substance with as many atoms or other units as there are in 12 grams of the carbon isotope C-12.

for propane is it  1 mole = 44.094g

[laxplayer]

so since there is 190 grams, you divide 190 by 44.094 to get 4.309 moles, right?

[Sev]

so since there is 190 grams, you divide 190 by 44.094 to get 4.309 moles, right?

But there is 1.9g of propane.

[billnotgatez]

is it not 1.9 grams

would carbon be 12 grams per mole according to the wiki

[laxplayer]

ooooh, I feel dumb :]

[billnotgatez]

could you write out your answer so i can get it right in my head

[laxplayer]

the answer is 4.1 atm.
you multiply (.04309)(.0821)(291.15) to get 1.0299, then you divide that by .25 to get 4.1199, 4.1 with two SF

[billnotgatez]

Thanks i was having trouble getting my head around how you got the moles

i assume you divided 1.9 by 44.094 to get (.04309)

[laxplayer]

yeah, I read the problem wrong, and calculated the moles from 190 g rather than 1.9 g