Was tryin' not to, but ok. So you've got a room, and in it you have three gases:
28.1% is nitrogen molecules
21.0% is oxygen molecules
0.9% is argon atoms
The barometer in the room reads 740 torr.
Find the partial pressure fo the nitrogen in the room.
Ok, you are going to use mass percentage to find the moles of each gas.
First, convert percentage to grams. So Nitrogen will be 28.1g, O will be 21.0g, and Ar will be .9g.
To get moles, you take the grams you just found, and divide them by the element's actual mass. Then you multiply by one mole to get an answer in moles.
So you will take :
28.1g N / 14.007g N * 1 mole = 2.0061 moles
21.0g O / 15.999g O * 1 mole = 1.3126 moles
.9g Ar / 39.948g Ar * 1 mole = .02253 moles
Add all of the moles together to get a total mole #. This = 3.34123 moles. Now you can find the mole fraction of each element (you only need Nitrogen's since it is what you are trying to find) by taking it's moles / total moles.
2.0061 moles N / 3.34123 total moles = .6004076343
.6004076343 is your mole fraction for N. To find the partial pressure of N, take the mole fraction of N and multiply it by the total pressure.
.6004076343 moles * .973984211 atm(same thing as 740 torr) = .5846074337 atm
or 444.30165 torr, which ever you need your answer to be in.