Mole fraction of unknown gas and molar mass of the gas

[catchtex]

A mixture of an unknown gas with nitrogen in a 5.00 L container exerts a pressure of 2.35 atm at 27.35°C. Before the unknown gas was added the container had a pressure 1.76 atm.

What is the mole fraction of the unknown gas?
(First time I tried this I got .119 by subtracting the two pressures and using PV=nRT)

If 2.42 g of the unknown gas was added to the container, what is the molar mass of the gas?
(Couldn't do this one without the first one)

Any help would be greatly appreciated!!!

[kingchemist]

Mole N2 = PV/RT = 1.76 x 5.00/0.082?057 x 300.35 = 0.3571 moles N2

When gas is added total moles = PV/RT = 2.35 x 5.00/0.082?057 x 300.35 = 0.4768 moles

So moles of unknown gas = 0.4768 moles - 0.3571 moles N2 = 0.1197 moles (same as you)

So mole fraction of unknown gas = 0.1197/0.4768 = 0.2510

0.1197 moles of unknown gas weigh 2.42g
1 mole   of unknown gas weighs (2.42/0.1197)  x 1 = 20.2172 g

20.2172 g/mol is the molar mass

[catchtex]

Wow  Thank you!  That was very helpful. I forgot the last step on the first part of the problem...