Molar mass of ideal gas... IDEAL GAS LAW?

Question

What is the molar mass of an ideal gas if 9.50 g occupies 3.73 L at 1418.35 torr and 12.8 degrees C ?

Answer 1

PV=nRT

molar mass = m/n

m= 9.50 g

n=PV/RT

R=0.082 atm-L/gmol-K
P=1418.35 torricelli =1418.35/760 atm = 1.86625 atm
T=12.8 C = 285.95 K
V= 3.73 L

n= (1.86625 atm x 3.73 L)/(0.082 atm-L/gmol-K x 285.95)=0.2969 gmol

molar mass = 9.5 g / 0.2969 gmol = 31.997 g/gmol

HEY! The gas might be diatomic oxygen.

Answer 2

You already asked this question... and I already answered it, but here you go again:

To get molar mass, we need mass/moles. We're given the mass (9.50g) and enough info about the gas to use ideal gas law to get the number of moles.
Ideal gas law is PV=nRT. Rearrange to solve for number of moles, n:
n = PV/RT

Given:
P = 1418.35 torr
V = 3.53 L
T = 12.8 C = 285.95 K

Wikipedia was kind enough to give the ideal gas constant, R, in the proper units:
R = 62.3637 torr*L/(mol*K)

Plugging in the values we find,
n = 0.297 mol

Molar mass, M, is mass / moles, so
M = 9.50 g/0.297 mol
M = 32.0 g/mol (to 3 sig.figs.)

Answer 3

because pv=nrt n is number of moles, not the mass. you dont know what the sample of gas consists of. if you did, then you could find the number of moles