A) A 2.49 sample of gas occupies a volume of 752 mL at 1.98 atm and 62oC. What is the gas?
a. SO2
b. SO3
c. NH3
d. NO2
e. Ne
B) why? (show work)
3 Answers

V = 0.752 L
T = 62 + 273 =335 K
n = pV / RT = 1.98 x 0.752 / 0.0821 x 335 = 0.0541
MM = 2.49 g / 0.0541 =46 g/mol
NO2 has molar mass = 14 + (16 x 2) = 46 g/mol

Why not fold the mass right into the ideal gas equation?
Use PV = nRT and n=m/M where m is the mass and M is the molar mass.
You get:
PV = mRT / M
Then solve for M
M = mRT / PV
M = (2.49 g) (0.0821 L atm / mol K) (335 K) / 1.98 atm / 0.752 L
M = 46.0 g / mol
The gas is NO2. It has a molar mass of 46.0 g/mol

PV = nRT
n = PV/RT
n = (1.98)(0.752)/(0.0821)(335)
n = 0.054 mol
2.4 g / 0.054 mol = 44.3 g/mol
NO2 with a molecular weight of 46 g/mol comes closest to 44.3 g/mol, so pick it.