can someone help me with this? calculate the relative molecular mass of a gas if a 500cm3 sample at 20c and 1atm has a mass of 0.66g?

For the hydrocarbon gas :
Volume, V = 500 cm³ = 0.5 L
Mass, m = 0.66 g
Temperature = (273 + 20) K = 293 K
Pressure, P = 1 atm
Gas constant, R = 0.0821 L atm / (mol K)
Molar maass, M = ? g/mol

PV = nRT and n = m/M
Then, PV = (m/M)RT
Hecne, M = mRT/(PV)

Molar mass, M = 0.66 × 0.0821 × 293 / (1 × 0.5) g/mol = 32 g/mol
Relative molecular mass = 32

well.. you could use
.. PV = nRT
.. n = PV / RT
to calculate moles, then divide mass by moles to get molar mass

but, let's do some algebra first
.. PV = nRT.... ideal gas law
.. . n = mass / molar mass.... mass / mw... by definition of molar mass

subbing
.. PV = (mass / mw) RT

rearranging
.. mw = mass x RT / (PV)

solving
.. mw = 0.66g x (0.08206 Latm/molK) x (293.15k) / ((1 atm) x (0.500L)) = 32 g/mol

********
now.. watch what happens if I do this
.. PV = nRT
.. PV = (mass / mw) RT
.. mw = (mass / V) x RT / P

see that term "mass / V"... that's density
.. mw = density x RT / P

remember this. You will see problems where you're given density, T, and P and asked to find molar mass.

Use the Ideal Gas Eq'n , which is PV = nRT
Hence by algebraic rearrangement
n = PV/RT

Next convert the units to S.I. values.
P = 1 atm = 101325 Pa
V = 500 cm^3 = 0.5 dm^3 = 0.0005 m^3
R = 8.314 (The Gas Constant)
T = 20c = 293 K

Substituting in
n = (101325 Pa X 0.0005)/ ( 8.314 X 293)
n = 0.0208 moles.

Remember also
moles = mass(g) / Mr
Mr = mass(g) / n(moles)
Mr = 0.66 / 0.0208
Mr = 31.73 ~ 32
From Periodic Table atomic Oxygen(O) = 16, Then molecular oxygen (O2) = 16 x 2 = 32 , the Mr.
Hence the unknown gas is molecular oxygen (O2) .

NB
When using the Ideal Gas Eq'n, make sure you convert to the correct units (as I have done)!!!!