If you can not use the ideal gas law I am guessing there is one more fact you left out. The volume of one mole of any gas at standard temperature and pressure is 22.4 liters. Given this fact you can use Charles' Law and Boyle's Law to get the molar mass.
Using Charles' Law you first need to find the volume of one liter of your gas given at standard temperature (273 K or 0 deg C).
V1/T1 = V2/T2
V1 = 1 liter T1 = 546 K (given as 273 C)
V2 = ? and T2 = 273 K (standard temperature)
Solving for V2:
V2 = (V1 * T2)/T1 = (1 liter * 273 K)/546 K = 0.5 liters
So far, the 2.4 grams that fit into 1 liter (from your given density) will fit into a 0.5 liter container if we cut the temperature in half (going from 546 K to 273 K). Next we have to figure out what changing the pressure will do.
Using Boyle's Law
P1 * V1 = P2 * V2
P1 = 1140 mm Hg V1 = 0.5 liters (answer from above)
P2 = 760 mm Hg (standard pressure) V2 = ?
Solve for V2
V2 = (P1 * V1)/P2 = (1140 mm Hg * 0.5 liters)/760 mm Hg
V2 = .75 liters
That means, the 2.4 grams that fit into the 1 liter container under the conditions given will fit int a 0.75 liter container at STP (standard temperature and pressure). Since 1 moles of any gas will occupy 22.4 liters at STP you can use proportions to show you only have .0335 moles of a gas.
x moles / 1 mole = 0.75 liters / 22.4 liters
solve for x to get .0335 moles.
Remeber that the original mass will not change so, the .0335 moles of gas has a mass of 2.4 grams.
2.4 grams /.0335 moles = 71.6 grams/mole