Verified answer

Molar mass....

Use a variation on the ideal gas equation:

PV = nRT ........... ideal gas equation

PV = mRT / M .... n = m / M ..... m = mass, M = molar mass

M = mRT / (PV) .. solve for M (the molar mass)

M = 1.35g x 0.08206 Latm/molK x 336K / 1.05 atm / 2.45L

M = 14.5 g/mol

The catch is finding a gas with a molar mass of 14.5 g/mol.

========== Follow up ==========

David's solution is a bit confusing, and Trinity is puzzled by the "22.4". I can see why. There are no units or formulas to guide the novice chemistry student, only a jumble of numbers and some mis-matched brackets.

The idea rests on the fact that 1 mol of an ideal gas occupies a volume of 22.414L at standard temperature and pressure (STP). The ratio of the mass of a gas to the volume is the key to the solution.

mass of 1 mol / volume of 1 mol = given mass / calc. volume ... at STP

The mass of 1 mol is the molar mass, the unknown.

The volume of 1 mol is 22.414 L (at STP)

The calculated volume is what the gas will have at STP

To find the calculated volume we can use the combined gas law

..... P1V1 / T1 = P2V2 / T2 .......... and find V2 at STP

..... 1.05 atm x 2.45L / 336K = 1.00 atm x V2 / 273K

..... V2 = 2.09L ............ This is the volume the 1.35g of gas will have at STP

Now go back to this equation and solve for the mass of 1 mol (which is the molar mass)

mass of 1 mol / volume of 1 mol = given mass / calc. volume

mass of 1 mol = given mass x volume of 1 mol / calc. volume

mass of 1 mol = 1.35g x 22.4L/mol / 2.09L = 14.5 g/mol

And there you have it. This is what David's jumble of numbers is supposed to do.

At STP, this same quantity of gas would occupy

(2.45 L)(273.15/336.15)*1.05 = 2.0904 liters.

That is (2.0904 L)/(22.414 L/mol) = 0.09326 mol.

So the molar mass of the gas is

(1.35 g)/(0.09326 mol) = 14.5 gram/mol.

22.4 (1.35) / [2.45 X (1.05/1) X [(273)/(273+63)] = molar mass

... use a calculator