Asample of gas at 310 kelvin and 1.2 atmospheres has a density of 2.78 g/l. what is the molar mass of this gas? explain how you got your answer

Ideal gas law PV = nRT    or  PM = dRt ( M is molecular mass of gas ) and d is gas density 
where m/ M    ( m is mass of gas sample ) 

m = dRT / P  ( 2.78 g/L )   ( 0.082 L atm/ K mole )( 310K )  /  1.2 atm  =
58.9 g/mol

gas must be butane ( C4H10) has molecular mass = 58.9 g/mol

 They belong to the same kingdom but different phyla.

Explanation: