Question

A steel cylinder contains 6.812×10⁻³ kg of an unknown gas. Combustion analysis indicates that the gas has the empirical formula CH4. The volume of the cylinder is 2.20×103mL, and the pressure inside the cylinder is 3.6×103mmHg. The cylinder is stored in a closet at 25.0 ∘C.

What is the molecular formula of the gas?

Answer 1

P = 3.6*10^3 mmHg   = 3600mmHg    = 3600/760   = 4.74atm

V   = 2.2*10^3ml      = 2200ml   = 2.2L

T = 25+273 = 298K

W   = 6.812*10^-3Kg   = 6.812g

PV    = nRT

n   = W/G.M.Wt   = W/M

PV   = WRT/M

M   = WRT/PV

     = 6.812*0.0821*298/4.74*2.2    = 16g/mole

The empirical formula   = CH4  

the empirical formula weight = 16g/mole

molecular formula   = (empirical formula)n

                   n         = M.Wt/E.F.wt

                             = 16/16   = 1

molecular formula = (CH4)1 = CH4 >>>answer