Question

A gaseous mixture of CO and CO2 has a density of 0.9967 g/L at 14.24°C and 490.1 mm Hg. What is the mass percent of CO?

Answer 1

Molar mass of CO = 12 + 16 = 28 g/mol

Molar mass of CO2 = 12 + (2x16) = 44 g/mol

We know that PV = nRT

                             = ( m/M ) RT

                         P = ( m/V ) * ( RT / M )

                         P = dRT /M

Where

d = density = 0.9967 g / L

T = Temperature = 14.24 oC = 14.24+273 = 287.24 K

P = pressure = 490.1 mm Hg = 490.1 / 760 atm                  Since 1 atm = 760 mm Hg

                      = 0.64 atm

M = Molar mass of the mixture= ?

R = gas constant = 0.0821 L-atm /(mol - K)

m = mass of the gaseous mixture

V= Volume of the gas

Plug the values we get P = dRT / M

                                   M = dRT / P

                                      = ( 0.9967 x 0.0821 x 287.24) / 0.64

                                     = 36.73 g/mol

So the mass percentage of CO is = molar mass of the mixture x ( correction factor)

                                                     = 36.73 x (molar mass of CO / total mass of CO &CO₂ )

                                                    = 36.73 x ( 28 / (28+44) )

                                                    = 14.28 %