Question

Air is a mixture of several gases. The 10 most abundant of these gases are listed here along with their mole fractions and molar masses.

Component	Mole fraction	Molar mass (g/mol)
Nitrogen	0.78084	28.013
Oxygen	0.20948	31.998
Argon	0.00934	39.948
Carbon dioxide	0.000375	44.0099
Neon	0.00001818	20.183
Helium	0.00000524	4.003
Methane	0.000002	16.043
Krypton	0.00000114	83.80
Hydrogen	0.0000005	2.0159
Nitrous oxide	0.0000005	44.0128

What mass of carbon dioxide is present in 1.00 m3 of dry air at a temperature of 25 ∘C and a pressure of 737 torr ?

Calculate the mass percentage of oxygen in dry air.

Answer 1

a)

mass % of Oxygen in dry air

x-O2 = mass of O2 / total mass

assume a basis of 100 mol of air..

then...

mol of O2 = y-O2 * total mol of air = 0.20948 * 100 = 20.948 mol of O2

mass of O2 = mol*MW = 20.948 * 32 = 670.336 g

MW of air = 28.966 g/mol approx

mass of air = 100*28.966 = 2896.6 gof air

now...

x-O2 = 670.336 / (2896.6 ) = 0.23142 for O2 per mass

b)

mass of CO2 in V = 1 m3... at T = 25°C, 298K at P = 737 torr

so...

let us calculate mol of CO2:

total mol:

PV = nRT

P = 737 / 760 = 0.96973 atm

V =1 m3 = 1000 L

R = 0.082 Latm/molK

T = 298 K

n = PV/(RT) = 0.96973 *(1000)/(0.082*298)

n = 39.6844 mol of air

of which, we know that

y-CO2 = 0.000375

so

mol of CO2 = y-CO2 * total mol = 0.000375*39.6844 = 0.01488 moles of CO2

mass of CO2 = mol*W = 0.01488*44.0099 = 0.6548673 g = 0.6548673*10^3 mg = 654.8673 mg