A mixture containing nitrogen and hydrogen weighs 3.48 g and occupies a volume of 7.47 L at 296 K and 1.02 atm. Calculate the mass percent of these two gases. Assume ideal-gas behavior.

there is 2% of hydrogen and 98% of nitrogen (mass percent)

Explanation:

assuming ideal gas behaviour

P*V=n*R*T

n = P*V / (R*T)

where P = pressure=1.02 atm, V=volume=7.47 L, T=absolute temperature = 296 K and R = ideal gas constant = 0.082 atm*L / (mole*K)

thus

n = P*V / (R*T) = 1.02 atm*7.47 L / (296 K * 0.082 atm*L / (mole*K)) = 0.314 moles

since the number of moles is related with the mass m through the molecular weight M

n=m/M

thus denoting 1 as hydrogen and 2 as nitrogen

m₁+m₂ = mt (total mass)

m₁/M₁+m₂/M₂ = n

dividing one equation by the other and denoting mass fraction w₁ = m₁/mt, w₂ = m₂/mt, w₂ = 1 - w₁

w₁/M₁+w₂/M₂ = n/mt

w₁/M₁ + (1-w₁) / M₂ = n/mt

w₁ * (1/M₁ - 1/M₂) + 1/M₂ = n/mt

w₁ = (n/mt - 1/M₂) / (1/M₁ - 1/M₂)

replacing values

w₁ = (n/mt - 1/M₂) / (1/M₁ - 1/M₂) = (0.314 moles/3.48 g - 1 / (14 g/mole)) / (1 / (1 g/mole) - 1 / (14 g/mole)) = 0.02 (%)

and w₂ = 1-w₁ = 0.98 (98%)

thus there is 2% of hydrogen and 98% of nitrogen