The mass percent of each gas in a mixture and the partial pressure of gases at a certain pressure.

Question

The composition in volume percent of dry air at sea level is approximately 78.1% N₂, 20.9% O₂, 0.93% Ar, and 0.031% CO₂. Calculate:

a-     The mass percent of each gas in the mixture.

b-    The partial pressure of these gases at a total pressure of 1.0 atm.

Answer 1

If you have, for example, 1 liter of air, according to the % volume composition, you would have...

.781 L N2

0.209 L O2

0.0093 L Ar

0.00031 L CO2

Since the pressure, volume and temperature are all constant, the % or liters can be equated to moles b/c the ratio of # L is the same as the ratio of # of moles. In addition, we can assume any volume, and compute relative number of moles and relative mass. For example, let us assume we have 1 L volume. Then...

PV = nRT and P,V,R and T are constant, so moles, n = volume

mass N2 = 0.781 L = 0.781 mols x 28 g/mol = 21.9 g N2

mass O2 = 0.209 L = 0.209 mols x 32 g/mol = 6.69 g O2

mass Ar = 0.0093 L = 0.0093 mols x 40 g/mol = 0.372 g Ar

mass CO2 = 0.00031 L = 0.00031 mols x 44 g/mol = 0.0136 g CO2

Total mass = 28.98 g

mass % composition:

N2 = 21.9 g/28.98 g x100% = 75.6%

O2 = 6.69g/28.98 g x 100% = 23.1%

Ar = 0.372 g/28.98 g x 100% = 1.28%

CO2 = 0.0136 g/28.98 g x100% = 0.047%

Partial pressures:

P_gas = X_gas X P_total (P_gas = partial pressure; X = mol fraction; P_total = total pressure)

P_N2 = 0.781 X 1.0 atm = 0.781 atm

P_O2 = 0.209 X 1.0 atm = 0.209 atm

P_Ar = 0.0093 X 1.0 atm = 0.0093 atm

P_CO2 = 0.00031 X 1.0 atm = 0.00031 atm