Answer to Question #52339 in Inorganic Chemistry for Candidate

Question

Dry air is a mixture, by volume, of 78% nitrogen (N2), 21% Oxygen (O2) and Argon (Ar 40 g).
a) Compute the gram molecular weight of dry air
b) Assuming water vapor saturation the new mixture being at 20 deg.Celius contains 2.4% water vapor by volume. Compute the new gram molecular weight of the mixture
c) In this last case compute the percentage of Oxygen, by weight, of the mixture

Answer 1

a) The gram molecular weight of dry air equels its molecular weight. Lets find molecular weight of each gas in air:
M(N₂) = (78% x 28)/100% = 21.84 g/mol
M(O₂) = (21% x 32)/100% = 6.72 g/mol
M(Ar) = (1% x 40)/100% = 0.4 g/mol
Total molecular weight of air is 21.84 + 6.72 + 0.4 = 28.96 g/mol

b) Water is lighter than other gases in air so water will replace them. Now gases will occupy only 100%-2.4%=97.6% of volume.
M(H₂O) = (2.4% x 18)/100% = 0.43 g/m
M(N_2,O_2,Ar) = (97.6% x 28.96)/100% = 28.265 g/mol
Total molecular weight of the mixture is 28.265 + 0.43 = 28.7 g/mol

c) Percentage of Oxygen by weight in last mixture is:
1) molecular weight of Oxygen in new mixture is M(O₂) = (6.72 x 97.6%)/100% = 6.56 g/mol
2) percentage of Oxygen is 6.56 x 100%/32 = 20.5%