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Here’s one way of doing it.

You know that 1 mol of a gas occupies 22.4 L at 1 atm and 0 C.

If that gas is air 78 % of its volume (17.5 L) is ##N_2## and 22 % of its volume (4.9 L) is ##O_2##.

The molar mass of ##N_2## is 28.0 g/mol.

If the mass of 22.4 L of ##N_2## is 28.0 g the mass of 17.5 L of ##N_2## is

##17.5 color(red)(cancel(color(black)(L N_2))) (28.0 g N_2)/(22.4 color(red)(cancel(color(black)(L N_2)))) = 21.9 g N_2##

The molar mass of ##O_2## is 32.0 g/mol.

If the mass of 22.4 L of ##O_2## is 32.0 g the mass of 4.9 L of ##O_2## is

##4.9 color(red)(cancel(color(black)(L O_2))) (32.0 g O_2)/(22.4 color(red)(cancel(color(black)(L O_2)))) = 7.0 g O_2##

The total mass of the two gases in 22.4 L is

##m_N + m_O = 21.9 g + 7.0 g = 28.9 g##

The molar mass of dry air is 28.9 g/mol.