Calculate mass of "air" (in grams) in the container, given that air is approximately 21% O2 and 79% N2.

Question

During an experiment, a 25.0mL "empty" sealed container is placed on a balance and found to have a mass of 12.730g. The pressure and temperature in the room is measured to be 0.997 atm and 19.5°C, respectively. Calculate mass of "air" (in grams) in the container, given that air is approximately 21% O₂ and 79% N₂. 

Answer 1

Ideal gas law:

PV = nRT (assuming air acts as an ideal gas)

P = pressure in atm = 0.997 atm

V = volume in L = 25.0 ml x 1 L / 1000 ml = 0.0250 L

n = moles of gas = ?

R = gas constant = 0.0821 Latm/Kmol

T = temperature in K = 19.5C + 273 = 293K

Solving for n, we have ...

n = PV/RT = (0.997)(0.025) / (0.0821)(293)

n = 0.00104 mols of air

mols N₂ = 0.79 x 0.00104 = 0.000822 mols N₂

mols O₂ = 0.21 x 0.00104 = 0.000218 mols O₂

mass N₂ = 0.000822 mols x 28 g / mol = 0.0230 g N₂

mass O₂ = 0.000218 mols x 32 g / mol = 0.00699 g O₂

Total mass of air = 0.0230 g + 0.00699 g = 0.0300 g of air