# Air is 78.1% nitrogen, 20.9% oxygen, and 0.934% argon by moles. What is the density of air at 22 °C and 760…?

Air is 78.1% nitrogen, 20.9% oxygen, and 0.934% argon by moles. What is the density of air at 22 °C and 760 torr? Assume ideal behavior.

• Somehow you need to deal with the fact that your percentages don’t add up to 100%. One way is like this:

Take a hypothetical sample consisting of 78.1 moles of N2, 20.9 moles of O2, and 0.934 moles of Ar:

78.1 mol + 20.9 mol + 0.934 mol = 99.934 mol total

V = nRT / P = (99.934 mol) x (62.36367 L Torr/K mol) x (22 + 273 K) /

(760 torr) = 2419 L total

(78.1 mol N2) x (28.01344 g N2/mol) = 2187.85 g N2

(20.9 mol O2) x (31.99886 g O2/mol) = 668.78 g O2

(0.934 mol Ar) x (39.9481 g Ar/mol) = 37.31 g Ar

2187.85 g + 668.78 g + 37.31 g = 2893.94 g total

(2893.94 g) / (2419 L ) = 1.20 g/L

• you also can assume the total volume is 1L with Ideal gas Laws

n=PV/RT=((760/7.6)kPa*1)/8.3145*295.15=0.0412mol

nN2=0.0412*78.1%=0.032mol, mN2=0,032*28.02=0.90g

nO2=0.0412*20.9%=0.0085mol, mO2=0.0085*32=0.272g

nAr=0.0412*0.934%=0.00038mol, mAr=0.00038*40=0.0152g

mT=mO2+mN2+mAr= 0.9+0.272+0.0152=1.19g

d=mT/V=1.19/1=1.19g/L

• Hello Julia : Apply the Ideal Gas Law after getting the average molecular mass, Mbar , of the air:

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For a gas mixture:

Mbar = SUM ( y sub j )( M sub j )

Mbar = ( 0.781 )( 28.01 ) + (0.209 ) ( 32.00 0 + ( 0.00934 ) ( 39.95 )

Mbar = 21.8758 + 6.6880 + 0.3731 = 28.94 g / gmol

Apply Ideal Gas Law:

PV = mRT / M

d = m / v = PM / RT

d = ( 760 /760 atm ) ( 28.94 g /gmol ) / ( 0.08205 atm – L/ gmol – K ) ( 22.0 + 273.2 )

d = 1.195 g / L <—————–