Gases

Molar Mass and Gas Densities


Molar Mass and Gas Densities

Density

Example:

What is the density of carbon tetrachloride vapor at 714 torr and 125°C?

The molar mass of CCl4 is 12.0 + (4*35.5) = 154 g/mol. 125°C in degrees Kelvin would be (273+125) = 398K. Since we are dealing with torr, the value of the gas constant, R, would be 62.36 L torr/mol K.


Caesar's Las Breath

Of the molecules in Caesar's last gasp, how many of them are in the breath you just took?

Given:

 

1. Number of moles in Caesar’s last breathe:

n = PV/RT = (0.96 atm)(2L)/(0.0821 L atm/mol K)(310 K)

n = 0.075 mol

2. Number of moles in the atmosphere:

a. Surface area of earth:

Area = (4)(p)(r2)

Area = 5.10 x 1014 square meters

b. Pressure of the atmosphere on the earth’s surface:

Pressure = 760 mm Hg = 1.01 x 105 Pascals = 1.01 x 105 Newtons/square meter

Pressure = 1.01 x 105 kg/m s2

c. Force of the atmosphere on the earth

Pressure = Force/Area

Therefore

Force = (Pressure)(Area)

Force = (1.01 x 105 kg/m s2)(5.10 x 1014 m2)

Force = 5.15 x 1019 kg m /s2

d. Mass of the atmosphere

Force = (mass)(acceleration)

therefore

mass = Force/acceleration

mass = (5.15 x 1019 kg m/s2)/(9.8 m/s2) note: this is the acceleration due to gravity

mass = 5.26 x 1018 kg or 5.26 x 1021 g

e. Moles in the atmosphere

mol = (5.26 x 1021 g)(1 mol/29 g)

mol = 1.81 x 1020 mol

3. Fraction of atmosphere which represents molecules from Caesar’s last breath:

(0.075 mol)/(1.81 x 1020 mol) = 4.14 x 10-22

4. Moles of Caesar’s last breath in your last breath:

Assume your breath holds 0.075 mol:

(0.075 mol)(4.14 x 10-22) = 3.11 x 10-23mol

5. Number of molecules:

(6.022 x 1023 molecules/mol)(3.11 x 10-23mol) = 18.7 molecules

 

(An even more disconcerting fact is that he probably had flatulence as well.)


1996 Michael Blaber