The Molar Volume
Chapter 10 – Blue Book
HW: #1-17, 19-20 & 23-24
#1-11 DUE NEXT CLASS!!!
I.
Avogadro’s Principle &
Molar Volume
A. Relationship between the mass of a gas & its
volume:
1. Equal volumes of all gases, measured under the same
conditions of P & T, contain the same number of
particles.
a. 1 mole of any gas @ STP contains 6.02 x 1023
particles
b. 1 mole of any gas has a mass = to its molecular
mass
*EXAMPLES:* ** 1 mole N2 = 28.0 g N2 = 6.02 x 1023
molecules of N2
** 1 mole of CO2 = 44.0 g CO2= 6.02 x
1023 molecules of CO2
c. 1 mole of any gas (@STP) = molecular mass = 22.4
dm3 of the gas
Examples
1. A sample of gas has a mass of 1.248 g
and occupies 300.0 cm3 at STP. What is
the molecular mass of this gas?
Solving Process: The molecular mass of the
gas is equal to one mole of the gas. Start
with the relationship between volume &
mass & calculate the mass of one mole of
the gas. All units must be divided out
except g/mol.
Examples
1. How many grams of CO2 will occupy a
volume of 500 cm3 at STP?
Remember:
1 mol CO2 = 22.4 dm3 CO2 at STP
1 mol CO2 = 44.0 g CO2
II. Molar Volume & Gases
Collected Over Water
A. We can apply what we know about
molar volume to lab situations involving
gases collected over water
Example:
A reaction produces 200 cm3 of oxygen
over water and measured at 22 oC and
99.2 kPa. How many grams of the gas
are produced? Assume water vapor
pressure of 2.6 kPa at 22 oC.
III. Ideal Gas Equation
A. Combo of the four physical variables:
pressure, volume, temperature & number of
particles.
1. Equation: PV = nRT
P = pressure in kPa
V = volume dm3
T = temperature in K
n = number of moles of a gas
R = 8.31 (dm3 x kPa) / (mol x K)*
(other values for R depending on the
units)
2. n = m/M
m = mass
M = molecular mass
Other ways to manipulate the Ideal Gas equation:
3. PV = (m/M)RT
Example
3. A flask has a volume of 258 cm3. A gas
with a mass of 1.475 g is introduced
into the flask at a temperature of 300K
and a pressure of 98.6 kPa. Calculate
the molecular mass of the gas using the
ideal gas equation.
IV Gas Concentration
A. Can be expressed many ways:
1. Percent by volume – volume of the given
compound contained in 100 volumes of air
(example air contains about 20% O2 by volume)
2. Parts per million (ppm) - # of mg of a compound
per one kg of the whole.
3. Parts per billion (ppb) - # of μg (micrograms) per
kg of the whole.
Examples: 1 ppm is $.01 in $10,000
1ppb is $.01 in $10,000,000
1 ppm is 1mm in 1km
1 ppb is 1mm in 1000 km