Chapter 11
Molecular Composition of Gases
Section 11.7
volume – mass relationships of gases
Comparing volumes of reacting
gases:
• Recall the balanced equation for the
synthesis of water:
2H2 (g) + O2 (g) → 2 H2O (g)
• We have interpreted the coefficients in this
equation in terms of molecules of
reactants & products and moles of
reactants & products.
Gay-Lussac’s Law of Combining
Volumes of Gases:
• Gay-Lussac made the discovery that,
when the reactants & products are gases,
these coefficients can also be interpreted
in terms of volumes of reactants &
products.
2H2 (g) + O2 (g) → 2 H2O (g)
•
•
•
•
2 molecules
2 moles
2 liters
2 volumes
1 molecule
1 mole
1 liter
1 volume
2 molecules
2 moles
2 liters
2 volumes
Avogadro’s Law (1811)
• Equal volumes of gases at the same
temperature and pressure contain equal
numbers of particles (atoms or molecules).
Molar Volume of Gases
• At standard temperature (0°C, 273K) and
pressure (1atm) the volume of 1 mole of
gas is 22.4 L.
• 1 mole of gas at STP contains 6.02 x 1023
particles (atoms or molecules) and has a
mass = molar mass (grams)
Add a road to the mole map!!!!
More conversion factor problems!!!!
Mole Map
known
particles
unknown
particles
6.02 x 1023
6.02 x 1023
Balanced Equation Blvd.
moles-----------------------------moles
Mole ratio
molar
mass
grams
22.4L at STP
If not, use
PV=nRT
liters
volume ratio
22.4L at STP
If not use
PV=nRT
molar
mass
liters
grams
Sample Problem 1
• A chemical reaction produces 0.0680 mol
of oxygen gas. What volume in liters is
occupied by this gas at STP?
Sample Problem 1
• A chemical reaction produces 0.0680 mol
of oxygen gas. What volume in liters is
occupied by this gas at STP?
• Solution:
0.0680mol x
Sample Problem 1
• A chemical reaction produces 0.0680 mol
of oxygen gas. What volume in liters is
occupied by this gas at STP?
• Solution:
0.0680mol x 22.4L =
1 mol
Sample Problem 1
• A chemical reaction produces 0.0680 mol
of oxygen gas. What volume in liters is
occupied by this gas at STP?
• Solution:
0.0680mol x 22.4L = 1.52 liters
1 mol
Sample problem 2
• A chemical reaction produced 98.0 mL of
sulfur dioxide gas, SO2 at STP. What was
the mass (in grams) of the gas produced?
Sample problem 2
• A chemical reaction produced 98.0 mL of
sulfur dioxide gas, SO2 at STP. What was
the mass (in grams) of the gas produced?
• Solution:
98.0 mL x 1 L
1000mL
Sample problem 2
• A chemical reaction produced 98.0 mL of
sulfur dioxide gas, SO2 at STP. What was
the mass (in grams) of the gas produced?
• Solution:
98.0 mL x 1 L
1 mol
x
1000mL
22.4 L
Sample problem 2
• A chemical reaction produced 98.0 mL of
sulfur dioxide gas, SO2 at STP. What was
the mass (in grams) of the gas produced?
• Solution:
98.0 mL x 1 L
1 mol x 64.07 g =
x
1000mL
22.4 L 1 mol
Sample problem 2
• A chemical reaction produced 98.0 mL of
sulfur dioxide gas, SO2 at STP. What was
the mass (in grams) of the gas produced?
• Solution:
98.0 mL x 1 L
1 mol x 64.07 g =
x
1000mL
22.4 L 1 mol
= 0.280g SO2
Homework
• Complete the “Gases and the Mole”
worksheet