Molecular Composition of Gases

Presentation on theme: "Molecular Composition of Gases"— Presentation transcript:

Molecular Composition of Gases
Chapter 12 Molecular Composition of Gases

Molar Volume of a Gas One mole of a gas has the same volume at STP as any other gas 22.4 L / mole at STP

Practice Problem What volume would mol of oxygen gas occupy at STP? What about mol of nitrogen gas at STP?

Ideal Gas Law PV = nRT n = Number of Moles R = Ideal Gas Constant
L . atm/ mol . K V = Volume (must be in liters) P = Pressure (must be in atmospheres) T = Temperature (must be in Kelvin)

Practice Problem What is the P (in atm) exerted by a mol sample of nitrogen gas in a 10.0 L container at 298 K? PV = nRT rearranges to P= nRT/ V (0.500 mol)( L .atm/mol .K)(298K) 10.0 L Pressure = atm

Gas Stoichiometry Uses volume-volume calculations (e.g. L  L)
Liters can be used just like mole to mole ratios in a factor label problem

C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)
Practice Problem How many L of oxygen are required for the complete combustion of L of propane? C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g) Ratio: 5 L O2 makes 1 L C3H8 0.250 L C3H8 x 5 L O2 1 L C3H8 = 1.25 L O2

CaCO3(s)  CaO(s) + CO2(g)
Practice Problem How many grams of calcium carbonate must be decomposed to produce 2.00 L of CO2 at STP? CaCO3(s)  CaO(s) + CO2(g) 2.00 L CO2 x 1 mol CO2 x 1 mol CaCO3 22.4 L CO mol CO2 x g CaCO3 1 mol CaCO3 8.94 g CaCO3

Graham’s Law Diffusion Effusion

Graham’s Law of Effusion or Diffusion
The rate of effusion or diffusion is inversely proportional to the square roots of their molar masses Rate A = MB Rate B MA

Practice Problem Compare the rate of effusion of hydrogen and nitrogen gas rate H2 = = = 3.7 rate N Hydrogen effuses 3.7 times faster than nitrogen