Ch. 11 Molecular Composition of Gases

11-1 Volume-Mass Relationships of Gases  Gay-Lussac’s law of combining volumes of gases-at constant temperature and pressure, the volumes of gaseous reactants and products can be expressed as ratios of small whole numbers  Hydrogen + oxygen ->water vapor 2L 1L 2L 2 volumes 1 volume 2 volumes

Avogadro’s Law  Equal volumes of gases at the same temperature and pressure contain equal numbers of molecules (doesn’t matter which gas) Fig. 11-1  He discovered that some molecules can have 2 atoms or more (diatomic molecules)

 Avogadro’s law also indicates gas volume (L) directly proportional to the amount of a gas (n)  V = kn

Standard molar volume of a gas  Avogadro’s constant = 6.022 X 10 23 molecules = 1 mole  Standard molar volume of a gas-the volume occupied by one mole of a gas at STP (22.4 L)  Fig. 11-3 1 mole of each gas occupies 22.4 L but different masses

Avogadro’s Law Sample problem 11-1  A chemical reaction produces 0.0680 mol of oxygen gas. What volume in liters is occupied by this gas sample at STP?

 0.0680 mol X 22.4 L = 1.52 L 1 mol

Avogadro’s Law Practice  A sample of hydrogen gas occupies 14.1 L at STP. How many moles of the gas are present?

Converting to grams   Sample problem 11-2 A chemical reaction produced 98.0 mL of sulfur dioxide gas, SO 2 , at STP. What was the mass in grams of the gas produced?

.098 L X 1 mol X 64.07 g SO 2 = 0.280 g 22.4 L 1 mol

Converting to grams practice  What is the volume of 77 g of nitrogen dioxide gas at STP?

11-2: The Ideal Gas Law  Mathematical relationship among pressure, volume, temperature, and the number of moles of a gas.

 Combination of Boyle’s, Charles’s, Gay Lussac’s, and Avogadro’s Laws  PV = nRT

 Ideal gas constant (R), is derived by plugging in all standard values into the equation:  R = PV = 0.0821

nT

Ideal gas law sample  What is the pressure in atmospheres exerted by a 0.500 mol sample of nitrogen gas in a 10 L container at 298 K?

Answer = 1.22 atm

More ideal gas law practice  What is the volume, in liters, of 0.250 mol of oxygen gas at 20°C and 0.974 atm pressure?

Answer = 6.17 L

Sample problem 11-5  What mass of chlorine gas, Cl 3.5 atm of pressure?

2 , in grams, is contained in a 10 L tank at 27°C and Answer = 101 g

Finding molar mass or density  PV = mRT M M = mRT M = DRT PV P D = MP RT

Sample Problem  At 28°C and 0.974 atm, 1.00 L of gas has a mass of 5.16 g. What is the molar mass of this gas?

11-3 Stoichiometry of Gases  Coefficients can be used as volume ratios:  2CO + O 2 -> 2CO 2 2 volumes CO 1 volume O 2

Sample Problem 11-7 volume volume

Sample problem 11-8 volume-mass

Sample problem 11-9

11-4 Effusion and Diffusion  Graham’s Law of Effusion-rates of diffusion and effusion depend on the relative velocities of gas molecules  Rates of effusion of gases at the same temperature and pressure are inversely proportional to the square roots of their molar masses.

Graham’s Law formula  Rate of effusion A = √M B Rate of effusion B √M A Molar masses can also be replaced by densities of the gases:  Rate of effusion A = √density B Rate of effusion B √denisty A

Graham’s Law Problem   Sample problem 11-10 Compare the rates of effusion of hydrogen and oxygen at the same temperature and pressure. (smaller molar mass gas will diffuse faster-how much faster?)  Smaller molar mass goes on bottom

Diffusion Quicklab pg. 353