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 1023 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, SO2, at STP. What was the mass in grams of the gas produced? .098 L X 1 mol X 64.07 g SO2 = 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, Cl2, in grams, is contained in a 10 L tank at 27°C and 3.5 atm of pressure? 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 + O2 -> 2CO2 2 volumes CO 1 volume O2

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 = √MB Rate of effusion B √MA Molar masses can also be replaced by densities of the gases: Rate of effusion A = √densityB Rate of effusion B √denistyA

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