Molecular Composition of Gases Chapter 11 Molecular Composition of Gases

Gay Lussac’s Law of Combining Volumes When measured at the same temperature and pressure, the ratio of the volumes of reacting gases are small whole numbers. N2 1 volume + 3 H2 3 volumes → 2 NH3 2 volumes 1. Students write balanced equation for reaction. Note balancing coefficients are same as volume in Liters.

Gay Lussac’s Law of Combining Volumes When measured at the same temperature and pressure, the ratio of the volumes of reacting gases are small whole numbers.

Avogadro’s Law Equal volumes of different gases at the same temperature and pressure contain the same number of molecules.

hydrogen + chlorine → hydrogen chloride H2 + Cl2 → 2 HCl 1 volume 1 volume 2 volumes 1 molecule 1 molecule 2 molecules 1 mol 1 mol 2 mol hydrogen + chlorine → hydrogen chloride Each molecule of hydrogen and each molecule of chlorine contains 2 atoms.

Mole-Mass-Volume Relationships

22.4 L at STP is known as the molar volume of any gas. Mole-Mass-Volume Relationships Volume of one mole of any gas at STP = 22.4 L. 22.4 L at STP is known as the molar volume of any gas.

Find the mass in grams of 2.80 L of carbon dioxide? (2.8 L CO2)(1 mol CO2/22.4 L CO2)(44 g CO2 /1 mol CO2 ) = 5.5 g CO2

Density of Gases grams liters

Density of Gases depends on T and P

1 mole of any gas occupies 22.4 L at STP The molar mass of SO2 is 64.07 g/mol. Determine the density of SO2 at STP. 1 mole of any gas occupies 22.4 L at STP

nT V a P Ideal Gas Equation Mathematical relationship of pressue, volume, temperature and number of moles of a gas

atmospheres nT V a P Number of molecules or moles affects other three quatities. Increase #, increase collision rate, increase pressure.

liters nT V a P

moles nT V a P

Kelvin nT V a P

nT V a P Ideal Gas Constant R= ideal gas constant. has the same value for all gases whose behavior approaches that of an ideal gas. Solve for R at STP. P= 1 atm, V= 22.4, n= 1 mol, T= 273 K

A balloon filled with 5.00 moles of helium gas is at a temperature of 25oC. The atmospheric pressure is 0.987 atm. What is the balloon’s volume? Step 1. Organize the given information. Convert temperature to kelvins. K = oC + 273 K = 25oC + 273 = 298K

Step 3. Substitute the given data into the equation and calculate. A balloon filled with 5.00 moles of helium gas is at a temperature of 25oC. The atmospheric pressure is .987 atm. What is the balloon’s volume? Step 2. Write and solve the ideal gas equation for the unknown. Step 3. Substitute the given data into the equation and calculate.

Determination of Molecular Weights Using the Ideal Gas Equation

Calculate the molar mass of an unknown gas, if 0 Calculate the molar mass of an unknown gas, if 0.020 g occupies 250 mL at a temperature of 305 K and a pressure of 0.045 atm. V = 250 mL = 0.250 L g = 0.020 g P = 0.045 atm T = 305 K

Determination of Density Using the Ideal Gas Equation Density = mass/volume Density varies directly with molar mass and pressure and inversely with Kelvin temp D = MP/ RT

The density of a gas was measured at 1 The density of a gas was measured at 1.50 atm and 27 ºC and found to be 1.95 g/L. Calculate the molar mass of the gas. M = dRT P (1.95g/L)(0.08206 L atm/K mol)(300K) 1.50 atm = 32.0 g/mol

Gases are assumed to behave as ideal gases. Gas Stoichiometry deals with the quantitative relationships among reactants and products in a chemical reaction. All calculations are done at STP. Gases are assumed to behave as ideal gases. A gas not at STP is converted to STP.

Gas Stoichiometry Primary conversions involved in stoichiometry. Coefficients for gaseous reactants or products indicate molar amounts, mole ratios and volume ratios. Primary conversions involved in stoichiometry.

What volume of oxygen (at STP) can be formed from 0 What volume of oxygen (at STP) can be formed from 0.500 mol of potassium chlorate? Step 1 Write the balanced equation 2 KClO3  2 KCl + 3 O2 Step 2 The starting amount is 0.500 mol KClO3. The conversion is moles KClO3  moles O2  liters O2

What volume of oxygen (at STP) can be formed from 0 What volume of oxygen (at STP) can be formed from 0.500 mol of potassium chlorate? 2 KClO3  2KCl + 3 O2 Step 3. Calculate the moles of O2, using the mole-ratio method. Step 4. Convert moles of O2 to liters of O2

The problem can also be solved in one continuous calculation. What volume of oxygen (at STP) can be formed from 0.500 mol of potassium chlorate? The problem can also be solved in one continuous calculation. 2 KClO3  2KCl + 3 O2

2 Al(s) + 6 HCl(aq)  2AlCl3(aq) + 3 H2(g) What volume of hydrogen, collected at 30.oC and 0.921 atm, will be formed by reacting 50.0 g of aluminum with hydrochloric acid? 2 Al(s) + 6 HCl(aq)  2AlCl3(aq) + 3 H2(g) Step 1 Calculate moles of H2. grams Al  moles Al  moles H2

Step 2 Calculate liters of H2. What volume of hydrogen, collected at 30.oC and 0.921 atm, will be formed by reacting 50.0 g of aluminum with hydrochloric acid? 2 Al(s) + 6 HCl(aq)  2AlCl3(aq) + 3 H2(g) Step 2 Calculate liters of H2. Convert oC to K: 30.oC + 273 = 303 K

Solve the ideal gas equation for V What volume of hydrogen, collected at 30.oC and 700. torr, will be formed by reacting 50.0 g of aluminum with hydrochloric acid? Solve the ideal gas equation for V PV = nRT

For reacting gases at constant temperature and pressure: Volume-volume relationships are the same as mole-mole relationships. H2(g) + Cl2(g)  2HCl(g) 1 mol H2 1 mol Cl2 2 mol HCl 22.4 L STP 22.4 L STP 2 x 22.4 L STP 1 volume 1 volume 2 volumes

What volume of nitrogen will react with 600 What volume of nitrogen will react with 600. mL of hydrogen to form ammonia? What volume of ammonia will be formed? N2(g) + 3H2(g)  2NH3(g)

Mass- volume problem

Graham’s Law of Effusion Effusion – the process whereby the molecules of a gas confined in a container randomly pass through a tiny opening in the container. Rate (like diffusion) depends on the relative velocities of gas molecules. Lighter molecules move faster than heavier molecules at the same temperature.

Graham’s Law of Effusion The rates of effusion of gases at the same temperature and pressure are inversely proportional to the square roots of their molar masses. Rate of effusion of A = MB Rate of effusion of B MA M = molar masses, density of a gas varies directly with molar mass, can replace molar mass with density

Calculate the ratio of the effusion rates of H2 & UF6, a gas used in the enrichment process to produce fuel for nuclear reactors. Rate of effusion of H2 Rate of effusion of UF Square root of molar mass of UF6 Square root of molar mass of H2 The effusion rate of the very light H2 molecules is about 13 times that of the massive UF6 molecules. Square root of 352.02/2.016 = 13.2 13.2