Figure 5.14: A gas whose density is greater than that of air.

1. Figure 5.14: A gas whose density is greater than that of air.

2. Figure 5.15: Finding the vapor density of a substance. (Example 5.8) Sample boils Excess sample escapes When liquid is gone, submerge flask in ice water, Vapor condenses to solid Determine mass Determine density D= m/v

3. Density Determination • If we look again at our derivation of the molecular mass equation, we can solve for m/V, which represents density.

4. A Problem to Consider • Calculate the density of ozone, O3 (Mm = 48.0g/mol), at 50 oC and 1.75 atm of pressure.

6. or Molecular Weight Determination • In Chapter 3 we showed the relationship between moles and mass.

7. Substitute into Ideal Gas Law: PV = nRT n = m/Mm Mm = molecular mass PV= mRT Mm PMm = mRT =dRT V

8. If we solve this equation for the molecular mass, we obtain Molecular Weight Determination • If we substitute this in the ideal gas equation, we obtain

9. A Problem to Consider • A 15.5 gram sample of an unknown gas occupied a volume of 5.75 L at 25 oC and a pressure of 1.08 atm. Calculate its molecular mass.

10. Stoichiometry!

11. Stoichiometry Problems Involving Gas Volumes • Consider the following reaction, which is often used to generate small quantities of oxygen. • Suppose you heat 0.0100 mol of potassium chlorate, KClO3, in a test tube. How many liters of oxygen can you produce at 298 K and 1.02 atm?

12. Stoichiometry Problems Involving Gas Volumes • First we must determine the number of moles of oxygen produced by the reaction.

13. Stoichiometry Problems Involving Gas Volumes • Now we can use the ideal gas equation to calculate the volume of oxygen under the conditions given. O2

14. Partial Pressures of Gas Mixtures • The composition of a gas mixture is often described in terms of its mole fraction. • Themole fraction, c , of a component gas is the fraction of moles of that component in the total moles of gas mixture.

15. Partial Pressures of Gas Mixtures • The partial pressure of a component gas, “A”, is then defined as • Applying this concept to the ideal gas equation, we find that each gas can be treated independently.

16. Partial Pressures of Gas Mixtures • Dalton’s Law of Partial Pressures: the sum of all the pressures of all the different gases in a mixture equals the total pressure of the mixture. (Figure 5.16)

17. Figure 5.17: An illustration of Dalton’s law of partial pressures before mixing.

18. Figure 5.17: An illustration of Dalton’s law of partial pressures after mixing.

20. A Problem to Consider • Given a mixture of gases in the atmosphere at 760 torr, what is the partial pressure of N2 (c = 0 .7808) at 25 oC?

21. A Problem to Consider • Suppose a 156 mL sample of H2 gas was collected over water at 19 oC and 769 mm Hg. What is the mass of H2 collected? • Table 5.6 lists the vapor pressure of water at 19 oC as 16.5 mm Hg.

22. A Problem to Consider • Now we can use the ideal gas equation, along with the partial pressure of the hydrogen, to determine its mass.

23. A Problem to Consider • From the ideal gas law, PV = nRT, you have • Next,convert moles of H2 to grams of H2.

24. Collecting Gases “Over Water” • A useful application of partial pressures arises when you collect gases over water. (see Figure 5.17) • As gas bubbles through the water, the gas becomes saturated with water vapor. • The partial pressure of the water in this “mixture” depends only on the temperature. (see Table 5.6)

25. Figure 5-18 Collection of a gas over water.

27. A Problem to Consider • Suppose a 156 mL sample of H2 gas was collected over water at 19 oC and 769 mm Hg. What is the mass of H2 collected? • First, we must find the partial pressure of the dry H2.

28. Conceptual Problem 5.27

29. Figure 5.27: The hydrogen fountain.Photo courtesy of American Color. Return to Slide 44

30. Figure 5.26: Model of gaseous effusion. Return to Slide 45