1. Applying the ideal gas equation Copyright © 2011 Pearson Canada Inc. Slide 1 of 19General Chemistry: Chapter 6

2. Copyright © 2011 Pearson Canada Inc. Slide 2 of 19General Chemistry: Chapter 6 Using the Gas Laws

3. Avogadro’ Hypothesis: Equal volumes of different gases at the same T and P contain equal numbers of molecules (or, equal numbers of moles of gas). Neglecting history (Avogadro’s elegant experiments!), we can apply the Ideal Gas Law Equation to two gases (Gas 1 and Gas 2). n 1 = P 1 V 1 /RT 1 and n 2 = P 2 V 2 /RT 2 If P 1 =P 2 and V 1 =V 2 and T 1 =T 2 then n 1 =n 2

4. Class Example – Avogadro’s Hypothesis: At a given T and P, 8.00 g of oxygen gas (O 2 (g)) has a volume of 8.00 L. At the same T and P 10.0 L of a gas having the molecular formula XO 2 has a mass of 20.0 g. Identify element X. Partial Solution: Apply Avogadro’s Hypothesis Here: Number of moles of = Number of moles of O 2 (g) per liter XO 2 (g) per liter # Moles O 2 (g) per liter = 0.250 mol/8.00L = 0.0313 mol∙L -1 “Aside”: Moles O 2 = 8.00g/(32.0 g. mol -1 ) = 0.250mol

5. Completion of “XO 2 example” in class

6. Ideal Gas Law and Molecular Formulas: In high school you used % composition data for compounds to derive corresponding empirical formulas. The Ideal Gas Law Eqtn can be used to determine molar masses. Combining an empirical formula with a molar mass allows a molecular formula to be determined. Empirical formulas specify relative numbers of atoms of each element. Knowing “too much chemistry” Can lead you astray. How?

7. Molecular Formulas using PV=nRT Elemental analysis shows that a compound containing carbon, hydrogen and fluorine, C x H y F z, is 63.17 % carbon, 3.53 % hydrogen and 33.30 % fluorine by mass. At 44.2 o C a 2.400 g sample of this substance is completely evaporated in a previously empty 2.50 L container and a gas pressure of 22.2 kPa is observed. Determine (a) the empirical formula of the compound and (b) the molecular formula of the compound.

8. A Step at a Time? One possible strategy is: Step 1: Use mass % composition data to determine the empirical formula for C x H y F z. Step 2: Use the ideal gas law equation to get (a) the number of moles of C x H y F z in 2.400g of compound and (b) the molar mass of C x H y F z. Step 3: Combine the results to Step 1 and Step 2 to find the molecular formula of C x H y F z.

9. Empirical Formula of C x H y F z “Knowing too much chemistry” one might be tempted to use 2.016 g. mol -1 for the molar mass of H. Why is this absolutely wrong? We’ll complete this problem in class. Mass of C atoms Mass of H atoms Mass of F atoms Moles of C atoms Moles of H atoms Moles of F atoms

10. 6-5 Gases in Chemical Reactions Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 6Slide 10 of 41 Stoichiometric factors relate gas quantities to quantities of other reactants or products. Ideal gas equation relates the amount of a gas to volume, temperature and pressure. Law of Combining Volumes can be developed using the gas law.

11. Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 6Slide 11 of 41 6-6 Mixtures of Gases Partial pressure – Each component of a gas mixture exerts a pressure that it would exert if it were in the container alone. Gas laws apply to mixtures of gases. Simplest approach is to use n total, but....

12. Dalton’s law of partial pressures illustrated Figure 6-12 Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 6Slide 12 of 41 The total pressure of a mixture of gases is the sum of the partial pressures of the components of the mixture.

13. Partial Pressure Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 6Slide 13 of 41 P tot = P a + P b +… V a = n a RT/P tot and V tot = V a + V b +… VaVa V tot n a RT/P tot n tot RT/P tot = = nana n tot PaPa P tot n a RT/V tot n tot RT/V tot = = nana n tot nana = aa Recall