Accelerated Chemistry Molecular Composition of Gases 5/20/2019 Chapter 11 Molecular Composition of Gases Chapter 12 - Ideal Gas Law NotesChapter 12 - Ideal Gas Law Notes

First, Let’s Review Chapter 9-11 Accelerated Chemistry 5/20/2019 First, Let’s Review Chapter 9-11 Stoichiometric Calculations The coefficients in the balanced equation give the ratio of moles of reactants and products Chapter 12 - Ideal Gas Law NotesChapter 12 - Ideal Gas Law Notes

Stoichiometric Calculations Accelerated Chemistry 5/20/2019 Stoichiometric Calculations From the mass of Substance A you can use the ratio of A and B to calculate the mass of Substance B formed (if it’s a product) or used (if it’s a reactant) Mole Grams Liters Molecules Coefficients Molecules Chapter 12 - Ideal Gas Law NotesChapter 12 - Ideal Gas Law Notes

Stoichiometric Calculations Accelerated Chemistry 5/20/2019 Stoichiometric Calculations C6H12O6 + 6 O2  6 CO2 + 6 H2O How many grams of H2O can be made from 1.00g C6H12O6? Chapter 12 - Ideal Gas Law NotesChapter 12 - Ideal Gas Law Notes

11.1 Volume - Mass Relationships of Gases Accelerated Chemistry 5/20/2019 11.1 Volume - Mass Relationships of Gases Last chapter, we related the volume and the mass of a gas: Reaction Stoichiometry Note: a “stoich” problem can be recognized by the fact that information from one substances is given and the problem is asking about a different substance. The Combined Gas Law Let’s compare the volumes of gases in two example problems… Chapter 12 - Ideal Gas Law NotesChapter 12 - Ideal Gas Law Notes

…Reaction Stoichiometry at standard conditions Accelerated Chemistry 5/20/2019 …Reaction Stoichiometry at standard conditions Example 1 - How many L of Oxygen are needed to react with 50.0L of Hydrogen at STP? 2H2(g) + O2(g) ---> 2 H2O(g) x 1 mole H2 22.4 L H2 1 mole O2 2 moles H2 22.4 L O2 50.0 L H2 = 25.0 L O2 We are comparing the mass and volume of a gas at standard conditions (STP) using “stoich” 1 3 2 But...we can skip steps 1 and 3. Why? All gases take up the same amount of space at STP. Chapter 12 - Ideal Gas Law NotesChapter 12 - Ideal Gas Law Notes

…The Combined Gas Law at nonstandard conditions Accelerated Chemistry 5/20/2019 …The Combined Gas Law at nonstandard conditions Example 2 - Calculate the volume of one mole of H2 at 20ºC and 1000 torr. P1 = 760 torr P2 = 1000 torr T1 = 273K T2 = 293K V1 = 22.4 L V2 = ? V2 = 18.3 L H2 We can compare the volume and mass of a gas at non-standard conditions using the Combined Gas Laaw Chapter 12 - Ideal Gas Law NotesChapter 12 - Ideal Gas Law Notes

Accelerated Chemistry 5/20/2019 The Ideal Gas Law We solved example 2 using the combined gas law – we could have also used the ideal gas law A relationship between pressure, volume, temperature and the # of moles of a gas PV = nRT n = # of moles in Liters Volume R = ideal gas constant Kelvin Temp. 1 atm 1 atm = 760 torr Pressure at STP, R = (1atm)(22.4L / 1mol)(273K) = .0821 L · atm / mol· K Crash Course Ideal Gas Law Chapter 12 - Ideal Gas Law NotesChapter 12 - Ideal Gas Law Notes

Deriving the Ideal Gas Law Accelerated Chemistry 5/20/2019 Deriving the Ideal Gas Law Let’s derive the ideal gas law and gas constant... Volume is proportional to 1/P (as P is reduced, the V increases) V is proportional to T (as T increases, the V increases) V is proportional to n (as more moles are added, the V increases) Chapter 12 - Ideal Gas Law NotesChapter 12 - Ideal Gas Law Notes

Accelerated Chemistry 5/20/2019 In general… The combined gas law, P1V1 = P2V2 T1 T2 is used for changing conditions. The Ideal Gas Law, can be introduced when you have problems containing: one set of conditions solving for grams solving for moles calculating molecular weight (molar mass) calculating density involving stoichiometry and non-STP conditions Chapter 12 - Ideal Gas Law NotesChapter 12 - Ideal Gas Law Notes

Accelerated Chemistry 5/20/2019 One Set of Conditions Example: Calculate the volume of 1.00 mole of Hydrogen at 20.0 ˚C and 1000. torr Chapter 12 - Ideal Gas Law NotesChapter 12 - Ideal Gas Law Notes

Accelerated Chemistry 5/20/2019 Solving for Moles Example: A sample of CO2 in a 10.0 L container at 293K exerts a pressure of 50,000. torr. How many moles of CO2 are in your sample? Chapter 12 - Ideal Gas Law NotesChapter 12 - Ideal Gas Law Notes

Accelerated Chemistry 5/20/2019 Solving for Grams Example: Calculate the number of grams of helium in a 6.0 liter cylinder at 27˚C and 800. torr. Chapter 12 - Ideal Gas Law NotesChapter 12 - Ideal Gas Law Notes

Accelerated Chemistry 5/20/2019 Calculating M.W. Example: If 18.0 grams of a gas at 380 torr and 546.0 K occupies 44.8 L, what is the molecular weight of the gas? Step 1: Identify the m.w. formula. Step 2: Now, use PV = nRT to solve for n (the number of moles). Step 3: Plug this value into the molecular weight (m.w.) equation. Chapter 12 - Ideal Gas Law NotesChapter 12 - Ideal Gas Law Notes

Calculating Density at STP Accelerated Chemistry 5/20/2019 Calculating Density at STP Example: Find the density of carbon dioxide at STP. Step 1: Identify the density formula. Step 2: If it is assumed that one mole of CO2 is present, then the mass can be calculated from the periodic table and the volume at STP is 22.4 L. Chapter 12 - Ideal Gas Law NotesChapter 12 - Ideal Gas Law Notes

Calculating Density (not at STP) Accelerated Chemistry 5/20/2019 Calculating Density (not at STP) Example: Find the density of carbon dioxide at 546.0 K and 4.00 atm. Step 1: Identify the density formula. D = m/V Step 2: Assume one mole of CO2. Thus, CO2 weighs 44.0 grams. Plug this into the Density formula. Step 3: Because the conditions are non-STP values, 22.4 L can’t be used. So, use PV=nRT and solve for V (assume one mole). Step 4: Plug this value into the density equation. Chapter 12 - Ideal Gas Law NotesChapter 12 - Ideal Gas Law Notes

Deviations from Ideal Behavior Accelerated Chemistry 5/20/2019 Deviations from Ideal Behavior Real gases do not behave according to the KMT - Why? Real gases have molecules that occupy space Real gases have attractive and repulsive forces Ideal gases conform exactly to the KMT no such gas exists gases only behave close to ideally at low P and high T. At low T and high P, gases deviate greatly from ideal behavior. Some gases are close to ideal (if they are small and nonpolar): H2, He, Ne (these are small and nonpolar!!!) O2 and N2 are not too bad NH3, H2O are not even close to ideal Chapter 12 - Ideal Gas Law NotesChapter 12 - Ideal Gas Law Notes

Stoichiometry of Gases Accelerated Chemistry 5/20/2019 Stoichiometry of Gases At STP, there is nothing new here. The only thing new is that 1 mole = 22.4 L must be adjusted if not at STP. Problems for Stoichiometry of Gases include converting: Grams to Liters Liters to Grams Liter to Liter Chapter 12 - Ideal Gas Law NotesChapter 12 - Ideal Gas Law Notes

Accelerated Chemistry 5/20/2019 Grams to Liters Example: How many liters of O2 are generated when 50.0 grams of sodium chlorate decomposes at 0.950 atm and 20.0˚C? 2NaClO3(s) + heat ---> 2NaCl(s) + 3O2(g) Chapter 12 - Ideal Gas Law NotesChapter 12 - Ideal Gas Law Notes

Accelerated Chemistry 5/20/2019 Liter to Grams Example: If a lawn mower engine generates 555.0 L CO2 on a lovely Sunday afternoon (.996 atm and 37.00 C0 )- how many grams of octane were consumed? 2C8H18(l) + 25O2(g) ---> 16CO2(g) + 18H2O(g) Chapter 12 - Ideal Gas Law NotesChapter 12 - Ideal Gas Law Notes

Liter to Liter (not at STP) Accelerated Chemistry 5/20/2019 Liter to Liter (not at STP) Example: How many L of carbon dioxide can be made from the combustion of 2.00 L of propane (C3H8) at 500. K and 3.00 atm? C3H8(l) + 5O2(g) ---> 3CO2(g) + 4H2O(g) Chapter 12 - Ideal Gas Law NotesChapter 12 - Ideal Gas Law Notes