The Molar Volume Chapter 10 – Blue Book Chapter 13.2 – Red Book HW: #1-17, & Crash Course Intro Crash Course Intro

I.Avogadro’s Principle & Molar Volume A.Relationship between the mass of a gas & its volume: 1.Equal volumes of all gases, measured under the same conditions of P & T, contain the same number of particles. a.1 mole of any STP contains 6.02 x particles

Avogadro’s Principle & Molar Volume b.1 mole of any gas has a mass = to its molecular mass *EXAMPLES:* ** 1 mole N 2 = 28.0 g = 6.02 x molecules of N 2 ** 1 mole of CO 2 = 44.0 g = ** 1 mole of CO 2 = 44.0 g = 6.02 x molecules of CO 2 c. 1 mole of any gas = molecular mass = 22.4 dm 3 of gas

Examples 1.A sample of gas has a mass of g and occupies cm 3 at STP. What is the molecular mass of this gas? Solving Process: The molecular mass of the gas is equal to one mole of the gas. Start with the relationship between volume & mass & calculate the mass of one mole of the gas. All units must be divided out except g/mol. Answer: 93.2 g/mol

Examples 1.How many grams of CO 2 will occupy a volume of 500 cm 3 at STP? Remember: 1 mol CO 2 = 22.4 dm 3 CO 2 at STP 1 mol CO 2 = 44.0 g CO 2 Answer: g CO 2

Examples 2. How many grams of NH 3 will occupy a volume of 350 cm 3 at STP? Remember: 1 mol NH 3 = 22.4 dm 3 NH 3 at STP 1 mol NH 3 = 17.0 g NH 3 Answer: g NH 3

II. Molar Volume & Gases Collected Over Water A.We can apply what we know about molar volume to lab situations involving gases collected over water

Example: A reaction produces 200 cm 3 of oxygen over water and measured at 22 o C and 99.2 kPa. How many grams of the gas are produced? Assume water vapor pressure of 2.6 kPa at 22 o C. 1.Correct to dry pressure 2.Use Old/New chart to convert volume to Standard conditions 3.Once Volume is at STP you can use 22.4 dm 3 /mol, the mol  grams of O 2

III. Ideal Gas Equation A.Combo of the four physical variables: pressure, volume, temperature & number of particles. 1.Equation: PV = n RT P = pressure in kPa V = volume dm 3 T = temperature in K n = number of moles of a gas R = 8.31 (dm 3 x kPa) / (mol x K) * (other values for R depending on the units)

III. Ideal Gas Equation 1.Break down of n n = m/M n = m/M m = massM = molecular massm = massM = molecular mass (g/mol) (g/mol) Other ways to manipulate the Ideal Gas equation: PV = (m/M)RT m = PVM M = (mRT) RT PV RT PV

Example 3.A flask has a volume of 258 cm 3. A gas with a mass of g is introduced into the flask at a temperature of 300K and a pressure of 98.6 kPa. Calculate the molecular mass of the gas using the ideal gas equation. Use M = mRT/PV then “plug & chug” plug in the values & solve.Use M = mRT/PV then “plug & chug” plug in the values & solve. Answer: 145 g/mol

A Video…  FYI: Mark Rosengarten is using atm instead of kPa… Remember it’s OK to have other values for R depending on the units, he’ll explain that at the end, so hang on Remember it’s OK to have other values for R depending on the units, he’ll explain that at the end, so hang on Crash Course Click on movie projector to play video