The Molar The Molar VolumeVolumeChapter 10 – Blue BookChapter 10 – Blue Book

Chapter 13.2 – Red Book Chapter 13.2 – Red Book

HW: #1-17, 19-20 & 23-24HW: #1-17, 19-20 & 23-24Crash Course Intro

I.I. Avogadro’s Avogadro’s Principle & Principle & Molar VolumeMolar Volume

A.A. Relationship between the mass of Relationship between the mass of a gas & its volume:a gas & its volume:

1.1. Equal volumes of all gases, Equal volumes of all gases, measured under the same measured under the same conditions of P & T, contain the conditions of P & T, contain the same number of particles.same number of particles.

a.a. 1 mole of any gas @ STP 1 mole of any gas @ STP contains 6.02 x 10contains 6.02 x 1023 23 particlesparticles

Avogadro’s Principle & Avogadro’s Principle & Molar VolumeMolar Volumeb.b. 1 mole of any gas has a mass = 1 mole of any gas has a mass =

to its molecular mass to its molecular mass *EXAMPLES:**EXAMPLES:* ** 1 mole N ** 1 mole N22 = 28.0 g = = 28.0 g =

6.02 x 106.02 x 1023 23 molecules of Nmolecules of N22

** 1 mole of CO** 1 mole of CO22 = 44.0 g = = 44.0 g = 6.02 x 106.02 x 1023 23 molecules of COmolecules of CO22

c. 1 mole of c. 1 mole of anyany gas (@STP) = gas (@STP) = molecular mass = 22.4 dmmolecular mass = 22.4 dm3 3 of gasof gas

ExamplesExamples1.1. A sample of gas has a mass of A sample of gas has a mass of

1.248 g and occupies 300.0 cm1.248 g and occupies 300.0 cm33 at at STP. What is the molecular mass STP. What is the molecular mass of this gas?of this gas?

Solving Process: The molecular mass of Solving Process: The molecular mass of the gas is equal to one mole of the gas. the gas is equal to one mole of the gas. Start with the relationship between Start with the relationship between volume & massvolume & mass & calculate the mass & calculate the mass of one mole of the gas. All units must of one mole of the gas. All units must be divided out except g/mol.be divided out except g/mol.

Answer: 93.2 g/molAnswer: 93.2 g/mol

ExamplesExamples

1.1. How many grams of COHow many grams of CO22 will will occupy a volume of 500 cmoccupy a volume of 500 cm33 at at STP?STP?

Remember: Remember:

1 mol CO1 mol CO22 = 22.4 dm = 22.4 dm33 CO CO22 at STP at STP

1 mol CO1 mol CO22 = 44.0 g CO = 44.0 g CO22

Answer: 0.982 g COAnswer: 0.982 g CO22

ExamplesExamples

2. How many grams of NH2. How many grams of NH33 will will occupy a volume of 350 cmoccupy a volume of 350 cm33 at STP? at STP?

Remember: Remember:

1 mol NH1 mol NH33 = 22.4 dm = 22.4 dm33 NH NH33 at STP at STP

1 mol NH1 mol NH33 = 17.0 g NH = 17.0 g NH33

Answer: 0.266 g NHAnswer: 0.266 g NH33

II. Molar Volume & Gases II. Molar Volume & Gases Collected Over WaterCollected Over Water

A.A. We can apply what we We can apply what we know about molar volume know about molar volume to lab situations involving to lab situations involving gases collected over watergases collected over water

Example: Example: A reaction produces 200 cmA reaction produces 200 cm33 of oxygen of oxygen

over water and measured at 22 over water and measured at 22 ooC and C and 99.2 kPa. How many grams of the gas 99.2 kPa. How many grams of the gas are produced? Assume water vapor are produced? Assume water vapor pressure of 2.6 kPa at 22 pressure of 2.6 kPa at 22 ooC.C.

1.1. Correct to dry pressureCorrect to dry pressure

2.2. Use Old/New chart to convert volume to Use Old/New chart to convert volume to Standard conditionsStandard conditions

3.3. Once Volume is at STP you can use Once Volume is at STP you can use 22.4 dm22.4 dm33/mol, the mol /mol, the mol grams of O grams of O22

III. Ideal Gas EquationIII. Ideal Gas EquationA.A. Combo of the four physical variables: Combo of the four physical variables:

pressure, volume, temperature & number of pressure, volume, temperature & number of particles.particles.

1.1. Equation: PV = Equation: PV = nnRTRT

P = pressure in kPaP = pressure in kPa

V = volume dmV = volume dm33

T = temperature in KT = temperature in K

nn = number of moles of a gas = number of moles of a gas

R = 8.31 (dmR = 8.31 (dm33 x kPa) / (mol x K) x kPa) / (mol x K)* *

(other values for R depending on the units)(other values for R depending on the units)

III. Ideal Gas EquationIII. Ideal Gas Equation

1.1. Break down of Break down of nn

• n n = m/M= m/M

• m = massm = mass M = molecular massM = molecular mass

(g/mol)(g/mol)

Other ways to manipulate the Ideal Other ways to manipulate the Ideal Gas equation: PV = (m/M)RTGas equation: PV = (m/M)RT

m = PVM m = PVM M = (mRT) M = (mRT)

RT PVRT PV

ExampleExample3.3. A flask has a volume of 258 cmA flask has a volume of 258 cm33. A gas with . A gas with

a mass of 1.475 g is introduced into the flask a mass of 1.475 g is introduced into the flask at a temperature of 300K and a pressure of at a temperature of 300K and a pressure of 98.6 kPa. Calculate the molecular mass of 98.6 kPa. Calculate the molecular mass of the gas using the ideal gas equation.the gas using the ideal gas equation.

• Use M = mRT/PV then “plug & chug” plug in Use M = mRT/PV then “plug & chug” plug in the values & solve.the values & solve.

Answer: 145 g/molAnswer: 145 g/mol

A Video…A Video…

FYI: Mark Rosengarten is using atm FYI: Mark Rosengarten is using atm instead of kPa… instead of kPa…

Remember it’s OK to have other values Remember it’s OK to have other values for R depending on the units, he’ll for R depending on the units, he’ll explain that at the end, so hang on explain that at the end, so hang on

CrashCrash

CourseCourseClick on movie projector to play video