Ch. 10 – The Mole Molar Conversions

A. What is the Mole? A counting number (like a dozen) Avogadro’s number (NA) 1 mole = 6.022  1023 representative particles

B. Mole/Particle Conversions 6.022  1023 NA MOLES NUMBER OF PARTICLES (particles/mol) Particles = atoms, molecules, formula units, ions, etc NA atoms/mol NA molecules/mol

B. Mole/Particle Conversion Examples How many molecules are in 2.50 moles of C12H22O11? 6.02  1023 molecules C12H22O11 1 mol C12H22O11 2.50 mol C12H22O11 = 1.51  1024 molecules C12H22O11

B. Mole/Particle Conversion Examples If you have 2.23 x 1018 atoms of sodium, how many moles is that? 2.23  1018 atoms Na 1 mole Na 6.02  1023 atoms Na 3.70 x 10-6 moles Na =

B. Mole/Particle Conversion Examples How many molecules is 3.75 moles of calcium hydroxide? 6.02  1023 molecules Ca(OH)2 1 mol Ca(OH)2 3.75 mol Ca(OH)2 = 2.26  1024 molecules Ca(OH)2

C. Molar Mass Avogadro discovered the relationship between number of particles and volume of a gas This was used to find the relationship for particles in a mole

Representative Particles & Moles Substance Chemical Formula Representative Particle Rep Particles in 1.00 mole Carbon C Atom 6.02 x 1023 Nitrogen gas N2 Molecule Calcium ion Ca2+ Ion Magnesium fluoride MgF2 Unit Cell

C. Molar Mass Mass of 1 mole of an element or compound Atomic mass (on the PT) tells the... mass of each atom (amu) grams per mole (g/mol)

C. Molar Mass Examples carbon aluminum 12.01 g/mol 26.98 g/mol

C. Molar Mass Examples water H2O 2(1.01) + 16.00 sodium chloride NaCl 22.99 + 35.45 = 18.02 g/mol = 58.44 g/mol

Particles = atoms, molecules, formula units, ions, etc D. Molar Conversions molar mass 6.02  1023 NA MASS IN GRAMS MOLES NUMBER OF PARTICLES (g/mol) (particles/mol) Particles = atoms, molecules, formula units, ions, etc NA atoms/mol NA molecules/mol

D. Molar Conversion Examples How many moles of carbon are in 26 g of carbon? 26 g C 1 mol C 12.01 g C = 2.2 mole C

D. Molar Conversion Examples Find the mass of 2.1  1024 molecules of NaHCO3. 2.1  1024 Molecules NaHCO3 1 mol NaHCO3 6.02  1023 Molecules NaHCO3 84.01 g NaHCO3 1 mol NaHCO3 = 290 g NaHCO3

D. Molar Conversion Examples How many atoms are in 22.5 grams of potassium? 6.02  1023 atoms K 1 mol K 22.5 g K 1 mol K 39.10 g K = 3.46 x 1023 atoms K

Gases Many of the chemicals we deal with are gases. They are difficult to weigh. Need to know how many moles of gas we have. Two things effect the volume of a gas Temperature and pressure Compare at the same temp. and pressure.

Standard Temperature and Pressure 0ºC and 1 atm pressure abbreviated STP At STP 1 mole of gas occupies 22.4 L Called the molar volume Avogadro’s Hypothesis - at the same temperature and pressure equal volumes of gas have the same number of particles.

Molar Volume Avogadro's Theory: Two gases containing equal numbers of molecules occupy equal volumes under similar conditions. Standard Temperature and Pressure: 0OC and 1 atm. Molar Volume at STP = 22.4 L

Mole Calculations 48 grams of O3 1 mol O3 1 mol O3 22.4 L O3 What is the mass of 3.36 L of ozone gas, O3, at STP? 48 grams of O3 1 mol O3 3.36 L O3 1 mol O3 22.4 L O3 = 7.2 grams of O3

Mole Calculations 1 mol H2 22.4 L H2 = 1.34 x 1022 molecules H2 How many molecules of hydrogen gas, H2, occupy 0.500 L at STP? 6.02  1023 molecules of H2 1 mol H2 0.50 L H2 1 mol H2 22.4 L H2 = 1.34 x 1022 molecules H2

atomic 1 mole mass Mass 22.4 L Representative Particles Volume of Gas at STP 6.02x10 23 Mole

II. Formula Calculations Ch 10 The Mole - Formula Calculations Ch. 10 – The Mole II. Formula Calculations Chilton

Ch 10 The Mole - Formula Calculations A. Percent Composition the percentage by mass of each element in a compound Chilton

A. Percent Composition  100 = 79.85% Cu %Cu =  100 = 20.15% S %S = Find the % composition Copper and Sulfur in Cu2S. Known: Mass of Cu in Cu2S = 2 (63.55g) = 127.10 g Cu Mass of S in Cu2S = 32.07 g S Molar Mass (total) = 127.10 g + 32.07 g = 159.17 g/mol 127.10 g Cu 159.17 g Cu2S  100 = 79.85% Cu %Cu = 32.07 g S 159.17 g Cu2S  100 = 20.15% S %S =

Gases Physical Properties

Kinetic Molecular Theory Particles in an ideal gas… have no volume have elastic collisions are in constant, random, straight-line motion don’t attract or repel each other average kinetic energy is directly proportional to the absolute temperature To change pressure you can: Change temperature Change volume Change amount

Characteristics of Gases Gases expand to fill any container random motion, no attraction Gases are fluid (like liquids) no attraction Gases have very low densities no volume = lots of empty space

Characteristics of Gases Gases can be compressed no volume = lots of empty space Gases undergo diffusion & effusion random motion State Changes

Describing Gases K atm L # Gases can be described by their: Temperature Pressure Volume Number of molecules/moles K atm L #

K = ºC + 273 Temperature ºF ºC K -459 32 212 -273 100 273 373 Temperature: Every time temperature is used in a gas law equation it must be stated in Kelvin ºF ºC K -459 32 212 -273 100 273 373 K = ºC + 273

Pressure Pressure of a gas is the force of its particles exerted over a unit area (number of collisions against a container’s walls) Which shoes create the most pressure?

Pressure Pressure may have different units: torr, millimeters of mercury (mm Hg), atmospheres (atm), Pascals (Pa), or kilopascals (kPa). KEY EQUIVALENT UNITS 760 torr 760 mm Hg 1 atm 101,325 Pa 101.325 kPa (kilopascal) 14.7 psi

Standard Temperature & Pressure STP STP Standard Temperature & Pressure 0°C 273 K 1 atm 101.325 kPa The volume of 1 mole of gas at STP is 22.4 L -OR-

A) Pressure Problem 1 101.325 kPa 0.830 atm = 84.1 kPa 1 atm The average pressure in Denver, Colorado, is 0.830 atm. Express this in kPa. 101.325 kPa 0.830 atm = 84.1 kPa 1 atm

B) Pressure Problem 2 14.7 psi 1.75 atm = 25.7 psi 1 atm Convert a pressure of 1.75 atm to psi. 14.7 psi 1.75 atm = 25.7 psi 1 atm

The Behavior of Gases The Gas Laws P V T

Boyle’s Law Investigated the relationship between pressure and volume on a gas. He found that pressure and volume of a gas are inversely related at constant mass & temp P V

P1V1 = k Boyle’s Law As pressure increases Volume decreases

Boyle’s Law

Gas Law Problems BOYLE’S LAW P V A gas occupies 100. mL at 150. kPa. Find its volume at 200. kPa. BOYLE’S LAW GIVEN: V1 = 100. mL P1 = 150. kPa V2 = ? P2 = 200. kPa P V WORK: P1V1 = P2V2 (150.kPa)(100.mL)=(200.kPa)V2 V2 = 75.0 mL

Charles’ Law Jacques Charles investigated the relationship of temperature & volume in a gas. He found that the volume of a gas and the Kelvin temperature of a gas are directly related. V T

Charles’ Law As volume increases, temperature increases Charles Law: V

Charles’ Law

Gas Law Problems CHARLES’ LAW T V A gas occupies 473 cm3 at 36°C. Find its volume at 94°C. CHARLES’ LAW GIVEN: V1 = 473 cm3 T1 = 36°C = 309K V2 = ? T2 = 94°C = 367K T V WORK: V1T2 = V2T1 (473 cm3)(367 K)=V2(309 K) V2 = 562 cm3

Gas Law Problems CHARLES’ LAW T V GIVEN: V1 = 43.5 L T1 = 0°C = 273K Example: The volume of a sample of gas is 43.5 L at STP. What will the volume of the gas be if the temperature is increased to 35.4°C? CHARLES’ LAW GIVEN: V1 = 43.5 L T1 = 0°C = 273K V2 = ? T2= 35.4°C= 308.4K T V WORK: V1T2 = V2T1 (43.5 L)(308.4 K)=V2(273 K) V2 = 49.1 L

Gay-Lussac’s Law Since temperature and volume are related, it would make sense that temperature and pressure are related. The equation for this will resemble Charle’s law: P T

Gay-Lussac’s Law The pressure and absolute temperature (K) of a gas are directly related at constant mass & volume P T

Gas Law Problems GAY-LUSSAC’S LAW P T A gas’ pressure is 765 torr at 23°C. At what temperature will the pressure be 560. torr? GAY-LUSSAC’S LAW GIVEN: P1 = 765 torr T1 = 23°C = 296K P2 = 560. torr T2 = ? P T WORK: P1T2 = P2T1 (765 torr)T2 = (560. torr)(296K) T2 = 217 K = -56°C

PV T V T P T PV = k P1V1 T1 = P2V2 T2 P1V1T2 = P2V2T1 Combined Gas Law PV T V T P T PV = k P1V1 T1 = P2V2 T2 P1V1T2 = P2V2T1

Gas Law Problems COMBINED GAS LAW P T V V1 = 7.84 cm3 P1 = 71.8 kPa A gas occupies 7.84 cm3 at 71.8 kPa & 25°C. Find its volume at STP. COMBINED GAS LAW GIVEN: V1 = 7.84 cm3 P1 = 71.8 kPa T1 = 25°C = 298 K V2 = ? P2 = 101.325 kPa T2 = 273 K P T V WORK: P1V1T2 = P2V2T1 (71.8 kPa)(7.84 cm3)(273 K) =(101.325 kPa) V2 (298 K) V2 = 5.09 cm3

Gases II. Ideal Gas Law Ideal Gas Law and Gas Stoichiometry

Ideal Gas Law Part 1

Avogadro’s Law The volume of a gas increases or reduces, as its number of moles is being increased or decreased. The volume of an enclosed gas is directly proportional to its number of moles.

Avogadro’s Principle Equal volumes of gases contain equal numbers of moles at constant temp & pressure true for any gas n = number of moles V n

UNIVERSAL GAS CONSTANT Ideal Gas Law Merge the Combined Gas Law with Avogadro’s Principle: V n PV nT PV T = k = R UNIVERSAL GAS CONSTANT R=0.08206 Latm/molK R=8.315 dm3kPa/molK

UNIVERSAL GAS CONSTANT Ideal Gas Law PV=nRT UNIVERSAL GAS CONSTANT R=0.08206 Latm/molK R=8.315 dm3kPa/molK

Ideal Gas Law Problems P = ? atm n = 0.412 mol T = 16°C = 289 K Calculate the pressure in atmospheres of 0.412 mol of Heat 16°C & occupying 3.25 L. GIVEN: P = ? atm n = 0.412 mol T = 16°C = 289 K V = 3.25 L R = 0.08206Latm/molK WORK: PV = nRT P(3.25)=(0.412)(.08206)(289) L mol Latm/molK K P = 3.01 atm

Ideal Gas Law Problems V = ? n = 85 g T = 25°C = 298 K P = 104.5 kPa Find the volume of 85 g of O2 at 25°C and 104.5 kPa. GIVEN: V = ? n = 85 g T = 25°C = 298 K P = 104.5 kPa R = 8.315 dm3kPa/molK WORK: 85 g 1 mol O2 = 2.7 mol 32.00 g O2 = 2.7 mol PV = nRT (104.5)V=(2.7) (8.315) (298) kPa mol dm3kPa/molK K V = 64 dm3