1. Ch. 10 – The Mole
Molar Conversions

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

3. 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

4. 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

5. 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 =

6. 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

7. 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

8. 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

9. 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)

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

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

12. 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

13. 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

14. 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

15. 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

16. 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.

17. 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.

18. 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

19. 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

20. Mole Calculations 1 mol H2 22.4 L H2 = 1.34 x 1022 molecules H2
How many molecules of hydrogen gas, H2, occupy 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

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

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

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

24. 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) = g Cu
Mass of S in Cu2S = g S
Molar Mass (total) = g g = g/mol
g Cu
g Cu2S
 100 =
79.85% Cu
%Cu =
32.07 g S
g Cu2S
 100 =
20.15% S
%S =

25. Gases
Physical Properties

26. 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

27. 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

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

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

30. 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

31. 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?

32. 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
kPa (kilopascal)
14.7 psi

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

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

35. 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

36. The Behavior of Gases
The Gas Laws P V T

37. 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

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

39. Boyle’s Law

40. 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

41. 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

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

43. Charles’ Law

44. 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

45. 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

46. 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

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

48. 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

49. 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

50. 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 = kPa
T2 = 273 K
P T V
WORK:
P1V1T2 = P2V2T1
(71.8 kPa)(7.84 cm3)(273 K)
=( kPa) V2 (298 K)
V2 = 5.09 cm3

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

52. Ideal Gas Law
Part 1

53. 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.

54. 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

55. 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= Latm/molK
R=8.315 dm3kPa/molK

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

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

58. 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 kPa.
GIVEN:
V = ?
n = 85 g
T = 25°C = 298 K
P = kPa
R = 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