Ch. 10 – The Mole
I. Molar Conversions
I
II
III
IV
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
NUMBER
MOLES
OF
PARTICLES
(particles/mol)
NA atoms/mol
NA molecules/mol
Particles =
atoms,
molecules,
formula units,
ions, etc
B. Mole/Particle Conversion Examples
 How
many molecules are in
2.50 moles of C12H22O11?
2.50 mol
C12H22O11
6.02  1023
molecules
C12H22O11
1 mol
C12H22O11
= 1.51  1024
molecules
C12H22O11
B. Mole/Particle Conversion Examples
you have 2.23 x 1018 atoms of
sodium, how many moles is
that?
 If
2.23  1018 1 mole Na
atoms 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
3.75
molecules
mol Ca(OH)2 Ca(OH)2
= 2.26  1024
1 mol Ca(OH)2 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 Representative Rep Particles
Formula
Particle
in 1.00 mole
Carbon
C
Atom
6.02 x 1023
Nitrogen
gas
N2
Molecule
6.02 x 1023
Calcium ion
Ca2+
Ion
6.02 x 1023
Magnesium
fluoride
MgF2
Unit Cell
6.02 x 1023
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
12.01 g/mol

aluminum
26.98 g/mol
C. Molar Mass Examples

water
 H2O
 2(1.01) + 16.00 = 18.02 g/mol

sodium chloride
 NaCl
 22.99 + 35.45 = 58.44 g/mol
D. Molar Conversions
6.02  1023
NA
molar
mass
MASS
NUMBER
MOLES
IN
GRAMS
(g/mol)
OF
(particles/mol)
PARTICLES
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
the mass of 2.1  1024
molecules of NaHCO3.
2.1  1024
84.01 g
Molecules 1 mol
NaHCO3
NaHCO3
NaHCO3
 Find
6.02  1023 1 mol
Molecules
NaHCO3
NaHCO3
= 290 g NaHCO3
D. Molar Conversion Examples
 How
many atoms are in 22.5
grams of potassium?
22.5 g K
1 mol K
6.02  1023
atoms K
39.10 g K 1 mol 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

What is the mass of 3.36 L of
ozone gas, O3, at STP?
3.36 L O3
1 mol O3
48 grams
of O3
22.4 L O3 1 mol O3
= 7.2 grams of O3
Mole Calculations

How many molecules of hydrogen
gas, H2, occupy 0.500 L at STP?
0.50 L H2
1 mol H2
6.02  1023
molecules of H2
22.4 L H2
1 mol H2
= 1.34 x 1022 molecules H2
Ch. 10 – The Mole
II. Formula
Calculations
I
II
III
IV
A. Percent Composition

the percentage by mass of each
element in a compound
total mass of element
% mass of element 
 100
total mass of compound
A. Percent Composition

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
%Cu =
%S =
127.10 g Cu
159.17 g Cu2S
32.07 g S
159.17 g Cu2S
 100 = 79.85% Cu
 100 = 20.15% S
Gases
Physical
Properties
I
II
III
IV

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

Gases can be described by their:
 Temperature
•K
 Pressure
• atm
 Volume
•L
 Number of molecules/moles
•#
Temperature

Temperature: Every time temperature is
used
ºF in a gas law equation it must be
stated
-459 in Kelvin
32
212
ºC
-273
K
0
0
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)
force
pressure 
area
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
STP
STP
Standard Temperature & Pressure
0°C
273 K
-OR-
1 atm
101.325 kPa
The volume of 1 mole of gas at STP is 22.4 L
A) Pressure Problem 1

The average pressure in Denver,
Colorado, is 0.830 atm. Express this
in kPa.
0.830 atm 101.325 kPa= 84.1
1 atm
kPa
B) Pressure Problem 2

Convert a pressure of 1.75 atm
to psi.
1.75 atm
14.7 psi
1 atm
= 25.7 psi
The Behavior of Gases
The Gas
Laws
I
II
III
IV
Boyle’s Law
 Investigated
P
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
V
Boyle’s Law

As pressure increases

Volume decreases
Boyles LAW:
P1V1 = k
P
V
Boyle’s Law
Gas Law Problems

A gas occupies 100. mL at 150. kPa.
Find its volume at 200. kPa.
BOYLE’S LAW
GIVEN: P V
V1 = 100. mL
P1 = 150. kPa
V2 = ?
P2 = 200. kPa
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
V1
k
T1
Charles’ Law
V
T

As volume increases,
temperature increases

Charles Law:
V1
k
T1
Charles’ Law
Gas Law Problems

A gas occupies 473 cm3 at 36°C.
Find its volume at 94°C.
CHARLES’ LAW
GIVEN: T V
V1 = 473 cm3
T1 = 36°C = 309K
V2 = ?
T2 = 94°C = 367K
WORK:
V1T2 = V2T1
(473 cm3)(367 K)=V2(309 K)
V2 = 562 cm3

Gas Law Problems
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: T V
V1 = 43.5 L
T1 = 0°C = 273K
V2 = ?
T2= 35.4°C= 308.4K
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:
P1
k
T1
P
T
Gay-Lussac’s Law

The pressure and absolute
temperature (K) of a gas are
directly related
 at constant mass & volume
P1
k
T1
P
T
Gas Law Problems

A gas’ pressure is 765 torr at 23°C.
At what temperature will the pressure
be 560. torr?
GAY-LUSSAC’S LAW
GIVEN: P T WORK:
P1 = 765 torr
P1T2 = P2T1
T1 = 23°C = 296K (765 torr)T2 = (560. torr)(296K)
P2 = 560. torr
T2 = 217 K = -56°C
T2 = ?
Combined Gas Law
P
V
PV
PV = k
T
P 1V 1 P 2V 2
=
T1
T2
P 1 V 1T 2 = P 2V 2 T 1
Gas Law Problems

A gas occupies 7.84 cm3 at 71.8 kPa &
25°C. Find its volume at STP.
COMBINED GAS LAW
GIVEN: P T V WORK:
V1 = 7.84 cm3
P1V1T2 = P2V2T1
P1 = 71.8 kPa
(71.8 kPa)(7.84 cm3)(273 K)
T1 = 25°C = 298 K
=(101.325 kPa) V2 (298 K)
V2 = ?
P2 = 101.325 kPa V2 = 5.09 cm3
T2 = 273 K
Gases
II. Ideal Gas
Law
Ideal Gas Law and Gas
Stoichiometry
I
II
III
IV
Ideal Gas Law
Part 1
I
II
III
IV
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
V1
k
n1
V
n
Ideal Gas Law
Merge the Combined Gas Law with Avogadro’s Principle:
PV
V
=R
k
nT
T
n
UNIVERSAL GAS
CONSTANT
R=0.08206 Latm/molK
3
R=8.315 dm kPa/molK
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

GIVEN:
Calculate the pressure in atmospheres
of 0.412 mol of Heat 16°C & occupying
3.25 L.
WORK:
P = ? atm
PV = nRT
n = 0.412 mol
P(3.25)=(0.412)(.08206)(289)
L
mol Latm/molK K
T = 16°C = 289 K
V = 3.25 L
P = 3.01 atm
R = 0.08206Latm/molK
Ideal Gas Law Problems

GIVEN:
Find the volume of 85 g of O2 at 25°C
and 104.5 kPa.
WORK:
V=?
85 g 1 mol O2 = 2.7 mol
n = 85 g = 2.7 mol
32.00 g
O2
T = 25°C = 298 K PV = nRT
P = 104.5 kPa
(104.5)V=(2.7) (8.315) (298)
kPa
mol
dm3kPa/molK K
R = 8.315 dm3kPa/molK
V = 64 dm3