Chapter 10
The Mole
Objectives

Explain how a mole is used to indirectly count the number of
particles of matter

Relate the mole to common everyday counting unit

Calculate the molar mass of a compound

Convert between moles and number of representative particles

Convert between number of moles and the mass of an element

Convert between number of moles and number of atoms of an
element
Chemical Quantities
Mole – represents a number of items
- is the SI base unit used to measure the amount of a
substance
 Just like a dozen = 12, gross = 144, ream =500
 Mole has 6.02 x1023 particles of any substance
Avogadro’s Constant
 6.02 x1023 particles = number of particles in 1 mole of
substance
 We use Avogadro’s number as a grouping
 Particles can be anything – Cars, people, atoms,
molecules
How big is a mole?


100 average-sized peas
 Roughly the volume of a cubic inch
1,000,000 peas (106)


Billion (109) peas


Fill a refrigerator
Fill a house basement to attic
Trillion (1012) peas

Fill a 1000 houses
How big is a Mole?

Quadrillion (1015) peas


Quintillion (1018) peas


Blanket of peas 4 feet deep all over ND
Sextillion (1021) peas


Fill all the buildings in a larger city like Minneapolis
Earth covered in 4 ft
Septillion (1024) peas

250,000 planets covered 4ft deep
Molecular Weight
Molar mass – of any substance is the mass (in grams) of 1
mole of substance
3 different types Molar Masses

Each of the following represents 1 mole:
1) Gram Atomic mass (gam)

1 mole of atoms (single element)

Can get from periodic table

write values to the tenth
Examples:
Oxygen
1 mole O = 16.0 g
Iron
1 mole Fe = 55.9 g
Cu+2
1 mole Cu+2 = 63.6 g
Molecular Weight
2)
Gram molecular mass (gmm)

1 mole of molecules –> used with molecular compounds

Add the molar mass of each element multiplied by its
subscript
Example:
What is the molar mass or Gram molecular mass of CO2 ?
C
O
12.0g x 1 =
16.0g x 2 =
12.0g
32.0g
44.0g
1 mole of CO2 = 44.0g
Molecular Weight
3) Gram formula mass (gfm)

1 mole of formula units –>ionic compounds

Add the molar mass of each element multiplied by
its subscript
Examples:
What is the molar mass or Gram Formula mass of NaCl ?
Na
Cl
23.0g x 1 = 23.0g
35.5g x 1 = 35.5g
58.5 g
1 mole of NaCl = 58.5 g
Molecular Mass
What is the molar mass or Gram Formula mass of Na2SO4 ?
Na
S
O
23.0g x 2 = 46.0g
32.1g x 1 = 32.1g
16.0g x 4 = 64.0g
142.1g
1 mole of Na2SO4 = 142.1g
How many molecules of Na2SO4 is there in 1 mole?
6.02 x 1023 molecules of Na2SO4 = 1 mole Na2SO4
Moles to mass conversions
1 mole of substance is equal to its molar mass


1 mole of NaCl = 58.5 g NaCl
2.54 moles of NaCl contains how many grams?
1
(
(
2.54 moles NaCl
(
58.5 g NaCl
1 mole NaCl
(

=
149 g NaCl
252 grams of NaCl is how many moles?
(
1
(
1 mole NaCl
58.5 grams NaCl
(
(
252 grams NaCl
=
4.31 moles NaCl
Conversions – Mass/grams/particles
1
(
252 grams NaCl
(
1 mole NaCl
58.5 grams NaCl
(
(
How many particles (Formula Units) of NaCl are there in
252 grams of NaCl
(
6.02 x 1023 FU of NaCl
1 mole NaCl
(

=
2.59 x 1024 FU NaCl
1
(
3.50 x 1024 molecules of NO3
1 mole NO3
( 6.02 x 10
23
(
(
3.50 x 1024 molecules of NO3 has a mass of how many
grams?
molecules of NO3
=
(
62.0 grams NO3
1 mole NO3
360. grams NO3
(

Flow chart for conversions
Mass of
Compound
Number of Moles
of compound
Molar
Mass
Avagadro’s
Number
Number of particles

Atoms

Molecules

Formula Units
Volume of a Mole of a Gas



Volume of 1 mole of gas varies with
 Δ in temperature
 Δ in pressure
Gases usually measured @ Standard temp and pressure
(STP)
 0° and 101.3 KPa or 1 ATM
1 Mole of any gas occupies a volume of 22.4L
 22.4 L Know as Molar Volume of Gas
Mole to a gas
Example:
25 g of Neon = ? L
20.2 g Ne
(
(
1
(
1 mole Ne
(
22. 4 L Ne
1 mole Ne
(
(
25 g Ne
=
28 L Ne
Percent Composition of a Compound
Percent composition – amounts, by % mass, of each
element in a compound
2 types of problems
1) Percent composition – relative amounts
 you are provided amounts
 % mass of element = grams of element x 100
grams of compound
Example:
Find the percent composition of the compound that is
formed from 128.0 g of sulfur and 192.0 g of oxygen.
Example:
Find the percent composition of the compound that is
formed from 128.0 g of sulfur and 192.0 g of oxygen.
192.0 g
+128.0 g
320.0 g Total mass of compound
% mass of S =
128.0 g x 100 = 40.00 % S
320.0 g
% mass of O =
192.0 g x 100 =
320.0 g
60.00 % O
Percent Composition of a Compound
2)
Percent composition of known compound
You are given the composition
Step 1 – use the chemical formula to find molar mass
Step 2 – for each element
 Find the mass


%mass
Divide it grams of element by molar mass of
compound
Multiply by 100
= mass of 1 mole of element x 100
molar mass of compound
Example

Find the percent composition of Cu, SO4 and H2O in
CuSO42H2O
Empirical Formula
Empirical Formula – lowest whole-number ratio of elements in
a compound
Example
 C3 H 6 O
 C6H12O2
Empirical formula
Molecular formula
The percentage composition of diborane is 78.1% B and 21.9%
H Determine the empirical formula.
Empirical Formula


Step 1
 Change % to grams
 78.1% B = 78.1 g B
21.9% H = 21.9 g H
Step 2
 Change the grams to moles for each element
(78.1 g B ) (1 mole B ) = 7.23 mole B
1
(10.8 g B)
(21.9 g H) (1 mole H ) = 22 mole H
1
(1.0 g H )
Empirical Formula

Step 3
 Divide each amount by the smallest amount of moles
 Use these Ratios to determine the Empirical Formula
7.23 moles = 1.00 B
7.23 moles
BH3
22 moles = 3.0 H
7.23 moles