The document discusses the mole concept in chemistry. Some key points: - A mole is the amount of substance containing Avogadro's number (6.022x1023) of elementary entities like atoms, molecules, formula units. - One mole of any substance has a mass in grams equal to its formula/molar mass. For example, 1 mole of iron (Fe) has a mass of 55.85 g. - The mole can be used to convert between the number of particles/formula units and the mass of a substance using molar mass and the definition of 1 mole. - Common calculations include determining moles from mass or vice versa using molar mass, as well as

1. The Mole Concept
(The counting unit of chemistry)
Chapter 1

2. • Atoms and molecules are too small to
keep track of individually! It is easier to
count them in packages.
Counting
• We count pieces in MOLES
• That’s why chemists created their own
counting unit called the mole.

3. Avogadro’s Number and The Mole
• Atoms come in moles
6.02 x 1023 = 1 mole = NA
NA is known as Avogadro’s number

4. Mole
Definition:
• A mole is the number of atoms in exactly 12g of the pure
carbon-12 isotope.
• The mole is NOT just a counting unit, like the dozen or a
gross, which identifies only the number of objects. The
definition of a mole signifies the number of objects in a
fixed mass of substance.
• Mass spectrometrically, the mass of a carbon-12 atom is
1.9926×10-23 g.
No. of carbon-12 atoms = atomic mass (g)
mass of one atom (g)
= 12 g _
1.9926×10-23 g
= 6.022 ×1023 atoms

5. Other definitions of the Mole
One mole contains Avogadro’s Number (6.022 x 1023)
A mole is the amount of a substance of a system which
contains as many elementary entities as there are atoms in
0.012kg (or12g) of Carbon-12
A mole is the quantity of a substance whose mass in grams is
the same as its formula weight
For example: Fe (55.85 g)
Iron has an atomic mass or 55.85 g mol-1, so one mole of iron
has a mass or 55.85 g.

6. Formodernsystems,12Cistakenasastandard.
So,12Cisassumedtobe12a.m.u.
112Catom=12a.m.u.
1 a.m.u.= 1/12 of the mass of a single 12C atom
= 1/12 (12/6.02 X 1023)
= 1/6.02 X 1023 grams
= mass of 1 1H atom
Then by comparing the masses of each atom with the mass of 12C,
atomic masses of other atoms can be easily calculated as “RELATIVE
ATOMIC MASSES”. Relative atomic masses easily calculated by MASS
SPECTROMETER.
1 16O = 16 a.m.u
1 40Ca = 40 a.m.u.
1 1H = 1 a.m.u.

7. Onemoleofanyobjectalwaysmeans6.022×1023 unitsofthoseobjects.
Forexample, 1molofH2Ocontains6.022×1023 molecules
1molofNaClcontains6.022×1023 formulaunits
Avogadro’s number is used to convert between the number of moles
and the number of atoms, ions or molecules.
Example
0.450 mol of iron contains how many atoms?
Number of atoms = number of moles × Avogadro’s number (NA)
Therefore,
No. of atoms = (0.450 mol) × (6.022 × 1023)
= 2.7 × 1023 atoms
Calculating the number of particles

8. Example
How many molecules are there in 4 moles of hydrogen peroxide
(H2O2)?
No. of molecules = no. of moles × Avogadro’s number (NA)
= 4 mol × (6.022 × 1023 mol-1)
= 24 ×1023 molecules
= 2.4 × 1024 molecules
Question
 How many atoms are there in 7.2 moles of gold (Au)?
Answer: 4.3 × 1024 atoms

9.  How many calcium atoms are in 0.250 moles of calcium?
Answer: 1.51 x 1023 Ca atoms
 Calculate the moles of 3.75 x 1027 molecules of oxygen gas?
Answer : 6230 mol O2
 How many Cu atoms are in 0.50 mole of Cu?
Answer: 3.0 x 1023 Cu atoms
 How many moles of CO2 are in 2.50 x 1024 molecules of CO2?
Answer: 4.15 moles of CO2

10. Mass and the Mole
• The mass in grams of one mole of any pure substance is called
its molar mass.
• The molar mass of any element is numerically equal to its
atomic mass and has the units g/mol.
• molar mass = mass of 1 mole of substance
• Molar mass can be determined by adding up the atomic
masses from the periodic table (atomic mass goes to 1
decimal place).

11. Q. Find the molar mass of CH4.
= 1C + 4H
= 1 x (12.0) + 4 x (1.0)
= 16.0 g/mol
Q. Find the molar mass of Mg(OH)2.
=1Mg + 2O + 2H
=1 x (24.3) + 2 x (16.00) + 2 x (1.0)
=58.3 g/mol
Q. Find the molar mass of MgSO4•7H2O.
=1Mg + 1S + 4O + 7(H2O)
=1 x (24.3) + 1 x (32.1) + 4 x (16.00) + 7 x [2 x (1.0) + 16.00]
=246.4 g/mol

12. Converting between mass and moles
In the laboratory, we measure the mass of reactants in grams using a
balance. However, when these react they do so in a ratio of moles.
Therefore, we need to convert between the mass we measure and the
number of moles we require.
The expression relating mass and number of moles is:
Mass of sample (g) = no. of moles (mol) × molar mass (gmol-1)
Example
Calculate the mass in grams in 0.75 mol of sodium hydroxide, NaOH
Step 1: Find the molar mass of the compound
Na: 22.99 gmol-1
O: 16.00 gmol-1
H: 1.008 gmol-1
Molar mass of NaOH: 40.00 gmol-1
Step 2: Substitute into the above expression
Mass of sample = 0.75mol × 40.00 gmol-1
= 30g

13. Questions
Calculate the mass in grams present in:
(a) 0.57mol of potassium permanganate (KMnO4)
Answer: Molar mass KMnO4 = 158.03 gmol-1
Mass in grams = 0.57mol × 158.03 gmol-1
= 90.07 g
(b) 1.16mol of oxalic acid (H2C2O4)
Answer: Molar mass H2C2O4 = 90.04 gmol-1
Mass in grams = 1.16mol × 90.04 gmol-1
= 104.44 g
(c) 2.36mol of calcium hydroxide Ca(OH)2
Answer: Molar mass Ca(OH)2 = 74.1 gmol-1
Mass in grams = 2.36mol × 74.1 gmol-1
= 174.87 g

14. Convertingbetweenmolesandmass
Number of moles = mass of sample (g)
molar mass (gmol-1)
Example
Convert 25.0g of KMnO4 to moles
Step 1: Calculate the molar mass
K
Mn
O
1 × 39.10 gmol-1
1 × 54.93 gmol-1
4 × 16.00 gmol-1
= 39.10 gmol-1
= 54.93 gmol-1
= 64.00 gmol-1
MM = 158.03 gmol-1
Step 2: Substitute into above expression
No. of moles =
25.0g .
158.03gmol-1
= 0.158 mol

15. Questions
Calculate the number of moles in:
(a) 1.00g of water (H2O)
Answer: Molar mass water = 18.02 gmol-1
1.00g H2O = 0.055mol
(b) 3.0g of carbon dioxide (CO2)
Answer: Molar mass carbon dioxide = 44 gmol-1
3.0g CO2 = 0.068mol
(c) 500g of sucrose (C12H22O11)
Answer: Molar mass sucrose = 342.30 gmol-1
500g C12H22O11 = 1.46mol
(d) 2.00g of silver chloride (AgCl)
Answer: Molar mass silver chloride = 143.38 gmol-1
2.00g AgCl = 0.014mol

16. Important formulaeso far….
Number of moles = mass of sample (g)
molar mass (gmol-1)
No. of carbon-12 atoms = atomic mass (g)
mass of one atom (g)
No. of atoms = No. of moles × Avogadro’s number (NA)
No. of molecules = No. of moles × Avogadro’s number (NA)
Mass of one molecule = Molar mass
Avogadro’s no.
Mass of sample (g) = no. of moles (mol) × molar mass (gmol-1)
Defining the mole:
Calculating the number of atoms or molecules, given the number of moles:
Most important equation:
Calculating the mass of an individual molecule:

17. Calculatingmasspercentagefromachemicalformula
Many elements in the periodic table occur in combination with other elements to
form compounds.
A chemical formula of a compound expresses the composition of that compound in
terms of the number of atoms of each element present in it.
The mass percentage composition allows you to determine the fraction of the total mass
each element contributes to the compound.
Example
Ammonium nitrate (NH4NO3) is an important compound in the fertiliser industry.
What is the mass % composition of ammonium nitrate?
Step 1: Calculate the molar mass of ammonium nitrate
Molar mass NH4NO3 = 80.05 gmol-1
Two N atoms: 28.016 gmol-1
Four H atoms: 4.032 gmol-1
Three O atoms: 48.00 gmol-1

18. Step2:Determinethemass%compositionforeachelement
Nitrogen: 28.016g N in one mol of ammonium nitrate
Mass fraction of N = 28.016g
80.05g
Mass % composition of N = 28.016g × 100%
80.05g
= 34.99% ≈ 35%
Hydrogen: 4.032g H in one mol of ammonium nitrate
Mass fraction of N = 4.032g
80.05g
Mass % composition of H = 4.032g × 100%
80.05g
= 5.04% ≈ 5%
Oxygen: 48.00g O in one mol of ammonium nitrate
As above, the mass % composition of O is found to be 60%

19. Therefore,themass%compositionofammoniumnitrate(NH4NO3)is:
% Nitrogen: 35%
% Hydrogen: 5%
% Oxygen: 60%
To check your answer, make sure it adds up to 100%
Question
What is the mass % composition of C12H22O11?
Answer: % Carbon: 42.1%
% Hydrogen: 6.5%
% Oxygen: 51.4%

20. Determiningempiricalformulafrommass
The empirical formula of a compound refers to the relative number of
atoms of each element present in that compound. It gives the simplest ratio
of the elements in the compound.
For example, the empirical formula of glucose (C6H12O6) is CH2O, giving the
C:H:O ratio of 1:2:1
If you know the mass % composition and the molar mass of elements present
in a compound, you can work out the empirical formula1
Example
What is the empirical formula of a compound which has a mass %
composition of 50.05% S and 49.95% O?
Step 1: Find the atomic masses of the elements present
Sulfur (S) : 32.066 gmol-1 Oxygen (O) : 16.000 gmol-1

21. Step2:Determinethenumberofmolesofeachelementpresent
Since we are dealing with percentages, we can express the mass % as
grams if we assume we have 100g of the compound.
21
Therefore, 100g of our compound contains 50.05g of sulfur and 49.95g
of oxygen.
Convert number of grams to number of moles
Number of mol Sulfur = mass of sulfur in sample (g)
atomic mass of sulfur (gmol-1)
= 50.05g .
32.066 gmol-1
= 1.56 mol
Similarly, the no. of mol of Oxygen is found to be 3.12mol
Step 3: Determining the ratios of elements
Sulfur: 1.56mol
Oxygen: 3.12mol
Ratio 1.56 : 3.12
Ratio must be in whole numbers. Here we must divide across by 1.56
Therefore, we have a ratio of 1:2 giving an empirical formula of SO2

22. Question
Determine the empirical formula of a compound that contains 27.3
mass% Carbon and 72.7 mass% Oxygen.
Answer: No. of mol Carbon = 2.27mol
No. of mol Oxygen = 4.54mol
Ratio 1:2 Empirical formula CO2
Monosodium glutamate (MSG) has the following mass percentage
composition: 35.51% C, 4.77 % H, 37.85% O, 8.29% N, and 13.60% Na.
What is its molecular formula if its molar mass is 169 gmol-1?
Answer: C5H8O4NNa

23. Molarity
Some chemical reactions involve aqueous solutions of reactants
The concentration of a solution is the amount of solute present in a
given quantity of solvent or solution
This concentration may be expressed in terms of molarity (M) or
molar concentration:
M = Molarity = no. of moles
volume in Litres
Molarity is the number of moles of solute in 1 Litre (L) of solution

24. Whatismolarityofan85.0mLethanol(C2H5OH)solutioncontaining1.77gof
ethanol?
Step 1: Determine the number of moles of ethanol
Example
Molar mass of ethanol, C2H5OH:
2 × carbon atoms
1 × oxygen atom
6 × hydrogen atoms
2 × 12.01 gmol-1
1 × 16.00 gmol-1
6 × 1.008 gmol-1
24.02 gmol-1
16.00 gmol-1
6.048 gmol-1
46.07 gmol-1
No. of moles = mass in g
molar mass
No. of moles ethanol = 1.77g .
46.07 gmol-1
= 0.038 mol

25. Step 2: Convert to molarity
Have 85.0mL ethanol
1 L = 1000mL
 Have 0.085 L of ethanol
Molarity = no. of moles
volume in L
= 0.038 mol
0.085 L
= 0.45 molL-1 ≡ 0.45 M
Questions
Calculate the molarities of each of the following solutions:
(a) 2.357g of sodium chloride (NaCl) in 75mL solution
Answer: 0.5378 M
(b) 1.567mol of silver nitrate (AgNO3) in 250mL solution
Answer: 6.268 M
(c) 10.4g of calcium chloride (CaCl2) in 2.20 × 102 mL of solution
Answer: 0.426 M

26. And if it is a gas at STP, 1 mole = 22.4 L
1 mole = 6.02 x 1023 = molar mass= 22.4 L
•Gases are mostly empty space so they
ALL have the same volume despite
different mass
•22.4 L per 1 mole of any gas at STP
• STP is Standard Temperature and Pressure
• 0°C and 1 atm

27. Example 1:
Determine the volume, in liters, of 0.600 mol of SO2
gas at STP.
Answer: 13.4 L SO2
Example 2:
Determine the number of moles in 33.6 L of He gas
at STP.
Answer: 1.50 moles

28. Combustion Stoichiometry
• Balancing any chemical reaction requires equating the
number of atoms on both the reactant and product side of
the reaction equation.
• In combustion reactions, one of the reactants is air
• Air is approximately 20.9% O2 and 79.1% N2 by volume air
also has argon, CO2 and trace amounts of many other
species but we will ignore these for now
• This means 79.1/20.9 = 3.78 moles of N2 per mole of O2
• Consider the (unbalanced) combustion of methane:
CH4 +a (O2 +3.78 N2) CO2 + bH2O + 3.78a N2

29. • In this representation, N is inert, so why bother
including it?
• There are two reasons.
a. If you want to calculate the mole fraction of each
combustion product, you need to know how much N2 is
present since it will be the predominant species.
b. During actual combustion, atmospheric N2 can form
NOx and hence nitrogen is NOT always inert - more
about NOx formation later.
The complete combustion of methane and air is
now balanced.
CH4 + 2 (O2 +3.78 N2) CO2 + 2 H2O + 2 x 3.78 N2

30. • In general, a simple HC fuel with the composition CnHm will
undergo complete oxidation to form CO2 and H2O.
CnHm + (n +m/4)(O2+3.78 N2) n CO2 +m/2 H2O + 3.78 (n+m/4)N2
• For each mole of fuel burned, (n + m/4) x (1 + 3.78) = 4.78
x (n + m/4) moles of O2 and N2 are involved, and moles of
combustion products are generated is given as:
Moles of combustion product generated
=n + m/2 + 3.78n + 3.78m/4
=4.78n + m/4 + m/4 + 3.78m/4
=4.78(n + m/4) + m/4
The molar stoichiometric fuel-to-air ratio is
1 / [4.78 x (n + m/4) ]

31. The product mole fractions for complete combustion of this
simple HC are
4
/
)
4
/
(
78
.
4
)
4
/
(
78
.
3
4
/
)
4
/
(
78
.
4
2
/
4
/
)
4
/
(
78
.
4
2
2
2
m
m
n
m
n
x
m
m
n
m
x
m
m
n
n
x
N
O
H
CO











32. 066
.
0
1516
100
)
28
78
.
3
32
(
11
100













s
a
f
m
m
%
4
.
73
734
.
0
58
.
56
5
.
41
%
1
.
14
141
.
0
58
.
56
8
%
4
.
12
124
.
0
58
.
56
7
2
2
2









N
O
H
CO
x
x
x
Example: Calculate the stoichiometric fuel/air mass ratio and product gas
composition for the combustion of heptane in air.
Solution
Heptane is C-C-C-C-C-C-C or C7H16.
C7H16 + 11 (O2 +3.78 N2) ===> 7 CO2 + 8 H2O + 11 x 3.78 N2
For each mole of heptane burned, 11 x (1 + 3.78) = 52.5 moles of O2 and N2
are involved.
The molar mass of heptane is 7 x 12 + 16 x 1 = 100. Hence, the fuel/air
mass ratio is
The total number of moles of combustion products is 7 + 8 + 11 x 3.78
= 7 + 8 + 41.5 = 56.58
The product gas composition, on a mole fraction basis, is

33. References
1. General chemistry- Principles, patterns and Applications. Bruce
Averill, Strategic Energy Security Solutions. Patricia Eldredge, R.H.
Hand, LLC ISBN 13: 9781453322307. Saylor Foundation. Pages:
186-222