SCH3U1
In Unit 1, you learned how to use isotopic abundances and isotopic masses to find the average atomic mass of an element. The average atomic mass (found on the periodic table) describes the average mass of an atom in a large sample of matter.
Why is relating average atomic mass to the mass of large samples important? In a laboratory, as in everyday life, we deal with macroscopic samples (i.e. one we can directly measure with a balance). These samples contain incredibly large numbers of atoms or molecules. Imagine if a procedure asked for six septillion molecules
(6 x 10 24 or 6,000,000,000,000,000,000,000,000,000) of water to be added! The numbers involved would be ridiculously inconvenient. Chemists needed a way to deal with the large number of particles present, and the answer was the Mole.
The mole is a convenient unit to measure the very large number of particles present in a substance. a pair of shoes; a dozen eggs; a mole of particles)
The mole was based on the number of atoms in 12 grams of carbon-12:
1 .
994

12
10

24
/

6 .
022

10
23
This number, 6.022 x 10 23 particles, was defined as a mole and is known as
Avogadro’s Constant. The number of moles of a substance can be determined using this formula: n

N
N
A
Where: n = number of moles.
N = number of particles
N
A
= Avogadro’s Constant = 6.022 x 10 23

When you purchase a pharmaceutical from a drugstore, you are confident that each tablet contains the correct amount of the active ingredient. If the drug is present in too small a quantity, the product will be ineffective. If it is present in too high an amount, it may cause unintended side effects. When testing a product, chemist need to be sure the correct number of molecules (or moles of molecules) of the active ingredient are present. On its own, the mass of an active ingredient is not very useful; it does not tell him or her to number of molecules present. Yet the effect of a drug depends on the number of molecules, not their mass. Chemists need a way to relate the nass of a substance to the number of particles that are present.
The molar mass is the mass of 1 mole (or 6.022 x 10 23 particles) of any substance.
Therefore:
M
 m
Where: n
m = mass of a substance (g)
n = number of moles of a substance (mol)
M = molar mass (g/mol)
Calculating Molar Masses
The mass of 6.022 x 10 23 particles of any chemical can easily be calculated from the sum of the atomic masses from the periodic table. e.g.1 What is the molar mass (M) of magnesium chloride, MgCl
M
Mg
= 24.31 g/mol M
Cl
= 35.45 g/mol
2
? e.g. 2
Therefore: M
MgCl2
= M
MgCl2
= 24.31 +(2x35.46) = 95.23 g/mol
What is the molar mass of Fe(OH)
3
?
M
Fe(OH)3
= M
Fe
+ 3xM
O
+ 3xM
H
= 55.85 + (3x16.00) + (3x1.01)
= 106.88 g/mol e.g. 3 e.g. 4
How many moles of MgCl number of moles
 n
 m
M

2
are there in 250 g of this salt?
250
95 .
23 g / g mol

2 .
63
How many actual “molecules” are present? mol
* MgCl2 is actually ionic so they are called “formula units”!
number of particles

N
 n

N
A
N

( 2 .
63

6 .
022

10
23
)

1 .
58

10
24 formula units

SCH3U1
THE MOLE
1. One litre of air contains 2 x 10 22 molecules of nitrogen gas. How many moles of nitrogen are there in one litre of air?
2. A cup of water (250 mL) contains approximately 13.9 mol of H
2
O. How many molecules of water are present in a cup of water?
3. a) Calculate the molar mass of carbon dioxide (CO
2
)? b) Calculate the number of moles in 110 g of carbon dioxide. c) How many molecules are there in 110 g of carbon dioxide? d) How many atoms are there in 110 g of carbon dioxide?
4. A grain of sand has a mass of 0.012 g. If the sand is composed of silicon dioxide, how many molecules of SiO
2
are there in the grain of sand?

M OLES AND MOLAR MASS WORKSHEET
1. Using the periodic table, calculate the molar mass of the following substances:
XeF
4
BaSO
4
I
2
SiO
2
________________
________________
________________
________________
H
Hg
S
2
8
CH
SO
4
3
________________
________________
________________
________________
NH
4
NO
3
________________
2. Calculate the mass of the following:
Fe
3
[Fe(CN)
6
]
2
______________
4 mol H
2
SO
4
___________________________________
2.5 mol MgCO
3
___________________________________
0.5 mol FeCl
2
___________________________________
0.01 mol CaCO
3
___________________________________
0.25 mol KF ___________________________________
6.0 mol NaOH ___________________________________
1.8 mol Al
2
(SO
4
)
3
___________________________________
3. Calculate the number of moles in the following:
68 g H
2
S ______________ 23.2 g Zn(ClO
3
)
2
____________
1.7 g NH
3
______________ 1.7 g AgNO
3
______________
5.6 g KOH ______________ 174 g MnO
2
______________
0.58 g Li
3
PO
4
____________ 11.0 g CO
2
_______________
4. A sample of the mineral hematite ( iron (III) oxide ) has a mass of 12.4 g. a) How many moles of the mineral are present? b) How many molecules of the mineral are present? c) How many atoms of iron and oxygen are present?
4. One sample of gold contains 9.0 x 10 21 atoms of gold sells for $50. Another sample of gold with a mass of 3.6 g of gold also sells for $50. \Which sample contains more gold? Hint: convert both values to moles to compare them!

HOW MUCH IS A MOLE?
1. Observe the substances at this station and complete the table.
Substance 1 Substance 2 Substance 3
Name
Chemical
Formula
Mass (g)
Molar Mass
(g/mol)
Number of
Moles (mol)
Number of
Particles
Useful Formulas:
N n

N
A
n = number of moles (mol)
N = number of particles
N
A
= Avogadro’s number = 6.022 x 10 23
M
 m
M = molar mass (g/mol)
m = mass (g) n n = number of moles (mol)
Remember that the mass of 1 mole af any element is equal to the atomic mass in grams (e.g. M helium
= 4.0 g/mol).