Atomic mass and the mole
An equation for a chemical reaction can provide us with a lot of useful information. It tells us what the reactants and the products are in the reaction, and it also tells us the ratio in which the reactants combine to form products. Look at the equation below:
$\text{Fe}+\text{S}\to \text{FeS}$
In this reaction, every atom of iron ($\text{Fe}$) will react with a single atom of sulphur ($\text{S}$) to form iron sulphide ($\text{FeS}$). However, what the equation does not tell us, is the quantities or the amount of each substance that is involved. You may for example be given a small sample of iron for the reaction. How will you know how many atoms of iron are in this sample? And how many atoms of sulphur will you need for the reaction to use up all the iron you have? Is there a way of knowing what mass of iron sulphide will be produced at the end of the reaction? These are all very important questions, especially when the reaction is an industrial one, where it is important to know the quantities of reactants that are needed, and the quantity of product that will be formed. This chapter will look at how to quantify the changes that take place in chemical reactions.
The mole
Sometimes it is important to know exactly how many particles (e.g. atoms or molecules) are in a sample of a substance, or what quantity of a substance is needed for a chemical reaction to take place.
The amount of substance is so important in chemistry that it is given its own name, which is the mole.
 Definition 1: Mole
The mole (abbreviation “mol”) is the SI (Standard International) unit for “amount of substance”.
The mole is a counting unit just like hours or days. We can easily count one second or one minute or one hour. If we want bigger units of time, we refer to days, months and years. Even longer time periods are centuries and millennia. The mole is even bigger than these numbers. The mole is 602 204 500 000 000 000 000 000 or 6,022 × 10^{23} particles. This is a very big number! We call this number Avogadro's number.
 Definition 2: Avogadro's number
The number of particles in a mole, equal to 6,022 × 10^{23}.
If we had this number of cold drink cans, then we could cover the surface of the earth to a depth of over 300 km! If you could count atoms at a rate of 10 million per second, then it would take you 2 billion years to count the atoms in one mole!
Interesting Fact:
The original hypothesis that was proposed by Amadeo Avogadro was that “equal volumes of gases, at the same temperature and pressure, contain the same number of molecules”. His ideas were not accepted by the scientific community and it was only four years after his death, that his original hypothesis was accepted and that it became known as “Avogadro's Law”. In honour of his contribution to science, the number of particles in one mole was named Avogadro's number.
We use Avogadro's number and the mole in chemistry to help us quantify what happens in chemical reaction. The mole is a very special number. If we measure 12,0 g of carbon we have one mole or 6,022 × 10^{23} carbon atoms. 63,5 g of copper is one mole of copper or 6,022 × 10^{23} copper atoms. In fact, if we measure the relative atomic mass of any element on the periodic table, we have one mole of that element.
Exercise 1: Moles and mass
How many atoms are there in:

1 mole of a substance

2 moles of calcium

5 moles of phosphorus

24,3 g of magnesium

24,0 g of carbon
The relative atomic mass for nitrogen is the value written on the periodic table. To work out the number of moles in the sample we note that when we have one mole of a substance we have the same sample mass and relative atomic mass.
For hydrogen we have 1,01 g of sample and 1,01 u for relative atomic mass, so we have one mole of hydrogen. For carbon we have 24,0 g of sample and 12,0 u for relative atomic mass, so we have two moles of substance ($12,01\times 2=24,02$)
Doing this for the rest of the elements in the table we get:
Element  Relative atomic mass (u)  Sample mass (g)  Number of moles in the sample 

Hydrogen  1,01  1,01  1 
Magnesium  24,3  24,3  1 
Carbon  12,0  24,0  2 
Chlorine  35,45  70,9  2 
Nitrogen  14,0  42,0  3 
Complete the following table:

a) The number of atoms is given by Avogadros number. To determine the number of atoms in a sample we only need the number of moles, we do not need to know what the substance is. So the number of atoms in 1 mole of a substance is:
$1\times (6,02\times {10}^{23})=6,02\times {10}^{23}$
b) In 2 moles of calcium there are:
$2\times (6,02\times {10}^{23})=1,204\times {10}^{24}$
c) In 5 moles of phosphorus there are:
$5\times (6,02\times {10}^{23})=3,01\times {10}^{24}$
d) Before we can work out the number of atoms, we must work out the number of moles. The relative atomic mass of magnesium is 24,31 u.
So we have 1 mole of magnesium. (Relative atomic mass = sample mass)
The number of atoms are:
$1\times (6,02\times {10}^{23})=6,02\times {10}^{23}$
e) The relative atomic mass of carbon is 12,01 u
The number of moles in the sample is 2. (Relative atomic mass = half sample mass)
The number of atoms is:
$2\times (6,02\times {10}^{23})=1,204\times {10}^{24}$
Molar mass
 Definition 3: Molar mass
Molar mass (M) is the mass of 1 mole of a chemical substance. The unit for molar mass is grams per mole or g·mol^{−1}.
You will remember that when the mass, in grams, of an element is equal to its relative atomic mass, the sample contains one mole of that element. This mass is called the molar mass of that element.
Note:
You may sometimes see the molar mass written as ${M}_{m}$. We will use M in this book, but you should be aware of the alternate notation.
It is worth remembering the following: On the periodic table, the relative atomic mass that is shown can be interpreted in two ways.

The mass (in grams) of a single, average atom of that element relative to the mass of an atom of carbon.

The average atomic mass of all the isotopes of that element. This use is the relative atomic mass.

The mass of one mole of the element. This third use is the molar mass of the element.

Example 1: Calculating the number of moles from mass
Question
Calculate the number of moles of iron (Fe) in an 11,7 g sample.
Answer
Find the molar mass of iron
If we look at the periodic table, we see that the molar mass of iron is 55,8 g·mol^{−1}. This means that 1 mole of iron will have a mass of 55,8 g.
Find the mass of iron
If 1 mole of iron has a mass of 55,8 g, then: the number of moles of iron in 111,7 g must be:
There are 2 moles of iron in the sample.
Example 2: Calculating mass from moles
Question
You have a sample that contains 5 moles of zinc.

What is the mass of the zinc in the sample?

How many atoms of zinc are in the sample?
Answer
Find the molar mass of zinc
Molar mass of zinc is 65,4 g·mol^{−1}, meaning that 1 mole of zinc has a mass of 65,4 g.
Find the mass
If 1 mole of zinc has a mass of 65,4 g, then 5 moles of zinc has a mass of: $65,4\text{g}\times 5\text{mol}=327\text{g}$ (answer to a)
Find the number of atoms
$5\text{mol}\times 6,022\times {10}^{23}\text{atoms}\xb7{\text{mol}}^{1}=3,011\times {10}^{23}\text{atoms}$ (answer to b)
Exercise 2: Moles and molar mass
Give the molar mass of each of the following elements:

hydrogen gas

nitrogen gas

bromine gas
We simply find the element on the periodic table and read off its molar mass.
a) $1,01g\cdot mo{l}^{1}$
b) $14,01g\cdot mo{l}^{1}$
c) $79,9g\cdot mo{l}^{1}$
Calculate the number of moles in each of the following samples:

21,6 g of boron ($\text{B}$)

54,9 g of manganese ($\text{Mn}$)

100,3 g of mercury ($\text{Hg}$)

50 g of barium ($\text{Ba}$)

40 g of lead ($\text{Pb}$)
a) Molar mass of boron is: $10g\cdot mo{l}^{1}$
If one mole of boron has a mass of 10 g then the number of moles of boron in 21,62 g must be: $\frac{21,62g}{10,00g\cdot mo{l}^{1}}=2,162mols$
b) Molar mass of manganese is: $59,908g\cdot mo{l}^{1}$
If one mole of manganese has a mass of 59,908 g then the number of moles of manganese in 54,94 g must be:
$\frac{54,94g}{59,908g\cdot mo{l}^{1}}=0,917mols$
c) Molar mass of mercury is:$200,59g\cdot mo{l}^{1}$
If one mole of mercury has a mass of 200,59 g then the number of moles of manganese in 100,03 g must be:
$\frac{100,3g}{200,59g\cdot mo{l}^{1}}=0,50mols$
d) Molar mass of barium is: $137,33g\cdot mo{l}^{1}$
If one mole of barium has a mass of 137,33 g then the number of moles of barium in 50 g must be:
$\frac{50g}{137,33g\cdot mo{l}^{1}}=0,36mols$
e) Molar mass of lead is: $207,2g\cdot mo{l}^{1}$
If one mole of lead has a mass of 207,2 g then the number of moles of lead in 40 g must be:
$\frac{40g}{207,2g\cdot mo{l}^{1}}=0,193mols$
An equation to calculate moles and mass
We can calculate molar mass as follows: $\text{molar mass (M)}={\displaystyle \frac{\text{mass (g)}}{\text{mole (mol)}}}$
This can be rearranged to give the number of moles:
The following diagram may help to remember the relationship between these three variables. You need to imagine that the horizontal line is like a division sign and that the vertical line is like a multiplication sign. So, for example, if you want to calculate M, then the remaining two letters in the triangle are m and n and m is above n with a division sign between them. Your calculation will then be $\text{M}=\frac{\text{m}}{\text{n}}$
Tip:
Remember that when you use the equation $\text{n}=\frac{\text{m}}{\text{M}}$, the mass is always in grams (g) and molar mass is in grams per mol (g·mol^{−1}). Always write the units next to any number you use in a formula or sum.
Example 3: Calculating moles from mass
Question
Calculate the number of moles of copper there are in a sample that with a mass of 127 g.
Answer
Write down the equation
Find the moles
There are 2 moles of copper in the sample.
Example 4: Calculating atoms from mass
Question
Calculate the number of atoms there are in a sample of aluminium that with a mass of 81 g.
Answer
Find the number of moles
Find the number of atoms
Number of atoms in 3 mol aluminium $=3\times 6,022\times {10}^{23}$
There are 1,8069 × 10^{24} aluminium atoms in a sample of 81 g.
Exercise 3: Some simple calculations
Calculate the number of moles in each of the following samples:

5,6 g of calcium

0,02 g of manganese

40 g of aluminium
a) Molar mass of calcium: $40,08\text{g}\cdot {\text{mol}}^{1}$
$n=\frac{m}{M}=\frac{5,6}{40,08}=0,140\text{mol}$
b) Molar mass of manganese: $54,94\text{g}\cdot {\text{mol}}^{1}$
$n=\frac{m}{M}=\frac{0,02}{54,94}=0,00036\text{mol}$
c) Molar mass of aluminium: $26,98\text{g}\cdot {\text{mol}}^{1}$
$n=\frac{m}{M}=\frac{40}{26,98}=1,48\text{mol}$
A lead sinker has a mass of 5 g.

Calculate the number of moles of lead the sinker contains.

How many lead atoms are in the sinker?
a) Molar mass of lead: $207,2\text{g}\cdot {\text{mol}}^{1}$
$n=\frac{m}{M}=\frac{5}{207,2}=0,024\text{mol}$
b) $0,024\times (6,02\times {10}^{23})=1,44\times {10}^{22}$
Calculate the mass of each of the following samples:

2,5 mol magnesium

12 mol lithium

$4,5\times {10}^{25}$ atoms of silicon
a) Molar mass of magnesium: $24,31g\cdot mo{l}^{1}$
$n=\frac{m}{M}$ 
$m=nM$ 
$m=(2,5)(24,31)$ 
$m=60,775g$ 
b) Molar mass of lithium: $6,94g\cdot mo{l}^{1}$
$m=nM$ 
$m=\left(12\right)(6,94)$ 
$m=83,28g$ 
c) Number of mols of silicon
$4,5\times {10}^{25}\xf76,02\times {10}^{23}=74,751mols$
Silicon has a mass of $28g\cdot mo{l}^{1}$.
The mass of the sample is:
$m=n\times M$ 
$m=74,751\times 28$ 
$m=2093g$ 
Compounds
So far, we have only discussed moles, mass and molar mass in relation to elements. But what happens if we are dealing with a compound? Do the same concepts and rules apply? The answer is yes. However, you need to remember that all your calculations will apply to the whole compound. So, when you calculate the molar mass of a covalent compound, you will need to add the molar mass of each atom in that compound. The number of moles will also apply to the whole molecule. For example, if you have one mole of nitric acid (${\text{HNO}}_{3}$) the molar mass is 63,01 g·mol^{−1} and there are 6,022 × 10^{23} molecules of nitric acid. For network structures we have to use the formula mass. This is the mass of all the atoms in one formula unit of the compound. For example, one mole of sodium chloride ($\text{NaCl}$) has a formula mass of 63,01 g·mol^{−1} and there are 6,022 × 10^{23} molecules of sodium chloride in one formula unit.
In a balanced chemical equation, the number that is written in front of the element or compound, shows the mole ratio in which the reactants combine to form a product. If there are no numbers in front of the element symbol, this means the number is '1'.
Khan academy video on the mole
e.g. ${\text{N}}_{2}+3{\text{H}}_{2}\to 2\text{N}{\text{H}}_{3}$
In this reaction, 1 mole of nitrogen molecules reacts with 3 moles of hydrogen molecules to produce 2 moles of ammonia molecules.
Example 5: Calculating molar mass
Question
Calculate the molar mass of ${\text{H}}_{2}{\text{SO}}_{4}$.
Answer
Give the molar mass for each element
Hydrogen = 1,01 g·mol^{−1}
Sulphur = 32,1 g·mol^{−1}
Oxygen = 16,0 g·mol^{−1}
Work out the molar mass of the compound
Example 6: Calculating moles from mass
Question
Calculate the number of moles in 1 kg of ${\text{MgCl}}_{2}$.
Answer
Convert mass into grams
Calculate the molar mass
Find the number of moles
There are 10,5 moles of magnesium chloride in a 1 kg sample.
Group discussion 1: Understanding moles, molecules and Avogadro's number
Divide into groups of three and spend about 20 minutes answering the following questions together:

What are the units of the mole? Hint: Check the definition of the mole.

You have a 46 g sample of nitrogen dioxide (${\text{NO}}_{2}$)

How many moles of ${\text{NO}}_{2}$ are there in the sample?

How many moles of nitrogen atoms are there in the sample?

How many moles of oxygen atoms are there in the sample?

How many molecules of ${\text{NO}}_{2}$ are there in the sample?

What is the difference between a mole and a molecule?


The exact size of Avogadro's number is sometimes difficult to imagine.

Write down Avogadro's number without using scientific notation.

How long would it take to count to Avogadro's number? You can assume that you can count two numbers in each second.

Exercise 4: More advanced calculations
Calculate the molar mass of the following chemical compounds:

$\text{KOH}$

${\text{FeCl}}_{3}$

$\text{Mg}{\left(\text{OH}\right)}_{2}$
a) Molar mass = molar mass K + molar mass O + molar mass H
$39,10+16,00+1,01=56,11g\cdot mo{l}^{1}$
b) Molar mass = molar mass Fe + 3 molar mass Cl
$55,85+3(35,45)=162,2g\cdot mo{l}^{1}$
c) Molar mass = molar mass Mg + 2 molar mass O + 2 molar mass H (note that the OH part is multiplied by 2.)
$24,31+2(16,00)+2(1,01)=58,33g\cdot mo{l}^{1}$
How many moles are present in:

10 g of ${\text{Na}}_{2}{\text{SO}}_{4}$

34 g of $\text{Ca}{\left(\text{OH}\right)}_{2}$

2,45 × 10^{23} molecules of ${\text{CH}}_{4}$?
a) Molar mass of $N{a}_{2}S{O}_{4}$ is:
$2(22,99)+32,06+4(16,00)=142,04g\cdot mo{l}^{1}$
$n=\frac{m}{M}$ 
$n=\frac{10}{142},04$ 
$n=0,07mols$ 
b) Molar mass of $Ca{\left(OH\right)}_{2}$:
$40,08+2(16,00)+2(1,01)=74.1g\cdot mo{l}^{1}$
$n=\frac{m}{M}$ 
$n=\frac{34}{74},1$ 
$n=0,46mols$ 
c) We divide by Avogadro's number to get the number of moles:
$2,45\times {10}^{23}\xf76,02\times {10}^{23}=0,41\text{mols}$
For a sample of 0,2 moles of magnesium bromide (${\text{MgBr}}_{2}$), calculate:

the number of moles of ${\text{Mg}}^{2+}$ ions

the number of moles of ${\text{Br}}^{}$ ions
a) 0,2 mols
b) 0,2 mols
You have a sample containing 3 mol of calcium chloride.

What is the chemical formula of calcium chloride?

How many calcium atoms are in the sample?
a) $CaC{l}_{2}$
b) There are 3 mols of calcium atoms. The number of calcium atoms is:
$3\times (6,023\times {10}^{23})=1,81\times {10}^{24}$
Calculate the mass of:

3 moles of ${\text{NH}}_{4}\text{OH}$

4,2 moles of $\text{Ca}{\left({\text{NO}}_{3}\right)}_{2}$
a) Molar mass of $N{H}_{4}OH$ is:
$14,0+4(1,01)+16,00+1,01=35,05\text{g}\cdot {\text{mol}}^{1}$
Mass is:
$m=nM$ 
$m=\left(3\right)(35,05)$ 
$m=105,18g$ 
b) Molar mass of $Ca{\left(N{O}_{3}\right)}_{2}$ is:
$40,1+2(14,0)+6(16,0)=164,1\text{g}\cdot {\text{mol}}^{1}$
Mass:
$m=nM$ 
$m=(4,2)(164,1)$ 
$m=689,22g$ 