Chapter 19: Quantitative aspects of chemical change

19.1 Atomic mass and the mole (ESAFW)

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:

\[\text{Fe} + \text{S} \rightarrow \text{FeS}\]

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 (ESAFX)

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.

Mole ($n$): 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 $\text{602 204 500 000 000 000 000 000}$ or $\text{6,022} \times \text{10}^{\text{23}}$ particles. This is a very big number! We call this number Avogadro's number.

Avogadro's number ($N_{\text{A}}$): The number of particles in a mole, equal to $\text{6,022} \times \text{10}^{\text{23}} \text{ mol}^{-1}$.

If we had this number of cold drink cans, then we could cover the surface of the earth to a depth of over $\text{300}$ $\text{km}$! 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!

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 $\text{12,0}$ $\text{g}$ of carbon we have one mole or $\text{6,022} \times \text{10}^{\text{23}}$ carbon atoms. $\text{63,5}$ $\text{g}$ of copper is one mole of copper or $\text{6,022} \times \text{10}^{\text{23}}$ 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.

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Moles and mass

Textbook Exercise 19.1

How many atoms are there in:

1 mole of a substance
2 moles of calcium
5 moles of phosphorus
$\text{24,3}$ $\text{g}$ of magnesium
$\text{24,0}$ $\text{g}$ of carbon

Solution not yet available

Complete the following table:

Element	Relative atomic mass (u)	Sample mass (g)	Number of moles in the sample
Hydrogen	$\text{1,01}$	$\text{1,01}$
Magnesium	$\text{24,3}$	$\text{24,3}$
Carbon	$\text{12,0}$	$\text{24,0}$
Chlorine	$\text{35,45}$	$\text{70,9}$
Nitrogen	$\text{14,0}$	$\text{42,0}$

Solution not yet available

Molar mass (ESAFY)

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 $\text{g·mol$^{-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.

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.

Element	Relative atomic mass (u)	Molar mass ($\text{g·mol$^{-1}$}$)	Mass of one mole of the element (g)
Magnesium	$\text{24,3}$	$\text{24,3}$	$\text{24,3}$
Lithium	$\text{6,94}$	$\text{6,94}$	$\text{6,94}$
Oxygen	$\text{16,0}$	$\text{16,0}$	$\text{16,0}$
Nitrogen	$\text{14,0}$	$\text{14,0}$	$\text{14,0}$
Iron	$\text{55,8}$	$\text{55,8}$	$\text{55,8}$

Table 19.1: The relationship between relative atomic mass, molar mass and the mass of one mole for a number of elements.

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Worked example 1: Calculating the number of moles from mass

Calculate the number of moles of iron (Fe) in an $\text{111,7}$ $\text{g}$ sample.

Find the molar mass of iron

If we look at the periodic table, we see that the molar mass of iron is $\text{55,8}$ $\text{g·mol$^{-1}$}$. This means that 1 mole of iron will have a mass of $\text{55,8}$ $\text{g}$.

Find the mass of iron

If 1 mole of iron has a mass of $\text{55,8}$ $\text{g}$, then: the number of moles of iron in $\text{111,7}$ $\text{g}$ must be:

\begin{align*} n & = \frac{\text{111,7}\text{ g}}{\text{55,8}\text{ g·mol$^{-1}$}} \\ & = \frac{\text{111,7}\text{ g·mol}}{\text{55,8}\text{ g}} \\ & = \text{2,0017...}\text{ mol} \\ & \approx \text{2}\text{ mol} \end{align*}

There are 2 moles of iron in the sample.

Worked example 2: Calculating mass from moles

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?

Find the molar mass of zinc

Molar mass of zinc is $\text{65,4}$ $\text{g·mol$^{-1}$}$, meaning that 1 mole of zinc has a mass of $\text{65,4}$ $\text{g}$.

Find the mass

If 1 mole of zinc has a mass of $\text{65,4}$ $\text{g}$, then 5 moles of zinc has a mass of: $\text{65,4}\text{ g} \times \text{5}\text{ mol} = \text{327}\text{ g}$ (answer to a)

Find the number of atoms

$\text{5}\text{ mol} \times \text{6,022} \times \text{10}^{\text{23}}\text{ atoms·mol$^{-1}$} = \text{3,011} \times \text{10}^{\text{24}}\text{ atoms}$ (answer to b)

Moles and molar mass

Textbook Exercise 19.2

Give the molar mass of each of the following elements:

hydrogen gas
nitrogen gas
bromine gas

Solution not yet available

Calculate the number of moles in each of the following samples:

$\text{21,6}$ $\text{g}$ of boron ($\text{B}$)
$\text{54,9}$ $\text{g}$ of manganese ($\text{Mn}$)
$\text{100,3}$ $\text{g}$ of mercury ($\text{Hg}$)
$\text{50}$ $\text{g}$ of barium ($\text{Ba}$)
$\text{40}$ $\text{g}$ of lead ($\text{Pb}$)

Solution not yet available

An equation to calculate moles and mass (ESAFZ)

We can calculate molar mass as follows: $\text{molar mass } (M) = \frac{\text{ mass (g)}}{\text{mole (mol)}}$

This can be rearranged to give the number of moles:

\[n = \frac{m}{M}\]

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 $M = \frac{m}{n}$

Remember that when you use the equation $n=\frac{m}{M}$, the mass is always in grams (g) and molar mass is in grams per mol ($\text{g·mol$^{-1}$}$). Always write the units next to any number you use in a formula or sum.

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Worked example 3: Calculating moles from mass

Calculate the number of moles of copper there are in a sample that with a mass of $\text{127}$ $\text{g}$.

Write down the equation

\[n = \frac{m}{M}\]

Find the moles

\begin{align*} n & = \frac{\text{127}\text{ g}}{\text{63,5}\text{ g·mol$^{-1}$}} \\ & = \text{2}\text{ mol} \end{align*}

There are 2 moles of copper in the sample.

Worked example 4: Calculating atoms from mass

Calculate the number of atoms there are in a sample of aluminium that with a mass of $\text{81}$ $\text{g}$.

Find the number of moles

\begin{align*} n & = \frac{m}{M} \\ & = \frac{\text{81}\text{ g}}{\text{27,0}\text{ g·mol$^{-1}$}} \\ & = \text{3}\text{ mol} \end{align*}

Find the number of atoms

Number of atoms in $\text{3}$ $\text{mol}$ aluminium $= 3 \times \text{6,022} \times \text{10}^{\text{23}}$

There are $\text{1,8066} \times \text{10}^{\text{24}}$ aluminium atoms in a sample of $\text{81}$ $\text{g}$.

Some simple calculations

Textbook Exercise 19.3

Calculate the number of moles in each of the following samples:

$\text{5,6}$ $\text{g}$ of calcium
$\text{0,02}$ $\text{g}$ of manganese
$\text{40}$ $\text{g}$ of aluminium

Solution not yet available

A lead sinker has a mass of $\text{5}$ $\text{g}$.

Calculate the number of moles of lead the sinker contains.
How many lead atoms are in the sinker?

Solution not yet available

Calculate the mass of each of the following samples:

$\text{2,5}$ $\text{mol}$ magnesium
$\text{12}$ $\text{mol}$ lithium
$\text{4,5} \times \text{10}^{\text{25}}$ atoms of silicon

Solution not yet available

Compounds (ESAGA)

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 $\text{63,01}$ $\text{g·mol$^{-1}$}$ and there are $\text{6,022} \times \text{10}^{\text{23}}$ 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 $\text{63,01}$ $\text{g·mol$^{-1}$}$ and there are $\text{6,022} \times \text{10}^{\text{23}}$ 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'.

Video: VPezc

e.g. $\text{N}_{2} + 3\text{H}_{2} \rightarrow 2\text{NH}_{3}$

In this reaction, 1 mole of nitrogen molecules reacts with 3 moles of hydrogen molecules to produce 2 moles of ammonia molecules.

Worked example 5: Calculating molar mass

Calculate the molar mass of $\text{H}_{2}\text{SO}_{4}$.

Give the molar mass for each element

Hydrogen = $\text{1,01}$ $\text{g·mol$^{-1}$}$

Sulphur = $\text{32,1}$ $\text{g·mol$^{-1}$}$

Oxygen = $\text{16,0}$ $\text{g·mol$^{-1}$}$

Work out the molar mass of the compound

\begin{align*} M_{\text{H}_{2}\text{SO}_{4}} &= 2(\text{1,01}\text{ g·mol$^{-1}$}) + (\text{32,1}\text{ g·mol$^{-1}$}) + 4(\text{16,0}\text{ g·mol$^{-1}$}) \\ & = \text{98,12}\text{ g·mol$^{-1}$} \end{align*}

Worked example 6: Calculating moles from mass

Calculate the number of moles in $\text{1}$ $\text{kg}$ of $\text{MgCl}_{2}$.

Convert mass into grams

\[m = \text{1}\text{ kg} \times \frac{\text{1 000}\text{ g}}{\text{1}\text{ kg}} = \text{1 000}\text{ g}\]

Calculate the molar mass

\[M_{\text{MgCl}_{2}} = \text{24,3}\text{ g·mol$^{-1}$} + 2(\text{35,45}\text{ g·mol$^{-1}$}) = \text{95,2}\text{ g·mol$^{-1}$}\]

Find the number of moles

\begin{align*} n & = \frac{\text{1 000}\text{ g}}{\text{95,2}\text{ g·mol$^{-1}$}} \\ & = \text{10,5042...}\text{ mol} \\ & \approx \text{10,5}\text{ mol} \end{align*}

There are $\text{10,5}$ $\text{moles}$ of magnesium chloride in a $\text{1}$ $\text{kg}$ sample.

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 $\text{46}$ $\text{g}$ sample of nitrogen dioxide ($\text{NO}_{2}$)
1. How many moles of $\text{NO}_{2}$ are there in the sample?
2. How many moles of nitrogen atoms are there in the sample?
3. How many moles of oxygen atoms are there in the sample?
4. How many molecules of $\text{NO}_{2}$ are there in the sample?
5. What is the difference between a mole and a molecule?
The exact size of Avogadro's number is sometimes difficult to imagine.
1. Write down Avogadro's number without using scientific notation.
2. How long would it take to count to Avogadro's number? You can assume that you can count two numbers in each second.

More advanced calculations

Textbook Exercise 19.4

Calculate the molar mass of the following chemical compounds:

$\text{KOH}$
$\text{FeCl}_{3}$
$\text{Mg}(\text{OH})_{2}$

Solution not yet available

How many moles are present in:

$\text{10}$ $\text{g}$ of $\text{Na}_{2}\text{SO}_{4}$
$\text{34}$ $\text{g}$ of $\text{Ca}(\text{OH})_{2}$
$\text{2,45} \times \text{10}^{\text{23}}$ molecules of $\text{CH}_{4}$?

Solution not yet available

For a sample of $\text{0,2}$ $\text{moles}$ of magnesium bromide ($\text{MgBr}_{2}$), calculate:

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

Solution not yet available

You have a sample containing $\text{3}$ $\text{mol}$ of calcium chloride.

What is the chemical formula of calcium chloride?
How many calcium atoms are in the sample?

Solution not yet available

Calculate the mass of:

$\text{3}$ $\text{moles}$ of $\text{NH}_{4}\text{OH}$
$\text{4,2}$ $\text{moles}$ of $\text{Ca}(\text{NO}_{3})_{2}$

Solution not yet available

End of chapter exercises

Table of Contents

19.2 Composition

Element	Relative atomic mass (u)	Sample mass (g)	Number of moles in the sample
Hydrogen	\(\text{1,01}\)	\(\text{1,01}\)
Magnesium	\(\text{24,3}\)	\(\text{24,3}\)
Carbon	\(\text{12,0}\)	\(\text{24,0}\)
Chlorine	\(\text{35,45}\)	\(\text{70,9}\)
Nitrogen	\(\text{14,0}\)	\(\text{42,0}\)

Element	Relative atomic mass (u)	Molar mass (\(\text{g·mol$^{-1}$}\))	Mass of one mole of the element (g)
Magnesium	\(\text{24,3}\)	\(\text{24,3}\)	\(\text{24,3}\)
Lithium	\(\text{6,94}\)	\(\text{6,94}\)	\(\text{6,94}\)
Oxygen	\(\text{16,0}\)	\(\text{16,0}\)	\(\text{16,0}\)
Nitrogen	\(\text{14,0}\)	\(\text{14,0}\)	\(\text{14,0}\)
Iron	\(\text{55,8}\)	\(\text{55,8}\)	\(\text{55,8}\)