The mole and Avogadro's constant

In a nutshell

The mole is a unit for measuring the amount of a substance. A mole is a certain number of particles in a substance. The number of particles in one mole is known as Avogadro's constant $(6.02 \, \times \, 10^{23})$ . The number of moles is calculated using the mass and molar mass or using the concentration and volume.

Equations

word equation	symbol equation
$number \space of \space moles \, = \, \frac{number \space of \space particles}{Avogadro's \space constant}$	$n \, = \, \frac{number \space of \space particles}{L}$
$number \space of \space moles \, = \, \frac{mass}{molar \space mass}$	$n \, = \, \frac{m}{M}$
$number \space of \space moles \, = \, concentration \, \times \, volume$	$n \, = \, c \, \times \, V$
$mass \, = \, concentration \, \times \, volume$	$m \, = \, c \, \times \, V$

Constants

Constant	Constant value
$Avogadro's \space constant$	$6.02 \, \times \, 10^{23}$

Variable units

quantity name	symbol	unit name	unit
$number \space of \space moles$	$n$	$mole$	$mol$
$Avogadro's \space constant$	$L$	$no \space unit$	$no \space unit$
$mass$	$m$	$gram$	$g$
$molar \space mass$	$M$	$gram \space per \space mole$	$g \, mol^{-1}$
$concentration$	$c$	$mole \space per \space cubic \space decimeter \space or \newline gram \space per \space cubic \space decimeter$	$mol \, dm^{-3} \space or \newline g \, dm^{-3}$
$volume$	$V$	$cubic \space decimeter$	$dm^3$

The mole and Avogadro's constant

The mole is a unit that is used to measure the amount of a substance. A mole is a certain number of particles in a substance.

Avogadro's constant ( $6.02 \, \times \, 10^{23}$ ) is the value for the number of particles in one mole of a substance. The equation to work out the number of moles given the number of particles and Avogadro's constant is:

$number \space of \space moles \, = \, \frac{number \space of \space particles}{Avogadro's \space constant} \\ \ \\ n \, = \, \frac{number \space of \space particles}{L}$

This equation can be rearranged to find the number of particles given the number of moles.

$number \space of \space particles \, = \, number \space of \space moles \, \times \, Avogadro's \space constant \\ \ \\ number \space of \space particles \, = \, n \, \times \, L$

Example

How many moles is there in $3.75 \, \times \, 10^{18}$ lithium atoms?

First, write down the equation you will use:

$n \, = \, \frac{number \space of \space atoms}{L}$

Write the value of Avogadro's constant:

$6.02 \, \times \, 10^{23}$

Substitute the values into the equation:

$n \, = \, \frac{3.75 \, \times \, 10^{18}}{6.02 \, \times \, 10^{23}} \, = \, 6.23 \, \times \, 10^{-6} \,mol$ 

There is $\underline{6.23 \, \times \, 10^{-6} \,moles}$ in $3.75 \, \times \, 10^{18}$ lithium atoms.

Moles

Molar mass is the mass per mole of a substance. The molar mass can be used alongside the mass to calculate the number of moles of a substance:

$number \space of \space moles \, = \, \frac{mass}{molar \space mass} \newline n \, = \, \frac{m}{M}$

This equation can be rearranged to find either the mass or molar mass of a substance using for formula triangle.

Chemistry; Formulae, equations and amounts of substances; KS5 Year 12; The mole and Avogadro's constant

Rearrangement using the formula triangle:

$mass \, = number \space of \space moles \, \times \, molar \space mass \\ \ \\ m \, = \, n \, \times \, M \\ \ \\ molar \space mass \, = \, \frac{mass}{number \space of \space moles} \\ \ \\ M \, = \, \frac{m}{n}$

Example

How many moles are there in $4.12 \, g$ of $Al_2(CO_3)_3$ ?

First, write down the equation you will use:

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

Next, work out the molar mass of $Al_2(CO_3)_3$ :

$M \, (Al_2(CO_3)_3) \, = \, (27 \, \times \, 2)\, + \,(12 \, \times \, 3) \, + \, (16 \, \times \, 9)\, = 234 \, g \, mol^{-1}$

Substitute the value for the mass and the molar mass into the equation to find the number of moles:

$n \, = \, \frac{4.12}{234} \, = 0.018 \, mol$

There is $\underline{0.018 \, moles}$ in $4.12 \, g$ of $Al_2(CO_3)_3$ .

Finding the number of atoms

Both the equations to find the number of moles and the number of particles can be linked together to find the number of atoms of a substance.

procedure

1.

Work out the molar mass of the given compound.

2.

Use the molar mass with the given mass to calculate the number of moles using the equation below:

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

3.

Using the number of moles and Avogadro's constant, find the number of atoms.

$number \space of \space atoms \, = \, n \, \times \, L$

Concentration

The number of moles can also be calculated using the concentration of a substance along with the volume.

$number \space of \space moles \, = \, concentration \, \times \, volume \newline n \, = \, c \, \times \, V$

Note: Using the number of moles, the mass can be calculated using the molar mass ( $M$ ).

This equation can be rearranged to find either the concentration or volume of a substance using the formula triangle.

Chemistry; Formulae, equations and amounts of substances; KS5 Year 12; The mole and Avogadro's constant

Rearrangement using the formula triangle:

$concentration \, = \, \frac{number \space of \space moles}{volume} \\ \ \\ c \, = \, \frac{n}{V} \\ \ \\ volume \, = \, \frac{number \space of \space moles}{concentration} \\ \ \\ V \, = \, \frac{n}{c}$

Example

What is the mass of $150 \, cm^3$ of $HCl$ with a concentration of $0.5 \, mol \, dm^{-3}$ ?

First, write down your quantities in the correct units:

$V \, = \, 150 \, cm^3 \, = \, \frac{150 \, cm^3}{1000} \, = \, 0.15 \, dm^3$

$c \, = \, 0.5 \, mol \, dm^{-3}$ 

Next, write down the equation you will use to find the number of moles:

$n \, = \, c \, \times \, V$

Substitute your values into the equation to find the number of moles:

$n \, = \, 0.15 \, \times \, 0.5 \, = \, 0.075 \, mol$ 

Write down the equation you will use to find the mass of $HCl$ :

$m \, = \, n \, \times \, M$ 

Work out the molar mass ( $M_r$ ) of $HCl$ :

$M \,(HCl) \, = \, 1 \, + \, 35.5 \, = \, 36.5 \, g \, mol^{-1}$ 

Substitute your values into the equation to find the mass:

$m \, = 0.075 \, \times \, 36.5 \,= \, 2.74 \, g$ 

The mass of $150 \, cm^3$ $HCl$ ( $0.5 \, mol \, dm^{-3}$ ) is $\underline{2.74 \, g}$ .

Concentration in grams

The concentration can also be measured using the mass.

$mass \, = \, concentration \, \times \, volume \newline m \, = \, c \, \times \, V$

As with the equation involving the number of moles, this equation can be rearranged to find either the concentration or volume of a substance.

Rearrangement:

$concentration \, = \, \frac{mass}{volume} \\ \ \\ c \, = \, \frac{m}{V} \\ \ \\ volume \, = \, \frac{mass}{concentration} \\ \ \\ V \, = \, \frac{m}{c}$

Note: Instead of concentration having the unit $mol \, dm^{-3}$ , when mass is used to calculate concentration the unit is $g \, dm^{-3}.$

Example

What is the mass of $5000 \, cm^3$  of sodium hydroxide with a concentration of $1.25 \, g \, dm^{-3}$ ?

First, write down your quantities in the correct units:

$V \, = \, 5000 \, cm^3 \, = \, \frac{5000 \, cm^3}{1000} \, = 5 \, dm^3$

$c \, = \, 1.25 \, g \, dm^{-3}$

Next, write down the equation you will use to find the mass of sodium hydroxide:

$m \, = \, c \, \times \, V$

Substitute your values into the equation to find the mass:

$m \, = \, 5 \, \times \, 1.25 \, = \, 6.25 \ g$

There is a mass of $\underline{6.25 \, g}$ sodium hydroxide ( $1.25 \, g \, dm^{-3}$ ) in a volume of $5000 \, cm^3$ .