MYP Chemistry Topic 5.4 the Mole Concept and Chemical…

The Mole – Counting the Uncountable

Atoms and molecules are tiny – far too small to count individually.
Chemists needed a way to:
1. talk about numbers of particles
2. but work with measurable amounts like grams and litres.
That’s why we use the mole.

What is a mole?

Definition

Mole

The amount of substance that contains $N_A=6.022×10^{23}$ particles (atoms, molecules, ions, etc.).

That big number in the definition of the mole is called Avogadro’s constant.

Analogy

Think of a mole as a chemist’s dozen:

A dozen = 12 of something
A mole = $6.022×10^{23}$ of something

Example

1 mol of carbon atoms = $6.022×10^{23}$ $C$ atoms
1 mol of water molecules = $6.022×10^{23}$ $H_2O$ molecules
This is how we connect the microscopic world (particles) to the macroscopic world (mass, volume).

Molar Mass – Linking Moles to Mass

Using the Periodic Table to Find Ar and Mr

To use the mole in calculations, you start with the periodic table.
On a typical periodic table each element has:
1. Atomic number (Z) – number of protons
2. Relative atomic mass (Aᵣ) – the average mass of its atoms, compared to carbon-12
We use Aᵣ values to calculate relative molecular mass (Mᵣ) and molar mass.

Example

Step 1 – Pick up Aᵣ from the periodic table

The most frequently used values:

H ≈ 1
C ≈ 12
O ≈ 16
Na ≈ 23
Cl ≈ 35.5
Ca ≈ 40
N ≈ 14

These numbers have no units when used as Aᵣ, but the molar mass has units of g mol⁻¹ and is numerically equal to Mᵣ.

Step 2 – Use Aᵣ to find Mᵣ (relative molecular / formula mass)

For molecules, add up the Aᵣ of all atoms in the formula
For ionic compounds, do the same (it’s often called “relative formula mass” instead)

Water – H₂O

$$M_r\left(\mathrm{H}_2 \mathrm{O}\right)=2 \times A_r(\mathrm{H})+A_r(\mathrm{O})=2 \times 1+16=18$$

So molar mass = 18 g mol⁻¹

Sodium chloride – NaCl

$$M_r(\mathrm{NaCl})=A_r(\mathrm{Na})+A_r(\mathrm{Cl})=23+35.5=58.5$$

Molar mass = 58.5 g mol⁻¹

Calcium hydroxide – Ca(OH)₂

Be careful with brackets – (OH)₂ means two OH groups:

$$M_r\left(\mathrm{Ca}(\mathrm{OH})_2\right)=A_r(\mathrm{Ca})+2 \times\left[A_r(\mathrm{O})+A_r(\mathrm{H})\right]=40+2 \times(16+1)=$$ $$=40+2 \times 17=40+34=74$$

Molar mass = 74 g mol⁻¹

Diatomic gases
Some elements exist as diatomic molecules in nature: H₂, N₂, O₂, F₂, Cl₂, Br₂, I₂.

Example for O₂: $$M_r\left(\mathrm{O}_2\right)=2 \times 16=32$$

Tip

When you see O₂, H₂, Cl₂ in an equation, always use the molecular Mr, not the atomic Ar.

Molar Mass

The molar mass of a substance is the mass of 1 mole of that substance.
Unit: g mol⁻¹
Numerically equal to the relative atomic/molecular mass on the periodic table.

Example

Carbon (C): Ar ≈ 12.0 → molar mass = 12.0 g mol⁻¹
Water (H₂O): $$M_r=2 \times 1.0+16.0=18.0 \Rightarrow \text { molar mass }=18.0 \mathrm{~g} \mathrm{~mol}^{-1}$$

Key formula

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

Where:

$n$ = moles (mol)
$m$ = mass (g)
$M$ = molar mass (g mol⁻¹)

Example question

How many moles are in 36.0 g of water?

Solution

$$M\left(\mathrm{H}_2 \mathrm{O}\right)=18.0 \mathrm{~g} \mathrm{~mol}^{-1} \Rightarrow n=\frac{36.0}{18.0}=2.0 \mathrm{~mol}$$

Example question

What mass is 2.0 mol of H₂O?

Solution

$$m=n M=2.0 \times 18.0=36.0 \mathrm{~g}$$

Particles and Moles

Once you know the moles, you can find the number of particles using Avogadro’s constant: $$N=n \times N_A$$

where:

$N$ = number of particles
$n$ = moles
$N_A=6.022 \times 10^{23} \mathrm{~mol}^{-1}$

Example

1.0 mol of carbon atoms contains: $$N=1.0 \times 6.022 \times 10^{23}=6.022 \times 10^{23} \text { atoms }$$

1.5 mol of carbon atoms: $$N=1.5 \times 6.022 \times 10^{23} \approx 9.03 \times 10^{23} \text { atoms }$$

Moles and Gases – Avogadro’s Law

Avogadro’s law:

Equal volumes of gases at the same temperature and pressure contain the same number of particles.

This leads to the idea of molar volume.

At STP (standard temperature and pressure): $$1 \mathrm{~mol} \text { of any gas } \approx 22.4 \mathrm{dm}^3$$

Gas volume formula

$$V=n×V_m$$

where:

$V$ = gas volume (dm³)
$n$ = moles of gas (mol)
$V_m$= molar volume (e.g. 22.4 dm³ mol⁻¹ at STP)

Example question

What volume is 0.50 mol of O₂ at STP?

Solution

$$V=0.50 \times 22.4=11.2 \ \mathrm{dm}^3$$

Example question

What volume is 64 g of O₂ at STP?

Solution

$$M\left(\mathrm{O}_2\right)=32.0 \mathrm{~g} \mathrm{~mol}^{-1} \Rightarrow n=\frac{64}{32}=2.0 \mathrm{~mol}$$

$$V=2.0 \times 22.4=44.8 \ \mathrm{dm}^3$$

Stoichiometry: Turning Equations into Numbers

Once an equation is balanced, its coefficients give you ratios of moles.
With the mole and molar mass, you can turn those ratios into masses, gas volumes, or numbers of particles.

Example question

Magnesium reacts with hydrochloric acid: $$\mathrm{Mg}+2 \mathrm{HCl} \rightarrow \mathrm{MgCl}_2+\mathrm{H}_2$$

What mass of magnesium chloride is formed from 4.8 g of Mg, assuming excess acid?

Solution

Step 1 - Moles of $Mg$

$$
M_r(\mathrm{Mg})=24 \quad \Rightarrow \quad n(\mathrm{Mg})=\frac{4.8}{24}=0.20 \mathrm{~mol}
$$

Step 2 - Use mole ratio
Equation shows:

$$
1 \mathrm{Mg} \rightarrow 1 \mathrm{MgCl}_2
$$

So:

$$
n\left(\mathrm{MgCl}_2\right)=0.20 \mathrm{~mol}
$$

Step 3 - Convert moles of $\mathrm{MgCl}_2$ to mass

$$
\begin{gathered}
M_r\left(\mathrm{MgCl}_2\right)=24+2 \times 35.5=95 \\
m\left(\mathrm{MgCl}_2\right)=n M=0.20 \times 95=19 \mathrm{~g}
\end{gathered}
$$

Answer: $\mathbf{1 9 ~ g}$ of $\mathrm{MgCl}_2$

Example question

Calcium carbonate decomposes as follows:

$$
\mathrm{CaCO}_3 \rightarrow \mathrm{CaO}+\mathrm{CO}_2
$$

What volume of $\mathrm{CO}_2$ (at STP, $22.4 \mathrm{dm}^3 \mathrm{~mol}^{-1}$ ) is produced from 10.0 g of $\mathrm{CaCO}_3$ ?

Solution

Step 1 - Moles of $\mathrm{CaCO}_3$

$$
\begin{gathered}
M_r\left(\mathrm{CaCO}_3\right)=40+12+3 \times 16=100 \\
n\left(\mathrm{CaCO}_3\right)=\frac{10.0}{100}=0.10 \mathrm{~mol}
\end{gathered}
$$

Step 2 - Use mole ratio

$$
1 \mathrm{CaCO}_3 \rightarrow 1 \mathrm{CO}_2
$$

So:

$$
n\left(\mathrm{CO}_2\right)=0.10 \mathrm{~mol}
$$

Step 3 - Moles to volume

$$
V\left(\mathrm{CO}_2\right)=n V_m=0.10 \times 22.4=2.24 \ \mathrm{dm}^3
$$

Answer: $2.24 \ \mathrm{dm}^3$ of $\mathrm{CO}_2$ at STP

Using Moles in Chemical Equations (Stoichiometry)

Balanced equations tell you mole ratios.
Those coefficients are literally “moles of each substance”.

Example

Haber process:

$$\mathrm{N}_2+3 \mathrm{H}_2 \rightarrow 2 \mathrm{NH}_3$$

This means:

1 mol N₂ reacts with 3 mol H₂ to form 2 mol NH₃.

General stoichiometry steps

Write and balance the equation.
Convert the known quantity → moles.
Use the mole ratio from the equation to get moles of the unknown.
Convert moles of the unknown → mass, gas volume, or number of particles.

Example question

How many moles of H₂ are needed to make 2.50 mol of NH₃?

Solution

From the equation:

$$
\mathrm{N}_2+3 \mathrm{H}_2 \rightarrow 2 \mathrm{NH}_3
$$

Mole ratio $\mathrm{H}_2: \mathrm{NH}_3=3: 2$

$$
n\left(\mathrm{H}_2\right)=2.50 \times \frac{3}{2}=3.75 \mathrm{~mol}
$$

Example question

How much ammonia from 4.0 mol of H₂?

Solution

Same equation:

$$
\mathrm{N}_2+3 \mathrm{H}_2 \rightarrow 2 \mathrm{NH}_3
$$

Mole ratio $\mathrm{NH}_3: \mathrm{H}_2=2: 3$

$$
n\left(\mathrm{NH}_3\right)=4.0 \times \frac{2}{3} \approx 2.67 \mathrm{~mol}
$$

(Assuming excess nitrogen is present)

Limiting Reactants – When One Reactant Runs Out

In real reactions, reactants are often not in perfect ratio.
The one that runs out first is the limiting reactant – it limits how much product you can make.
The other reactant(s) are in excess.

Definition

Limiting reactant

Substance in a chemical reaction that gets completely used up first, thereby "limiting" or stopping the reaction and determining the maximum amount of product that can be formed, with any leftover reactants being in excess.

Example

Limiting reactant in Haber

$$\mathrm{N}_2+3 \mathrm{H}_2 \rightarrow 2 \mathrm{NH}_3$$

Suppose you mix:

1.0 mol N₂
4.0 mol H₂

From the equation:

1.0 mol N₂ needs 3.0 mol H₂ → you actually have 4.0 mol.
So N₂ is limiting, H₂ is in excess.

Product moles (from limiting reagent):

1.0 mol N₂ → 2.0 mol NH₃

The extra 1.0 mol H₂ is left over.

Example

Limiting reactant in water formation

$$2 \mathrm{H}_2+\mathrm{O}_2 \rightarrow 2 \mathrm{H}_2 \mathrm{O}$$

If you have:

5.0 mol H₂
2.0 mol O₂

Required ratio H₂ : O₂ = 2 : 1

For 2.0 mol O₂ you need $2×2.0=4.0$ mol H₂.

You have 5.0 mol H₂ → H₂ is in excess
O₂ is limiting

Product:

From 2.0 mol O₂ → 4.0 mol H₂O

Theoretical Yield and Percentage Yield

Definition

Theoretical yield

The maximum possible amount of product, calculated from the limiting reactant (assuming the reaction goes to completion with no losses).

Definition

Actual yield

Measured mass or amount of product actually obtained from a chemical reaction in an experiment. This is usually less than the calculated maximum (theoretical yield) due to incomplete reactions or material loss during processes like filtering or transferring

Definition

Percentage yield

Ratio of the actual yield (product obtained in an experiment) to the theoretical yield (maximum possible product from stoichiometry), expressed as a percentage, showing reaction efficiency.

$$\% \text { yield }=\frac{\text { actual yield }}{\text { theoretical yield }} \times 100$$

Low yield might be caused by:

incomplete reaction
side reactions
loss of product during separation/purification.

Example

If the theoretical yield of $\mathrm{NH}_3$ is 2.0 mol but the actual yield is 1.8 mol:

$$
\% \text { yield }=\frac{1.8}{2.0} \times 100=90 \%
$$

Why Do Mole Calculations Matter in the Real World?

Mole calculations aren’t just exam exercises – they are core to real chemistry.

Efficiency and Cost

Using correct mole ratios avoids wasting expensive reactants.
Industry always wants maximum yield from minimum input.

Example

In the Haber process, conditions and catalyst are chosen to get a good compromise between:

high yield (enough NH₃ formed)
reasonable rate (not too slow)
acceptable cost (pressure, temperature, catalyst).

All of that is designed using mole-based calculations.

Safety and Waste Reduction (Green Chemistry)

Knowing how many moles of gas will be produced helps design safe equipment (so it doesn’t over-pressurise).
Mole calculations help reduce:
1. toxic by-products
2. waste chemicals
3. energy use (e.g. not heating 10× more than needed).

Note

This supports green chemistry goals: less waste, less energy, fewer harmful emissions.

Correct Doses in Medicine

Drug doses are based on moles of active ingredient, not just mass.
Too little → ineffective.
Too much → toxic.

Example

Pharmacists use moles to ensure safe and effective concentrations in tablets, injections and drips.

Active recall

What is Avogadro’s constant, and what does it represent?
How would you calculate the mass of 0.50 mol of NaCl if its molar mass is 58.5 g mol⁻¹?
How many moles of H₂ are required to react completely with 4.0 mol of N₂ in the Haber reaction?
In the reaction $$2 \mathrm{H}_2+\mathrm{O}_2 \rightarrow 2 \mathrm{H}_2 \mathrm{O}$$ you have 3.0 mol H₂ and 2.0 mol O₂.
Which reactant is limiting?
How many moles of H₂O can form?
Why is it important for industry to know which reactant is limiting and what the theoretical yield is?

The Mole – Counting the Uncountable

Atoms and molecules are tiny – far too small to count individually.
Chemists needed a way to:
1. talk about numbers of particles
2. but work with measurable amounts like grams and litres.
That’s why we use the mole.

What is a mole?

Definition

Mole

The amount of substance that contains $N_A=6.022×10^{23}$ particles (atoms, molecules, ions, etc.).

That big number in the definition of the mole is called Avogadro’s constant.

Analogy

Think of a mole as a chemist’s dozen:

A dozen = 12 of something
A mole = $6.022×10^{23}$ of something

Example

1 mol of carbon atoms = $6.022×10^{23}$ $C$ atoms
1 mol of water molecules = $6.022×10^{23}$ $H_2O$ molecules
This is how we connect the microscopic world (particles) to the macroscopic world (mass, volume).

Molar Mass – Linking Moles to Mass

Using the Periodic Table to Find Ar and Mr

To use the mole in calculations, you start with the periodic table.
On a typical periodic table each element has:
1. Atomic number (Z) – number of protons
2. Relative atomic mass (Aᵣ) – the average mass of its atoms, compared to carbon-12
We use Aᵣ values to calculate relative molecular mass (Mᵣ) and molar mass.

Example

Step 1 – Pick up Aᵣ from the periodic table

The most frequently used values:

H ≈ 1
C ≈ 12
O ≈ 16
Na ≈ 23
Cl ≈ 35.5
Ca ≈ 40
N ≈ 14

These numbers have no units when used as Aᵣ, but the molar mass has units of g mol⁻¹ and is numerically equal to Mᵣ.

Step 2 – Use Aᵣ to find Mᵣ (relative molecular / formula mass)

For molecules, add up the Aᵣ of all atoms in the formula
For ionic compounds, do the same (it’s often called “relative formula mass” instead)

Water – H₂O

$$M_r\left(\mathrm{H}_2 \mathrm{O}\right)=2 \times A_r(\mathrm{H})+A_r(\mathrm{O})=2 \times 1+16=18$$

So molar mass = 18 g mol⁻¹

Sodium chloride – NaCl

$$M_r(\mathrm{NaCl})=A_r(\mathrm{Na})+A_r(\mathrm{Cl})=23+35.5=58.5$$

Molar mass = 58.5 g mol⁻¹

Calcium hydroxide – Ca(OH)₂

Be careful with brackets – (OH)₂ means two OH groups:

$$M_r\left(\mathrm{Ca}(\mathrm{OH})_2\right)=A_r(\mathrm{Ca})+2 \times\left[A_r(\mathrm{O})+A_r(\mathrm{H})\right]=40+2 \times(16+1)=$$ $$=40+2 \times 17=40+34=74$$

Molar mass = 74 g mol⁻¹

Diatomic gases
Some elements exist as diatomic molecules in nature: H₂, N₂, O₂, F₂, Cl₂, Br₂, I₂.

Example for O₂: $$M_r\left(\mathrm{O}_2\right)=2 \times 16=32$$

Tip

When you see O₂, H₂, Cl₂ in an equation, always use the molecular Mr, not the atomic Ar.

Molar Mass

The molar mass of a substance is the mass of 1 mole of that substance.
Unit: g mol⁻¹
Numerically equal to the relative atomic/molecular mass on the periodic table.

Example

Carbon (C): Ar ≈ 12.0 → molar mass = 12.0 g mol⁻¹
Water (H₂O): $$M_r=2 \times 1.0+16.0=18.0 \Rightarrow \text { molar mass }=18.0 \mathrm{~g} \mathrm{~mol}^{-1}$$

Key formula

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

Where:

$n$ = moles (mol)
$m$ = mass (g)
$M$ = molar mass (g mol⁻¹)

Example question

How many moles are in 36.0 g of water?

Solution

$$M\left(\mathrm{H}_2 \mathrm{O}\right)=18.0 \mathrm{~g} \mathrm{~mol}^{-1} \Rightarrow n=\frac{36.0}{18.0}=2.0 \mathrm{~mol}$$

Example question

What mass is 2.0 mol of H₂O?

Solution

$$m=n M=2.0 \times 18.0=36.0 \mathrm{~g}$$

Particles and Moles

Once you know the moles, you can find the number of particles using Avogadro’s constant: $$N=n \times N_A$$

where:

$N$ = number of particles
$n$ = moles
$N_A=6.022 \times 10^{23} \mathrm{~mol}^{-1}$

Example

1.0 mol of carbon atoms contains: $$N=1.0 \times 6.022 \times 10^{23}=6.022 \times 10^{23} \text { atoms }$$

1.5 mol of carbon atoms: $$N=1.5 \times 6.022 \times 10^{23} \approx 9.03 \times 10^{23} \text { atoms }$$

Moles and Gases – Avogadro’s Law

Avogadro’s law:

Equal volumes of gases at the same temperature and pressure contain the same number of particles.

This leads to the idea of molar volume.

At STP (standard temperature and pressure): $$1 \mathrm{~mol} \text { of any gas } \approx 22.4 \mathrm{dm}^3$$

Gas volume formula

$$V=n×V_m$$

where:

$V$ = gas volume (dm³)
$n$ = moles of gas (mol)
$V_m$= molar volume (e.g. 22.4 dm³ mol⁻¹ at STP)

Example question

What volume is 0.50 mol of O₂ at STP?

Solution

$$V=0.50 \times 22.4=11.2 \ \mathrm{dm}^3$$

Example question

What volume is 64 g of O₂ at STP?

Solution

$$M\left(\mathrm{O}_2\right)=32.0 \mathrm{~g} \mathrm{~mol}^{-1} \Rightarrow n=\frac{64}{32}=2.0 \mathrm{~mol}$$

$$V=2.0 \times 22.4=44.8 \ \mathrm{dm}^3$$

Stoichiometry: Turning Equations into Numbers

Once an equation is balanced, its coefficients give you ratios of moles.
With the mole and molar mass, you can turn those ratios into masses, gas volumes, or numbers of particles.

Example question

Magnesium reacts with hydrochloric acid: $$\mathrm{Mg}+2 \mathrm{HCl} \rightarrow \mathrm{MgCl}_2+\mathrm{H}_2$$

What mass of magnesium chloride is formed from 4.8 g of Mg, assuming excess acid?

Solution

Step 1 - Moles of $Mg$

$$
M_r(\mathrm{Mg})=24 \quad \Rightarrow \quad n(\mathrm{Mg})=\frac{4.8}{24}=0.20 \mathrm{~mol}
$$

Step 2 - Use mole ratio
Equation shows:

$$
1 \mathrm{Mg} \rightarrow 1 \mathrm{MgCl}_2
$$

So:

$$
n\left(\mathrm{MgCl}_2\right)=0.20 \mathrm{~mol}
$$

Step 3 - Convert moles of $\mathrm{MgCl}_2$ to mass

$$
\begin{gathered}
M_r\left(\mathrm{MgCl}_2\right)=24+2 \times 35.5=95 \\
m\left(\mathrm{MgCl}_2\right)=n M=0.20 \times 95=19 \mathrm{~g}
\end{gathered}
$$

Answer: $\mathbf{1 9 ~ g}$ of $\mathrm{MgCl}_2$

Example question

Calcium carbonate decomposes as follows:

$$
\mathrm{CaCO}_3 \rightarrow \mathrm{CaO}+\mathrm{CO}_2
$$

What volume of $\mathrm{CO}_2$ (at STP, $22.4 \mathrm{dm}^3 \mathrm{~mol}^{-1}$ ) is produced from 10.0 g of $\mathrm{CaCO}_3$ ?

Solution

Step 1 - Moles of $\mathrm{CaCO}_3$

$$
\begin{gathered}
M_r\left(\mathrm{CaCO}_3\right)=40+12+3 \times 16=100 \\
n\left(\mathrm{CaCO}_3\right)=\frac{10.0}{100}=0.10 \mathrm{~mol}
\end{gathered}
$$

Step 2 - Use mole ratio

$$
1 \mathrm{CaCO}_3 \rightarrow 1 \mathrm{CO}_2
$$

So:

$$
n\left(\mathrm{CO}_2\right)=0.10 \mathrm{~mol}
$$

Step 3 - Moles to volume

$$
V\left(\mathrm{CO}_2\right)=n V_m=0.10 \times 22.4=2.24 \ \mathrm{dm}^3
$$

Answer: $2.24 \ \mathrm{dm}^3$ of $\mathrm{CO}_2$ at STP

Using Moles in Chemical Equations (Stoichiometry)

Balanced equations tell you mole ratios.
Those coefficients are literally “moles of each substance”.

Example

Haber process:

$$\mathrm{N}_2+3 \mathrm{H}_2 \rightarrow 2 \mathrm{NH}_3$$

This means:

1 mol N₂ reacts with 3 mol H₂ to form 2 mol NH₃.

General stoichiometry steps

Write and balance the equation.
Convert the known quantity → moles.
Use the mole ratio from the equation to get moles of the unknown.
Convert moles of the unknown → mass, gas volume, or number of particles.

Example question

How many moles of H₂ are needed to make 2.50 mol of NH₃?

Solution

From the equation:

$$
\mathrm{N}_2+3 \mathrm{H}_2 \rightarrow 2 \mathrm{NH}_3
$$

Mole ratio $\mathrm{H}_2: \mathrm{NH}_3=3: 2$

$$
n\left(\mathrm{H}_2\right)=2.50 \times \frac{3}{2}=3.75 \mathrm{~mol}
$$

Example question

How much ammonia from 4.0 mol of H₂?

Solution

Same equation:

$$
\mathrm{N}_2+3 \mathrm{H}_2 \rightarrow 2 \mathrm{NH}_3
$$

Mole ratio $\mathrm{NH}_3: \mathrm{H}_2=2: 3$

$$
n\left(\mathrm{NH}_3\right)=4.0 \times \frac{2}{3} \approx 2.67 \mathrm{~mol}
$$

(Assuming excess nitrogen is present)

Limiting Reactants – When One Reactant Runs Out

In real reactions, reactants are often not in perfect ratio.
The one that runs out first is the limiting reactant – it limits how much product you can make.
The other reactant(s) are in excess.

Definition

Limiting reactant

Example

Limiting reactant in Haber

$$\mathrm{N}_2+3 \mathrm{H}_2 \rightarrow 2 \mathrm{NH}_3$$

Suppose you mix:

1.0 mol N₂
4.0 mol H₂

From the equation:

1.0 mol N₂ needs 3.0 mol H₂ → you actually have 4.0 mol.
So N₂ is limiting, H₂ is in excess.

Product moles (from limiting reagent):

1.0 mol N₂ → 2.0 mol NH₃

The extra 1.0 mol H₂ is left over.

Example

Limiting reactant in water formation

$$2 \mathrm{H}_2+\mathrm{O}_2 \rightarrow 2 \mathrm{H}_2 \mathrm{O}$$

If you have:

5.0 mol H₂
2.0 mol O₂

Required ratio H₂ : O₂ = 2 : 1

For 2.0 mol O₂ you need $2×2.0=4.0$ mol H₂.

You have 5.0 mol H₂ → H₂ is in excess
O₂ is limiting

Product:

From 2.0 mol O₂ → 4.0 mol H₂O

Theoretical Yield and Percentage Yield

Definition

Theoretical yield

The maximum possible amount of product, calculated from the limiting reactant (assuming the reaction goes to completion with no losses).

Definition

Actual yield

Definition

Percentage yield

Ratio of the actual yield (product obtained in an experiment) to the theoretical yield (maximum possible product from stoichiometry), expressed as a percentage, showing reaction efficiency.

$$\% \text { yield }=\frac{\text { actual yield }}{\text { theoretical yield }} \times 100$$

Low yield might be caused by:

incomplete reaction
side reactions
loss of product during separation/purification.

Example

If the theoretical yield of $\mathrm{NH}_3$ is 2.0 mol but the actual yield is 1.8 mol:

$$
\% \text { yield }=\frac{1.8}{2.0} \times 100=90 \%
$$

Why Do Mole Calculations Matter in the Real World?

Mole calculations aren’t just exam exercises – they are core to real chemistry.

Efficiency and Cost

Using correct mole ratios avoids wasting expensive reactants.
Industry always wants maximum yield from minimum input.

Example

In the Haber process, conditions and catalyst are chosen to get a good compromise between:

high yield (enough NH₃ formed)
reasonable rate (not too slow)
acceptable cost (pressure, temperature, catalyst).

All of that is designed using mole-based calculations.

Safety and Waste Reduction (Green Chemistry)

Knowing how many moles of gas will be produced helps design safe equipment (so it doesn’t over-pressurise).
Mole calculations help reduce:
1. toxic by-products
2. waste chemicals
3. energy use (e.g. not heating 10× more than needed).

Note

This supports green chemistry goals: less waste, less energy, fewer harmful emissions.

Correct Doses in Medicine

Drug doses are based on moles of active ingredient, not just mass.
Too little → ineffective.
Too much → toxic.

Example

Pharmacists use moles to ensure safe and effective concentrations in tablets, injections and drips.

Active recall

What is Avogadro’s constant, and what does it represent?
How would you calculate the mass of 0.50 mol of NaCl if its molar mass is 58.5 g mol⁻¹?
How many moles of H₂ are required to react completely with 4.0 mol of N₂ in the Haber reaction?
In the reaction $$2 \mathrm{H}_2+\mathrm{O}_2 \rightarrow 2 \mathrm{H}_2 \mathrm{O}$$ you have 3.0 mol H₂ and 2.0 mol O₂.
Which reactant is limiting?
How many moles of H₂O can form?
Why is it important for industry to know which reactant is limiting and what the theoretical yield is?

1 Matter

2 Pure and impure substances

3 Periodic table

4 Bonding and periodic properties

5 Types of chemical reaction

6 International Union of Pure and Applied Chemistry (IUPAC)

7 The atmosphere

1 Matter

2 Pure and impure substances

3 Periodic table

4 Bonding and periodic properties

5 Types of chemical reaction

6 International Union of Pure and Applied Chemistry (IUPAC)

7 The atmosphere

MYP Chemistry 5.4 the Mole Concept and Chemical Calculations Notes

MYP Chemistry 5.4 the Mole Concept and Chemical Calculations Notes

The Mole – Counting the Uncountable

What is a mole?

Molar Mass – Linking Moles to Mass

Using the Periodic Table to Find Ar and Mr

Step 1 – Pick up Aᵣ from the periodic table

Step 2 – Use Aᵣ to find Mᵣ (relative molecular / formula mass)

Molar Mass

Key formula

Particles and Moles

Moles and Gases – Avogadro’s Law

Gas volume formula

Stoichiometry: Turning Equations into Numbers

Using Moles in Chemical Equations (Stoichiometry)

General stoichiometry steps

Limiting Reactants – When One Reactant Runs Out

Limiting reactant in Haber

Limiting reactant in water formation

Theoretical Yield and Percentage Yield

Why Do Mole Calculations Matter in the Real World?

Efficiency and Cost

Safety and Waste Reduction (Green Chemistry)

Correct Doses in Medicine

The Mole – Counting the Uncountable

What is a mole?

Molar Mass – Linking Moles to Mass

Using the Periodic Table to Find Ar and Mr

Step 1 – Pick up Aᵣ from the periodic table

Step 2 – Use Aᵣ to find Mᵣ (relative molecular / formula mass)

Molar Mass

Key formula

Particles and Moles

Moles and Gases – Avogadro’s Law

Gas volume formula

Stoichiometry: Turning Equations into Numbers

Using Moles in Chemical Equations (Stoichiometry)

General stoichiometry steps

Limiting Reactants – When One Reactant Runs Out

Limiting reactant in Haber

Limiting reactant in water formation

Theoretical Yield and Percentage Yield

Why Do Mole Calculations Matter in the Real World?

Efficiency and Cost

Safety and Waste Reduction (Green Chemistry)

Correct Doses in Medicine