R2.1.2 Using the mole ratio

Mole Ratios and Their Applications in Chemical Calculations

Understanding Mole Ratios

Definition

Mole ratio

A mole ratio is the ratio of the amounts (in moles) of reactants and products in a balanced chemical equation. These ratios are determined by the stoichiometric coefficients in the equation.

Example

Let’s take the combustion of methane as an example:

${CH}_{4} (g) + 2 O_{2} (g) \to {CO}_{2} (g) + 2 H_{2} O (g)$

From the coefficients, we can see:

1 mole of methane reacts with 2 moles of oxygen.
This produces 1 mole of carbon dioxide and 2 moles of water.

These relationships allow us to calculate the amounts of any reactant or product, provided we know the quantity of at least one substance in the reaction.

Tip

Always ensure your chemical equation is balanced before using mole ratios. An unbalanced equation gives incorrect ratios and leads to errors in calculations.

Applications of Mole Ratios

1. Calculating Masses of Reactants and Products

To find the mass of a substance in a reaction, we use the formula:

$Mass = Moles \times Molar Mass$

Example question

How much carbon dioxide is produced when 16 g of methane ${CH}_{4}$ is completely combusted?

Solution

Write the balanced equation:
${CH}_{4} (g) + 2 O_{2} (g) \to {CO}_{2} (g) + 2 H_{2} O (g)$
Calculate the moles of methane:
$n ({CH}_{4}) = \frac{Mass}{Molar Mass} = \frac{16 g}{16.04 g/mol} = 1.00 mol$
Use the mole ratio ${CH}_{4} : {CO}_{2} = 1 : 1$ to find the moles of ${CO}_{2}$ :
$n ({CO}_{2}) = 1.00 mol$
Calculate the mass of ${CO}_{2}$ :
$Mass ({CO}_{2}) = n \times Molar Mass = 1.00 mol \times 44.01 g/mol = 44.01 g$

Answer: 44.01 g of carbon dioxide is produced.

2. Calculating Volumes of Gases

Definition

Avogadro's law

Avogadro's law states that equal volumes of all gases, measured under the same conditions of temperature and pressure, contain equal numbers of molecules.

At STP (273 K and 100 kPa), 1 mole of any gas occupies 22.7 $d m^{3}$ . This is invaluable for gas-related calculations.

Common Mistake

Many students forget to convert gas volumes to moles using the molar volume of 22.7 $d m^{3} m o l^{- 1}$ . Always check whether your problem involves gases at STP!

Example question

What volume of oxygen gas is required to completely combust 2.00 mol of propane $C_{3} H_{8}$ ?

Solution

Write the balanced equation:
$C_{3} H_{8} (g) + 5 O_{2} (g) \to 3 {CO}_{2} (g) + 4 H_{2} O (g)$
Use the mole ratio $C_{3} H_{8} : O_{2} = 1 : 5$ to find the moles of $O_{2}$ :
$n (O_{2}) = 2.00 mol \times 5 = 10.00 mol$
Calculate the volume of $O_{2}$ at STP:
$V (O_{2}) = n \times 22.7 {dm}^{3} / mol = 10.00 mol \times 22.7 {dm}^{3} / mol = 227 {dm}^{3}$

Answer: $227 d m^{3}$ of oxygen gas is required.

3. Reactions in Solution: Concentration and Volume

In solution-based reactions, the relationship between concentration $C$ , moles $n$ , and volume $V$ is given by:

$C = \frac{n}{V}$

This formula is especially useful for titration calculations.

Example question

A 25.0 $c m^{3}$ solution of 0.100 $m o l d m^{3}$ hydrochloric acid $HCl$ reacts with sodium hydroxide $NaOH$ according to:
$HCl (a q) + NaOH (a q) \to NaCl (a q) + H_{2} O (l)$

What volume of 0.200 $m o l d m^{3}$ $NaOH$ is required for complete neutralization?

Solution

Calculate moles of $HCl$ :
$n (HCl) = C \times V = 0.100 {mol dm}^{- 3} \times 0.0250 {dm}^{3} = 0.00250 mol$
Use the mole ratio $HCl : NaOH = 1 : 1$ to find moles of $NaOH$ :
$n (NaOH) = 0.00250 mol$
Calculate the volume of $NaOH$ :
$V (NaOH) = \frac{n}{C} = \frac{0.00250 mol}{0.200 {mol dm}^{- 3}} = 0.0125 {dm}^{3} = 12.5 {cm}^{3}$

Answer: 12.5 $c m^{3}$ of $NaOH$ is required.

Tip

When working with solutions, always ensure volumes are in dm $^{3}$ for consistency with mol dm $^{3}$ units.

Reflection and Challenges

Theory of Knowledge

To what extent does the universality of mole ratios and Avogadro’s law reflect the interconnectedness of scientific principles across languages and cultures?

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R2.1.2 Using the mole ratio

Mole Ratios and Their Applications in Chemical Calculations

Understanding Mole Ratios

Applications of Mole Ratios

1. Calculating Masses of Reactants and Products

2. Calculating Volumes of Gases

3. Reactions in Solution: Concentration and Volume

Reflection and Challenges

You've read 2/2 free chapters this week.

Introduction to Mole Ratios