Chemistry (9202) | Using molar concentrations of solutions and amount of substance in relation to volumes of gases

Introduction: The Chemist's Recipe Book

Hello future chemists! Welcome to a crucial chapter in Quantitative Chemistry. Don’t worry if the titles sound complicated—this section is all about being a precise chemist, like following a perfect recipe.

We are going to learn how to measure the exact amount of stuff (moles) when that stuff is either dissolved in a liquid (a solution) or floating around as a gas. These calculations are fundamental to laboratory work, helping us determine how much product we can make or how strong a solution truly is!

Get ready to master two key concepts: molar concentration and the molar volume of gases.

1. Understanding Molar Concentration of Solutions

1.1 What is Concentration?

Think about making a glass of orange squash. If you add only a tiny splash of concentrate to a lot of water, the drink is weak. If you add a lot of concentrate, the drink is strong. This "strength" is the concentration!

In Chemistry, molar concentration (often just called concentration) tells us exactly how much solute (the chemical dissolved) is present in a specific volume of solvent (usually water).

We measure the amount of solute in moles (\(n\)) and the volume of the solution in cubic decimetres (\(dm^3\)).

The units for molar concentration are moles per cubic decimetre (\(mol/dm^3\)).

1.2 The Crucial Step: Volume Conversion

In the lab, we often measure volumes using small measuring cylinders or pipettes, giving us readings in cubic centimetres (\(cm^3\)). However, our formula requires volume in \(dm^3\)!

Remember this conversion fact:
\(1 \text{ cubic decimetre } (dm^3) = 1000 \text{ cubic centimetres } (cm^3)\)

How to convert \(cm^3\) to \(dm^3\):

1. Start with the volume in \(cm^3\).
2. Divide by 1000.

Example: \(250 \text{ cm}^3 \div 1000 = 0.25 \text{ dm}^3\)

1.3 The Molar Concentration Formula

We can link concentration (C), moles (\(n\)), and volume (V) using the following equation:

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

Where:
\(C\) = Concentration (\(mol/dm^3\))
\(n\) = Amount of substance (moles)
\(V\) = Volume of solution (\(dm^3\))

Memory Aid: The Chemistry Triangle

This triangle helps you rearrange the formula easily. Cover the value you want to find:

\( \begin{array}{|c|} \hline \quad n \quad \\ \hline C \quad | \quad V \\ \hline \end{array} \)

This means:

• To find Moles (\(n\)): \(n = C \times V\) (Concentration times Volume)
• To find Volume (\(V\)): \(V = \frac{n}{C}\) (Moles divided by Concentration)

Key Takeaway for Solutions: Molar concentration links the amount of substance (moles) to the volume of the liquid, but only when the volume is measured in \(dm^3\).

2. Amount of Substance in Relation to Volumes of Gases

Gases behave differently from solutions, but the great news is that calculating moles of gas is often simpler because of one magical number!

2.1 The Molar Gas Volume (MGV)

Imagine you have one mole of *any* gas—be it tiny hydrogen gas or heavy carbon dioxide gas. If you measure their volume under the exact same conditions, they will take up the same amount of space!

This is based on Avogadro's Law. For simplicity, chemists use standard conditions known as Room Temperature and Pressure (RTP).

At RTP, one mole of any gas occupies a volume of \(24 \text{ dm}^3\).

This value, \(24 \text{ dm}^3/mol\), is called the Molar Gas Volume (MGV). You must know and use this specific value for your calculations at International GCSE level.

2.2 Calculating Gas Volumes and Moles

Since 1 mole = \(24 \text{ dm}^3\), we can create a simple relationship to find the volume of any amount of gas.

$$\text{Volume of Gas } (dm^3) = \text{Moles } (n) \times 24$$

Or, if you need to find the moles of gas you have:

$$n = \frac{\text{Volume of Gas } (dm^3)}{24}$$

Volume Units for Gases

Just like with solutions, gas volume is often measured in \(cm^3\).

• If the volume is given in \(dm^3\): Use the value 24 in the formula.
• If the volume is given in \(cm^3\): You must convert it to \(dm^3\) first (divide by 1000), OR you can use 24000 directly: \(n = \frac{V (cm^3)}{24000}\).

Did you know? Even though a Helium atom is much lighter than a Carbon Dioxide molecule, one mole of Helium gas takes up the exact same volume (\(24 \text{ dm}^3\)) as one mole of Carbon Dioxide gas at RTP! This is because gas particles are so far apart that the size of the individual particle hardly matters.

Key Takeaway for Gases: The volume of any gas at RTP is directly proportional to the number of moles. Always use the standard volume of \(24 \text{ dm}^3\) per mole.

3. Linking Solutions, Gases, and Stoichiometry

Now we put it all together! Most exam questions will require you to use the molar concentration formula AND the molar gas volume formula in the same problem, using the mole ratio from a balanced chemical equation.

3.1 Step-by-Step Calculation Strategy

This process applies whenever you have information about one substance (A) and need to find an unknown quantity for another substance (B).

Scenario Example: Calculating the volume of \(CO_2\) gas produced when an acid reacts with a carbonate solution.

Equation: \(\text{HCl}(aq) + \text{Na}_2\text{CO}_3(aq) \rightarrow 2\text{NaCl}(aq) + \text{H}_2\text{O}(l) + \text{CO}_2(g)\)

Step 1: Find the Moles of the Known Substance (A).
If substance A is a solution, use the concentration formula: \(n_A = C \times V\). (Don't forget to convert volume to \(dm^3\)!)

Step 2: Use Stoichiometry (Mole Ratio) to Find Moles of the Unknown Substance (B).
Look at the balanced equation. Use the coefficients to find the mole ratio between A and B.
Example: The ratio of \(Na_2CO_3\) to \(CO_2\) is 1:1. So, \(n_B = n_A \times \frac{\text{Coefficient of B}}{\text{Coefficient of A}}\).

Step 3: Calculate the Final Quantity of the Unknown Substance (B).
If substance B is a gas, use the molar gas volume formula:
Volume of gas = \(n_B \times 24\) (Volume in \(dm^3\)).

If substance B were a solution and you needed its concentration, you would use \(C_B = \frac{n_B}{V}\).

3.2 Practicing the Flow

The flow of calculations almost always moves through moles (\(n\)):

$$ \text{Concentration/Volume (Solution A)} \rightarrow \text{Moles of A} \rightarrow \text{Moles of B} \rightarrow \text{Volume of Gas B} $$

3.3 Example Walkthrough

Question: \(100 \text{ cm}^3\) of a \(0.5 \text{ mol/dm}^3\) solution of hydrochloric acid (\(\text{HCl}\)) reacts completely with excess metal. Calculate the volume of hydrogen gas (\(H_2\)) produced at RTP. (Assume the ratio of \(\text{HCl}\) to \(\text{H}_2\) is 2:1).

Solution:

Step 1: Find the moles of \(\text{HCl}\) (the known solution).
1. Convert volume: \(100 \text{ cm}^3 \div 1000 = 0.10 \text{ dm}^3\).
2. Calculate moles (\(n = C \times V\)):
\(n_{\text{HCl}} = 0.5 \text{ mol/dm}^3 \times 0.10 \text{ dm}^3 = 0.05 \text{ moles}\).

Step 2: Find the moles of \(\text{H}_2\) (the unknown gas).
1. The mole ratio (\(\text{HCl}\) : \(\text{H}_2\)) is 2:1.
2. Divide the moles of \(\text{HCl}\) by 2:
\(n_{\text{H}_2} = 0.05 \text{ moles} \div 2 = 0.025 \text{ moles}\).

Step 3: Calculate the volume of \(\text{H}_2\) gas.
1. Use the Molar Gas Volume (24 \(dm^3/mol\)):
Volume = \(n \times 24\)
Volume \(= 0.025 \text{ moles} \times 24 \text{ dm}^3/\text{mol} = 0.60 \text{ dm}^3\).

Answer: The volume of hydrogen gas produced is \(0.60 \text{ dm}^3\).

4. Final Review: Essential Formulas to Master

To succeed in this chapter, make sure these three relationships are memorised and understood:

1. General Moles Formula (Prerequisite)

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

2. Molar Concentration Formula (Solutions)

$$n = C \times V \quad (\text{Volume must be in } dm^3)$$

3. Molar Gas Volume Formula (Gases at RTP)

$$V_{\text{gas}} = n \times 24 \quad (\text{Volume will be in } dm^3)$$

Keep practising these steps, and you'll become a quantitative chemistry whiz! Good luck!