Mole concept and stoichiometry (KCSE Chemistry Form 3)

TL;DR

The mole is a central unit in chemistry, allowing us to count atoms and molecules by weighing them. Stoichiometry uses these mole relationships to predict the amounts of reactants and products in chemical reactions. Mastering these concepts helps you understand chemical formulas and balance equations.

1. The Mental Model

Think of the mole as a "chemist's dozen" – it's just a specific number of particles. This number is so huge that we can't count individual atoms, so we weigh them instead. Stoichiometry is like following a recipe, but for chemical reactions, making sure you have the right amounts of ingredients.

2. The Core Material

What is a Mole?

A mole (mol) is the amount of substance that contains as many elementary entities (atoms, molecules, ions, electrons, etc.) as there are atoms in 12 grams of carbon-12. This number is Avogadro's constant, approximately 6.022 x 10^23.

So, 1 mole of anything contains 6.022 x 10^23 particles of that thing.
* 1 mole of carbon atoms = 6.022 x 10^23 carbon atoms
* 1 mole of water molecules = 6.022 x 10^23 water molecules

Molar Mass

The molar mass of a substance is the mass of one mole of that substance, usually expressed in grams per mole (g/mol).
* For elements, the molar mass is numerically equal to its relative atomic mass (Ar) in grams. For example, the Ar of Carbon is 12, so its molar mass is 12 g/mol.
* For compounds, the molar mass is the sum of the relative atomic masses of all the atoms in its chemical formula, expressed in grams. This is also called the relative molecular mass (Mr) or relative formula mass.

Example: Calculate the molar mass of H₂O.
Ar(H) = 1, Ar(O) = 16
Molar mass of H₂O = (2 x Ar(H)) + (1 x Ar(O)) = (2 x 1) + (1 x 16) = 2 + 16 = 18 g/mol.

Converting Between Mass, Moles, and Number of Particles

You'll often need to convert between these three quantities. Here's how:

graph TD
    A[Mass (g)] -->|Divide by Molar Mass (g/mol)| B(Moles)
    B -->|Multiply by Molar Mass (g/mol)| A
    B -->|Multiply by Avogadro's Constant (6.022 x 10^23)| C[Number of Particles]
    C -->|Divide by Avogadro's Constant (6.022 x 10^23)| B

Example 1: Mass to Moles
How many moles are in 49 g of H₂SO₄?
Ar(H)=1, Ar(S)=32, Ar(O)=16
Molar mass of H₂SO₄ = (2x1) + 32 + (4x16) = 2 + 32 + 64 = 98 g/mol
Moles = Mass / Molar Mass = 49 g / 98 g/mol = 0.5 mol

Example 2: Moles to Number of Particles
How many molecules are in 0.25 mol of CO₂?
Number of molecules = Moles x Avogadro's Constant
Number of molecules = 0.25 mol x 6.022 x 10^23 molecules/mol = 1.5055 x 10^23 molecules

Molar Gas Volume

At Standard Temperature and Pressure (STP: 0°C or 273 K, 1 atmosphere or 101325 Pa), one mole of any gas occupies a volume of 22.4 dm³ (or 22400 cm³).
At Room Temperature and Pressure (RTP: 25°C or 298 K, 1 atmosphere), one mole of any gas occupies a volume of 24 dm³ (or 24000 cm³).

Example: What is the volume occupied by 0.1 mol of oxygen gas at STP?
Volume = Moles x Molar Gas Volume = 0.1 mol x 22.4 dm³/mol = 2.24 dm³

Stoichiometry and Chemical Equations

Stoichiometry is about the quantitative relationships between reactants and products in a balanced chemical equation. The coefficients in a balanced equation represent the mole ratio of the substances involved.

Example: 2H₂(g) + O₂(g) → 2H₂O(l)
This equation tells us:
* 2 moles of hydrogen react with 1 mole of oxygen to produce 2 moles of water.
* The mole ratio H₂ : O₂ : H₂O is 2 : 1 : 2.

You can use these mole ratios to calculate amounts of reactants or products.

Steps for Stoichiometry Problems:
1. Write a balanced chemical equation. This is crucial!
2. Convert the given quantity to moles. (Using molar mass or molar gas volume).
3. Use the mole ratio from the balanced equation to find the moles of the desired substance.
4. Convert the moles of the desired substance to the required units. (Mass, volume, or number of particles).

3. Worked Example

Problem: What mass of carbon dioxide (CO₂) is produced when 4.8 g of methane (CH₄) is completely combusted?
(Ar: C=12, H=1, O=16)

Solution:

1. Write a balanced chemical equation:
   CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)

2. Convert the given quantity (mass of CH₄) to moles:
   Molar mass of CH₄ = 12 + (4 x 1) = 16 g/mol
   Moles of CH₄ = Mass / Molar Mass = 4.8 g / 16 g/mol = 0.3 mol

3. Use the mole ratio from the balanced equation to find the moles of CO₂:
   From the equation, the mole ratio CH₄ : CO₂ is 1 : 1.
   So, if 0.3 mol of CH₄ reacts, 0.3 mol of CO₂ will be produced.

4. Convert the moles of CO₂ to the required units (mass):
   Molar mass of CO₂ = 12 + (2 x 16) = 12 + 32 = 44 g/mol
   Mass of CO₂ = Moles x Molar Mass = 0.3 mol x 44 g/mol = 13.2 g

Therefore, 13.2 g of carbon dioxide is produced.

4. Key Takeaways

• A mole is a specific number of particles (Avogadro's constant, 6.022 x 10^23).
• Molar mass is the mass of one mole of a substance in grams per mole.
• You can convert between mass, moles, and number of particles using molar mass and Avogadro's constant.
• One mole of any gas occupies 22.4 dm³ at STP and 24 dm³ at RTP.
• Balanced chemical equations provide the essential mole ratios for stoichiometric calculations.
• Always balance the chemical equation before starting any stoichiometry calculation.
• Stoichiometry allows you to predict the amounts of reactants and products in a reaction.

Common mistakes to avoid:
- Not balancing the chemical equation correctly before starting calculations.
- Using the wrong molar mass for a substance (e.g., using Ar for a compound).
- Confusing molar mass with relative atomic mass or relative molecular mass.
- Forgetting to use Avogadro's constant when converting to or from the number of particles.
- Mixing up STP and RTP molar gas volumes.

5. Now Try It

Calculate the mass of oxygen gas (O₂) required to react completely with 11.2 dm³ of hydrogen gas (H₂) at STP to form water.
(Ar: H=1, O=16)

What to do:
1. Write the balanced chemical equation for the formation of water from hydrogen and oxygen.
2. Convert the given volume of hydrogen gas to moles.
3. Use the mole ratio from your balanced equation to find the moles of oxygen gas needed.
4. Convert the moles of oxygen gas to its mass.

What success looks like: You should get an answer around 8 grams.