Chemical Reactions and Stoichiometry

TL;DR

Chemical reactions are just atoms rearranging themselves into new substances, and we represent these changes with balanced chemical equations. Stoichiometry is the math that lets us predict exactly how much of each substance is involved in a reaction. It's crucial for understanding how much reactant you need or how much product you'll get.

1. The Mental Model

Think of chemical reactions like building with LEGOs: you take apart existing structures and use the same blocks to build new ones. Stoichiometry is like knowing exactly how many blocks of each color you'll need for your new design.

2. The Core Material

Chemical reactions are fundamental processes where one or more substances, called reactants, are converted into different substances, called products. This involves breaking old chemical bonds and forming new ones.

Chemical Equations

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We use chemical equations to represent these reactions. A chemical equation shows the reactants on the left, the products on the right, and an arrow indicating the direction of the reaction.

For example, when hydrogen gas (H₂) burns in oxygen gas (O₂) to form water (H₂O):

H₂ + O₂ → H₂O

This equation isn't complete yet because it doesn't follow the Law of Conservation of Mass, which states that matter cannot be created or destroyed. This means the number of atoms of each element must be the same on both sides of the equation.

Balancing Chemical Equations

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To obey the Law of Conservation of Mass, we must balance chemical equations by placing coefficients (numbers) in front of the chemical formulas. You can only change coefficients, never the subscripts within a chemical formula (because that would change the substance itself!).

Let's balance the water formation reaction:
1. Start with H₂ + O₂ → H₂O
2. Count atoms:
- Reactants: H=2, O=2
- Products: H=2, O=1
3. We need more oxygen on the product side. Put a '2' in front of H₂O:
H₂ + O₂ → 2H₂O
4. Now count again:
- Reactants: H=2, O=2
- Products: H=4, O=2
5. Now hydrogen is unbalanced. Put a '2' in front of H₂ on the reactant side:
2H₂ + O₂ → 2H₂O
6. Final count:
- Reactants: H=4, O=2
- Products: H=4, O=2
It's balanced!

Stoichiometry: The Math of Reactions

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Stoichiometry is about using the mole concept and balanced equations to calculate amounts of reactants and products. The coefficients in a balanced equation represent the mole ratio (and also the molecule ratio) between substances.

The general pathway for stoichiometry calculations looks like this:

graph TD
    A["Known Mass of Substance A"] --> B["Moles of Substance A"];
    B --> C["Moles of Substance B (using mole ratio from balanced equation)"];
    C --> D["Desired Mass of Substance B"];
    style A fill:#f9f,stroke:#333,stroke-width:2px;
    style B fill:#bbf,stroke:#333,stroke-width:2px;
    style C fill:#bbf,stroke:#333,stroke-width:2px;
    style D fill:#f9f,stroke:#333,stroke-width:2px;

You'll need:
1. Molar Mass: The mass of one mole of a substance (g/mol). You calculate this from the atomic masses on the periodic table.
2. Balanced Chemical Equation: To get the mole ratio.

Types of Stoichiometry Problems

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• Mole-to-Mole: Given moles of A, find moles of B. (Use mole ratio).
• Mole-to-Mass: Given moles of A, find mass of B. (Mole ratio, then molar mass of B).
• Mass-to-Mole: Given mass of A, find moles of B. (Molar mass of A, then mole ratio).
• Mass-to-Mass: Given mass of A, find mass of B. (Molar mass of A, mole ratio, then molar mass of B). This is the most common and involves all steps in the diagram.

3. Worked Example

Let's say we burn 100.0 g of propane (C₃H₈) in oxygen. How many grams of carbon dioxide (CO₂) are produced?

1. Write and balance the chemical equation:
   C₃H₈(g) + O₂(g) → CO₂(g) + H₂O(g)
   Balancing:
   C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
   (Check: C:3 on both sides, H:8 on both sides, O:10 on both sides. It's balanced!)

2. Identify knowns and unknowns:
   Known: Mass of C₃H₈ = 100.0 g
   Unknown: Mass of CO₂ = ? g

3. Calculate molar masses:
   Molar mass of C₃H₈ = (3 × 12.01 g/mol C) + (8 × 1.01 g/mol H) = 44.11 g/mol
   Molar mass of CO₂ = (1 × 12.01 g/mol C) + (2 × 16.00 g/mol O) = 44.01 g/mol

4. Convert known mass to moles:
   Moles of C₃H₈ = 100.0 g C₃H₈ × (1 mol C₃H₈ / 44.11 g C₃H₈) = 2.267 mol C₃H₈

5. Use the mole ratio from the balanced equation to find moles of CO₂:
   From the balanced equation, 1 mol C₃H₈ produces 3 mol CO₂.
   Moles of CO₂ = 2.267 mol C₃H₈ × (3 mol CO₂ / 1 mol C₃H₈) = 6.801 mol CO₂

6. Convert moles of CO₂ to mass of CO₂:
   Mass of CO₂ = 6.801 mol CO₂ × (44.01 g CO₂ / 1 mol CO₂) = 299.3 g CO₂

So, burning 100.0 g of propane produces 299.3 g of carbon dioxide.

4. Key Takeaways

• Chemical reactions involve rearranging atoms to form new substances.
• Balanced chemical equations follow the Law of Conservation of Mass, showing equal numbers of each type of atom on both sides.
• Coefficients in a balanced equation represent the mole ratios of reactants and products.
• Stoichiometry uses molar masses and mole ratios to predict exact amounts in reactions.
• The general path for stoichiometry is: grams A → moles A → moles B → grams B.

Common Mistakes

• Not balancing the equation first: All stoichiometry calculations depend on a correctly balanced equation.
• Confusing coefficients with subscripts: Never change subscripts; only change coefficients to balance.
• Incorrect molar mass calculations: Double-check your periodic table values and math.
• Forgetting units or cancelling them incorrectly: Units help you check if your setup is right.
• Rounding too early: Keep extra significant figures during intermediate steps and round only at the very end.

5. Now Try It

You're making ammonia (NH₃) from nitrogen gas (N₂) and hydrogen gas (H₂).

1. Write and balance the chemical equation for this reaction.
2. If you start with 28.0 grams of nitrogen gas, calculate how many grams of hydrogen gas you would need for a complete reaction.

What success looks like:
You'll have a balanced equation and a final answer for the mass of hydrogen gas, including units, showing the step-by-step conversion from grams of nitrogen to moles of nitrogen, then to moles of hydrogen using the mole ratio, and finally to grams of hydrogen.