Topic 4.5 Notes – Stoichiometry

1. Stoichiometry and Why It Works

At the core is conservation of atoms. In a chemical reaction, atoms are rearranged, not created or destroyed. That’s why we must balance equations.

Take this reaction:

$$\ce{2Al(s) + 3Cl2(g) -> 2AlCl3(s)}$$

The coefficients (2, 3, 2) tell you the proportional relationship between substances.

• 2 moles Al react with 3 moles Cl₂
• 2 moles AlCl₃ are formed

Those numbers are not random. They represent mole ratios, and because 1 mole = $6.022 \times 10^{23}$ particles, the ratio works at the particle level too.

If a balanced equation looks like:

$$aA + bB \rightarrow cC + dD$$

then $a : b : c : d$ is a mole ratio.

That’s why stoichiometry works. The equation already contains the quantitative relationships.

Balanced equation → mole ratio → measurable quantities.

2. Mole Ratios and What They Represent

A mole ratio comes directly from coefficients.

Using the aluminum reaction:

$$\ce{2Al + 3Cl2 -> 2AlCl3}$$

Possible mole ratios include:

• $\frac{2 \text{ mol Al}}{3 \text{ mol Cl2}}$
• $\frac{2 \text{ mol AlCl3}}{2 \text{ mol Al}}$
• $\frac{3 \text{ mol Cl2}}{2 \text{ mol AlCl3}}$

You can form a ratio between any two substances in the equation.

Two reminders students mess up:

• Coefficients relate moles, not grams.
• Subscripts never change. They are part of the compound identity.

If you need to convert from Al to AlCl₃, the mole ratio is your bridge. Flip it so the starting unit cancels.

On tests, they often hide this in words. “How many moles of product form from…” just means use the coefficient ratio.

3. The Stoichiometry Roadmap

Almost every problem follows the same flow.

Core Strategy

1. Write and balance the equation.
2. Start with the given amount.
3. Convert to moles (if needed).
4. Use a mole ratio to switch substances.
5. Convert to the final unit.

The path usually looks like:

Given → moles (given) → moles (wanted) → desired unit

Here’s a quick example.

How many grams of $\ce{AlCl3}$ form from 5.00 g Al?

Balanced equation:
$$\ce{2Al + 3Cl2 -> 2AlCl3}$$

Step 1. Convert grams Al to moles:

$$5.00 \text{ g Al} \times \frac{1 \text{ mol Al}}{26.98 \text{ g Al}} = 0.185 \text{ mol Al}$$

Step 2. Use mole ratio (2:2 simplifies to 1:1):

$$0.185 \text{ mol Al} = 0.185 \text{ mol AlCl3}$$

Step 3. Convert to grams:

$$0.185 \text{ mol} \times 133.34 \text{ g/mol} = 24.7 \text{ g AlCl3}$$

Units cancel at each step. If they don’t, something is wrong.

Conversions You Must Know

• Grams ↔ moles using molar mass (g/mol)
• Particles ↔ moles using $6.022 \times 10^{23}$
• Gas volume at STP ↔ moles using 22.4 L/mol
• Molarity:
  $$M = \frac{\text{mol}}{\text{L}}$$
  So moles = $M \times L$

These are tools you combine with mole ratios.

4. Stoichiometry with Gases and Solutions

Gas Stoichiometry

If a gas is at STP:

• 1 mol = 22.4 L

If not at STP, use the ideal gas law:

$$PV = nRT$$

Solve for $n$, then use the mole ratio.

Flow:
Gas data → moles (via PV = nRT or 22.4 L/mol) → mole ratio → target

On the AP exam, they often give pressure and temperature that are not STP to force you into PV = nRT.

Solution Stoichiometry

If given molarity and volume:

1. Convert volume to liters.
2. Use $n = M \times L$.
3. Apply mole ratio.
4. Convert to final unit.

Example setup:

25.0 mL of 0.200 M $\ce{Na2CO3}$

$$0.0250 \text{ L} \times 0.200 \frac{\text{mol}}{\text{L}} = 0.00500 \text{ mol}$$

That gives moles of reactant. Then use the balanced equation.

AP questions love combining molarity and stoichiometry in one chain. Don’t try to shortcut around moles. Always go through them.

5. What Stoichiometry Tells You About a Reaction

From a balanced equation, you can determine:

• How much product forms from a given amount of reactant.
• How much reactant is required for a target amount of product.
• The proportional consumption of reactants.
• Quantitative relationships across solids, gases, and aqueous species.

Every calculation traces back to one fact. The equation is balanced because atoms are conserved. Stoichiometry is just using that conservation mathematically with the mole as the counting unit.