Stoichiometry — Algebra Balances Every Reaction
Stoichiometry
Every chemical reaction obeys a strict accounting system: atoms in = atoms out. The math that keeps the books? Ratios, proportions, and algebra.
One mole = $6.022 \times 10^{23}$ particles (Avogadro's number). This enormous number bridges the atomic world to the lab scale.
$$n = \frac{m}{M}$$
where $n$ is moles, $m$ is mass (grams), and $M$ is molar mass (g/mol).
Balancing Equations — Solving for Unknowns
Consider the combustion of methane:
$$\text{CH}_4 + a\,\text{O}_2 \rightarrow b\,\text{CO}_2 + c\,\text{H}_2\text{O}$$
Atom balance gives us a system of equations:
- Carbon: $1 = b$
- Hydrogen: $4 = 2c \Rightarrow c = 2$
- Oxygen: $2a = 2b + c = 2(1) + 2 = 4 \Rightarrow a = 2$
$$\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}$$
This is solving simultaneous linear equations — the same algebra you learn in class.
How many grams of O₂ are needed to completely burn 100 g of methane (CH₄)? Molar masses: CH₄ = 16 g/mol, O₂ = 32 g/mol.
Step 1: Moles of CH₄:
$$n_{\text{CH}_4} = \frac{100}{16} = 6.25 \text{ mol}$$
Step 2: From the balanced equation, 1 mol CH₄ requires 2 mol O₂:
$$n_{\text{O}_2} = 6.25 \times 2 = 12.5 \text{ mol}$$
Step 3: Mass of O₂:
$$m_{\text{O}_2} = 12.5 \times 32 = 400 \text{ g}$$
You need 400 grams of oxygen — found using nothing but division, multiplication, and ratios.
Every chemical calculation — from cooking recipes to industrial production — uses ratios and proportions. When you solve $\frac{a}{b} = \frac{c}{d}$, you're doing exactly what chemists do to scale reactions.