Stoichiometry and the Mole
Stoichiometry and the Mole
Every chemical reaction obeys strict bookkeeping: atoms in must equal atoms out. The math behind this is ratios, proportions, and algebra.
One mole = $6.022 \times 10^{23}$ particles (Avogadro's number). Moles connect atomic masses to lab-scale masses:
$$n = \frac{m}{M}$$
where $n$ = moles, $m$ = mass in grams, $M$ = molar mass (g/mol).
Balancing a chemical equation means solving a system of linear equations — one equation per element — so that atoms are conserved.
Balance: $\text{Fe} + \text{O}_2 \rightarrow \text{Fe}_2\text{O}_3$
Let the equation be $a\text{Fe} + b\text{O}_2 \rightarrow c\text{Fe}_2\text{O}_3$.
Fe: $a = 2c$. O: $2b = 3c$.
Let $c = 2$: then $a = 4$, $b = 3$.
$$4\text{Fe} + 3\text{O}_2 \rightarrow 2\text{Fe}_2\text{O}_3$$
How many grams of O₂ are needed to burn 64 g of methane (CH₄)?
CH₄ + 2 O₂ → CO₂ + 2 H₂O. Molar masses: CH₄ = 16, O₂ = 32.
Moles of CH₄: $n = 64/16 = 4$ mol.
From the equation, 1 mol CH₄ requires 2 mol O₂, so we need $4 \times 2 = 8$ mol O₂.
Mass of O₂: $8 \times 32 = 256$ g.
A 50.0 g sample of CaCO₃ decomposes: CaCO₃ → CaO + CO₂. How many grams of CO₂ are produced? (Molar masses: CaCO₃ = 100, CO₂ = 44)
Moles of CaCO₃: $50/100 = 0.50$ mol. The 1 : 1 ratio gives 0.50 mol CO₂.
Mass: $0.50 \times 44 = 22.0$ g CO₂.
Practice Problems
Show Answer Key
1. $10$ mol
2. $\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3$
3. 1 mol N₂ → 2 mol NH₃, so 34 g
4. $2 \times 6.022 \times 10^{23} = 1.204 \times 10^{24}$
5. 1 mol C → 1 mol CO₂, so 44 g
6. $0.25 \times 18 = 4.5$ g