Training Chemistry Math Stoichiometry and the Mole
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Stoichiometry and the Mole

24 min Chemistry Math

Stoichiometry is the quantitative heart of chemistry — it tells you exactly how much of each substance is consumed and produced in a chemical reaction. The mole concept bridges the microscopic world of atoms and molecules to the macroscopic world of grams and liters that you can measure in a laboratory. By balancing chemical equations and converting between moles, mass, and number of particles using Avogadro’s number and molar masses, you can predict yields, identify limiting reagents, and scale reactions from a test tube to an industrial reactor.

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.

The Mole

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 Equations

Balancing a chemical equation means solving a system of linear equations — one equation per element — so that atoms are conserved.

Example 1

Balance: $\text{Fe} + \text{O}_2 \rightarrow \text{Fe}_2\text{O}_3$

  1. Let the equation be $a\text{Fe} + b\text{O}_2 \rightarrow c\text{Fe}_2\text{O}_3$.
  2. Fe:
  3. $a = 2c$.   O: $2b = 3c$.
  4. Let $c = 2$: then $a = 4$, $b = 3$.
  5. $$4\text{Fe} + 3\text{O}_2 \rightarrow 2\text{Fe}_2\text{O}_3$$
Example 2

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.

  1. Moles of CH₄:
  2. $n = 64/16 = 4$ mol.
  3. From the equation, 1 mol CH₄ requires 2 mol O₂
  4. so we need $4 \times 2 = 8$ mol O₂.
  5. Mass of O₂: $8 \times 32 = 256$ g.
Example 3

A 50.0 g sample of CaCO₃ decomposes: CaCO₃ → CaO + CO₂. How many grams of CO₂ are produced? (Molar masses: CaCO₃ = 100, CO₂ = 44)

  1. Moles of CaCO₃: $50/100 = 0.50$ mol.
  2. The 1 : 1 ratio gives 0.50 mol CO₂.
  3. Mass:
  4. $0.50 \times 44 = 22.0$ g CO₂.

Practice Problems

1. How many moles are in 180 g of water (H₂O, M = 18)?
2. Balance: $\text{N}_2 + \text{H}_2 \rightarrow \text{NH}_3$
3. How many grams of NH₃ are produced from 28 g of N₂? (M: N₂ = 28, NH₃ = 17)
4. How many molecules are in 2 moles of CO₂?
5. If 12 g of carbon reacts with excess O₂ to form CO₂, how many grams of CO₂ form? (C = 12, CO₂ = 44)
6. A reaction produces 0.25 mol of H₂O. What mass is this?
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

⚗ Mole ↔ Mass ↔ Particles Converter
Moles (n = m / M)
Particles (× 6.022×10²³)
Volume at STP (L, if gas)
The pH Scale and Logarithms