Stoichiometry

A molecular formula states the actual number of each type of atom in one molecule, e.g. C₆H₁₂O₆ for glucose. An empirical formula gives the simplest whole-number ratio, e.g. CH₂O for glucose.

To write the formula of an ionic compound, balance the charges so the total is zero:

State symbols: (s) solid · (l) liquid · (g) gas · (aq) in aqueous solution.

A balanced equation has equal numbers of each type of atom on both sides. The coefficients in a balanced equation give the mole ratio of reactants and products.

Relative atomic mass (Aᵣ) — the average mass of the isotopes of an element relative to 1/12 of the mass of carbon-12. It is a dimensionless ratio.

Key values to know by heart:

Chlorine's Aᵣ is 35.5 because it is a weighted average of Cl-35 (75%) and Cl-37 (25%): (0.75 × 35) + (0.25 × 37) = 35.5.

Relative molecular mass (Mᵣ) — sum of the Aᵣ values for all atoms in the formula. For ionic compounds this is called the relative formula mass but the symbol Mᵣ is still used.

Examples:

• H₂O: (2 × 1) + 16 = 18
• CO₂: 12 + (2 × 16) = 44
• H₂SO₄: (2 × 1) + 32 + (4 × 16) = 98
• CaCO₃: 40 + 12 + (3 × 16) = 100

One mole (mol) is the amount of substance that contains 6.02 × 10²³ particles (atoms, ions, molecules — whatever the formula unit is). This number is the Avogadro constant.

The molar mass of a substance (in g/mol) is numerically equal to its Mᵣ.

Quick examples:

At room temperature and pressure (r.t.p.), one mole of any gas occupies 24 dm³ (24 000 cm³).

Volume (dm³) = moles × 24 or moles = volume (dm³) ÷ 24

Remember: 1 dm³ = 1000 cm³. To convert cm³ → dm³, divide by 1000.

Example: What volume does 0.25 mol CO₂ occupy at r.t.p.? → 0.25 × 24 = 6 dm³

Concentration can be expressed in mol/dm³ (molarity) or g/dm³.

c (mol/dm³) = n (mol) ÷ V (dm³)

To use volumes in cm³, divide by 1000 first.

Example: 0.5 mol NaOH in 250 cm³ → V = 0.25 dm³ → c = 0.5 ÷ 0.25 = 2 mol/dm³

Use the balanced equation mole ratio to scale between substances:

1. Write the balanced equation.
2. Convert given mass to moles (n = m/M).
3. Use the mole ratio to find moles of the other substance.
4. Convert moles back to mass (m = n × M).

The limiting reactant is the one completely consumed first — it determines the maximum amount of product. Any excess of the other reactant is left over.

From percentage composition:

1. Write the percentage as a mass in grams (assume 100 g sample).
2. Divide each mass by the element's Aᵣ to get moles.
3. Divide all by the smallest mole value.
4. Round to simple whole-number ratios → empirical formula.

To find molecular formula from empirical formula:

• Calculate the Mᵣ of the empirical formula.
• Divide the actual Mᵣ by the empirical Mᵣ to get the multiplier n.
• Multiply all subscripts by n.

Example: Empirical formula CH₂O (Mᵣ = 30), actual Mᵣ = 180 → n = 6 → molecular formula C₆H₁₂O₆

Percentage yield: % yield = (actual yield / theoretical yield) × 100%

Percentage purity: % purity = (mass of pure substance / mass of impure sample) × 100%

Percentage composition by mass: % of element = (Aᵣ × number of atoms of that element / Mᵣ of compound) × 100%

Example: % O in H₂O = (16 ÷ 18) × 100% = 88.9%

In an acid–base titration:

1. Calculate moles of the known solution: n = c × V (convert V to dm³).
2. Use the equation mole ratio to find moles of the unknown.
3. Calculate the concentration of the unknown: c = n ÷ V.

Example: 25 cm³ of 0.1 mol/dm³ NaOH neutralises 12.5 cm³ of HCl. Equation: NaOH + HCl → NaCl + H₂O (1:1 ratio).

• Moles NaOH = 0.1 × 0.025 = 0.0025 mol
• Moles HCl = 0.0025 mol (1:1)
• Concentration HCl = 0.0025 ÷ 0.0125 = 0.2 mol/dm³