Remember: 1 dm³ = 1000 cm³. Convert cm³ to dm³ by dividing by 1000.
EXAMPLES:
0.5 mol of NaOH dissolved in 250 cm³:
V = 250 ÷ 1000 = 0.25 dm³
c = 0.5 ÷ 0.25 = 2 mol/dm³
How many moles in 100 cm³ of 2 mol/dm³ HCl?
V = 100 ÷ 1000 = 0.1 dm³
n = 2 × 0.1 = 0.2 mol
USING TITRATION DATA:
If 25.0 cm³ of NaOH is neutralised by 20.0 cm³ of 0.1 mol/dm³ HCl:
n(HCl) = 0.1 × 0.020 = 0.002 mol
NaOH + HCl → NaCl + H₂O (1:1 ratio)
n(NaOH) = 0.002 mol
c(NaOH) = 0.002 ÷ 0.025 = 0.08 mol/dm³
Limiting Reactants
In a reaction, the LIMITING REACTANT is the reactant that is completely used up first — it determines how much product forms.
The other reactant is in EXCESS — some of it remains unreacted.
IDENTIFYING THE LIMITING REACTANT:
Calculate moles of each reactant.
Compare to the molar ratio from the balanced equation.
The reactant that runs out first (relative to the ratio) is the LIMITING reactant.
EXAMPLE:
Mg + 2HCl → MgCl₂ + H₂
2.4 g of Mg and 3.65 g of HCl are mixed.
n(Mg) = 2.4 ÷ 24 = 0.1 mol
n(HCl) = 3.65 ÷ 36.5 = 0.1 mol
For complete reaction: 0.1 mol Mg needs 0.2 mol HCl.
Only 0.1 mol HCl available → HCl is the LIMITING REACTANT.
Product is determined by limiting reactant:
n(H₂) = n(HCl) ÷ 2 = 0.05 mol
mass(H₂) = 0.05 × 2 = 0.1 g
Using Moles to Balance Equations
EMPIRICAL FORMULA from masses or percentages:
Convert % or masses to moles (divide by Ar).
Find the simplest whole number ratio.
Write the empirical formula.
EXAMPLE:
4.0 g sulfur + 4.0 g oxygen react to form an oxide.
n(S) = 4.0 ÷ 32 = 0.125 mol
n(O) = 4.0 ÷ 16 = 0.25 mol
Ratio S:O = 0.125 : 0.25 = 1 : 2
Empirical formula = SO₂
MOLECULAR FORMULA from empirical formula:
Divide molecular mass by empirical formula mass.
Multiply empirical formula by this factor.
Example: empirical formula CH₂ (mass 14). Molecular mass = 56.
Factor = 56 ÷ 14 = 4. Molecular formula = C₄H₈.
⚠️ Common Mistake
The LIMITING REACTANT is the one that runs out — not the one in smaller quantity. You must compare the RATIO of moles available to the RATIO NEEDED from the equation. A large excess of one reactant does not make it the limiting reactant — only the molar comparison matters.
📐 Variables
cConcentration (c) is measured in mol/dm³ (mol/dm³)
nMoles (n) is measured in mol (mol)
VVolume (V) is measured in dm³ (dm³)
📐 Key Equations
c (mol/dm³) = n (mol) ÷ V (dm³)
n = c × V
Empirical formula: divide masses by Ar → find simplest ratio
📌 Key Note
c (mol/dm³) = n ÷ V (V in dm³). Limiting reactant: convert to moles → compare to equation ratio → whichever runs out first limits yield. Empirical formula: masses → moles → simplest ratio. Molecular formula: divide Mr by empirical mass → multiply.
🎯 Matching Activity — Moles Calculations Match
Match each calculation to the correct answer. — drag the symbols on the right to match the component names on the left.