Normality expresses concentration in gram equivalents per litre of solution (eq/L, written N). An equivalent is the amount of a substance that supplies or reacts with one mole of reactive units: one mole of H⁺ in an acid–base reaction, one mole of OH⁻, or one mole of electrons in a redox reaction. So 1 M sulfuric acid, which can give up two protons, is 2 N, while 1 M hydrochloric acid is 1 N.
The advantage of normality is that equivalents always react one-to-one. At the end point of a titration, N₁V₁ = N₂V₂ regardless of how many protons or electrons each molecule carries. That is why normality is still common in water treatment (alkalinity and hardness in meq/L), clinical chemistry (electrolytes in mEq/L) and classical titrimetric analysis.
The catch is that the number of equivalents per mole depends on the reaction: phosphoric acid is 1, 2 or 3 N per mole depending on how far it is neutralised, and potassium permanganate supplies 5 electrons in acid but only 3 in neutral solution. IUPAC therefore discourages normality in favour of molarity, but it remains widely used. Choose a substance below or enter your own values.
How to use the Normality Calculator
- Calculate from
- Substance
- Molar mass
- Equivalents per mole (n-factor) — Protons donated (acids), hydroxide ions accepted/donated (bases) or electrons transferred per formula unit (redox).
- Equivalent weight of solute
- Mass of solute
- Volume of solution
- Molarity
Results update as you type, every field has its own unit menu, and everything is calculated in your browser.
How it works
Choose your method. With an equivalent weight, normality is mass ÷ (equivalent weight × volume). With a molar mass, the equivalent weight is first found as molar mass ÷ equivalents per mole. From molarity, multiply by the equivalents per mole.
Worked example
4.9 g of sulfuric acid (M = 98.072 g/mol, 2 acidic protons) is diluted to 500 mL. The equivalent weight is 98.072 ÷ 2 = 49.036 g/eq, so the solution contains 4.9 ÷ 49.036 = 0.09993 eq. Normality = 0.09993 eq ÷ 0.5 L = 0.1999 N, while the molarity is 4.9 ÷ 98.072 ÷ 0.5 = 0.0999 M – exactly half the normality.
Equivalents per mole of common reagents
| Reagent | Reaction | Equivalents per mole | Equivalent weight (g/eq) |
|---|---|---|---|
| HCl | acid–base | 1 | 36.458 |
| H₂SO₄ | acid–base | 2 | 49.036 |
| H₃PO₄ | full neutralisation | 3 | 32.665 |
| NaOH | acid–base | 1 | 39.997 |
| Ca(OH)₂ | acid–base | 2 | 37.046 |
| Na₂CO₃ | titrated to CO₂ | 2 | 52.994 |
| KMnO₄ | redox, acidic (→ Mn²⁺) | 5 | 31.606 |
| K₂Cr₂O₇ | redox, acidic (→ Cr³⁺) | 6 | 49.030 |
Tips
- Always start from a balanced equation – coefficients give the mole ratio.
- Convert everything to moles first, then use the ratio, then convert back.
- Every field has its own unit menu – g, mg, mol, mmol, L and mL all work.
- Everything is calculated in your browser.
Common mistakes to avoid
- Using mass ratios instead of mole ratios.
- Forgetting water of crystallisation in the molar mass of hydrates.
- Confusing molarity (per litre of solution) with molality (per kilogram of solvent).
Inputs explained
| Field | What to enter |
|---|---|
| Calculate from | Choose one: Mass, equivalent weight and volume, Mass, substance (molar mass and equivalents) and volume, Molarity and equivalents per mole |
| Substance | Choose one: Hydrochloric acid HCl (1 H⁺), Nitric acid HNO₃ (1 H⁺), Sulfuric acid H₂SO₄ (2 H⁺), Phosphoric acid H₃PO₄ (3 H⁺, full neutralisation) … |
| Molar mass | Units: g/mol, kg/mol, Da (u), kDa |
| Equivalents per mole (n-factor) | A number. Protons donated (acids), hydroxide ions accepted/donated (bases) or electrons transferred per formula unit (redox). |
| Equivalent weight of solute | Units: g/eq, mg/meq |
| Mass of solute | Units: g, mg, µg, kg, oz, lb |
| Volume of solution | Units: L, mL, µL, m³, dm³, cm³ … |
| Molarity | Units: mol/L (M), mmol/L (mM), µmol/L (µM), nmol/L (nM), mol/m³ |
What each result means
- Equivalents of solute = [Mass of solute]÷[Equivalent weight]
- Volume of a 1 N reagent that reacts with 1 L of this solution = [Normality]×1000
Formula
N = m ÷ (E × V) · E = M ÷ n_eq · N = c × n_eqNumber of equivalents = m ÷ E = n × n_eqWorked example
Inputs — Calculate from: Mass, substance (molar mass and equivalents) and volume; Substance: Sulfuric acid H₂SO₄ (2 H⁺); Mass of solute: 4.9 g; Volume of solution: 0.5 L.
Results — Molar mass = 98.072 g/mol; Equivalents per mole = 2 eq/mol; Equivalent weight = 49.036 g/eq; Normality = 0.199853 eq/L (N); Equivalents of solute = 0.099927 eq; Molarity = 0.099927 mol/L (M); Mass of solute per litre = 9.8 g/L; Volume of a 1 N reagent that reacts with 1 L of this solution = 199.85 mL.
Key terms
- Normality (N)
- Gram equivalents of solute per litre of solution.
- Equivalent
- Amount of substance that reacts with or supplies one mole of H⁺, OH⁻ or electrons.
- Equivalent weight
- Molar mass divided by the number of equivalents per mole, g/eq.
- n-factor
- Number of equivalents per mole of a substance in a specific reaction.
Frequently asked questions
What is normality in chemistry?
Normality is the number of gram equivalents of solute per litre of solution (eq/L, N). It equals molarity × the number of reactive units (H⁺, OH⁻ or electrons) per formula unit.
What is the normality of 1 M H₂SO₄?
2 N, because each molecule can donate two protons. 1 M HCl is 1 N and 1 M H₃PO₄ is up to 3 N.
How do I convert normality to molarity?
Divide by the equivalents per mole: M = N ÷ n. 0.5 N H₂SO₄ is 0.25 M.
What is an equivalent weight?
The mass that supplies one equivalent: molar mass ÷ equivalents per mole. For H₂SO₄ it is 98.072 ÷ 2 = 49.036 g/eq; for Na₂CO₃ titrated to CO₂ it is 105.988 ÷ 2 = 52.994 g/eq.
What is the normality of KMnO₄?
In acidic solution permanganate takes 5 electrons (MnO₄⁻ → Mn²⁺), so 0.02 M KMnO₄ is 0.1 N. In neutral solution (→ MnO₂) it takes only 3.
Why use normality in titrations?
Equivalents react 1 : 1, so N₁V₁ = N₂V₂ at the end point – 25.0 mL of 0.1 N acid needs exactly 25.0 mL of 0.1 N base, whatever the acid or base.
What is meq/L?
Milliequivalents per litre, 1/1000 of eq/L. Blood sodium of 140 mEq/L is 0.140 N (and 140 mmol/L, because Na⁺ has one charge).
What formula does the Normality Calculator use?
N = m ÷ (E × V) · E = M ÷ n_eq · N = c × n_eq.