Audited 29 Jul 2026·Last updated 28 Oct 2026·6 citations·Tier 1·0 uses

Normality Calculator (equivalents per litre)

Convert between normality and molarity, find the normality of a weighed mass or a titrated unknown, with equivalent weight — using the z of your reaction.

Normality Calculator

What do you want to work out?
Protons donated, hydroxides donated, or electrons transferred by one mole of the solute in the specific reaction you are running. H2SO4 fully neutralised is 2; H2SO4 to the first endpoint only is 1; KMnO4 is 5 in acid and 3 near neutral. Write the balanced half-equation and count. There is no universal value — this is why the field is required.
eq/mol
Default 98.072 g/mol = sulfuric acid, H2SO4, from IUPAC/CIAAW 2021 abridged standard atomic weights: 2(1.008) + 32.06 + 4(15.999). Supplier labels often print 98.08 from an older sulfur value — use whichever matches your certificate of analysis.
g/mol
Litres of solution you are preparing or analysing. Convert mL by dividing by 1000. Not used by the titration mode, which uses the aliquot volume instead.
L
Moles of the compound itself per litre — the SI-preferred figure. Used when solving for normality.
mol/L
Equivalents per litre, as printed on an older reagent label or specified in a titrimetric protocol. Used when solving for molarity.
eq/L
Grams actually placed on the balance. Used only by the 'normality from a mass' mode.
g
The standardised solution in the burette. Titration mode only.
eq/L
Burette reading at the endpoint, in litres. 21.20 mL is 0.02120 L. Titration mode only.
L
Volume of the unknown you pipetted into the flask, in litres. 25.00 mL is 0.02500 L. Titration mode only.
L
Normality
0.5
Equivalents of solute per litre of solution, in eq/L, for the reaction whose z you entered.
Molarity (SI-preferred)
0.25 mol/L
Equivalent weight
49.036 g/eq
Mass of solute
24.518 g
Equivalents present
0.5 eq

Background.

Older analytical methods are full of instructions such as "titrate with 0.1 N sodium hydroxide" or "prepare 0.5 N sulfuric acid". The N stands for normality: equivalents of reacting substance per litre of solution. For sodium hydroxide it changes nothing, because one mole supplies one hydroxide ion and 0.1 N is simply 0.1 mol/L. For sulfuric acid it halves the number you order. Each mole gives up two protons, so 0.5 N sulfuric acid is a 0.25 mol/L solution, and buying a bottle labelled 0.5 M would double the acid in every titration.

The multiplier is the equivalence factor z, and it belongs to the reaction rather than to the chemical. Sulfuric acid titrated through both protons has z = 2; stopped at the first endpoint, z = 1. Potassium permanganate takes up five electrons in acid but three near neutral, so a single 0.02 mol/L solution is 0.1 N for one method and 0.06 N for another. No formula can reveal which reaction you intend, so z is entered by hand in every mode, along with the molar mass of the exact compound on the balance.

That ambiguity is why NIST Special Publication 811 calls the term obsolete and asks for the concentration of a named entity instead — c[(1/2)H₂SO₄] = 0.5 mol/dm³ in place of 0.5 N — and why IUPAC's Green Book leaves normality out of its concentration table altogether. The arithmetic is untouched by the relabelling. Normality survives in pharmacopoeial volumetric solutions and water-analysis methods because it lets a titration be solved without any mole ratio: at the endpoint, equivalents of titrant equal equivalents of sample, whatever the two substances are. Every result here comes with the molarity beside the normality, so you can report whichever your audience expects.

What is normality calculator?

Normality, symbol N, is the number of equivalents of a solute per litre of solution, in eq/L. What counts as one equivalent depends on the reaction: one mole of hydrogen ions transferred in an acid–base reaction, one mole of electrons in a redox reaction, one mole of unit charge in a precipitation. If one mole of the compound supplies z equivalents in the reaction you are running, then N = z × c, where c is the ordinary molarity, and the equivalent weight — the mass that supplies a single equivalent — is the molar mass divided by z. Because z is a whole number of at least 1 in the reactions met in practice, normality is never smaller than molarity, and the two are equal only when z = 1, as for hydrochloric acid, sodium hydroxide or thiosulfate in an iodine titration. NIST's formal version of the same idea counts a fractional entity: if a solution holds 0.25 mol of H₂SO₄ per litre, it holds 0.5 mol of (1/2)H₂SO₄ per litre, and that second figure is exactly the normality.

How to use this calculator.

  1. Work out z before touching the calculator. Write the balanced equation or half-equation for your reaction and count what one mole of the solute supplies: HCl 1; H₂SO₄ 2, or 1 if you stop at the first endpoint; NaOH 1; Ca(OH)₂ 2; Na₂CO₃ 2 to the methyl-orange endpoint but 1 to phenolphthalein; KMnO₄ 5 in acid and 3 near neutral; K₂Cr₂O₇ 6.
  2. Enter the molar mass of the exact formula you weigh, water of crystallisation included. The default, 98.072 g/mol, is sulfuric acid from the 2021 standard atomic weights; supplier labels often print 98.08 from an older sulfur value.
  3. Choose the mode that matches your question: Normality when you have a molarity, Molarity when a method specifies a normality, Normality from a mass for a solution you made up, or the titration mode for an unknown.
  4. Enter every volume in litres. 25.00 mL is 0.02500 L and 21.20 mL is 0.02120 L. The titration mode ignores the solution-volume field and uses the aliquot volume instead.
  5. Copy the molarity into your notebook alongside the normality, or write the entity explicitly, such as c[(1/2)H₂SO₄] — either form tells the next reader which reaction the number assumed.

The formula.

N = z · c EW = M ⁄ z N₁V₁ = N₂V₂

Six relations cover every mode. Normality from molarity: N = z × c. Equivalent weight: EW = M ÷ z. Equivalents in a weighed mass: eq = m ÷ EW. Normality of a made-up solution: N = eq ÷ V. Mass for a given normality and volume: m = N × V × EW. And at a titration endpoint: N₁V₁ = N₂V₂.

The first line is a definition. NIST SP 811 gives the counting rule behind it: the amount of (1/x)B is x times the amount of B, so if 5 mol of H₂SO₄ is present, 15 mol of (1/3)H₂SO₄ is present. Applied per litre, the concentration of (1/z)B is z × c(B), which is N = z × c with the entity named.

The endpoint relation needs no separate assumption. An equivalent of acid is defined as the amount that reacts with one equivalent of base, so at the equivalence point the counts on both sides are equal by construction. Normality times volume turns each concentration into a count of equivalents; setting the two counts equal gives N₁V₁ = N₂V₂, with no stoichiometric coefficient anywhere. With molarities the same calculation needs the mole ratio from the balanced equation as an extra factor, and that factor is exactly what z has absorbed.

The equivalent weight is carried at full precision into every mass. Rounding 49.036 g/eq to 49.04 before multiplying would turn 24.518 g into 24.520 g, a 2 mg difference that a four-place analytical balance displays plainly.

Normality is per litre of solution, so it shares molarity's temperature dependence. Water expands by roughly 0.02 to 0.03 percent per degree Celsius near room temperature, and a dilute solution's normality falls by about that much as it warms, because the equivalents stay put while the volume grows. Volumetric glassware is calibrated at 20 °C for this reason.

z and the molar mass must be greater than zero, as must any volume that is divided by or used to report a mass. A normality or mass of exactly zero is accepted and simply returns zeros.

A worked example.

Example

A method calls for 0.5 N sulfuric acid in a titration where both protons react, and you need to know what to order and what to weigh for one litre. Choose Molarity mode, enter a normality of 0.5, z = 2, the molar mass 98.072 g/mol (2 × 1.008 + 32.06 + 4 × 15.999) and a volume of 1 L. The molarity is the normality divided by z: 0.5 ÷ 2 = 0.25 mol/L, so the bottle to buy is 0.25 M, not 0.5 M. The equivalent weight is 98.072 ÷ 2 = 49.036 g/eq. One litre at 0.5 eq/L holds 0.5 × 1 = 0.5 equivalents, and their mass is 0.5 × 49.036 = 24.518 g of pure H₂SO₄. The results read Normality 0.5 eq/L, Molarity 0.25 mol/L, Equivalent weight 49.036 g/eq, Mass of solute 24.518 g and Equivalents present 0.5 eq. This is the case NIST SP 811 uses when it retires the unit: rather than "a 0.5 N solution of H₂SO₄" it asks for "an amount-of-substance concentration of c[(1/2)H₂SO₄] = 0.5 mol/dm³". The half-molecule entity carries the reaction information that the bare N leaves out, and its value is the same 0.5. The 24.518 g is a mass of the pure acid; concentrated sulfuric acid is sold as a solution of stated strength, so in practice the mass or volume of concentrate is scaled from that assay, then added slowly to water.

volume1
equivalents Per Mole2
molar Mass98.072
normality0.5
solve Formolarity

Frequently asked questions.

Is a 1 N solution the same as a 1 M solution?
Only when z = 1. For hydrochloric acid, nitric acid, sodium hydroxide and potassium hydroxide, one mole supplies one equivalent and the two labels describe the same liquid. Sulfuric acid titrated through both protons is 0.5 mol/L at 1 N. Potassium dichromate, which takes six electrons when reduced to chromium(III), is 0.1667 mol/L at 1 N, with an equivalent weight of 49.0302 g/eq.
How do I decide the equivalence factor z for my reaction?
Count what one mole of your compound actually transfers under your conditions. For an acid, that is the protons neutralised up to the endpoint your indicator shows: phosphoric acid gives 1 at the first endpoint and 2 at the second, while its third proton is too weak to give a usable endpoint in water. For a base, count hydroxides supplied or protons accepted. For an oxidant or reductant, count electrons in the balanced half-equation.
Can the same bottle have two different normalities?
Yes. A 0.02 mol/L potassium permanganate solution is 0.1 N in acid, where MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O, and 0.06 N near neutral, where MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻. The liquid is identical; only the reaction changed. Its equivalent weight is 31.6064 g/eq in the first case and 52.6773 g/eq in the second, so a bottle labelled only in N is ambiguous outside the method that defined it.
How do I find the normality of an unknown from a titration?
Choose the titration mode and enter the titrant's normality, the burette volume at the endpoint and the aliquot volume. If 21.20 mL of 0.1000 N base neutralises a 25.00 mL aliquot of acid, the acid is 0.1 × 0.0212 ÷ 0.025 = 0.0848 eq/L, and the aliquot held 0.00212 equivalents. With z = 2 for a diprotic acid titrated completely, the molarity is 0.0424 mol/L.
Can it standardise a titrant against a weighed primary standard?
Yes, through the mass mode. Weigh the standard, titrate it, then enter its mass, molar mass and z, with the titrant volume in the volume field. For 0.5105 g of potassium hydrogen phthalate (204.222 g/mol, z = 1) consumed by 24.95 mL of sodium hydroxide, the result is 0.1002 eq/L — the normality of the base, because the equivalents of standard and titrant are equal at the endpoint.
Why do water-analysis methods like 0.02 N acid?
Because it turns burette readings straight into milligrams. Calcium carbonate's equivalent weight is 100.086 ÷ 2 = 50.043 g/eq, so one millilitre of 0.02 N acid neutralises 0.02 × 0.001 × 50.043 g, about 1.0009 mg of CaCO₃. Alkalinity expressed as CaCO₃ can then be read almost directly from the titre — the kind of convenience that kept normality alive in routine methods.
Why does NIST call normality obsolete?
NIST SP 811 states that the term normality and the symbol N should no longer be used, and gives c[(1/2)H₂SO₄] = 0.5 mol/dm³ as the replacement for 0.5 N sulfuric acid. The reasoning rests on the SI rule, restated in the BIPM's SI Brochure, that the elementary entities must be specified whenever the mole is used. Half a sulfuric acid molecule is a specified entity; a bare equivalent is not. Only the label changes, not the number.
Does temperature change a solution's normality?
Yes, exactly as it changes molarity, because both are counted per litre of solution. Warming expands the liquid while the equivalents stay fixed, so a dilute aqueous solution loses roughly 0.02 to 0.03 percent of its normality per degree near room temperature. Careful titrimetry records the temperature and works with glassware calibrated at 20 °C. Molality, which divides by the mass of solvent, has no such drift.
How do I dilute a stock solution to a lower normality?
Use N₁V₁ = N₂V₂, just as you would with molarities. Dilution changes only the volume, never the number of equivalents, so z cancels and need not even be known. To make 500 mL of 0.1 N from a 1 N stock, take 0.1 × 0.500 ÷ 1 = 0.050 L, or 50 mL, and make it up to the mark. With concentrated acids, always add the acid to the water, never the reverse.
What goes wrong if I use the anhydrous molar mass for a hydrate?
The normality comes out too high by the ratio of the two molar masses. Oxalic acid dihydrate, H₂C₂O₄·2H₂O, is 126.064 g/mol; 6.3032 g in one litre is 0.1 N with z = 2. Enter the anhydrous 90.034 g/mol by mistake and the same weighing reports 0.14 N, an error of 40 percent, because the crystal water was counted as acid.

References& sources.

  1. [1]NIST Special Publication 811, 2008 edition, 'Guide for the Use of the International System of Units (SI)', section 8.6.5 note 2: 'The term normality and the symbol N should no longer be used because they are obsolete. One should avoid writing, for example, "a 0.5 N solution of H2SO4" and write instead "a solution having an amount-of-substance concentration of c[(1/2)H2SO4] = 0.5 mol/dm3".' This is the worked example reproduced on this page. Retrieved 2026-07-29.
  2. [2]NIST Special Publication 811, 2008 edition, section 8.6.1 note 2: 'In general, n(xB) = n(B)/x, where x is a number. Thus, for example, if the amount of substance of H2SO4 is 5 mol, the amount of substance of (1/3)H2SO4 is 15 mol: n[(1/3)H2SO4] = 3n(H2SO4).' This is the identity c[(1/z)B] = z·c(B) that makes N = z·c formal. Retrieved 2026-07-29.
  3. [3]BIPM, 'The International System of Units (SI)', 9th edition, 2019, section 2.3.1 (definition of the mole): 'The amount of substance, symbol n, of a system is a measure of the number of specified elementary entities. An elementary entity may be an atom, a molecule, an ion, an electron, any other particle or specified group of particles.' Appendix 1 reproduces 14th CGPM (1971) Resolution 3 clause 2: 'When the mole is used, the elementary entities must be specified.' Consulted as an authority independent of NIST and IUPAC; it agrees. Retrieved 2026-07-29.
  4. [4]IUPAC, 'Quantities, Units and Symbols in Physical Chemistry' (Green Book), 3rd edition, 2nd printing 2012, section 2.10, composition-of-mixtures table. Searched for 'normality', 'equivalent concentration' and 'equivalent entity': no occurrences. IUPAC does not endorse or explicitly deprecate normality in this table — it lists amount concentration c_B = n_B/V and omits normality entirely. Recorded here as documented silence rather than agreement. Retrieved 2026-07-29.
  5. [5]IUPAC Commission on Isotopic Abundances and Atomic Weights (CIAAW), Standard Atomic Weights 2021, abridged values. Used to build the molar masses printed on this page: H 1.008, S 32.06, O 15.999 giving M(H2SO4) = 98.072 g/mol; K 39.098, Mn 54.938, O 15.999 giving M(KMnO4) = 158.032 g/mol. Supplier labels often print 98.08 for sulfuric acid, derived from an older sulfur value. Retrieved 2026-07-29.
  6. [6]NIST Special Publication 811, 2008 edition, front-matter check list item (18): 'The obsolete term normality and the symbol N, and the obsolete term molarity and the symbol M, are not used, but the quantity amount-of-substance concentration of B ... and its symbol cB and SI unit mol/m3 ... are used instead.' Confirms the deprecation is editorial policy for the whole document, not a footnote. Retrieved 2026-07-29.

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Method
N = z · c EW = M ⁄ z N₁V₁ = N₂V₂
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