# What is the normality?

Finds the normality of a solution (equivalents per liter) from its molarity and the equivalents per mole, or from grams of solute, molar mass and volume; or the molarity from a normality, or the grams to weigh for a normality.

- Page: https://www.acalculator.org/chemistry/normality-calculator
- JSON spec: https://www.acalculator.org/chemistry/normality-calculator.json
- Version: 531f665df248

## Default answer

Example with the default inputs (Find Normality, Molarity (mol/L) 0.1, Equivalents per mole (n) 2): Normality = 0.2 N (n = 2).

## Inputs

| Key | Label | Description |
| --- | --- | --- |
| find | Find | Normality from molarity, molarity from normality, normality from grams, or the grams to weigh. |
| m | Molarity (mol/L) | The molar concentration: moles of solute per liter of solution. |
| nn | Normality (eq/L) | Equivalents of solute per liter of solution (N). |
| g | Mass of solute (g) | The grams of solute dissolved. |
| mm | Molar mass (g/mol) | The formula weight of the solute, such as 98.08 for H₂SO₄. |
| v | Volume of solution | The final volume of the solution, in milliliters or liters. |
| eq | Equivalents per mole (n) | H⁺ given or taken (acid–base), electrons (redox), the ion’s charge (precipitation): 2 for H₂SO₄ giving both H⁺. |

## Outputs

| Key | Label | Description |
| --- | --- | --- |
| result | Answer | The normality, the molarity, or the grams to weigh. |
| normalityValue | Normality (eq/L) | Equivalents per liter: n × molarity. |
| molarityValue | Molarity (mol/L) | Moles per liter: normality ÷ n. |
| equivalents | Equivalents (eq) | The equivalents of solute: n × moles. |
| moles | Moles of solute (mol) | Grams ÷ molar mass. |
| gramsValue | Mass of solute (g) | The grams of solute. |
| ew | Equivalent weight (g/eq) | Molar mass ÷ n: the grams that give one equivalent. |

## Method

N = n × M; M = N ÷ n; M = grams ÷ (molar mass × liters); equivalent weight EW = molar mass ÷ n; N = grams ÷ EW ÷ liters; grams = N × liters × EW.

## Assumptions

- n is the equivalents per mole of the solute in the reaction you use it for, so one solution can have several normalities: 1 M H₂SO₄ is 2 N when both H⁺ react and 1 N when one does.
- The volume is the final volume of the solution, not of the solvent.

## Worked examples

1. find = n, m = 0.1, eq = 2 gives normalityValue = 0.2, result = Normality = 0.2 N. Source: Harvey, Analytical Chemistry 2.1, §16.1 Normality (N = n × M; EW = FW ÷ n; n = 2 for H₂SO₄ with NH₃), https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Analytical_Chemistry_2.1_(Harvey)/16%3A_Appendix/16.01%3A_Normality.
2. find = m, nn = 0.5, eq = 3 gives molarityValue = 0.166667, result = Molarity = 0.166667 M. Source: Harvey, Analytical Chemistry 2.1, §16.1 Normality (N = n × M; EW = FW ÷ n; n = 2 for H₂SO₄ with NH₃), https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Analytical_Chemistry_2.1_(Harvey)/16%3A_Appendix/16.01%3A_Normality: M = N ÷ n.
3. find = mass, g = 4.904, mm = 98.08, v = 0.0005, eq = 2 gives normalityValue = 0.2, molarityValue = 0.1, equivalents = 0.1, ew = 49.04. Source: Harvey, Analytical Chemistry 2.1, §16.1 Normality (N = n × M; EW = FW ÷ n; n = 2 for H₂SO₄ with NH₃), https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Analytical_Chemistry_2.1_(Harvey)/16%3A_Appendix/16.01%3A_Normality; OpenStax, Chemistry 2e, §3.3 Molarity (M = mol solute ÷ L solution; Example 3.17: 5.30 M NaCl, 0.250 L, 58.44 g/mol gives 77.4 g), https://openstax.org/books/chemistry-2e/pages/3-3-molarity.
4. find = grams, nn = 5.3, v = 0.00025, mm = 58.44, eq = 1 gives gramsValue = 77.433, moles = 1.325, result = Weigh 77.433 g. Source: OpenStax, Chemistry 2e, §3.3 Molarity (M = mol solute ÷ L solution; Example 3.17: 5.30 M NaCl, 0.250 L, 58.44 g/mol gives 77.4 g), https://openstax.org/books/chemistry-2e/pages/3-3-molarity: 77.4 g; Harvey, Analytical Chemistry 2.1, §16.1 Normality (N = n × M; EW = FW ÷ n; n = 2 for H₂SO₄ with NH₃), https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Analytical_Chemistry_2.1_(Harvey)/16%3A_Appendix/16.01%3A_Normality.

## FAQ

### How do I calculate normality?

Multiply the molarity by n, the equivalents per mole: N = n × M. Sulfuric acid gives two H⁺ to a base such as ammonia, so n = 2 and a 0.1 M solution is 0.2 N.

### What is n, the number of equivalents?

It depends on the reaction. In an acid–base reaction it is the H⁺ an acid gives or a base takes; in a redox reaction, the electrons a species gives or takes; in a precipitation, the charge of the ion; in a complexation, the electron pairs a metal takes or a ligand gives (Harvey).

### How do I convert normality to molarity?

Divide by n: M = N ÷ n. A 0.5 N phosphoric acid solution used to give all three H⁺ is 0.5 ÷ 3 = 0.166667 M.

### What is equivalent weight?

The molar mass divided by n: the grams of the substance that give one equivalent. For H₂SO₄ with n = 2, it is 98.08 ÷ 2 = 49.04 g/eq. Normality is then grams ÷ equivalent weight ÷ liters.

### How many grams do I need for a solution of a given normality?

Grams = normality × liters × equivalent weight. For 250 mL of 5.30 N sodium chloride (n = 1, 58.44 g/mol): 5.30 × 0.250 × 58.44 = 77.433 g, the same as OpenStax’s 77.4 g for 5.30 M.

### Why is normality not used much any more?

Because it depends on the reaction: the same 1 M sulfuric acid is 1 N or 2 N as one or both H⁺ react, and Harvey notes that one solution can have several normalities. Molarity has one value, so it is the usual unit; normality is still met in titration work.

## Sources

- Harvey, Analytical Chemistry 2.1 (LibreTexts), §16.1 Normality: N = n × M, EW = FW ÷ n, and n for acid–base (H₂SO₄ with NH₃: n = 2), redox, precipitation and complexation reactions. https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Analytical_Chemistry_2.1_(Harvey)/16%3A_Appendix/16.01%3A_Normality (retrieved 2026-10-19)
- OpenStax, Chemistry 2e, §3.3 Molarity: M = mol solute ÷ L solution; Example 3.17 (0.250 L of 5.30 M NaCl at 58.44 g/mol: 77.4 g). https://openstax.org/books/chemistry-2e/pages/3-3-molarity (retrieved 2026-10-19)