What is the normality?
Enter the molarity of a solution and the equivalents each mole gives in your reaction, or the grams of solute, molar mass and volume. The calculator gives the normality, the molarity, the equivalent weight, or the grams to weigh for a solution of a given normality.
- Answer
- Normality = 0.2 N
Normality = 0.2 N (n = 2).
- Normality (eq/L)
- 0.2
- Molarity (mol/L)
- 0.1
Answer: Normality = 0.2 N. Normality = 0.2 N (n = 2).
How to calculate
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.
Example with the default inputs (Find Normality, Molarity (mol/L) 0.1, Equivalents per mole (n) 2): Normality = 0.2 N (n = 2).
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.
- 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
Each example is checked against the calculator on every build.
- Find Normality, Molarity (mol/L) 0.1, Equivalents per mole (n) 2 gives Normality (eq/L) 0.2, Answer 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
- Find Molarity, Normality (eq/L) 0.5, Equivalents per mole (n) 3 gives Molarity (mol/L) 0.166667, Answer 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
- Find From grams, Mass of solute (g) 4.904, Molar mass (g/mol) 98.08, Volume of solution 500 mL, Equivalents per mole (n) 2 gives Normality (eq/L) 0.2, Molarity (mol/L) 0.1, Equivalents (eq) 0.1, Equivalent weight (g/eq) 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
- Find Grams needed, Normality (eq/L) 5.3, Volume of solution 250 mL, Molar mass (g/mol) 58.44, Equivalents per mole (n) 1 gives Mass of solute (g) 77.433, Moles of solute (mol) 1.325, Answer 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
How it works
Normality (Harvey §16.1).
N = n × M
N is in equivalents per liter (eq/L, written N), M in moles per liter, and n is the equivalents per mole of the solute in its reaction: H⁺ given or taken for an acid or a base, electrons for a redox reaction, the ion’s charge for a precipitation, electron pairs for a complexation. The equivalent weight is EW = molar mass ÷ n.
- Normality: N = n × M.
- Molarity: M = N ÷ n.
- From grams: moles = grams ÷ molar mass; M = moles ÷ liters; equivalents = n × moles; N = equivalents ÷ liters (the same as grams ÷ EW ÷ liters).
- Grams needed: grams = N × liters × EW.
The volume is typed in mL or L and read back exactly as typed (1 mL = 0.001 L). The modes with grams also show the moles, the equivalents, the molarity and the equivalent weight.
Rules the page checks. n is a whole number from 1 to 10. Molarity from 10⁻¹² to 100 mol/L; normality from 10⁻¹² to 1,000 eq/L; grams from 10⁻⁹ to 10⁹; molar mass from 0.001 to 1,000,000 g/mol; volume from 10⁻⁹ mL to 10⁹ L.
How results are written. Every result is worked exactly from the typed decimals and rounded once. The sentence reads “Normality = 0.2 N”, “Molarity = 0.166667 M” or “Weigh 77.433 g”, then “(n = 2).”: 6 significant figures, trailing zeros dropped, thousands commas, and m × 10ⁿ below 0.000001 or from 10²¹.
Assumptions. n is for the reaction the solution is used in, and the volume is the final volume of the solution.
Worked examples by hand
0.1 M H₂SO₄, both H⁺ (Harvey: n = 2). N = 2 × 0.1 = 0.2 N.
0.5 N H₃PO₄, all three H⁺. M = 0.5 ÷ 3 = 0.166667 M.
4.904 g of H₂SO₄ in 500 mL. Moles = 4.904 ÷ 98.08 = 0.05 mol; equivalents = 2 × 0.05 = 0.1 eq; N = 0.1 ÷ 0.5 L = 0.2 N (M = 0.1). EW = 98.08 ÷ 2 = 49.04 g/eq.
250 mL of 5.30 N NaCl (n = 1; OpenStax Example 3.17). EW = 58.44 g/eq; grams = 5.30 × 0.250 × 58.44 = 77.433 g (OpenStax: 77.4 g), 1.325 mol.
Other questions people ask
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.