Moles to Equivalents Converter
Common Conversions
| mol | eq |
|---|---|
| 0.01 | 0.01 |
| 0.1 | 0.1 |
| 0.5 | 0.5 |
| 1 | 1 |
| 2 | 2 |
| 5 | 5 |
| 10 | 10 |
| 0.5 | 1 |
| 1 | 2 |
| 1 | 3 |
| 1 | 5 |
Why this conversion matters in chemistry
Acid-base titration math runs on equivalents rather than moles because what the titrant consumes is one proton (or electron) per reactive site. 0.050 mol of H₂SO₄ gives 0.100 equivalents of acid — the base needed to reach the second endpoint. 0.050 mol of Na₂CO₃ gives 0.100 equivalents for full neutralization to CO₂. The mole-equivalent step normalizes different polyprotic acids, polyatomic bases, and multi-electron oxidants onto a common titration scale. The valence factor n is reaction-specific, not intrinsic to the substance.
Formula
eq = mol × n (where n = valence factor for the reaction)
Worked Examples
1 mol HCl = 1 eq
HCl donates 1 H⁺, so n = 1; 1 mol = 1 eq.
1 mol H₂SO₄ = 2 eq
H₂SO₄ donates 2 H⁺, so n = 2; 1 mol = 2 eq for full neutralization.
1 mol Ca(OH)₂ = 2 eq
Ca(OH)₂ provides 2 OH⁻, so n = 2.
1 mol KMnO₄ = 5 eq
In acidic solution MnO₄⁻ gains 5 e⁻, so n = 5 for the redox reaction.
Frequently Asked Questions
What is an equivalent?
An equivalent is the amount of substance that reacts with or supplies one mole of H⁺, OH⁻, electrons, or charges, depending on the reaction type. The definition depends on the reaction context, not on the substance alone.
How do I convert moles to equivalents?
Multiply by the valence factor n: for acids, n = number of H⁺ donated; for bases, n = OH⁻ provided; for redox, n = electrons transferred. The factor depends on the specific reaction.
Is the equivalent concept still used?
Yes, but less than before. IUPAC discourages equivalents in favor of moles with explicit stoichiometric context. The notation survives in clinical chemistry (mEq/L) and in some normality-based titration work.
How do equivalents relate to normality?
Normality (N) = equivalents per liter, equivalently N = M × n where M is molarity and n is the valence factor. So a 0.1 N H₂SO₄ solution is 0.05 M (since n = 2 for sulfuric acid in a full neutralization).