Normality (N) Calculator
Solve for normality, grams to weigh, molarity, volume, or total equivalents — with real compound presets, worked steps, and visuals that show exactly how mass, moles, equivalents, and normality connect.
Background
Normality is equivalents of solute per liter of solution, related to molarity by N = M × n, where n — the n-factor — is the number of reactive units each mole delivers: H⁺ ions for an acid, OH⁻ ions for a base, or electrons transferred in a redox reaction. The same compound can have a different n depending on which reaction you're using it in, so always pick n for your specific reaction.
How to use this calculator
- Pick a goal: normality, grams to weigh, molarity, volume, or total equivalents.
- Set the n-factor: the number of H⁺ (acid), OH⁻ (base), or electrons (redox) delivered per mole in your specific reaction.
- Use a quick pick to auto-fill real molar masses and n-factors for common titration reagents, or enter your own.
- Read the visual and steps: a chart or flow diagram plus a full worked solution appear with every result.
Tip: for acid–base problems, n is usually the number of H⁺ or OH⁻ delivered per mole. For redox, n equals electrons transferred per mole in the balanced half-reaction.
Formulas & Equations Used
- Normality from molarity: N = M × n
- Normality from mass: N = (m/Mm × n) / V(L)
- Grams for target N: m = N × V(L) × Mm / n
- Molarity from N: M = N / n
- Volume from N, m: V(L) = (m/Mm × n) / N
- Equivalents: Eq = N × V(L)
n-factor quick reference
| Compound | Molar mass | n-factor | Why |
|---|---|---|---|
| HCl | 36.46 g/mol | 1 | Monoprotic acid, 1 H⁺ |
| H₂SO₄ | 98.08 g/mol | 2 | Diprotic acid, 2 H⁺ |
| H₃PO₄ | 98.00 g/mol | 3 (or less) | Triprotic; n depends on which H⁺ react |
| NaOH | 40.00 g/mol | 1 | Monobasic, 1 OH⁻ |
| Ca(OH)₂ | 74.09 g/mol | 2 | Dibasic, 2 OH⁻ |
| Na₂CO₃ | 105.99 g/mol | 2 | Fully neutralized carbonate accepts 2 H⁺ |
| KMnO₄ (acidic medium) | 158.03 g/mol | 5 | Mn⁷⁺ → Mn²⁺, 5 electrons |
| Na₂S₂O₃ | 158.11 g/mol | 1 | Typical iodometric titration |
n-factor is reaction-specific — the same compound can have a different n in a different reaction. Always confirm against your balanced equation.
Example Problems & Step-by-Step Solutions
Example 1 — Normality from molarity
0.250 M H₂SO₄ (n = 2).
N = M × n = 0.250 × 2 = 0.500 N
Example 2 — Grams for a target N
Make 250 mL of 0.100 N HCl (Mm = 36.46, n = 1).
m = 0.100 × 0.250 × 36.46 / 1 = 0.9115 g
Example 3 — Normality from mass (redox)
Dissolve 0.790 g KMnO₄ (Mm = 158.03, n = 5) to make 500 mL, in acidic medium.
N = [(0.790/158.03) × 5] / 0.500 ≈ 0.0500 N
Example 4 — Volume needed
You have 2.5 g NaOH (Mm = 40.00, n = 1) and need 0.200 N.
mol = 0.0625, Eq = 0.0625, V = 0.0625/0.200 = 0.3125 L (312.5 mL)
Frequently Asked Questions
What is the n-factor?
Equivalents per mole for the specific reaction: number of ionizable H⁺ for acids, OH⁻ delivered for bases, or electrons transferred per mole for redox reactions.
Is normality always the same for a substance?
No — it depends on the reaction. The same compound can have different n-factors in different reactions, so normality changes accordingly even though molarity doesn't.
Why use normality instead of molarity at all?
Normality directly tracks reactive capacity — one liter of 1 N acid always neutralizes one liter of 1 N base, regardless of how many H⁺ or OH⁻ each molecule carries. That makes titration math simpler than converting through molarity every time.
Can I enter volume in mL?
Yes — choose mL or L wherever a volume field appears, and the calculator converts to liters internally before computing.
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