Dilution Calculator (C₁V₁ = C₂V₂)
Solve C₁V₁ = C₂V₂ for the final volume, the final concentration or the stock volume, with the solvent to add and the dilution factor.
Inputs
- Working
- C₂ = C₁V₁ ÷ V₂ = 5 × 0.85 ÷ 1.8 = 2.3611 mol/L
- Moles of solute (C₁V₁ = C₂V₂)
- 4.25 mol
- Dilution factor
- 2.1176×
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What each Dilution input means
- Solve for
- The unknown: the final volume, the final concentration, or the volume of stock to measure out.
- Stock concentration C₁
- The stock's concentration, in mol/L. Any unit works if C₂ uses the same one — %, g/L or ppm.
- Stock volume V₁
- The volume of stock used, in litres.
- Target concentration C₂
- The concentration you want after diluting, in the same unit as C₁.
- Final volume V₂
- The total volume after diluting, in litres — stock plus solvent.
Dilution formula
C₁ × V₁ = C₂ × V₂ (the moles of solute do not change) · V₂ = C₁V₁ ÷ C₂ · C₂ = C₁V₁ ÷ V₂ · V₁ = C₂V₂ ÷ C₁ · solvent to add = V₂ − V₁
Adding solvent to a solution spreads the same amount of solute through a larger volume, so the concentration falls in proportion. Because concentration × volume is the amount of solute, and that amount is unchanged, C₁V₁ before equals C₂V₂ after. Knowing any three of the four gives the fourth: how much stock to measure out to make a given volume at a target concentration, what concentration a stock becomes when made up to a volume, or what final volume a portion of stock must reach. The units just have to match on both sides — mL with mL, mol/L with mol/L, or even percent with percent — because they cancel. The page also gives the solvent to add (final minus stock volume) and the dilution factor.
Dilution worked example
OpenStax's Example 3.19: 0.850 L of 5.00 M copper nitrate is diluted to 1.80 L. C₂ = C₁V₁ ÷ V₂ = 5.00 × 0.850 ÷ 1.80 = 2.36 M, a dilution factor of 1.80 ÷ 0.850 = 2.12. The same solute, 4.25 mol, is in both. Asked the other way — how much 5.00 M stock makes 1.80 L of 2.36 M — V₁ = 2.36 × 1.80 ÷ 5.00 = 0.850 L, and 0.95 L of water is added.
More Dilution examples
500 mL of 1 M hydrochloric acid from 12 M concentrated stock: V₁ = 1 × 0.5 ÷ 12 = 0.0417 L — about 41.7 mL of acid, made up to 500 mL by adding it to about 458 mL of water (acid into water, never the reverse).
25 mL of a 2 M solution diluted to 250 mL in a volumetric flask: C₂ = 2 × 0.025 ÷ 0.25 = 0.2 mol/L, a tenfold dilution.
Where the Dilution calculator can give the wrong answer
- Solvent to add is final volume minus stock volume only when volumes are additive, which they nearly are for dilute aqueous solutions. Concentrated acids and alcohol–water mixes shrink on mixing; make the final volume up to the mark rather than adding a calculated amount.
- Add acid to water, never water to concentrated acid. The arithmetic is the same; the order is a safety rule.
- A target concentration above the stock's cannot be reached by dilution — the page says so instead of returning a negative solvent volume.
- Any consistent units work, but mixing them does not: a stock in mol/L and a target in mg/mL need converting first.
- Serial dilutions multiply: two 1 : 10 steps are 1 : 100. Run the page once per step, feeding each C₂ in as the next C₁.
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Dilution reference table
| Target concentration | Stock to measure | Water to make it up to 1 L (about) |
|---|---|---|
| 6 M | 500 mL | 500 mL |
| 3 M | 250 mL | 750 mL |
| 1 M | 83.3 mL | 917 mL |
| 0.5 M | 41.7 mL | 958 mL |
| 0.1 M | 8.3 mL | 992 mL |
Dilution FAQ
- Put the stock's concentration and volume on the left and the final concentration and volume on the right, then solve for the unknown. To make 500 mL of 0.1 M from a 2 M stock: V₁ = 0.1 × 500 ÷ 2 = 25 mL of stock, made up to 500 mL with 475 mL of water. Pick what to solve for above.
- Final volume minus the stock volume you started with. Diluting 100 mL of 1 M to 0.25 M needs a final volume of 1 × 100 ÷ 0.25 = 400 mL, so 300 mL of water. The page shows that line whenever it solves for a volume.
- The ratio of final volume to stock volume, or equivalently stock concentration to final concentration. A 1 : 10 dilution has a factor of 10 — 1 part stock in 10 parts total. Note that '1 : 10' in biology usually means 1 part in 10 total, not 1 part stock to 10 parts water.
- Yes — the equation only needs the same concentration unit on both sides. Diluting 10% bleach to 0.5% is a factor of 20, so 50 mL of bleach in 1 L of solution.
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