Dilution Calculator — C₁V₁ = C₂V₂

How it works

When you dilute a solution by adding solvent, the amount of solute doesn't change — only the volume does. Since moles of solute = concentration × volume, that conservation gives the dilution equation (OpenStax Chemistry):

C₁V₁ = C₂V₂

where C₁ and V₁ are the concentration and volume of the concentrated stock, and C₂ and V₂ are those of the final diluted solution. To find how much stock you need, rearrange:

V₁ = (C₂ × V₂) ÷ C₁

Then the amount of solvent to add is simply the final volume minus the stock volume. The concentrations can be in any consistent unit (molarity, %, ppm) as long as C₁ and C₂ match, and the volumes likewise — the units cancel. In the lab, you measure out V₁ of stock and then add solvent until the total reaches V₂, rather than adding a pre-measured amount of water, because volumes are not always perfectly additive.

Worked example

To make 1 L of 0.5 M solution from a 2 M stock:

• Stock needed: V₁ = (0.5 × 1) ÷ 2 = 0.25 L
• Solvent to add: 1 − 0.25 = 0.75 L

Check: C₁V₁ = 2 × 0.25 = 0.5, and C₂V₂ = 0.5 × 1 = 0.5 — the solute is conserved.

Frequently asked questions

How do you calculate a dilution?

Use V₁ = (C₂ × V₂) ÷ C₁ to find the stock volume, then add solvent up to the final volume V₂. For a 2 M stock diluted to 0.5 M in 1 L, you need 0.25 L of stock.

What is the dilution equation?

C₁V₁ = C₂V₂. It states that the concentration times volume (which equals moles of solute) is the same before and after dilution, because adding solvent doesn't change the amount of solute.

What units should I use?

Any consistent ones. C₁ and C₂ must be in the same concentration unit and V₁ and V₂ in the same volume unit; the units cancel in the equation.

Should I add the calculated amount of water exactly?

Not exactly — measure the stock volume, then add solvent up to the final volume V₂. Volumes of solution and solvent are not always perfectly additive, so topping up to the mark is more accurate.

Can I use this to concentrate a solution?

No. Dilution only lowers concentration. If you ask for a final concentration higher than the stock, the required stock volume exceeds the final volume (the solvent-to-add comes out negative), which is not physically achievable by adding solvent.

Does this work for serial dilutions?

Each step of a serial dilution follows C₁V₁ = C₂V₂. Apply it repeatedly, using each step's diluted solution as the stock for the next, to reach very low concentrations.