Dilution Calculator (C1V1 = C2V2)

Background.

You have a bottle of 10 mol/L sodium hydroxide and a protocol that wants a litre at 0.5 mol/L. How much of the bottle goes into the flask? The answer is 50 mL, topped up to the litre mark, and the reasoning behind it is the dilution law, C1 × V1 = C2 × V2: the concentration and volume of the stock you take, multiplied together, equal the concentration and volume of what you end up with.

The law is conservation, nothing more. Taking V1 of a stock at concentration C1 removes a fixed amount of solute, C1 × V1 of it, whether that amount is in moles, milligrams or colony-forming units. Adding solvent spreads the same amount through a bigger volume without creating or destroying any, so the new concentration is that amount divided by the new volume. David Harvey's Analytical Chemistry 2.1 derives it exactly this way, from the definition of concentration as amount per volume.

Because only that conservation is involved, the units do not matter so long as they match. The same equation prepares a 0.01 mol/L standard from a 1 mol/L stock, a 20 mg/L working solution from a 1,000 mg/L reference, a ppm dosing solution, or a tenfold step in a bacterial plate count. What does matter is that C1 and C2 are in the same unit, and V1 and V2 are in the same unit.

Choose which of the four quantities you are missing and enter the other three. Most often that is the stock volume, since a protocol usually fixes the target concentration and the final volume and leaves you to work out how much to pipette; but the same equation answers how far a fixed aliquot will stretch, what concentration a dilution produced, or what a stock must have been. Alongside the answer you get the volume of diluent, V2 − V1, and the dilution factor, V2 ÷ V1, the number that goes on the tube label.

What is dilution calculator?

Dilution lowers the concentration of a solute by adding more solvent while the amount of solute stays fixed. Concentration is amount per unit volume, the IUPAC definition c = n/V for amount-of-substance concentration, so the amount in any portion of solution is concentration times volume. Before dilution that amount is C1 × V1; after, the same amount is C2 × V2. Setting the two equal gives the dilution law.

Two derived quantities do most of the practical talking. The diluent volume, V2 − V1, is how much water, buffer or solvent goes in on top of the stock. The dilution factor, V2 ÷ V1, equal to C1 ÷ C2, says how many times weaker the result is: a factor of 20 means the final solution has one-twentieth of the stock's concentration. Laboratories usually write that as "1:20" or "1 in 20", meaning one part stock in twenty parts total.

Serial dilutions repeat the step on the output of the previous one, and their factors multiply. Three successive 1-in-10 dilutions give 1 in 1,000, which is how a microbiologist brings a dense culture down to a plate with a countable number of colonies and then multiplies back through the chain to recover the original count.

How to use this calculator.

1. 01Pick the unknown from Solve for: stock volume (V1), final volume (V2), final concentration (C2) or stock concentration (C1). The field you are solving for is ignored, so its contents do not matter.
2. 02Enter the two concentrations in the same unit, whatever it is: mol/L, mg/mL, % w/v, ppm or CFU/mL. If a reagent bottle states a mass percentage, convert it to an amount-per-volume unit first using the density on the label or safety data sheet.
3. 03Enter the volumes in litres, as labelled; 50 mL is 0.05 L. Because only the ratio of volumes matters, you can enter both in millilitres instead, as long as you then read every volume result in millilitres too.
4. 04Read the solved value, then the diluent to add and the dilution factor. If your target is stronger than the stock, the calculator refuses the inputs, because no amount of added solvent can achieve that.
5. 05At the bench, measure the stock into a volumetric flask partly filled with diluent, mix, let it return to room temperature, then make up to the mark, rather than adding the calculated diluent volume directly.

The formula.

Start from the definition c = n/V. The solute in the stock aliquot is n = C1 × V1. Dilute that aliquot to a total volume V2 and the solute is still n, now spread through V2, so C2 = n ÷ V2 = C1 × V1 ÷ V2, which rearranges to C1V1 = C2V2. The four modes are this identity solved for each unknown in turn: V1 = C2 × V2 ÷ C1, V2 = C1 × V1 ÷ C2, C2 = C1 × V1 ÷ V2 and C1 = C2 × V2 ÷ V1. Every input must be greater than zero, since a zero concentration or volume would put a zero in a denominator or describe no solution at all.

The equation contains only products and ratios of matched quantities, which is why it is indifferent to units. Concentrations appear as C1 ÷ C2 and volumes as V2 ÷ V1, so mol/L, mg/mL or CFU/mL all cancel, provided both concentrations share one unit and both volumes share another.

Two assumptions sit underneath, and both occasionally fail. The first is that concentration means amount per volume. A mass-fraction concentration such as 37% w/w hydrochloric acid is amount per mass, and treating it as per volume ignores the change in density between the concentrated reagent and the dilute result. The second is that the final volume is measured, not summed. Mixing liquids does not always produce a volume equal to the sum of the parts; ethanol and water, for instance, shrink noticeably on mixing. That is why V2 should be the mark on a volumetric flask, and why the diluent figure V2 − V1 is a guide to how much solvent you will need rather than an amount to pour blind.

The bare equation does not stop C2 from exceeding C1, but the calculator does. When C2 is above C1 the law would give a V2 smaller than V1, a negative diluent volume and a dilution factor below 1, which describes concentrating, not diluting. Adding solvent can only lower a concentration, so those inputs are refused.

A worked example.

Frequently asked questions.

References& sources.

1. [1]IUPAC Gold Book, 'amount-of-substance concentration, c' — the formal IUPAC definition of concentration, c = n/V, underlying the dilution law's conservation-of-solute derivation. International Union of Pure and Applied Chemistry.
2. [2]Harvey, D.T. Analytical Chemistry 2.1, §2.5 'Preparing Solutions' — derives the dilution law C_o × V_o = C_d × V_d from conservation of the amount of solute and gives worked dilution examples. LibreTexts / DePauw University.
3. [3]NIST Chemistry WebBook — searchable reference database of molar masses, concentrations, and physical property data used to cross-check dilution calculations for real compounds. National Institute of Standards and Technology, Standard Reference Database 69.
4. [4]Skoog, D.A., West, D.M., Holler, F.J. & Crouch, S.R. (2013). Fundamentals of Analytical Chemistry, 9th ed. Cengage. Chapter 4 covers concentration units, dilution calculations, and standard laboratory dilution practice. ISBN 978-0495558286.
5. [5]Atkins, P.W. & de Paula, J. (2014). Atkins' Physical Chemistry, 10th ed. Oxford University Press. Foundations section on amount of substance and molar concentration, the basis for the conservation argument behind C1V1 = C2V2.