Dilution Calculator
C1V1 = C2V2, with the solvent to add and a serial dilution view.
About
Dilution Calculator
Diluting a solution changes the volume the solute sits in and leaves the amount of solute alone. That single fact is the whole calculation, and it is why one short equation covers preparing a buffer, making up a standard and mixing a cleaning solution.
The equation
C1V1 = C2V2
C1 and V1 are the concentration and volume of the stock you start from. C2 and V2 are the concentration and total volume you end with. Both sides equal the moles of solute, which is what stays constant.
Rearranged for whichever value you need:
- V1 = C2V2 ÷ C1, the stock volume to measure out
- C2 = C1V1 ÷ V2, the concentration you end with
- V2 = C1V1 ÷ C2, the volume to make up to
- C1 = C2V2 ÷ V1, the stock concentration needed
A worked example
You need 100 mL of 0.5 M solution and have a 5 M stock.
- V1 = (0.5 M × 100 mL) ÷ 5 M = 10 mL of stock
- Solvent to add: 100 - 10 = 90 mL
- Dilution factor: 5 ÷ 0.5 = 1 in 10
Measure 10 mL of stock, then add solvent up to the 100 mL mark. The moles of solute are 0.05 either way; only the volume changed.
Final volume, not solvent added
V2 is the total volume you finish with. It is not the amount of solvent you pour in.
Adding 100 mL of water to 10 mL of stock gives 110 mL, and a concentration of 0.4545 M rather than 0.5 M. That is a 9% error, and it is the most common mistake in the whole calculation. Make up to the mark rather than adding to it.
There is a further subtlety with concentrated solutions. Volumes are not always additive, since the solute and solvent can pack together more tightly than they sat apart. Mixing 50 mL of ethanol with 50 mL of water gives about 96 mL rather than 100. Making up to a volumetric mark sidesteps this entirely, which is why the technique exists.
Units cancel, so only consistency matters
The concentrations must share a unit and the volumes must share a unit, but the two pairs are independent. Molar with millilitres works fine, since the units cancel from both sides of the equation.
What does not work is mixing millimolar with molar in the same equation, or microlitres with millilitres. Convert first. This calculator normalises everything internally, so the stock can be entered in molar and the target in millimolar without a problem.
Serial dilution
Large dilutions are done in steps rather than in one go, because pipetting a very small volume accurately is difficult. Diluting 1 in 1000 by taking 1 µL into 1 mL depends entirely on that 1 µL being right, and a 5% pipetting error becomes a 5% error in the result.
Three tenfold steps reach the same place with far better accuracy: 1 mL into 9 mL, three times over. Each step uses volumes large enough to measure well, and the errors partly cancel rather than compounding on one measurement.
The arithmetic is straightforward. The overall factor is the step factor raised to the number of steps, so four 1-in-10 steps give 1 in 10,000. Microbiology plate counts and antibody titrations are built on exactly this.
Percent solutions
The same equation works on percentage concentrations, with a caution about which percentage is meant.
| Notation | Meaning | Example |
|---|---|---|
| % w/v | grams per 100 mL | 0.9% saline is 0.9 g NaCl in 100 mL |
| % v/v | mL per 100 mL | 70% ethanol is 70 mL in 100 mL total |
| % w/w | grams per 100 g | 37% HCl is 37 g in 100 g of solution |
Diluting 70% ethanol to 30% for a 500 mL batch needs (30 × 500) ÷ 70 = 214.3 mL of stock, made up to 500 mL. The percentages cancel just as molarities do.
Safety with concentrated acids
Add acid to water, never water to acid. Diluting concentrated sulfuric acid releases a large amount of heat, and water poured onto acid sits on the dense acid layer, boils, and throws acid out of the vessel. Adding acid slowly into a larger volume of water spreads the heat through the water instead.
Making the stock in the first place
Before there is anything to dilute, a stock has to be made from solid. Moles equal mass divided by molar mass, and molarity is moles per litre, so:
mass = molarity × volume in litres × molar mass
A 500 mL bottle of 0.1 M sodium chloride needs 0.1 × 0.5 × 58.44 = 2.922 g. Weigh the solid, dissolve it in less than the final volume, then make up to the mark. Dissolving in the full volume first gives a solution slightly too dilute, since the solid adds volume of its own.
Dilutions outside the laboratory
The same equation covers most household mixing.
| Task | Working |
|---|---|
| Bleach at 8.25% down to 0.1% for 1 L | 0.1 × 1000 ÷ 8.25 = 12.1 mL of bleach, made up to 1 L |
| Squash at 1 part in 5 | 50 mL concentrate into 250 mL total |
| Antifreeze to a 50% mix | Equal volumes, made up together |
Product labels often give a ratio rather than a percentage, and the ratio convention is inconsistent. Some read 1:10 as one part concentrate to ten parts water, an eleven-fold dilution. Others read it as one part in ten total, a tenfold dilution. The difference is about 10%, which matters for a disinfectant with a specified contact concentration.
Checking the answer
Two quick tests catch most errors. The stock volume must be smaller than the final volume, since dilution adds solvent. And the ratio of the volumes should match the ratio of the concentrations in reverse: diluting tenfold means the stock volume is a tenth of the final volume.
In the worked example above, 10 mL into 100 mL is a tenth, and 5 M to 0.5 M is a tenth. The two agreeing is a sign the arithmetic went the right way round, and disagreement usually means a concentration and a volume were swapped.
Diluting by mass instead of volume
Volumetric work depends on temperature, since liquids expand when warm and a flask calibrated at 20 °C reads slightly low at 30 °C. Weighing does not have that problem, so precise work sometimes dilutes by mass.
The equation becomes m1C1 = m2C2 with concentrations in molal units, moles per kilogram of solvent, rather than molar. Molality is independent of temperature for exactly this reason, and it is the unit used for freezing-point and boiling-point calculations where the sample is heated or cooled well away from where it was prepared.
Common mistakes
Treating V2 as the solvent added. It is the total final volume.
Mixing concentration units. Millimolar and molar in the same equation gives an answer off by a thousand.
Trying to concentrate. If C2 comes out above C1, dilution cannot get you there. Evaporation or a stronger stock is needed.
Ignoring the stock volume in the total. The stock is part of the final volume, not additional to it.
Common questions
Frequently asked questions
It states that the amount of solute is the same before and after dilution. Concentration times volume gives moles, so if the moles do not change, the product of the two must be equal on both sides. Only the volume the solute occupies has changed.
Subtract the stock volume from the final volume. For 10 mL of stock made up to 100 mL, that is 90 mL. Better practice is to add solvent up to the mark rather than measuring it separately, since volumes are not always exactly additive.
Not in the raw equation, since the units must cancel. Convert both to the same unit first. This calculator handles the conversion internally, so you can enter a molar stock and a millimolar target directly.
The ratio of the starting concentration to the final one. Going from 5 M to 0.5 M is a factor of 10, written as 1 in 10 or 1:10. Confusingly, some fields write 1:10 to mean one part stock to ten parts solvent, which is a factor of 11, so check which convention is in use.
Because very small volumes are hard to pipette accurately. Reaching 1 in 1000 through three tenfold steps uses volumes that can be measured well, while a single step would depend entirely on getting 1 microlitre exactly right.
Yes, as long as both concentrations use the same kind of percentage. Weight per volume, volume per volume and weight per weight are different measures and cannot be mixed, but any one of them works in the equation because the units cancel.
Diluting a concentrated acid releases a lot of heat. Water poured onto acid floats on the denser acid, boils at the surface, and spits acid out of the container. Adding acid slowly into a larger volume of water spreads the heat safely.
No. Mixing 50 mL of ethanol with 50 mL of water gives about 96 mL, because the molecules pack together more tightly than they did apart. Making the solution up to a volumetric mark rather than adding measured volumes avoids the problem.