Free Dilution Calculator: How Much Stock, How Much Water
Enter any three of stock concentration, stock volume, final concentration, and final volume — 2.00 M down to 0.500 M in 250 mL — and get the fourth, the milliliters of solvent to add, and the rearrangement and substitution behind them, so the working is yours to reproduce on paper.
Dilution Calculator
Pick what to solve for, fill in the rest, and read the worked steps — every unit conversion, the rearranged formula, and the solvent you need to add are shown.
Diluting keeps the moles of solute the same and spreads them through more solvent, so M1 × V1 = M2 × V2. This page works one dilution from a stock solution, start to finish.
Which number is M1?
1 M = 1000 mM and 1 mM = 1000 µM — enter the unit you were given, the steps convert it.
1 M = 1000 mM and 1 mM = 1000 µM — enter the unit you were given, the steps convert it.
Total volume of the finished solution, not just the solvent added.
Volume of stock to measure out (V1)
0.025 L (25 mL)
- 1. Convert the final volume to litres250 mL ÷ 1000 = 0.25 L
- 2. Write the dilution equationM1 × V1 = M2 × V2
- 3. Rearrange M1 × V1 = M2 × V2 for the volume of stockV1 = (M2 × V2) ÷ M1
- 4. Substitute and solveV1 = (0.1 mol/L × 0.25 L) ÷ 1 mol/L = 0.025 L
- 5. Work out the solvent to add at the benchV(solvent) = V2 − V1 = 250 mL − 25 mL = 225 mL
What a dilution actually changes
Diluting a solution means adding more solvent to it. The amount of solute — the number of moles sitting in the flask — does not change at all; only the volume it is spread through gets bigger, so the concentration falls. Every dilution problem is an accounting of that one fact.
That is where M1V1 = M2V2 comes from. Moles equal concentration × volume, so M1V1 counts the moles you start with and M2V2 the moles you finish with. No solute was added or removed, so the two products must be equal. Subscript 1 is always the stock; subscript 2 the diluted solution.
The equation assumes the solute is spread out, never consumed, so it applies to one stock solution diluted once, not to a reaction. And it says nothing about technique: V2 is the volume the finished solution occupies, not the amount of solvent you poured in.
How to calculate a dilution: rearranging M1V1 = M2V2
One equation, four quantities: know any three and the fourth follows. Getting to it is ordinary algebra — divide both sides by whatever is sitting next to your unknown. Say you need V1, the volume of stock to measure out. Divide both sides of M1V1 = M2V2 by M1: on the left the M1 cancels and leaves V1 on its own, so V1 = M2 × V2 ÷ M1. Do that once with a pen and there is nothing to memorize.
The units look after themselves as long as the two concentrations share a unit and the two volumes share a unit — the FAQ below explains why. The four rearrangements, each with the division that produced it:
- Volume of stock to measure out — divide both sides by M1: V1 = M2 × V2 ÷ M1
- Concentration you end up with — divide both sides by V2: M2 = M1 × V1 ÷ V2
- Final volume to dilute up to — divide both sides by M2: V2 = M1 × V1 ÷ M2
- Concentration of the original stock — divide both sides by V1: M1 = M2 × V2 ÷ V1
Worked example: 250 mL of 0.500 M HCl
A lab sheet asks for 250 mL of 0.500 M hydrochloric acid, and the shelf holds a 2.00 M stock. The steps below are the ones the calculator prints.
Step 1 — label the four quantities. M1 = 2.00 M is the stock, M2 = 0.500 M the target, V2 = 250 mL the batch size, and V1, the stock volume, is the unknown.
Step 2 — rearrange for that unknown. Dividing both sides of M1V1 = M2V2 by M1 cancels the M1 on the left and gives V1 = M2 × V2 ÷ M1.
Step 3 — substitute, and let the units cancel. V1 = (0.500 M × 250 mL) ÷ 2.00 M = 125 ÷ 2.00 = 62.5 mL. The molarities divide out, so the answer arrives in milliliters, the unit V2 was already in.
Step 4 — work out the solvent. 250 − 62.5 = 187.5 mL of water. Put the water in first, add the acid to it, then fill to the mark: acid into water, never the reverse.
Mistakes that quietly change your concentration
- Adding V2 worth of solvent instead of diluting up to V2. V2 is the total final volume: here you add 187.5 mL of water, not 250 mL. Adding 250 mL leaves 312.5 mL at 0.400 M.
- Mixing units inside a pair. Writing V1 in milliliters and V2 in liters puts the answer out by a factor of 1000 — each pair has to agree with itself.
- Swapping the subscripts. The stock is always 1, so V1 can never exceed V2 and M2 can never exceed M1. If your entries say otherwise the calculator stops and says so, rather than reporting a negative solvent volume.
- Reusing M1V1 = M2V2 for a titration. The neutralization formula looks almost identical but carries stoichiometric coefficients, because there the solute is consumed rather than spread out.
- Assuming the unit must be molarity. M1V1 = M2V2 holds in any unit both sides share — 30% to 5% w/v is identical arithmetic — but the calculator is molar-only, so work percent or mg/mL yourself.
- Expecting volumes to add up exactly. Mixing contracts the volume slightly, so 62.5 + 187.5 mL may not read 250.0 mL; dilute to the mark in a volumetric flask.
How to check a dilution answer
A dilution answer audits itself, because the formula is really a statement about moles. Count them on both sides before anything else: 2.00 × 0.0625 = 0.125 mol going into the flask and 0.500 × 0.250 = 0.125 mol coming out of it. If those two products disagree, a number is in the wrong slot.
Next, check the direction. A dilution always ends less concentrated and larger in volume, so M2 < M1 and V2 > V1 with no exceptions; if either runs backwards, the subscripts are swapped. Then compare M1 ÷ M2 against V2 ÷ V1 — both are 4 here, and they must always match.
Finally, compare your working against the steps panel line by line; the first line where you disagree is where the error lives.
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Frequently Asked Questions
What is the dilution formula?
M1V1 = M2V2, where M1 and V1 are the concentration and volume of the stock solution and M2 and V2 are the concentration and volume of the diluted solution. It works because diluting adds solvent but no solute, so the moles on each side — concentration multiplied by volume — must come out the same. Knowing any three of the four values gives you the fourth by rearranging.
How much water do I need to add to dilute a solution?
Work out the volume of stock first, then subtract. For 250 mL of 0.500 M acid from a 2.00 M stock, V1 = (0.500 × 250) ÷ 2.00 = 62.5 mL of stock, so the solvent is 250 − 62.5 = 187.5 mL. The formula never hands you the water directly: V2 is the total volume of the finished solution, and the water is whatever is left once the stock is in.
Do I have to convert milliliters to liters first?
No, and that is the one place dilution is easier than molarity. Because concentration appears on both sides of M1V1 = M2V2, the concentration units cancel and so do the volume units — the only requirement is that both volumes share a unit and both concentrations share a unit. Enter 250 mL and 62.5 mL comes back. Molarity on its own is different: there the volume must be in liters.
What if my stock is given in grams, not molarity?
Turn it into a concentration first: M1V1 = M2V2 has no slot for grams. Divide the mass by the molar mass for moles, then by the stock volume in liters: 14.6 g of NaCl (58.44 g/mol) in 250 mL is 0.250 mol ÷ 0.250 L = 1.00 M, your M1. The bridge runs both ways: for grams in the diluted batch, multiply M2 × V2 in liters by the molar mass.
My textbook writes C1V1 = C2V2 — is that different?
No, it is the same equation with a different letter. M is used when the concentration is specifically molarity; C is the general symbol, chosen because the relationship holds for percent, ppm, mg/mL, or any other concentration unit. Some books write N1V1 = N2V2 for normality, which again is the same arithmetic. Whichever letter your course uses, subscript 1 is the stock and subscript 2 is the diluted solution.
How do I work out the dilution factor?
Divide the stock concentration by the final one, or the final volume by the stock volume; they agree by definition, and give 4 in the worked example. Ratio notation is where people slip: in general chemistry coursework, "1:4" means one part stock plus four parts solvent — five parts total, a dilution factor of 5, so 250 mL takes 50 mL of stock. Take that reading unless your lab sheet says otherwise.
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