How to Make and Dilute Aqueous Solutions: Calculations, Formulas, and Worked Examples

The four calculations solution preparation actually requires, and the three things that most often make the concentration come out wrong.

Written byMichelle Dotzert, PhD
Reviewed byTrevor J Henderson
Updated | 8 min read
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Editor’s note, September 2026: This article has been updated with an interactive calculator, worked examples using named reagents, and a concentration unit conversion table. The serial dilution section has been corrected: a dilution from 1 M to 1 mM is a thousandfold dilution requiring three tenfold steps, not six. Updated by [Trevor Henderson].

Preparing an aqueous solution is four calculations and one habit. The calculations are mass from molarity, dilution from a stock, dilution from a concentrated liquid reagent, and serial dilution. The habit is checking the formula weight on the bottle in front of you rather than the one in your head — because hydrates, assay percentages, and unit slips account for most of the solutions that come out at the wrong strength.

At a glance

To make a solution from a solid: mass (g) = concentration (mol/L) × volume (L) × formula weight (g/mol). Dissolve in roughly three-quarters of the final volume, then make up to volume — never add the full volume of water to the solid. To dilute a stock: C₁V₁ = C₂V₂, in any consistent units. To dilute a concentrated liquid acid or base, first convert its density and assay percentage into a molarity, then use C₁V₁ = C₂V₂. For a serial dilution, the number of steps is log(C_stock ÷ C_final) ÷ log(dilution factor) — and 1 M to 1 mM is a thousandfold dilution, which is three tenfold steps.

Before you calculate: three things that change the answer

Formula weight, not molecular weight of the anhydrous compound. This is the single most common source of a wrong-strength solution. Copper(II) sulfate is usually supplied as the pentahydrate, CuSO₄·5H₂O, at 249.68 g/mol — not as the anhydrous salt at 159.61. To make 250 mL of 0.1 M solution you need 6.242 g of the pentahydrate. Weigh out 3.990 g because you used the anhydrous figure and you get 0.064 M: 36 percent low. The number you want is the formula weight printed on that specific container, which already accounts for the waters of hydration.

Assay percentage, for anything that is not 100 percent pure. Reagent-grade solids are usually close enough to ignore. Technical grades, concentrated liquids, and anything hygroscopic are not. If the label says 95 percent, divide the calculated mass by 0.95.

What "concentration" means in the protocol you are following. Percent can mean three different things and they are not interchangeable. % w/v is grams per 100 mL — the usual meaning in a biological protocol. % v/v is millilitres per 100 mL, used for liquids in liquids. % w/w is grams per 100 g, used on reagent labels. A 10 percent solution made three ways gives three different concentrations. If the protocol does not say which, find out before you weigh anything.

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Making a solution from a solid

The calculation is mass (g) = concentration (mol/L) × volume (L) × formula weight (g/mol).

  1. Calculate the mass. Use the formula weight from the container, and convert your volume to litres before multiplying.
  2. Weigh the solid on an analytical balance, into a weigh boat. Wear the PPE the SDS calls for — our guide to ensuring a proper PPE fit covers the specifics, and powdered reagents are a respirable hazard as well as a contact one.
  3. Dispense roughly three-quarters of the final volume of purified water into a beaker. Purity grade matters more than people assume — see our guide to laboratory water purification, and for regulated work, USP purified water and WFI standards.
  4. Add a stir bar and start stirring, then add the solid. Adding solid to still water gives you a cake on the bottom that takes far longer to dissolve.
  5. Adjust the pH once the solid is fully dissolved, not before — dissolution changes it. Add dilute NaOH or HCl slowly. Our guides to choosing a pH meter and pH meter calibration and care cover the instrument side.
  6. Transfer to a volumetric flask and make up to the mark. This is the step the calculation depends on: q.s. to final volume, from quantum satis, "as much as is enough." Adding a litre of water to the solid instead of making up to a litre overshoots the volume and undershoots the concentration.

For a solution with more than one solute, calculate and weigh each one separately, add them all to the same beaker, then follow the same sequence from step 4.

Worked examples with real reagents

Formula weights below are typical published values. Confirm against your own container before weighing — this is the point of the section, not a formality.

Reagent

Formula weight

Target

Volume

Weigh out

Sodium chloride, physiological saline

58.44 g/mol

0.154 M (0.9% w/v)

500 mL

4.500 g

Tris base

121.14 g/mol

1 M

1 L

121.14 g

EDTA disodium dihydrate

372.24 g/mol

0.5 M

500 mL

93.06 g

Glucose, anhydrous

180.16 g/mol

0.1 M

1 L

18.02 g

Sodium phosphate dibasic

141.96 g/mol

0.1 M

500 mL

7.098 g

HEPES

238.30 g/mol

25 mM

1 L

5.958 g

Copper(II) sulfate pentahydrate

249.68 g/mol

0.1 M

250 mL

6.242 g

 

The last row is the hydrate trap, worked through. CuSO₄·5H₂O has a formula weight of 249.68 g/mol; the anhydrous salt is 159.61. Both numbers appear in reference tables and only one is on your bottle. Use the wrong one and 6.242 g becomes 3.990 g, and the solution you have made is 0.064 M rather than 0.1 M. EDTA disodium dihydrate at 372.24 is the same trap in the other direction, and it is common enough in buffer recipes to be worth memorising.

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Diluting from a stock solution

Where the mass required is too small to weigh accurately, make a concentrated stock and dilute it. The relationship is:

C₁V₁ = C₂V₂

C₁ and V₁ are the concentration and volume of the stock; C₂ and V₂ are the concentration and volume you want. Units cancel, so any consistent pair works — as long as both concentrations share a unit and both volumes share a unit.

Worked example. To prepare 100 mL of 0.5 M solution from a 5 M stock: V₁ = C₂V₂ ÷ C₁ = (0.5 × 100) ÷ 5 = 10 mL. Take 10 mL of stock and make up to 100 mL, which means adding 90 mL of diluent. As with a solid, add the stock first and bring up to volume — do not add 100 mL of diluent to 10 mL of stock.

For a stock containing several solutes at different concentrations, calculate each one independently, add each calculated volume to the flask, then make up to the final volume once.

Making a solution from a concentrated liquid reagent

Concentrated acids and bases are not sold by molarity. The label gives a density and an assay percentage, and you have to convert those into a molarity before C₁V₁ = C₂V₂ is any use:

stock molarity = (density × 1000 × assay% ÷ 100) ÷ formula weight

Worked example. Concentrated hydrochloric acid at density 1.18 g/mL and 37 percent assay, formula weight 36.46: (1.18 × 1000 × 0.37) ÷ 36.46 = 11.97 M. To make 500 mL of 1 M HCl: V₁ = (1 × 0.5) ÷ 11.97 = 0.0418 L, or 41.8 mL.

Reagent

Typical density

Typical assay

Formula weight

Approx. molarity

Hydrochloric acid, concentrated

1.18 g/mL

37%

36.46

11.97 M

Sulfuric acid, concentrated

1.84 g/mL

98%

98.08

18.38 M

Nitric acid, concentrated

1.42 g/mL

70%

63.01

15.77 M

Acetic acid, glacial

1.05 g/mL

99.5%

60.05

17.40 M

Ammonium hydroxide, concentrated

0.90 g/mL

28%

17.03

14.80 M

Phosphoric acid, concentrated

1.70 g/mL

85%

98.00

14.74 M

 

Two things about this table. Density and assay vary by supplier and by lot, so these are starting points and the label on your bottle is the authority. And the safety rule is not optional: add acid to water, never water to acid. Dilution of concentrated acids is strongly exothermic, and adding water to concentrated sulfuric acid can boil it back out of the vessel. Add the acid slowly to most of the final volume of water, with stirring, then make up to volume.

Serial dilutions

A serial dilution is a sequence of stepwise dilutions. It exists to avoid measuring volumes too small to pipette accurately — reaching a millionfold dilution in one step would mean pipetting 1 µL into 1 L — and to generate the concentration series a standard curve needs.

The number of steps is: steps = log(C_stock ÷ C_final) ÷ log(dilution factor).

Worked example, done carefully. To reach 1 mM from a 1 M stock using tenfold steps: 1 M ÷ 1 mM is a 1,000-fold dilution, because 1 M = 1,000 mM. steps = log(1,000) ÷ log(10) = 3. Three tubes, not six.

The unit conversion is where this goes wrong, so it is worth stating plainly: 1 M = 1,000 mM = 1,000,000 µM. Reaching 1 µM from a 1 M stock is the six-step version. Reaching 1 mM is three. Confusing the two produces a solution a thousandfold off target, and because every tube in the series looks identical there is nothing at the bench to catch it.

For a tenfold series with 1 mL in each tube: pipette 900 µL of diluent into every tube, transfer 100 µL of stock into tube 1 and vortex, transfer 100 µL from tube 1 into tube 2 and vortex, and continue. After three tubes you are at 1 mM.

Tube

Diluent

Transfer in

Concentration

Tube 1

900 µL

100 µL of 1 M stock

100 mM

Tube 2

900 µL

100 µL from tube 1

10 mM

Tube 3

900 µL

100 µL from tube 2

1 mM

Tube 4

900 µL

100 µL from tube 3

100 µM

Tube 5

900 µL

100 µL from tube 4

10 µM

Tube 6

900 µL

100 µL from tube 5

1 µM

 

Vortex every tube before drawing from it. Incomplete mixing is the most common error in a serial dilution and the most damaging, because the error compounds at every subsequent step — a 10 percent error in tube 1 is still there, multiplied, in tube 6. Change tips between transfers, and label the tubes before you start rather than during.

Concentration units and how to convert between them

The same solution gets described several ways depending on who is writing the protocol. Conversions between molar and mass-based units require the formula weight.

Unit

Definition

Conversion

Molarity (M)

Moles of solute per litre of solution

The reference unit. g/L = M × formula weight

Millimolar (mM)

Thousandths of a mole per litre

1 M = 1,000 mM

Micromolar (µM)

Millionths of a mole per litre

1 M = 1,000,000 µM; 1 mM = 1,000 µM

Normality (N)

Gram equivalents per litre

N = M × equivalents per mole. Equivalents depend on the reaction, not just the compound — sulfuric acid is 2 N per mole in a full neutralisation

mg/mL

Milligrams per millilitre

mg/mL = g/L = M × formula weight

% w/v

Grams per 100 mL of solution

% w/v = g/L ÷ 10. So 0.9% w/v NaCl = 9 g/L = 0.154 M

% v/v

Millilitres of liquid per 100 mL of solution

Not interchangeable with % w/v. Used for liquid-in-liquid

% w/w

Grams per 100 g of solution

Used on reagent labels. Converting to volume requires the density

ppm

Parts per million

For dilute aqueous solutions, ppm ≈ mg/L. The approximation holds where density is close to 1 g/mL

 

Normality is the one that causes trouble, because it is not a property of the compound alone — it depends on how many reactive equivalents the reaction consumes. The same 1 M sulfuric acid is 2 N in a titration against a strong base and something else in a redox reaction. Our guide to titrants and titrators covers standardisation, where this matters most.

Where the error actually creeps in

  • Graduated cylinder instead of a volumetric flask. A graduated cylinder is for approximate volumes. Final volume in a quantitative solution goes into a volumetric flask, which is calibrated to a single mark at a stated temperature.
  • Reading the meniscus from the wrong angle. Eye level, bottom of the meniscus, on the mark. A parallax error of a couple of millimetres in a 100 mL flask is a real percentage.
  • Making up to volume while the solution is warm. Dissolution can be exothermic and glassware is calibrated at a stated temperature, usually 20 °C. Let it equilibrate before the final addition.
  • Undissolved solute at the point of q.s. If it has not dissolved, it is not in solution, and topping up to the mark locks in a concentration you cannot recover.
  • Rinsing losses on transfer. Rinse the beaker into the flask two or three times and include the rinsings. Solute left on the beaker wall is solute missing from the solution.
  • Tap water. Ions in tap water participate in reactions, alter pH, and precipitate with phosphate and carbonate buffers. Purified water, always.

Labelling and storage

Every prepared solution needs, at minimum: the identity of the solute, the concentration and its unit, the solvent, the date prepared, the preparer’s initials, and any hazard information. An unlabelled beaker is an unknown, and an unknown is waste. Our guide to chemical labelling covers the convention and the secondary container question.

Add an expiry or re-check date for anything that degrades — buffers grow things, reducing agents oxidise, and light-sensitive compounds do not care that the label looks fine. Store in a container appropriate to the contents, and record the lot number of the reagent you used if the work is regulated or will be repeated.


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Frequently Asked Questions (FAQs)

  • How do you calculate how much solute you need?

    mass (g) = concentration (mol/L) × volume (L) × formula weight (g/mol). Convert the volume to litres first, and take the formula weight from the container rather than a reference table, because hydrated salts have a higher formula weight than their anhydrous equivalents.

  • What is the C₁V₁ = C₂V₂ formula used for?

    Calculating a dilution. C₁ and V₁ are the concentration and volume of the stock, C₂ and V₂ the concentration and volume you want. Units cancel, so any consistent pair works provided both concentrations share a unit and both volumes share a unit. Rearranged for the usual question: V₁ = C₂V₂ ÷ C₁.

  • How many steps does a serial dilution need?

    steps = log(C_stock ÷ C_final) ÷ log(dilution factor). Going from 1 M to 1 mM with tenfold steps is three tubes, because 1 M = 1,000 mM and a thousandfold dilution is three factors of ten. Going from 1 M to 1 µM is six. If the calculation does not give a whole number, either change the dilution factor or run one fewer step and finish with a single direct dilution.

  • How do you convert percent to molarity?

    For % w/v: multiply by 10 to get grams per litre, then divide by the formula weight. So 0.9% w/v sodium chloride is 9 g/L, and 9 ÷ 58.44 = 0.154 M. Note that % v/v and % w/w are different quantities and cannot be converted the same way — % w/w needs the solution density.

  • What is the difference between molarity and normality?

    Molarity is moles of solute per litre. Normality is gram equivalents per litre: N = M × equivalents per mole. The number of equivalents depends on the reaction rather than on the compound alone, so 1 M sulfuric acid is 2 N when fully neutralised by a strong base. Normality is mostly encountered in titration and standardisation.

  • How do you make a solution from concentrated acid?

    Convert the label information into a molarity first: stock molarity = (density × 1000 × assay% ÷ 100) ÷ formula weight. Concentrated hydrochloric acid at 1.18 g/mL and 37 percent works out to about 11.97 M. Then apply C₁V₁ = C₂V₂. Always add the acid to water rather than water to acid, because the dilution is strongly exothermic.

  • Why do you dissolve in less water than the final volume?

    Because dissolved solute occupies volume. If you add one litre of water to the solid, the resulting solution is more than one litre and the concentration is below target. Dissolve in roughly three-quarters of the final volume, then make up to the mark — q.s., from quantum satis, "as much as is enough."

  • Does it matter whether the salt is hydrated?

    Considerably. Copper(II) sulfate pentahydrate has a formula weight of 249.68 g/mol against 159.61 for the anhydrous salt. Using the anhydrous figure to prepare 250 mL of 0.1 M solution gives 3.990 g instead of 6.242 g, and the result is 0.064 M — 36 percent below target. The formula weight on the container already accounts for the waters of hydration.

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About the Reviewer

  • Trevor Henderson headshot

    Trevor Henderson BSc (HK), MSc, PhD (c), has more than two decades of experience in the fields of scientific and technical writing, editing, and creative content creation. With academic training in the areas of human biology, physical anthropology, and community health, he has a broad skill set of both laboratory and analytical skills. Since 2013, he has been working with LabX Media Group developing content solutions that engage and inform scientists and laboratorians. He can be reached at thenderson@labmanager.com.

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