Reconstitution Calculator
Calculate the volume of solvent needed to reconstitute a dry compound to a desired concentration. Supports mg, mM, uM, and M units.
Reconstitution Calculator
Calculator
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Formula: Volume = (mass / molar mass) / desired concentration
Worked example โ Add 416.7 uL of solvent
Formula
Volume = (mass / molar mass) / desired concentration
Calculate the moles of compound available from the mass and molar mass, then divide by the desired molar concentration to find the volume of solvent needed. The result gives the total volume of solution to prepare.
Worked Examples
Example 1: Peptide Reconstitution
Problem:You have 5 mg of a peptide (MW = 1200 g/mol). How much solvent is needed for a 10 mM stock solution?
Solution:Moles = 5 mg / 1000 / 1200 g/mol = 0.000004167 mol Volume = 0.000004167 / 0.01 M = 0.0004167 L Volume = 0.4167 mL = 416.7 uL
Result:Add 416.7 uL of solvent
Example 2: Drug Reconstitution
Problem:Dissolve 50 mg of a drug (MW = 350 g/mol) to make a 100 mM solution.
Solution:Moles = 50 / 1000 / 350 = 0.0001429 mol Volume = 0.0001429 / 0.1 = 0.001429 L Volume = 1.429 mL
Result:Add 1.429 mL of solvent
Frequently Asked Questions
What is reconstitution in chemistry?
Reconstitution is the process of dissolving a dry or lyophilized (freeze-dried) substance back into a liquid solvent to create a solution of a specific concentration. This is commonly done with peptides, proteins, antibodies, and pharmaceutical compounds that are stored in powder form for stability. The key calculation involves determining how much solvent to add to achieve the desired concentration based on the mass of the powder and its molecular weight.
How do you calculate the volume needed for reconstitution?
First, calculate the number of moles of substance you have by dividing the mass (in grams) by the molar mass. Then divide the number of moles by the desired concentration (in mol/L) to get the volume of solvent needed in liters. For example, if you have 5 mg of a compound with a molar mass of 500 g/mol and want a 10 mM solution, you have 0.00001 mol. Dividing by 0.01 mol/L gives 0.001 L or 1 mL of solvent needed.
What solvents are commonly used for reconstitution?
The choice of solvent depends on the compound being dissolved. Water or sterile saline is used for water-soluble compounds. DMSO (dimethyl sulfoxide) is commonly used for hydrophobic compounds and small molecules in research. Ethanol or methanol can be used for moderately hydrophobic substances. For proteins and peptides, buffered solutions like PBS (phosphate buffered saline) are preferred to maintain biological activity. Always check the compound datasheet for recommended reconstitution solvents.
How should reconstituted solutions be stored?
Reconstituted solutions generally have shorter shelf lives than their dry counterparts. Most reconstituted proteins and peptides should be stored at -20 or -80 degrees Celsius and used within a few weeks to months. Aliquoting into small single-use volumes prevents repeated freeze-thaw cycles that can degrade the compound. Some reconstituted drugs must be used within hours at room temperature. Always check the product documentation for specific storage recommendations and expiration guidelines after reconstitution.
How do you work directly in milligrams, millimolar and millilitres?
Combine the unit factors into one step: volume in millilitres = (mass in mg x 1000) divided by (molar mass in g/mol x concentration in mM). For 5 mg of a compound with molar mass 500 g/mol at 10 mM, that is (5 x 1000) / (500 x 10) = 5000 / 5000 = 1 mL. Working in mg and mM avoids the string of decimal-place conversions between grams and moles, which is where most reconstitution mistakes are made.
Why does the powder's own volume matter in a vial?
A lyophilized solid displaces volume once it dissolves, so the final solution volume is larger than the diluent volume you injected. For a few milligrams the difference is negligible, but for a heavily loaded vial it dilutes the stock below the intended concentration. Protocols therefore distinguish between adding a stated volume of diluent and bringing the contents to a stated final volume. Follow whichever the datasheet specifies, and note that Reconstitution Calculator returns the final solution volume.
What is the difference between reconstitution and dilution?
Reconstitution dissolves a dry solid to create a stock of known molar concentration, so it needs the mass and the molar mass. Dilution takes an existing liquid stock and weakens it using C1V1 = C2V2, needing no molar mass at all. Most bench work uses both in sequence: reconstitute a concentrated stock that can be aliquoted and frozen, then dilute an aliquot into buffer or medium on the day of the experiment.
How does peptide net content change the mass you weigh?
Supplied peptide mass is often gross weight that includes counter-ions such as trifluoroacetate plus bound water, so the net peptide content can be well below 100%. Divide the peptide mass you need by the net content fraction to get the amount to weigh. To obtain 5 mg of peptide at 80% net content, weigh 5 / 0.80 = 6.25 mg. Ignoring this makes every downstream concentration lower than the label states, which quietly shifts a dose-response curve.
Why are DMSO stocks prepared at high concentration?
A concentrated stock lets the final assay receive a large dilution factor, which keeps the solvent itself from perturbing the system. A 10 mM stock diluted one thousand-fold gives 10 micromolar compound and carries only 0.1% DMSO into the well. Preparing a weak stock forces a small dilution factor and a much higher solvent fraction. The trade-off is solubility, since not every compound dissolves at 10 mM, and cold DMSO stocks can precipitate on thawing.
How can you confirm that the solid actually dissolved?
Inspect the vial against light for haze, undissolved flecks or a film on the glass, since anything left as solid means the true concentration is lower than calculated. Gentle vortexing, brief warming, or sonication usually completes dissolution. If material remains, the solubility limit in that solvent has been exceeded and either a different solvent or a lower target concentration is needed. Filtering or centrifuging out the residue without re-measuring leaves an unknown concentration.
References
Background & Theory
History
Reviewed for accuracy by Manoj Kumar, Mathematics Educator ยท Editorial policy
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