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Molarity Calculator

Mass to weigh, molarity or volume — with a ready-to-follow preparation recipe.

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Solute

Molar mass NaCl: 58.440 g/mol

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%
From the label, for the mass to weigh.
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How to prepare it

      How it was worked out

      Next steps

      About the Molarity Calculator

      Solve c = n ÷ V for whichever you need: the mass of solute to weigh for a solution of a given molarity and volume, the molarity of a solution you made, or the volume that holds a given amount. Type the solute as a formula or name and its molar mass comes from IUPAC (CIAAW 2024) atomic weights — hydrates included, such as CuSO₄·5H₂O — or type a molar weight yourself, in kDa for a protein.

      Concentrations can be molar (M, mM, µM, nM, pM) or by mass (g/L, mg/mL, µg/mL, % w/v), and amounts can be grams or moles. A purity correction gives the mass of a less-than-pure reagent to weigh, and every result comes with the working and a short preparation recipe you can copy into a lab notebook.

      How to use it

      1. Type the solute (NaCl, Tris, CuSO4·5H2O) or choose Molar mass and type it (g/mol, or kDa for proteins).
      2. Choose what to solve for: Mass to weigh, Concentration or Volume.
      3. Type the two known values with their units; add the reagent’s purity if it is below 100 %.
      4. Read the answer, the same concentration in other units and the recipe; copy them if you need them.

      Examples

      Mass to weigh
      Input
      100 mL of 1 M NaCl
      Result
      Weigh 5.844 g of NaCl (58.440 g/mol)
      Molarity from mass
      Input
      25.2 g of acetic acid (CH3COOH) in 0.500 L
      Result
      0.8393 M

      OpenStax Chemistry 2e Example 3.16: 0.839 M.

      Molarity from moles
      Input
      0.133 mol of sucrose in 355 mL
      Result
      0.3746 M

      OpenStax Example 3.14: 0.375 M.

      Grams in a solution
      Input
      0.250 L of 5.30 M NaCl
      Result
      77.43 g

      OpenStax Example 3.17: 77.4 g.

      Volume for a mass
      Input
      75.6 g of acetic acid at 0.839 M
      Result
      1.500 L

      OpenStax Example 3.18.

      A buffer
      Input
      500 mL of 1 M Tris
      Result
      Weigh 60.57 g of Tris base, C₄H₁₁NO₃ (121.136 g/mol)
      Saline
      Input
      0.9 % w/v NaCl
      Result
      0.154 M (9 g/L)
      A protein
      Input
      1 mg/mL of a 66.5 kDa protein
      Result
      15.04 µM

      Common uses

      • Preparing stock solutions and buffers in the lab: how much to weigh.
      • Converting between mg/mL, % w/v and molar concentrations.
      • Protein and DNA work: micromolar and nanomolar concentrations from kDa.
      • Chemistry homework on molarity, with the working shown.

      Formulas and units

      • Molarity (amount concentration): c = n ÷ V, in mol per litre of solution, written M; 1 mM = 10⁻³ M, 1 µM = 10⁻⁶ M (IUPAC Green Book §2.10).
      • Mass of solute: m = c × V × M, where M is the molar mass.
      • Mass concentration: γ = m ÷ V (g/L, which is the same as mg/mL); molarity = γ ÷ M.
      • % w/v: grams per 100 mL of solution, so 1 % w/v = 10 g/L. Normal saline, 0.9 % NaCl, is 9 g/L = 0.154 M.
      • Purity: to get m grams of solute from a reagent that is p % pure, weigh m × 100 ÷ p grams.

      Preparing the solution

      Molarity counts the volume of the finished solution, not of the solvent you add (OpenStax Chemistry 2e §3.3). So weigh the solute, dissolve it in less solvent than the final volume, transfer it to a volumetric flask of that volume (rinsing the beaker into it), add solvent up to the mark and mix. Dissolving 5.844 g of NaCl in 100 mL of water gives slightly more than 100 mL of solution, and a concentration slightly below 1 M.

      Hydrates, salts and the right molar mass

      Use the formula of exactly what is on the label. Copper(II) sulfate pentahydrate, CuSO₄·5H₂O, is 249.677 g/mol, against 159.602 g/mol for anhydrous CuSO₄: weighing 24.97 g of the pentahydrate or 15.96 g of the anhydrous salt both give 1 L of 0.1 M copper sulfate. Buffers such as Tris and Tris hydrochloride, or disodium EDTA dihydrate, differ in the same way; their names work in the solute box.

      Sources

      • IUPAC, Quantities, Units and Symbols in Physical Chemistry (Green Book, 3rd ed., 2007), §2.10: amount concentration and mass concentration.
      • OpenStax, Chemistry 2e (CC BY 4.0), §3.3 Molarity (definition and worked examples).
      • IUPAC CIAAW, Standard Atomic Weights 2024; compound formulas behind the names checked against PubChem.
      • BIPM SI Brochure (2019): the Avogadro constant, 6.022 140 76 × 10²³ mol⁻¹, for the number of particles.

      Limitations

      • The calculator does not know solubility: it will happily compute a concentration that cannot dissolve. Check a data sheet for anything near saturation.
      • Molarity changes slightly with temperature because the volume of a solution does; volumetric flasks are calibrated at a stated temperature (often 20 °C).
      • For mass percentages by weight (% w/w) or molality you also need the density of the solution, which is not used here.
      • For diluting a stock solution (C₁V₁ = C₂V₂), the dilution is a separate step: the amount of solute stays the same while the volume changes.

      Privacy

      Everything happens in your browser. What you enter or open here is not uploaded or stored by MySmartCoPilot.

      Frequently asked questions

      How many grams do I need for a 1 M solution?

      Its molar mass in grams per litre of solution: for 1 L of 1 M NaCl weigh 58.44 g, for 100 mL weigh 5.844 g. In general m = c × V × M.

      How do I convert mg/mL to molarity?

      mg/mL equals g/L, so divide by the molar mass: 1 mg/mL of a 66.5 kDa (66,500 g/mol) protein is 1 ÷ 66,500 = 1.504 × 10⁻⁵ M, or 15.04 µM.

      What is the difference between molarity and molality?

      Molarity is moles per litre of solution (M); molality is moles per kilogram of solvent (mol/kg). Molality does not change with temperature, but converting between them needs the solution’s density.

      Should I use the hydrate or the anhydrous molar mass?

      The one that matches the bottle. If you weigh CuSO₄·5H₂O, type that formula (249.677 g/mol); the water in the crystals is part of what you weigh.

      What does % w/v mean?

      Grams of solute per 100 mL of solution: 5 % w/v glucose is 50 g/L, which is 50 ÷ 180.156 = 0.2775 M.

      How do I correct for purity?

      Type the purity from the label. For 5.844 g of NaCl from a 98 % pure reagent, weigh 5.844 ÷ 0.98 = 5.963 g.

      Quick answers and tool search

      Type to search tools or to get a quick answer, for example 18% of 2500. Use the up and down arrow keys to move through the results, Enter to choose, and Escape to close.