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Pick what to solve for, enter the values you know, and the calculator applies the right rearrangement of M = n / V, where n = mass ÷ molar mass. Finding molarity divides your moles by litres of solution; finding mass multiplies molarity × volume × molar mass; dilution mode uses M1V1 = M2V2. Every answer comes with its unit and the moles of solute as a cross-check, so the arithmetic is always transparent.
Molarity is the most common way chemists express concentration: the number of moles of solute dissolved in each litre of solution. Its unit is the molar, written M, so a 1 M solution contains 1 mole of solute per litre. The molarity formula is M = n / V, where n is moles of solute and V is the solution volume in litres. Molarity is always measured against the final solution volume, not the volume of solvent you started with, which is the single most misunderstood part of the definition.
Every molarity problem is one of four rearrangements. To find molarity: M = n / V, with n = mass in grams ÷ molar mass in g/mol. To find the mass you need: mass = M × V × molar mass. To find the volume a given mass makes: V = n / M. And for dilutions: C1V1 = C2V2, where C1 and V1 are the stock concentration and volume and C2 and V2 are the target. Our molarity calculator switches between these automatically based on the mode you pick, and it always shows the moles of solute as an intermediate so you can check its arithmetic.
The mode dropdown covers all four solve directions plus dilution. "Find molarity (mol/L)" takes mass, molar mass and volume and returns the concentration. "Find mass (g)" takes molarity, volume and molar mass and returns the grams to weigh out. "Find volume (L)" takes mass, molar mass and molarity and returns the litres that mass will make. "Find molar mass (g/mol)" works backwards from molarity, volume and mass to identify an unknown compound. "Dilution: find final volume (L)" uses M1V1 = M2V2 to tell you how much total solution a stock aliquot will make at a lower concentration.
Dilution problems conserve moles: the moles you take from the stock equal the moles in the final solution, so M1V1 = M2V2 (also written C1V1 = C2V2). Rearranged, the final volume is V2 = M1V1 / M2. For example, to dilute 0.5 L of 2 M stock to 0.5 M: V2 = 2 × 0.5 ÷ 0.5 = 2.0 L. The solvent to add is the difference, 2.0 - 0.5 = 1.5 L. The calculator reports both the final volume and the moles carried over, so you can confirm the dilution is correct before you touch a flask.
Molar mass is the sum of the atomic masses of every atom in the formula, in grams per mole. For sodium chloride (NaCl): Na = 22.99 + Cl = 35.45 = 58.44 g/mol. For water (H2O): 2 × 1.008 + 16.00 = 18.016 g/mol. Handy values: NaCl 58.44, HCl 36.46, NaOH 40.00, glucose C6H12O6 180.16, H2SO4 98.08, CaCO3 100.09, KOH 56.11, sucrose 342.30, ethanol 46.07 g/mol. Read atomic masses from the periodic table, watch out for subscripts and parentheses like Ca(OH)2, and add the waters of crystallisation for hydrated salts (more on that below).
Molarity spans many orders of magnitude, so prefixes do heavy lifting: 1 M = 1,000 mM (millimolar) = 1,000,000 µM (micromolar) = 1,000,000,000 nM (nanomolar). Match your mass and volume units: grams with litres, milligrams with millilitres, micrograms with microlitres, and the arithmetic stays clean. A classic slip is entering a volume in mL while the formula expects L; 250 mL is 0.250 L, and forgetting that factor of 1,000 is the most common error in molarity homework. Convert mL to L by dividing by 1,000 before you calculate.
Molarity (mol/L of solution) and molality (mol/kg of solvent) look similar but differ when temperature or density change, because solution volume expands with heat while solvent mass does not. Normality counts equivalents per litre, which matters for acids, bases and redox titrations: 1 M H2SO4 is 2 N because each mole releases two protons. Mass percent is grams of solute per 100 g of solution, common on reagent labels. Use molarity for volumetric lab work and dilutions, molality when temperature varies or for colligative properties like freezing-point depression, and normality when equivalents drive the reaction.
Molarity changes with temperature because liquids expand: warm a solution and its volume grows, so the same moles occupy more litres and the molarity drops slightly. That is why precise analytical work uses molality or standardises at a stated temperature. Solubility sets a hard ceiling: you cannot dissolve more solute than the solvent holds, no matter what the formula says. Hydrated salts carry water in their crystals (for example CuSO4·5H2O, molar mass 249.68 g/mol), so weigh the hydrate with its full molar mass; using the anhydrous mass would give you fewer moles than you planned.
The V in M = n/V is the final solution volume, not the volume of water you add. Dissolving 5.85 g of NaCl in 500 mL of water gives you more than 500 mL of solution, so the concentration is slightly less than 0.200 M. The correct procedure is to dissolve the solute in part of the solvent, transfer it to a volumetric flask, and then add solvent up to the mark. This "make up to volume" step is what makes the stated volume exact, and it is the difference between an approximate solution and a standard one.
Weigh your solute on an analytical balance, dissolve it completely in roughly two thirds of the final solvent volume, then transfer everything to a volumetric flask and rinse the beaker in. Add solvent until the bottom of the meniscus sits on the calibration mark. For concentrated acids, always add acid to water, never water to acid, because the dilution is strongly exothermic and water added to acid can boil explosively. Label every flask with the compound, concentration, date and your initials; an unlabelled clear liquid is a hazard, not a solution.
The classic stock case first: to make 1M NaCl solution at one litre, dissolve 58.44 g of sodium chloride in water and make up to 1.00 L. Example 1: make 250 mL of 0.5 M NaCl. Moles needed = 0.5 × 0.250 = 0.125 mol; mass = 0.125 × 58.44 = 7.305 g, so weigh 7.31 g and make up to 250 mL. Example 2: prepare 500 mL of 1 M HCl from 12 M stock. V1 = M2V2 / M1 = 1 × 0.5 ÷ 12 = 0.0417 L, so measure 41.7 mL of stock, add it to water, and make up to 500 mL. Example 3: 5.85 g NaCl (molar mass 58.44) in 500 mL. Moles = 5.85 ÷ 58.44 = 0.1001 mol; molarity = 0.1001 ÷ 0.5 = 0.2002 M.
Pick a mode from the dropdown, for example "Find molarity (mol/L)". Enter the values you know: mass of solute in grams, molar mass in g/mol, and solution volume in litres. Press calculate to get the answer with its unit plus the moles of solute as a cross-check. Switch to "Find mass (g)" when a protocol asks how many grams to weigh out, or to "Dilution: find final volume (L)" for M1V1 = M2V2 problems. Every input has sensible limits, and the result always shows four decimals so small concentrations stay readable.
Divide moles of solute by litres of solution: M = n / V. With mass instead of moles, use M = (mass ÷ molar mass) ÷ volume. For example, 5.85 g of NaCl (molar mass 58.44 g/mol) in 1 L: (5.85 ÷ 58.44) ÷ 1 = 0.1001 M. The calculator above does this in one step and shows the moles as a check.
First convert mass to moles: moles = mass in grams ÷ molar mass in g/mol. Then divide by the solution volume in litres. So 5.85 g NaCl in 500 mL is 0.1001 mol ÷ 0.5 L = 0.2002 M. Remember the volume is the final solution volume, measured after you make the solution up to the mark.
Rearrange the formula: mass = molarity × volume × molar mass. For 250 mL of 0.5 M NaCl: 0.5 × 0.250 × 58.44 = 7.31 g. Weigh that amount, dissolve it, and make the solution up to exactly 250 mL. Choose the "Find mass (g)" mode above to get the answer directly.
Use the "Find mass (g)" mode: enter the target molarity, the volume in litres, and the molar mass, and the calculator returns mass = M × V × molar mass. This is the reverse of the standard molarity calculation and the one you need most often at the bench, since protocols usually give concentration and volume.
M1V1 = M2V2 (also written C1V1 = C2V2) says moles are conserved when you dilute: stock concentration × stock volume = target concentration × target volume. Solve for the unknown: to find the final volume, V2 = M1V1 / M2. Example: 0.5 L of 2 M diluted to 0.5 M gives V2 = 2 × 0.5 ÷ 0.5 = 2.0 L, so add 1.5 L of solvent. The dilution mode above handles it, including the moles carried over.
Yes. The "Dilution: find final volume (L)" mode solves V2 = M1V1 / M2 from your stock molarity, stock volume and target molarity. It also shows the moles of solute transferred, so you can verify the dilution before preparing it. For the solvent to add, subtract the stock volume from the final volume.
Molarity is moles per litre of solution; molality is moles per kilogram of solvent. Molarity changes with temperature because liquids expand; molality does not, since mass is temperature independent. They are nearly equal in dilute aqueous solutions but diverge for concentrated solutions or non-aqueous solvents. See the comparison section above for when to choose each.
Choose molarity for volumetric lab work, titrations and dilutions at roughly constant temperature, since glassware measures volume. Choose molality for colligative properties (boiling-point elevation, freezing-point depression, osmotic pressure) and whenever the temperature varies during the experiment, because molality stays constant while molarity drifts.
Add the atomic masses of all atoms in the formula: NaCl = 22.99 + 35.45 = 58.44 g/mol, H2O = 2 × 1.008 + 16.00 = 18.016 g/mol. The common-chemicals table above lists handy values (HCl 36.46, NaOH 40.00, glucose 180.16, H2SO4 98.08). For hydrated salts like CuSO4·5H2O, include the water molecules: 249.68 g/mol.
Use the molar mass of the hydrate, waters included: CuSO4·5H2O is 249.68 g/mol, not 159.61. Weigh the hydrate directly; the crystal water dissolves with it and becomes part of the solvent. If a protocol specifies the anhydrous mass, convert first: multiply by (hydrate molar mass ÷ anhydrous molar mass).
Calculate the mass (mass = M × V × molar mass), weigh it, dissolve it in part of the solvent, transfer to a volumetric flask, and add solvent up to the calibration mark. Never just add the solute to the full final volume of solvent, because the dissolved solid changes the total volume. For acids, always add acid to water.
Because liquids expand when heated. The same moles of solute then occupy a larger volume, so moles per litre falls as temperature rises. The effect is small for water near room temperature but matters in precise analytical work, which is why standards are quoted at a stated temperature or use molality instead.
Moles per litre, mol/L, also written M (molar). Common subdivisions: mM (millimolar), µM (micromolar), nM (nanomolar), each a factor of 1,000 smaller than the last. Pair mass and volume units consistently: g with L, mg with mL, µg with µL.
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