How this works
Molarity (symbol M, units mol/L) is the workhorse concentration unit in wet-lab work: moles of solute per litre of solution. The arithmetic is small but the unit traps are real — molar mass in g/mol, volume in litres not millilitres, mass in grams not milligrams. Get one of those wrong and you'll be off by a factor of a thousand. The widget solves for any one of mass, volume or concentration given the other two and a molar mass; default molar mass is NaCl (58.44 g/mol) because it's the most-pipetted compound on earth.
The dilution mode runs the C₁V₁ = C₂V₂ identity — conservation of moles when you dilute a stock solution. Enter the stock concentration (C₁), the target concentration (C₂) and the final volume you want (V₂); the calculator returns the volume of stock to take (V₁). The rest of V₂ is solvent (usually water or buffer). This is the most common dilution question on a lab bench: "I have a 1 M stock, I want 100 mL of 0.1 M working solution, how much stock?" Answer: 10 mL stock + 90 mL solvent.
The maths is the same whether you're prepping antibiotics for a tissue culture, primers for a PCR, or sodium hydroxide for a titration. What changes is the units you reach for — micromolar (µM) for primers, millimolar (mM) for buffers, molar (M) for stock acids — so the calculator displays in mol/L and you scale to whichever unit is on your bottle.
The formula
n = moles, C = molar concentration (mol/L), V = volume (L), molar mass in g/mol. The dilution identity holds whenever moles are conserved — i.e. you're diluting a solution, not running a reaction. C₁ and C₂ must use the same units (both M, or both µM, etc.); same for V₁ and V₂. The widget asks for V₂ in mL because that's the bench-realistic unit; output V₁ is in the same.
Example calculation
- Make 1 L of 0.5 M NaCl. Molar mass 58.44 g/mol.
- Solve for mass: n = 0.5 × 1 = 0.5 mol → mass = 0.5 × 58.44 = 29.22 g.
- Dilute to 0.1 M: 100 mL of 0.1 M from 0.5 M stock → V₁ = 0.1 × 100 / 0.5 = 20 mL stock + 80 mL solvent.
Frequently asked questions
Why is the result in mol/L when my bottle says µM or mM?
mol/L (the SI base unit, also written M for molar) is the canonical concentration unit. The prefix forms are just scalings: 1 mM = 0.001 M, 1 µM = 0.000 001 M, 1 nM = 0.000 000 001 M. To convert, multiply or divide by the appropriate factor of 1,000. The widget shows mol/L because it doesn't make assumptions about which prefix you want — pick the one that gives a digit you'd actually write on a label.
How do I handle hydrates like CuSO₄·5H₂O?
Use the molar mass of the hydrated form, not the anhydrous one. CuSO₄ is 159.6 g/mol; CuSO₄·5H₂O is 249.7 g/mol because each formula unit carries five water molecules (5 × 18.0 = 90 g) in its crystal lattice. If you weigh out the pentahydrate but use the anhydrous molar mass, you'll under-concentrate by a factor of 249.7 / 159.6 ≈ 1.56. The label on the reagent bottle tells you which form you have — match it.
What if the solute changes the final volume noticeably?
The formula assumes the final volume is determined by what you measure out — that's why standard lab protocols dissolve the solute in a small amount of solvent first, then top up to the final volume in a volumetric flask, not the other way round. For dilute solutions (< 1 M) the volume contribution of the solute is usually negligible and you can shortcut by just measuring V mL of solvent and adding the solute. For concentrated solutions (e.g. > 5 M of a sugar or salt) always top-to-volume, otherwise you'll be 5 – 15 % off.