Compute molarity from mass, molar mass, and volume, or use C₁V₁ = C₂V₂ to find how much stock solution and diluent you need for a target concentration.
Molarity ties together mass, moles, and volume into a single concentration number, and the dilution equation lets you predict exactly how a concentration changes when you add solvent. Together they cover the two most common solution-chemistry calculations an engineer runs into — preparing a solution from scratch, and diluting a stock solution to a target strength.
Molarity (M) is defined as moles of solute per liter of solution: M = n/V. Since moles aren't usually what you measure on a scale, the first step is converting mass to moles using the solute's molar mass: n = mass / molar mass. Combining both steps, M = (mass / molar mass) / volume.
Diluting a solution adds solvent but doesn't change the total moles of solute present. Since moles = concentration × volume, that conservation gives C₁V₁ = C₂V₂ — the concentration-volume product before dilution equals the concentration-volume product after. Solving for the stock volume needed: V₁ = C₂V₂ / C₁.
Once V₁ (the volume of concentrated stock solution to measure out) is known, the amount of solvent to add is simply the difference between the final target volume and that stock volume: diluent volume = V₂ − V₁. Mix the stock volume with that much solvent and top up to exactly V₂ to hit the target concentration.
To make 1 L of a 1 mol/L solution of NaCl (molar mass 58.44 g/mol), you need 58.44 g of NaCl dissolved and diluted to exactly 1 L — moles = 58.44/58.44 = 1 mol, M = 1/1 = 1 mol/L. To then dilute a 5 mol/L stock down to 0.5 mol/L in a final volume of 1 L: V₁ = (0.5 × 1)/5 = 0.1 L of stock, topped up with 1 − 0.1 = 0.9 L of solvent.
The formula only cares about the final concentration and volume, but in lab practice you should always add concentrated stock to solvent (not the reverse) for safety, especially with acids or bases, since diluting can release heat. Mathematically, once mixed and topped up to V₂, the result is the same either way.
That's the mass-to-molarity direction, not dilution — rearrange M = (mass/molar mass)/volume to mass = M × volume × molar mass, and solve directly for the mass of pure solute needed in a fresh solution rather than diluting an existing one.
Yes — it works for any consistent concentration unit (molarity, normality, percent w/v, ppm) as long as C₁ and C₂ use the same units and V₁, V₂ use the same volume units, since the equation is really just a statement that total solute quantity is conserved.
Environmental engineers use the same dilution math to size chemical dosing for water/wastewater treatment (chlorine, coagulants), and process engineers use it for batch dilution steps in manufacturing — the underlying conservation principle is identical regardless of the specific chemical involved.