Concentration Calculator
Calculate solution molarity, the mass of solute needed for a target concentration, or the volume needed to dilute a known amount — with a visual for each, full steps, and a built-in converter between M, w/w%, w/v%, and ppm.
Background
Molarity (M) measures how many moles of solute are dissolved per liter of solution: M = n/V. The same relationship runs in three directions depending on what you already know and what you're solving for — the concentration itself, how much solute to weigh out, or how much to dilute to.
How to use this calculator
- Molarity finds concentration from a known amount of solute (mass or moles) dissolved in a known volume.
- Required Mass finds how much solute (in grams) to weigh out to reach a target molarity in a target volume.
- Required Volume finds how much solution volume a known amount of solute needs to reach a target molarity — the dilution question.
- Click Calculate to see the visual and full step-by-step working for whichever mode you picked.
How this calculator works
Molarity is moles of solute per liter of solution: M = n/V. Moles come either directly, or from mass divided by molar mass: n = m/MW.
Required mass rearranges the same relationship: first find the moles needed, n = M · V, then convert to grams with m = n · MW.
Required volume solves for V directly: V = n/M — with a fixed amount of solute, a larger volume always means a lower (more dilute) concentration.
The advanced converter relates M to lab-common percentage and ppm units using the same molar-mass relationship, adjusted for solution density where the unit is mass-based (w/w%) rather than volume-based (w/v%, ppm as commonly used for dilute aqueous solutions).
Formula & Equations Used
Molarity: M = n / V (mol·L⁻¹)
Moles from mass: n = m / MW
Required mass: m = M · V · MW
Required volume: V = n / M
Converter (approximate): w/v% ≈ (M·MW)/10; w/w% ≈ (M·MW·0.1)/ρ; ppm ≈ M·MW·1000 (ρ≈1)
Example Problems & Step-by-Step Solutions
These cover cases the Quick Examples chips above don't already demonstrate.
Example 1 — Diluting a stock solution
You have 2.50 mmol of KCl and need a 0.0500 M solution. What volume do you need?
Step: n = 0.00250 mol. V = n/M = 0.00250/0.0500 = 0.0500 L = 50.0 mL.
Example 2 — Weighing out a reagent
Prepare 500 mL of 0.250 M CaCl₂ (MW = 110.98 g/mol). How many grams are needed?
Step: n = M·V = 0.250 × 0.500 = 0.125 mol. m = n·MW = 0.125 × 110.98 = 13.87 g.
Example 3 — Finding molarity from mass
4.00 g of NaOH (MW = 40.00 g/mol) is dissolved to make 200 mL of solution. Find M.
Step: n = m/MW = 4.00/40.00 = 0.100 mol. M = n/V = 0.100/0.200 = 0.500 mol·L⁻¹.
Example 4 — Converting to w/v %
A 0.500 M glucose solution (MW = 180.16 g/mol) — what is its w/v %?
Step: w/v% ≈ (M·MW)/10 = (0.500 × 180.16)/10 ≈ 9.01%.
Frequently Asked Questions
Do I need the molar mass every time?
Only when working with mass instead of moles directly, or when using the converter — molar mass is what links grams to moles, and moles are what molarity is actually built from.
Why does entering both mass and moles use moles?
Moles are the more direct, exact quantity molarity is defined on, so when both are provided the calculator trusts the value that skips a conversion step.
Are the percentage/ppm conversions exact?
They're practical approximations assuming ideal mixing at 25 °C. w/w% and ppm depend on solution density; supply an accurate ρ for anything beyond a dilute aqueous solution.
What's the difference between w/w% and w/v%?
w/w% is grams of solute per 100 grams of solution (mass-based, needs density to relate to molarity); w/v% is grams of solute per 100 mL of solution (volume-based, and doesn't need density at all).
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