Prepare a solution

Type only the data you have: the mass, moles or volume of the solute, the mass or volume of the solvent, the mass, volume or density of the solution, or any concentration. The calculator finds everything else by solving one relation after another and shows each step; if data are missing it says exactly which.

What you can calculate

Molarity, molality, mass/mass percent, mass/volume percent, volume/volume percent, ppm, mg/L and mole fraction.
The mass of solute to weigh or the volume to measure to prepare a solution.
Conversions between concentrations with the density (for example, from mass percent to molarity).
Molar mass from the solute formula.
Dilute-solution approximation (ρ ≈ 1 g/mL, ppm ≈ mg/L) only when it is valid, with a notice; it can be turned off.

Guide: concentration units and preparing solutions

The same solution can be described in many ways: molarity, molality, mass percent, ppm, mole fraction… Each one is useful for something different, and converting between them requires knowing which data are needed, usually the molar mass and sometimes the density. This guide summarizes the definitions, explains when the density is needed and works two typical lab cases.

The most common units

Molarity, M = moles of solute / liters of solution (mol/L). It is the most used in the lab, but it changes slightly with temperature, because the volume does.
Molality, b = moles of solute / kilograms of solvent (mol/kg). It does not depend on temperature; it is used for colligative properties.
Mass percent, % w/w = 100 · mass of solute / mass of solution.
Mass/volume percent, % w/v = grams of solute per 100 mL of solution.
Parts per million, ppm (mass/mass) = mg of solute per kg of solution. In very dilute water it is close to mg/L, because 1 L of that solution weighs almost 1 kg.
Mole fraction, x = moles of solute / total moles.

When is the density needed?

Going from a mass-based unit (% w/w, molality, ppm) to a volume-based one (molarity, % w/v, mg/L) requires the density of the solution, which links its mass and its volume. Only for very dilute aqueous solutions can ρ ≈ 1 g/mL be used; the calculator uses that approximation only if needed, the solvent is water and the solution is very dilute, and it tells you so.

Example 1: weighing a solid

How much NaOH must be weighed to prepare 500 mL of a 0.1 M solution?

Moles needed: n = M · V = 0.1 mol/L · 0.500 L = 0.05 mol.
Molar mass of NaOH: 22.990 + 15.999 + 1.008 = 39.997 g/mol.
Mass to weigh: m = n · M = 0.05 · 39.997 ≈ 2.00 g.

The solid is dissolved in less water than needed, transferred to a 500 mL volumetric flask and filled to the mark. Molarity is defined with the final volume of the solution, not with the volume of water added.

Example 2: concentrated sulfuric acid

Commercial sulfuric acid is 98 % w/w H₂SO₄ with a density of 1.84 g/mL. What are its molarity and molality?

Basis: 1 L of solution, which weighs 1000 mL · 1.84 g/mL = 1840 g.
Mass of solute: 0.98 · 1840 = 1803 g of H₂SO₄; mass of water: 1840 − 1803 = 36.8 g.
Moles of solute: 1803 / 98.07 ≈ 18.39 mol, so the molarity is 18.39 M.
Molality: 18.39 mol / 0.0368 kg ≈ 500 mol/kg: there is very little water. The mole fraction of the acid is ≈ 0.90.

Dilution. To prepare 250 mL of 1.00 M H₂SO₄ from this acid, use the fact that the moles of solute do not change on dilution, C₁V₁ = C₂V₂: V₁ = 1.00 · 250 / 18.39 ≈ 13.6 mL of concentrated acid, made up to 250 mL. For safety, always add the acid to the water, slowly, never the other way around, because mixing releases a lot of heat.

The results of both examples are those the calculator gives with these data.

How to use the calculator

Type the solute's formula so its molar mass is computed, or type the molar mass directly.
Type only the data you have: for the solute, the solvent, the solution or a concentration.
Choose what you want to know. Everything possible is found step by step and, if data are missing, it tells you which.

Common mistakes

Confusing molarity and molality: liters of solution versus kilograms of solvent.
Using the volume of water instead of the final volume of the solution.
Assuming ρ = 1 g/mL for concentrated solutions. In the sulfuric acid example it would give a badly wrong result.
Forgetting the water of hydration if the solid is a hydrate: the molar mass to use is the hydrate's.

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