Molarity Calculator
Calculate the molarity (molar concentration) of a solution from the solute mass and molar mass, or directly from moles, and the total solution volume. Results are given in mol/L.
Find the Concentration of Your Solution
Molarity Formula
Where:
- \(c\) — molarity of the solution (mol/L)
- \(n\) — moles of solute (mol)
- \(V\) — total volume of the solution (L)
- \(m\) — mass of the solute (g)
- \(M\) — molar mass of the solute (g/mol)
The formula expresses the molar concentration: the number of moles of a substance dissolved per liter of solution. If you start from a mass, the moles are first obtained by dividing by the molar mass, \(n = m / M\).
Molarity – Calculation Example
Example: 58.44 g of sodium chloride (NaCl, molar mass 58.44 g/mol) is dissolved in water and the final solution volume is 0.5 L.
\[ n = \frac{m}{M} = \frac{58.44}{58.44} = 1 \text{ mol} \]
\[ c = \frac{n}{V} = \frac{1}{0.5} = 2 \text{ mol/L} \]
Result: the molarity is 2 mol/L (2 M), obtained from 1 mol of NaCl in 0.5 L of solution.
How to Use the Molarity Calculator
- Choose the Input Method: “Solute mass & molar mass” if you know how many grams you weighed out, or “Moles of solute” if you already know the amount in moles.
- If you selected the mass method, enter the Solute Mass (m) and pick its unit (g, mg, kg or lb), then enter the Molar Mass (M) in g/mol. For NaCl this is 58.44 g/mol; you can build molar masses from tabulated atomic masses.
- If you selected the moles method, enter the Moles of Solute (n) instead.
- Enter the Solution Volume (V) and select its unit (L, mL, µL, m³ or US gallons). This must be the total volume of the finished solution, not the volume of solvent added.
- Read the results: the Molarity in mol/L and the Moles of Solute in the solution.
What Is Molarity?
Molarity (symbol \(c\), often written as M) is the most common way to express the concentration of a chemical solution. It tells you how many moles of solute are contained in one liter of solution. A 1 M (“one molar”) solution contains exactly one mole of solute per liter; a 0.5 M solution contains half a mole per liter.
Because it is based on moles rather than mass, molarity connects directly to stoichiometry. When a reaction equation says 2 moles of A react with 1 mole of B, equal molarities and equal volumes mean the reactants are present in a simple stoichiometric ratio. That is why recipes in the lab, buffer preparations and titration calculations are nearly always written in mol/L.
A key point: molarity is defined per liter of solution, not per liter of solvent. In practice you dissolve the solute in somewhat less than the final volume, then top up to the exact volume in a volumetric flask. If you divided by the volume of water added instead, the result would be slightly wrong — dissolving a substance usually changes the total volume.
When you know a concentrated stock solution and need a dilute one, molarity combines naturally with the dilution relation \(C_1V_1 = C_2V_2\); our C1V1 = C2V2 calculator handles that step.
Assumptions, Limitations and Practical Tips
- Volume is the final solution volume. Always measure it after the solute is fully dissolved, ideally with volumetric glassware.
- The molar mass must match the actual substance. Use the molar mass of the form you weighed — anhydrous vs. hydrated salts differ a lot (e.g. CuSO₄ is 159.6 g/mol, but CuSO₄·5H₂O is 249.7 g/mol).
- Temperature matters for high precision. Volume expands slightly with temperature, so molarity is temperature-dependent; the related unit molality (moles per kg of solvent) is not.
- Molarity assumes the solute dissolves completely and unchanged. If the substance reacts with the solvent, dissociates or associates, the molarity you calculate refers to the formula unit you used, not to the actual species in solution. For acid–base work this distinction is important — our acetate buffer calculator shows how dissociation affects a prepared solution.
- Unit conversions are handled for you. Masses in mg, kg or pounds and volumes in mL, µL, m³ or gallons are converted internally, so the answer in mol/L is always on a consistent basis.
If you need to relate the amount of substance to the number of particles instead, the atoms-to-moles converter is the natural next step, since one mole always contains Avogadro's number of entities.
Frequently Asked Questions
How do I calculate molarity step by step by hand?
First convert the solute mass to moles by dividing by the molar mass: \( n = m / M \). Then divide the moles by the total solution volume in liters: \( c = n / V \). Example: 4 g of NaOH (M = 40 g/mol) in 500 mL gives \( n = 4/40 = 0.1 \) mol and \( c = 0.1/0.5 = 0.2 \) mol/L. If the volume is in mL, divide by 1000 first.
How do I convert molarity to other units like mM, µM or g/L?
Unit conversions are simple decimal shifts: 1 mol/L = 1000 mmol/L (mM) = 1,000,000 µmol/L (µM). To convert molarity to g/L, multiply by the molar mass: a 0.5 M NaCl solution contains 0.5 × 58.44 ≈ 29.22 g/L. For very dilute aqueous solutions, 1 mol/L of a substance with molar mass M roughly corresponds to \( 10^6/M \) ppm.
What is the difference between molarity, molality and normality?
Molarity (mol/L) counts moles per liter of solution; molality (mol/kg) counts moles per kilogram of solvent and is independent of temperature. Normality multiplies molarity by the number of equivalents — for H₂SO₄, a 1 M solution is 2 N in acid–base reactions because each molecule releases two protons. Pick the unit that matches your recipe or protocol.
Why is my calculated molarity wrong? Common mistakes
The most frequent error is dividing by the volume of solvent added instead of the final solution volume — dissolving solute changes the total volume, so always measure V after mixing. Other common mistakes: using the molar mass of the anhydrous salt when you weighed a hydrate (e.g. CuSO₄ vs. CuSO₄·5H₂O), forgetting to convert mL to L, or using a rounded molar mass with too few significant figures.
Does dissociation change the molarity of salts and acids?
Molarity always refers to the formula unit you dissolved. When NaCl dissociates, the solution is still 1 M in NaCl, but it contains 1 M Na⁺ ions and 1 M Cl⁻ ions. Similarly, 1 M CaCl₂ gives 1 M Ca²⁺ and 2 M Cl⁻. For stoichiometry and reaction calculations, decide whether you need the formula-unit concentration or the concentration of a specific ion.
How do I prepare a solution with a specific molarity?
Rearrange the formula to find the mass needed: \( m = c \cdot V \cdot M \). For example, to make 250 mL of a 0.1 M NaCl solution you need 0.1 × 0.25 × 58.44 ≈ 1.46 g. Dissolve the weighed solid in somewhat less than 250 mL of water, then top up to the mark in a volumetric flask — never just add 250 mL of water.
How can I dilute a stock solution to a lower molarity?
Use the dilution equation \( C_1V_1 = C_2V_2 \). If you need 100 mL of 0.5 M from a 2 M stock, then \( V_1 = (0.5 \times 100)/2 = 25 \) mL of stock, diluted with solvent to a final volume of 100 mL. The moles of solute stay constant; only the volume increases.