Molarity Calculator and Formula: Full Guide
Calculate molarity with a free calculator. Learn the molarity formula, worked examples, dilution, mg/mL and ppm conversions, and titration.
Giroscience Scientific Review Team
10/15/202615 min read
Summary
Molarity tells you how many moles of a substance are dissolved in one litre of solution, and it is the most common way chemists describe concentration. The molarity formula is short, M = n / V, yet small slips with units, molar mass, or volume cause most of the errors students and lab technicians make. This guide explains what molarity means, gives you a free molarity calculator, and walks through worked examples for grams to molarity, dilution with M1V1 = M2V2, mg/mL and ppm conversions, normality conversions, titration, and pH. It also compares molarity with molality and normality and lists the mistakes that quietly ruin results. Every example is calculated step by step so you can check your own work.
Key Points
Molarity (M) equals moles of solute divided by litres of solution, not litres of solvent.
To go from grams to molarity, first convert grams to moles with the molar mass.
For dilutions, moles stay constant, so M1V1 = M2V2.
1 mg/mL equals 1 g/L, so dividing by the molar mass gives molarity directly.
Molarity changes slightly with temperature because the volume of a liquid changes.
Table of Contents
What Is Molarity?
Molarity is a measure of concentration that counts how much of a dissolved substance is present in a given volume of solution. The "amount" is measured in moles, the chemist's counting unit. NIST describes the mole as the SI unit of amount of substance and fixes it at exactly 6.02214076 × 10²³ elementary entities, which is the Avogadro constant (NIST). When you say a solution is "2 M", you mean it contains 2 moles of solute in every litre of solution, so the unit of molarity is mol/L, often written as M.
The official name is a little different. The IUPAC Gold Book calls the quantity amount concentration, defines it as the amount of a constituent divided by the volume of the mixture, and notes that "molarity" survives as a synonym mainly in older literature (IUPAC Gold Book). In practice, both names are used, and "molarity" remains the everyday word in classrooms and laboratories. Strictly speaking, the SI coherent unit is mol/m³, and 1 mol/L equals 1,000 mol/m³, but nobody weighs reagents in cubic metres.
Three words matter here. The solute is the substance that dissolves, such as salt. The solvent is the liquid that does the dissolving, usually water. The solution is the finished mixture of both. Molarity always divides by the volume of the solution, which is why you dissolve the solute in a little solvent and then top up to a final mark rather than adding a fixed volume of water.
Molarity also tells you about the particles in the beaker, not just the formula on the bottle. Ionic compounds split into ions when they dissolve, so a 1 M solution of sodium chloride holds 1 M sodium ions and 1 M chloride ions, while 1 M calcium chloride holds 1 M calcium ions and 2 M chloride ions. If you want to review why ionic compounds behave this way, our guide to ionic vs covalent bonds explains the difference, and the rest of this article builds on that idea.
The Molarity Formula and the Molarity Triangle
The molarity formula is M = n / V, where M is the molarity in mol/L, n is the amount of solute in moles, and V is the volume of the solution in litres. Everything else in this article is a variation on that one line. If you know any two of the three quantities, you can find the third.
Rearranging gives two more forms. The number of moles is n = M × V, and the volume is V = n / M. Many students remember this as the molarity triangle: put n at the top and M and V side by side at the bottom, cover the quantity you want, and the remaining symbols show the operation. Cover n and you see M × V. Cover M and you see n over V. Cover V and you see n over M.
When the starting point is a mass rather than moles, the formula expands to M = m / (MM × V), where m is the mass in grams and MM is the molar mass in g/mol. This version is what most people mean when they search for a molarity calculator formula, because lab work almost always starts with a balance reading. Our guide on how to calculate molar mass covers the step that comes before it.
A useful shortcut saves unit conversions. Because a millimole per millilitre equals a mole per litre, you can enter moles in millimoles and volume in millilitres and still get molarity in mol/L. The same logic lets you use micromoles per microlitre. Smaller concentrations use prefixes: 1 mM is 0.001 M, and 1 µM is 0.000001 M. Always write the unit with the number, because "0.5" alone could mean 0.5 M, 0.5 mM, or 0.5 mol.
Free Molarity Calculator
The calculator below solves the molarity formula for whichever value you leave blank. Enter any two of the three quantities and it returns the third, and it shows each step so you can learn the method rather than just copy the answer. It works in four modes: molarity from mass, volume and molar mass; molarity from moles and volume; the mass you need to weigh for a target molarity; and the volume that contains a given amount of solute.
To use it, start by choosing what you want to find. If you pick molarity, enter the mass of solute, its molar mass (also called molecular weight), and the final volume of your solution. Select grams or milligrams and litres or millilitres, since mixing units is the commonest source of a wrong answer. The calculator converts everything internally, so you do not need to. If you are unsure of a molar mass, add up the atomic masses from the periodic table, remembering to include any water of crystallization.
Here is a quick check. Enter 5.00 g of sodium hydroxide, a molar mass of 40.00 g/mol, and a volume of 500 mL. The calculator converts the mass to 0.125 mol, converts the volume to 0.500 L, and divides to give 0.250 M. You can run the same numbers by hand with the steps in the next section.
Remember what the tool assumes. It treats the volume you enter as the final volume of the solution, not the volume of water you added, and it does not correct for temperature or for non-ideal behaviour in very concentrated solutions. For the vast majority of teaching and routine laboratory work, those assumptions are fine. For high-precision analytical chemistry, you would calibrate your glassware and report the temperature.
How to Calculate Molarity From Mass and Volume
Introductory chemistry courses define molarity as moles of solute per litre of solution, as the LibreTexts UBC introductory chemistry page does, and they teach the same grams-to-moles route used here. Calculating molarity from grams takes three steps. First, convert the mass of solute to moles by dividing by the molar mass, n = m / MM. Second, convert the volume of the solution to litres. Third, divide moles by litres. The pipeline is the same every time, so it helps to see it laid out.
Take 11.69 g of sodium chloride dissolved and made up to 250 mL. The molar mass of NaCl is 58.44 g/mol, so the amount is 11.69 / 58.44 = 0.2000 mol. The volume is 250 mL, which is 0.250 L. The molarity is 0.2000 / 0.250 = 0.800 M. If you had mistakenly used 250 for the volume in millilitres, you would have reported 0.0008 and been off by a factor of 1,000, which shows why the unit step matters.
The same method works in reverse when you need to prepare a solution. Suppose you want 500 mL of 0.250 M sodium hydroxide. The moles required are n = M × V = 0.250 × 0.500 = 0.125 mol. With a molar mass of 40.00 g/mol, the mass is 0.125 × 40.00 = 5.00 g. Weigh that out, dissolve it in less than 500 mL of water, transfer it to a 500 mL volumetric flask, and add water up to the mark. Dissolving sodium hydroxide releases heat, so let the solution cool before the final top-up, because warm liquid occupies more volume.
If you already know the number of moles, skip the first step. Dissolving 0.35 mol of a compound in enough water to make 2.0 L of solution gives 0.35 / 2.0 = 0.175 M. This is the case behind the common question of calculating molarity using solute moles, and it is the simplest form of the formula.
Once you can calculate concentrations, you can predict how much product a reaction in solution will make, because the volume and molarity of each reactant give you the moles to compare. That is the bridge to our stoichiometry guide with worked examples, where mole ratios from balanced equations take over from here.
Dilution: M1V1 = M2V2
Dilution means adding solvent to lower a concentration. The key insight is that adding solvent does not add any solute, so the number of moles stays the same before and after. Since moles equal molarity times volume, we can write M1V1 = M2V2, where the subscript 1 refers to the concentrated stock and 2 refers to the diluted solution. Any units work, as long as the volume units match on both sides.
Consider diluting 25.0 mL of 12.0 M hydrochloric acid to a final volume of 500 mL. Rearranging gives M2 = M1V1 / V2 = (12.0 × 25.0) / 500 = 0.600 M. The solution is twenty times more dilute because the volume grew twenty times. Now reverse the problem: how much 12.0 M stock do you need to make 250 mL of 1.00 M acid? Rearranging for V1 gives V1 = M2V2 / M1 = (1.00 × 250) / 12.0 = 20.8 mL. You measure out 20.8 mL of stock and add water to reach 250 mL in total.
Two details trip people up. First, V2 is the final volume of the whole solution, not the volume of water you add, so you fill to the mark rather than pouring in 229.2 mL of water. Mixing liquids does not always give perfectly additive volumes, which is another reason to use a volumetric flask. Second, when diluting concentrated acids, always add the acid slowly to the water and not the reverse, because the mixing releases heat and can spatter.
The formula also works for serial dilutions, which are repeated dilutions in steps. Each step multiplies the dilution factor, so three tenfold dilutions in a row give a thousandfold reduction. That trick is common in microbiology and analytical chemistry whenever a sample is too concentrated to measure directly.
Converting mg/mL, ppm and Normality to Molarity
Concentrations appear in many units, and the skill is knowing how to turn them into molarity. The easiest case is mg/mL. One milligram per millilitre is the same as one gram per litre, so molarity equals the concentration in g/L divided by the molar mass. A glucose solution at 18.0 mg/mL contains 18.0 g/L, and with a molar mass of 180.16 g/mol, the molarity is 18.0 / 180.16 = 0.0999 M, or about 0.100 M.
Parts per million works similarly for dilute aqueous solutions, where one ppm is roughly one milligram per litre. Convert milligrams to grams by dividing by 1,000, then divide by the molar mass. For 500 ppm sodium chloride, that is 0.500 g/L divided by 58.44 g/mol, which gives 0.00856 M, or 8.56 mM. This shortcut assumes the solution has a density close to water, so it is not reliable for concentrated or non-aqueous solutions.
Normality is concentration expressed in equivalents per litre, and the link to molarity is M = N / n, where n is the number of equivalents each mole supplies in the reaction at hand. A 2 N sulfuric acid solution used in an acid-base reaction is 1 M, because each mole supplies two protons. A 0.50 N calcium hydroxide solution is 0.25 M, because each mole supplies two hydroxide ions. Because the factor depends on the reaction, you must know what the solution is being used for before converting.
Commercial reagents are usually labelled as a percentage by mass with a density, and you can convert those to molarity with M = (10 × %w/w × density in g/mL) / molar mass. For hydrochloric acid at 37% and a density of about 1.18 g/mL, with a molar mass of 36.46 g/mol, the result is about 12.0 M. The same method gives roughly 18.4 M for 98% sulfuric acid at 1.84 g/mL and about 15.7 M for 70% nitric acid at 1.41 g/mL. Treat these as approximate and check the certificate of analysis on your own bottle, since density varies by batch.
Finally, remember the metric prefixes. One molar equals 1,000 mM, 1,000,000 µM, and 1,000,000,000 nM. When a result looks absurdly large or small, a prefix slip is the first thing to check.
Molarity in Titration and From pH
Titration is one of the most useful applications of molarity. In a titration, you add a solution of known molarity from a burette to a measured volume of an unknown solution until the reaction is complete, which is called the equivalence point. For a reaction with a 1:1 mole ratio, such as hydrochloric acid and sodium hydroxide, the moles are equal at the endpoint, so Ma × Va = Mb × Vb.
Suppose 25.00 mL of hydrochloric acid of unknown concentration needs 20.00 mL of 0.1000 M sodium hydroxide to reach the endpoint. The moles of base are 0.1000 × 0.02000 = 0.002000 mol. The acid reacted in a 1:1 ratio, so it also holds 0.002000 mol, and its molarity is 0.002000 / 0.02500 = 0.0800 M. For reactions that are not 1:1, you apply the mole ratio from the balanced equation. Sulfuric acid reacts with sodium hydroxide in a 1:2 ratio, so the moles of acid are half the moles of base before you divide by the volume.
You can also find molarity from pH, but only in simple cases. The pH is defined by the concentration of hydrogen ions, so [H⁺] = 10^(−pH) mol/L. For a strong monoprotic acid such as hydrochloric acid, which dissociates essentially completely, a pH of 2.00 means [H⁺] = 0.0100 M, so the acid is 0.0100 M. For a strong base, find the pOH from 14 − pH at 25 °C, then take 10^(−pOH).
The method has limits. Weak acids dissociate only partly, so you also need the acid dissociation constant to work backwards from pH to the original molarity. Polyprotic acids release protons in stages and need more care. And in extremely dilute solutions near pH 7, the hydrogen ions from water itself become significant. When in doubt, titration remains the more trustworthy way to find an unknown molarity.
Molarity vs Molality vs Normality
Three concentration terms sound similar but measure different things. Molarity is moles of solute per litre of solution. Molality is moles of solute per kilogram of solvent. Normality is equivalents per litre of solution. Choosing the right one depends on what you are doing and how sensitive your work is to temperature.
Molarity is convenient because you measure liquids by volume, but volume changes with temperature. A solution made at 20 °C has a slightly different molarity at 30 °C because the liquid expands, even though nothing has been added or lost. Molality uses mass, and mass does not change with temperature, so molality is preferred for properties that depend on temperature, such as boiling point elevation and freezing point depression.
Converting between them requires the density of the solution. For a solution with molarity M, density ρ in g/mL, and solute molar mass MM, the molality is m = 1000 M / (1000 ρ − M × MM). A 1.00 M sodium chloride solution has a density of roughly 1.04 g/mL, so its molality is about 1000 / (1040 − 58.44), or roughly 1.02 mol/kg. In dilute aqueous solutions the two values are close, but they drift apart as the solution gets more concentrated.
Normality is the odd one out. Because it counts reactive equivalents, the same bottle can have different normalities depending on the reaction, and that ambiguity is why modern practice increasingly avoids it in favour of molarity plus a balanced equation. You will still meet it in older methods and some industrial specifications, so it helps to know the conversion from the previous section.
For a deeper look at how a compound's formula determines its mass and therefore its molar concentration, see our reference list of polyatomic ions, which makes it easier to write correct formulas for salts and add up their masses.
Common Mistakes to Avoid
The most frequent error is using the volume of solvent instead of the volume of solution. Molarity divides by the final volume, so a flask filled to the mark is the right reference. Closely related is the unit slip, where millilitres go into the formula as if they were litres. Convert first, then divide, and sanity-check the answer: if a result is a thousand times smaller than you expected, a unit slip is the likely cause.
The second group of mistakes involves the molar mass. Hydrated salts carry water in the crystal, so copper(II) sulfate pentahydrate has a molar mass of about 249.7 g/mol, not the 159.6 g/mol of the anhydrous form. Weighing the hydrate but using the anhydrous mass gives a solution that is too dilute by more than a third. Write the full formula, including the water, before you add up the masses, and use the correct charges and formulas for polyatomic ions.
Third, forgetting dissociation. A 0.10 M solution of calcium chloride has 0.20 M chloride ions, which matters for conductivity, precipitation reactions and osmotic calculations. Fourth, treating volumes in a dilution as additive without a volumetric flask. And fifth, converting normality to molarity without knowing the reaction. Pause at each of these and the rest of the calculation almost always takes care of itself.
FAQ
What is the formula for molarity?
The molarity formula is M = n / V, where M is molarity in mol/L, n is the moles of solute, and V is the volume of the solution in litres. If you start with a mass, use M = m / (MM × V), where m is the mass in grams and MM is the molar mass in g/mol. For example, 0.35 mol dissolved to make 2.0 L of solution gives 0.175 M. Rearranged forms are n = M × V and V = n / M. Always use the final volume of the solution, not the volume of water you added, and keep the volume in litres unless you also use millimoles.
How do you calculate molarity from grams and volume?
Convert grams to moles by dividing by the molar mass, convert the volume to litres, then divide moles by litres. For 11.69 g of NaCl in 250 mL of solution, the moles are 11.69 / 58.44 = 0.2000 mol, the volume is 0.250 L, and the molarity is 0.800 M. The same approach works for any solute as long as you use the correct molar mass, including water of crystallization for hydrates. A molarity calculator that takes mass, molar mass and volume does the same steps for you.
How do you calculate molarity from moles and liters?
Divide the number of moles by the volume in litres. If a solution contains 0.50 mol of solute in 0.25 L, the molarity is 0.50 / 0.25 = 2.0 M. If your volume is in millilitres, either divide it by 1,000 first or use millimoles with millilitres, because mmol/mL gives the same number as mol/L. For instance, 125 mmol in 250 mL is 0.50 M. This is the simplest use of the formula because no molar mass is needed.
How do you convert mg/mL or ppm to molarity?
For mg/mL, remember that 1 mg/mL equals 1 g/L, so divide the value by the molar mass. A solution at 18.0 mg/mL of glucose (180.16 g/mol) is about 0.100 M. For ppm in dilute water-based solutions, treat 1 ppm as 1 mg/L, divide by 1,000 to get g/L, and then divide by the molar mass. So 500 ppm NaCl is about 8.56 mM. The ppm shortcut assumes a density close to water and does not suit concentrated or non-aqueous solutions.
How do you convert normality to molarity?
Divide the normality by the number of equivalents supplied per mole in the reaction: M = N / n. In an acid-base reaction, sulfuric acid supplies two protons per mole, so 2 N H2SO4 is 1 M. Calcium hydroxide supplies two hydroxide ions, so 0.50 N Ca(OH)2 is 0.25 M. For monoprotic acids such as HCl and for NaOH, normality and molarity are equal. Because the factor depends on the reaction, you need to know how the solution will be used before converting.
How do you calculate molarity after dilution (M1V1 = M2V2)?
Because dilution does not change the amount of solute, the moles before and after are equal, so M1V1 = M2V2. Solve for the unknown. Diluting 25.0 mL of 12.0 M HCl to 500 mL gives M2 = (12.0 × 25.0) / 500 = 0.600 M. To make 250 mL of 1.00 M acid from the same stock, you need V1 = (1.00 × 250) / 12.0 = 20.8 mL, then fill to 250 mL with water. Volume units can be anything, provided they match on both sides.
How do you find molarity from pH?
For a strong monoprotic acid, which dissociates completely, the molarity equals the hydrogen ion concentration, which is 10^(−pH). A pH of 2.00 therefore corresponds to 0.0100 M. For a strong base, calculate pOH as 14 − pH at 25 °C and use 10^(−pOH) for the hydroxide concentration. This shortcut fails for weak acids, which need the acid dissociation constant, and for polyprotic acids or extremely dilute solutions, where water's own ions matter. In those cases, titration is a more reliable route.
What is the difference between molarity and molality?
Molarity is moles of solute per litre of solution, while molality is moles of solute per kilogram of solvent. Molarity depends on temperature because liquid volume changes as it warms or cools. Molality depends on mass, so it stays constant, which is why it is used in boiling point and freezing point calculations. In dilute aqueous solutions the two values are almost equal, but they diverge in concentrated ones. Converting between them requires the density of the solution.
Conclusion
Molarity is a simple idea, moles per litre of solution, that supports a large share of everyday chemistry. Once you are confident with M = n / V, you can move between mass, moles and volume, plan dilutions with M1V1 = M2V2, convert from mg/mL, ppm and normality, and use titration to find unknown concentrations.
The habits that matter most are small ones: use the final volume of the solution, keep your units consistent, get the molar mass right, and sanity-check the size of your answer. Try the calculator above on your own numbers, then confirm the result by hand until the method feels automatic.
© 2025. All rights reserved.
Disclaimer: While browsing our website, you may encounter affiliate links, as well as direct links to our own platforms where you can purchase our digital illustrations, products featuring our designs, or other science-related products.If you click on an affiliate link and make a purchase, we may earn a small commission at no additional cost to you. We only recommend products and services that we genuinely believe in and that align with our brand and values. Thank you for your support—it helps us continue creating content!