Molarity Calculator
Example 1
A plant biologist prepares a potassium chloride (KCl) solution for use as a fertiliser feed. She dissolves 7.46 g of KCl in water and dilutes to a total volume of 250 mL. The molar concentration is needed to calculate the potassium ion concentration for the hydroponic nutrient formula.
Solution
KCl has a molar mass of 74.55 g/mol. The moles of KCl are:
The total volume is 250 mL, which is 0.250 L. Applying the molarity formula:
Since each KCl formula unit dissociates into one K⁺ and one Cl⁻ ion, the potassium ion concentration is also 0.400 M. This is within the typical range for hydroponic nutrient solutions.
Answer
0.400 M KCl
The potassium concentration is also 0.400 M, suitable for the hydroponic feed. For comparison, standard Hoagland solution uses approximately 6 mM K⁺, so this concentrate would be diluted further before use.
The potassium concentration is also 0.400 M, suitable for the hydroponic feed. For comparison, standard Hoagland solution uses approximately 6 mM K⁺, so this concentrate would be diluted further before use.
Example 2
A chemist needs exactly 0.50 mol of HCl for a neutralisation reaction. The stock bottle contains 2.0 M HCl. The chemist must calculate the volume to measure out.
Solution
The molarity formula relates moles, volume, and concentration:
The chemist knows the required moles (n = 0.50 mol) and the stock concentration (M = 2.0 M). Rearranging the formula to solve for volume:
Substituting the known values into the rearranged formula:
Converting litres to millilitres for practical measurement: 0.25 L × 1000 mL/L = 250 mL.
Answer
250 mL of 2.0 M HCl
Measure 250 mL of the stock acid using a graduated cylinder. For improved accuracy, use a 250 mL volumetric flask and a pipette rather than a graduated cylinder.
Measure 250 mL of the stock acid using a graduated cylinder. For improved accuracy, use a 250 mL volumetric flask and a pipette rather than a graduated cylinder.
Example 3
A biochemistry student needs to prepare 500 mL of 0.1 M NaCl for a protein purification protocol. The protocol specifies the concentration and volume precisely because the ionic strength of the buffer affects protein binding to the chromatography column.
Solution
The student needs 500 mL of 0.1 M NaCl. Rearranging the molarity formula to solve for mass gives:
Substituting the known values:
The units cancel as mol/L × g/mol × L = g, confirming the result is in grams.
Answer
2.922 g of NaCl
Weigh 2.922 g of NaCl, dissolve in about 400 mL of distilled water in a beaker, transfer to a 500 mL volumetric flask, and fill to the mark.
Weigh 2.922 g of NaCl, dissolve in about 400 mL of distilled water in a beaker, transfer to a 500 mL volumetric flask, and fill to the mark.
Example 4
A laboratory requires 500 mL of 0.5 M HCl for a series of titrations. The stock solution is 6.0 M HCl. The technician must determine how much stock acid to dilute.
Solution
The technician needs 500 mL of 0.5 M HCl from a 6.0 M stock. The dilution formula gives the relationship between stock and target concentrations:
Rearranging to solve for the stock volume:
Converting to millilitres: 0.0417 L × 1000 mL/L = 41.7 mL.
Answer
41.7 mL of 6.0 M HCl
Measure 41.7 mL of concentrated HCl, slowly add it to about 400 mL of distilled water, then dilute to 500 mL in a volumetric flask. Always add acid to water, never the reverse.
Measure 41.7 mL of concentrated HCl, slowly add it to about 400 mL of distilled water, then dilute to 500 mL in a volumetric flask. Always add acid to water, never the reverse.
Published:
Updated:
References
- [1]Pearson, Vogel's Textbook of Quantitative Chemical Analysis, 2000.
- [2]IUPAC, IUPAC Gold Book - Molar Concentration. https://goldbook.iupac.org/terms/view/M03988
Glossary
- Molarity (M) – The number of moles of solute per litre of solution, expressed in mol/L. It is the most frequently used concentration unit in chemistry and is temperature-dependent because volume changes with temperature.
- Mole (mol) – The SI base unit for amount of substance, defined as exactly 6.02214076 × 10²³ elementary entities (Avogadro’s number). One mole of any substance contains the same number of particles.
- Molar mass (M) – The mass of one mole of a substance, expressed in grams per mole (g/mol). It is numerically equal to the molecular weight of the substance and is calculated by summing the atomic masses of all atoms in the chemical formula.
- Solute – The component of a solution that is present in lesser amount and is dissolved by the solvent. In aqueous solutions, the solute is typically a solid, liquid, or gas that dissolves in water.
- Solvent – The component of a solution that is present in greater amount and dissolves the solute. Water is the most frequently used solvent in chemical laboratory work and is often called the universal solvent.
- Solution – A homogeneous mixture of two or more substances consisting of a solute dissolved in a solvent. Solutions have uniform composition and properties throughout the mixture.
- Avogadro’s number (NA) – The number of particles in one mole of any substance, equal to 6.02214076 × 10²³. It provides the fundamental link between the macroscopic scale (grams) and the atomic scale (atoms or molecules).
- Dilution – The process of reducing the concentration of a solution by adding more solvent without adding more solute. The relationship C₁V₁ = C₂V₂ governs dilution calculations.
- Stock solution – A concentrated solution of known concentration that is stored and used to prepare more dilute working solutions as needed. Stock solutions save time and improve reproducibility in the laboratory.
- Molality (m) – Moles of solute per kilogram of solvent, expressed in mol/kg. Unlike molarity, molality is temperature-independent because it depends on mass rather than volume. It is used in colligative property calculations and in situations where temperature fluctuates.
How to Use?
- 1
Select what you want to calculate
Choose from the dropdown: Molarity (M), Mass of solute (m), or Volume of solution (V). The other fields adjust automatically based on your selection.
- 2
Enter the known values
Type in the values you have. Use the unit dropdowns next to each field to switch between grams and milligrams, liters and milliliters, or molar and millimolar.
- 3
Use the chemical search (optional)
Search for any chemical compound by name, such as sodium chloride or glucose. The molar mass field fills in automatically using PubChem data, saving you a manual lookup.
- 4
Review and adjust units
Check that the units match your needs. The results panel shows the primary value plus alternative units like mM and uM so you always have the right format.
- 5
Click Calculate and interpret results
Press the Calculate button to see the answer instantly. The result includes the formula used and interactive graphs that visualize the relationship between variables.
Frequently Asked Questions
Use the standard formula M = (wt% × density × 1000) / molar mass. This formula converts weight-based concentration (grams of solute per 100 g solution) into moles per liter by using density to determine volume. Example (37% HCl): wt% = 37, density = 1.18 g/mL, molar mass = 36.46 g/mol. M = (37 × 1.18 × 1000) / 36.46 = 12.08 M. The calculator performs this conversion instantly and accurately.
Weight percent is mass-based while molarity is volume-based, so density is required for conversion. Formula: M = (wt% × density × 1000) / molar mass. Common conversions:
37% HCl → 12.08 M.
98% H₂SO₄ → 18.39 M.
50% NaOH → 19.06 M.
Glacial acetic acid → 17.45 M.
Use the Quick Fill feature for fast, pre-verified values.
37% HCl → 12.08 M.
98% H₂SO₄ → 18.39 M.
50% NaOH → 19.06 M.
Glacial acetic acid → 17.45 M.
Use the Quick Fill feature for fast, pre-verified values.
Concentrated reagents show wt% purity and density on the label. Apply the formula M = (wt% × density × 1000) / molar mass using the exact values provided on your bottle. Standard reference values:
HCl 37% → 12.08 M.
H₂SO₄ 98% → 18.39 M.
HNO₃ 70% → 15.67 M.
NaOH 50% → 19.06 M.
Always prefer the values printed on your specific reagent bottle for highest accuracy.
HCl 37% → 12.08 M.
H₂SO₄ 98% → 18.39 M.
HNO₃ 70% → 15.67 M.
NaOH 50% → 19.06 M.
Always prefer the values printed on your specific reagent bottle for highest accuracy.
The dilution formula C₁V₁ = C₂V₂ conserves the number of moles of solute. To find stock volume needed: V₁ = (C₂ × V₂) / C₁. Examples: 500 mL of 0.1 M HCl from 12.08 M stock → 4.14 mL stock. 1 L of 1 M H₂SO₄ from 18.39 M stock → 54.37 mL stock. The "Volume Needed" mode in this calculator applies the formula automatically.
Molarity (M) = moles of solute per liter of solution. Normality (N) = number of reactive equivalents per liter of solution. Relationship: N = M × n, where n is the equivalents factor. Common n values: HCl = 1, H₂SO₄ = 2, H₃PO₄ = 3, NaOH = 1. Normality is especially useful in titrations. Enable Advanced Mode to calculate both units simultaneously.
Determine stock molarity (use Quick Fill or the formula). Calculate required stock volume with C₁V₁ = C₂V₂. Add roughly half the final volume of distilled water to a volumetric flask. Slowly add the concentrated stock (acid to water only). Cool if necessary, then top up to the final volume and mix well. Label properly. This safe method is used for preparing 0.1 M, 1 M, and other common laboratory solutions.
Weight percent describes mass of solute per mass of solution, while molarity requires moles per volume (liter) of solution. Density bridges this gap by converting mass of solution into volume. Without density, accurate conversion is impossible. Even small variations in density can noticeably affect the calculated molarity.
Use the simple formula N = M × n, where n is the equivalents factor based on the number of reactive H⁺ or OH⁻ ions. Examples:
12.08 M HCl × 1 = 12.08 N.
18.39 M H₂SO₄ × 2 = 36.78 N.
19.06 M NaOH × 1 = 19.06 N.
This calculator computes normality automatically in Advanced Mode.
12.08 M HCl × 1 = 12.08 N.
18.39 M H₂SO₄ × 2 = 36.78 N.
19.06 M NaOH × 1 = 19.06 N.
This calculator computes normality automatically in Advanced Mode.
Dilution factor (DF) = initial concentration / final concentration = final volume / stock volume. Examples: 10-fold: 1 part stock + 9 parts diluent. 100-fold: 1 part stock + 99 parts diluent. For serial dilutions, multiply the individual factors. The calculator handles all dilution calculations instantly.
Calculate the required stock volume using V₁ = (C₂ × V₂) / C₁. Then follow safe lab practice: Start with half the final volume of water in the flask. Add the concentrated reagent slowly (acid to water). Cool if heat is generated. Make up to final volume and mix thoroughly. The calculator provides exact volumes for any target concentration.
Use the molar mass of the hydrated compound, not the anhydrous form. For CuSO4·5H2O, the molar mass is 249.69 g/mol (not 159.61 g/mol for anhydrous CuSO4). The water of crystallization adds to the mass you weigh but does not change the moles of CuSO4 in solution. Simply enter 249.69 as the molar mass in the calculator.
Molarity (M) is moles of solute per liter of solution. Molality (m) is moles of solute per kilogram of solvent. Molarity changes with temperature because volume expands. Molality is temperature-independent because mass does not change. Use molarity for most lab work, titrations, and spectrophotometry. Use molality for freezing point depression, boiling point elevation, and other colligative property calculations.
Always add concentrated acid to water slowly while stirring, never the reverse. Dilution of concentrated acids generates significant heat. Adding water to acid can cause rapid boiling and dangerous splashing. Wear safety goggles, a lab coat, and chemical-resistant gloves. Work in a fume hood for volatile acids like HCl and HNO3. Use this calculator to determine the exact volumes needed before you start.
For two solutions of the same solute mixed together, calculate total moles and divide by total volume. Total moles = (M1 × V1) + (M2 × V2). Total volume = V1 + V2 assuming volumes are additive. Mixture molarity = total moles / total volume. For different solutes, calculate each concentration independently.
Standard reference values:
Hydrochloric acid (37% wt) = 12.08 M.
Sulfuric acid (98%) = 18.39 M.
Nitric acid (70%) = 15.67 M.
Phosphoric acid (85%) = 14.63 M.
Glacial acetic acid (99.7%) = 17.45 M.
Ammonium hydroxide (28%) = 14.53 M.
Sodium hydroxide (50%) = 19.06 M.
Always verify the exact wt% and density printed on your specific reagent bottle, as values vary slightly between manufacturers.
Hydrochloric acid (37% wt) = 12.08 M.
Sulfuric acid (98%) = 18.39 M.
Nitric acid (70%) = 15.67 M.
Phosphoric acid (85%) = 14.63 M.
Glacial acetic acid (99.7%) = 17.45 M.
Ammonium hydroxide (28%) = 14.53 M.
Sodium hydroxide (50%) = 19.06 M.
Always verify the exact wt% and density printed on your specific reagent bottle, as values vary slightly between manufacturers.
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