Molarity Calculator
Calculate chemical solution molarity, solute mass requirements, solution volumes, and laboratory stock dilutions (C1V1 = C2V2).
| Molarity (mol/L) | 0.500 M (500 mM) |
| Total Solution Volume | 0.500 Liters (500.0 mL) |
| Formula Weight | 58.44 g/mol |
| Mass Concentration | 29.22 g/L (2.92% w/v) |
What is Molarity (M) in Chemistry?
Molarity (symbolized as M) is the standard scientific unit of concentration used in chemical, biochemical, and pharmaceutical laboratories worldwide.
Molarity is defined as the number of moles of solute per liter of total solution (mol/L). Because chemical reactions occur between individual molecules based on molar stoichiometric ratios rather than raw gram weights, preparing reagents by exact molarity ensures predictable chemical reactions.
The Mathematics of Molarity & Dilutions
Here are the fundamental formulas connecting mass, moles, volume, and concentration:
Moles of Solute (n) = Solute Mass in Grams (m) / Molecular Weight (g/mol)
Mass Required Formula:
Mass (grams) = Molarity (mol/L) * Volume (L) * Molecular Weight (g/mol)
Solution Dilution Formula (Conservation of Solute):
C1 * V1 = C2 * V2
Where C1 = Initial Stock Concentration, V1 = Volume of Stock to Pipette,
C2 = Target Concentration, V2 = Target Final Volume.
Volume of Solvent to Add = V2 - V1
Step-by-Step Worked Laboratory Example
Suppose you need to prepare 500 mL (0.500 L) of a 0.50 M Sodium Chloride (NaCl, MW = 58.44 g/mol) solution:
- Identify parameters:
M = 0.50 mol/L, V = 0.500 L, MW = 58.44 g/mol. - Calculate total moles:
0.50 mol/L * 0.500 L = 0.250 moles of NaCl. - Calculate required mass:
0.250 moles * 58.44 g/mol = 14.61 grams. - Preparation Protocol: Weigh 14.61 grams of NaCl on an analytical balance, transfer to a 500 mL volumetric flask, dissolve in ~400 mL deionized water, and bring to the 500 mL fill line.
Master Molecular Weight Reference Table
Formula weights and preparation requirements for 1.0 Liter of 1.0 M solution:
| Compound Name | Chemical Formula | Molecular Weight (g/mol) | Mass for 1 Liter of 1.0 M Solution | Common Laboratory Use |
|---|---|---|---|---|
| Sodium Chloride | NaCl | 58.44 g/mol | 58.44 grams | Saline buffers, cell culture, biochemistry |
| Sodium Hydroxide | NaOH | 39.997 g/mol | 40.00 grams | pH adjustment, base titrations |
| Hydrochloric Acid | HCl | 36.46 g/mol | 36.46 grams | pH adjustment, acid catalysis |
| Sulfuric Acid | H2SO4 | 98.08 g/mol | 98.08 grams | Dehydration reactions, battery acid |
| D-Glucose (Dextrose) | C6H12O6 | 180.16 g/mol | 180.16 grams | Cellular respiration, microbiology media |
| Sucrose (Table Sugar) | C12H22O11 | 342.30 g/mol | 342.30 grams | Density gradient centrifugation |
| Tris Base | C4H11NO3 | 121.14 g/mol | 121.14 grams | Tris-HCl DNA/protein electrophoresis buffers |
| Potassium Chloride | KCl | 74.55 g/mol | 74.55 grams | Electrolyte solutions, molecular biology |
Molarity (M) vs Molality (m) vs Normality (N)
- Molarity (M, mol/L): Moles of solute per liter of total solution. Most common in wet chemistry, but changes slightly with thermal expansion of liquids.
- Molality (m, mol/kg): Moles of solute per kilogram of solvent. Unaffected by temperature or pressure; standard in thermodynamic and colligative property experiments (boiling point elevation, freezing point depression).
- Normality (N, eq/L): Gram equivalent weights per liter of solution. Commonly used in acid-base acidimetric titrations and redox chemistry.
5 Essential Tips for Accurate Solution Preparation
Frequently Asked Questions
Molarity (symbolized as M) is a measure of concentration defined as the number of moles of solute dissolved in exactly one liter of solution (mol/L). A 1.0 M solution contains 1 mole of substance per liter.
To find the mass in grams: Mass (g) = Molarity (mol/L) * Volume (L) * Molecular Weight (g/mol).
When diluting a concentrated stock solution, the total amount of solute remains unchanged. The equation C1 * V1 = C2 * V2 allows you to calculate the exact volume of stock solution (V1) needed to prepare a target volume (V2) at a lower target concentration (C2).
1 Molar (1 M) equals exactly 1,000 millimolar (1,000 mM) and 1,000,000 micromolar (1,000,000 μM).
Because dissolving solid solute increases the total physical volume of the liquid (displacement). Adding 1 liter of water to 58 grams of salt yields roughly 1.02 liters of total solution, diluting the molarity below 1.0 M.