Pharmaceutical Chemistry
Molarity Calculator
Work out the molarity of a solution, or how much solid or liquid to use for a target molarity.
I want to calculate
Molar concentration
0.0171mol/L (M)
Moderately dilute solution
Inputs
Substance type
From the label or formula, e.g. NaCl 58.44, glucose 180.16.
The mass you weighed out.
The volume you make the solution up to.
In other units
Working
Your values substituted into the formula.
M = (1 g = 1.0000 g) / (58.44 × 1.0000)
Common solutions
Tap one to load it as a target molarity.
| Solution | MW (g/mol) | Amount |
|---|---|---|
| 58.44 | 58.44in 1L | |
| 36.46 | 8.3 mL*in 1L | |
| 40.00 | 20.00in 1L | |
| 372.24 | 3.7224in 1L | |
| 157.60 | 157.60in 1L |
*For 37% concentrated HCl (specific gravity 1.19)
For solids
M = (mass in g) / (MW × volume in L)
For liquids
M = (A × sg × purity%) / (MW × V)
M molarity (mol/L) · MW molecular weight (g/mol) · mass mass of solid, converted to grams · A amount of liquid (mL) · sg specific gravity (g/mL) · purity% percentage purity, as a decimal · V final volume, converted to litres.
To find the required amount
Mass (g) = M_target × MW × V(L)
Volume (mL) = (M_target × MW × V(L)) / (sg × purity%)
For a liquid, A × sg gives the grams of liquid and × purity gives the grams of the actual solute in it; dividing by MW turns grams into moles.
Worked example: weighing sodium bicarbonate for a 0.2 M solution
A practical asks for 250 mL of 0.2 M sodium bicarbonate (NaHCO₃, MW 84.01 g/mol) to use as a mild base in an acid–base experiment. How much solid do you weigh? (Illustrative values.)
- Mode / substance
- Amount from target molarity · Solid
- Target molarity
- 0.2 mol/L
- Molecular weight, MW
- 84.01 g/mol
- Final volume
- 250 mL
Write the formula
Mass (g) = M × MW × V(L)
Convert the volume to litres
V = 250 mL × 0.001 = 0.2500 L
Moles needed
n = M × V = 0.2 mol/L × 0.2500 L = 0.05 mol
Convert moles to grams
Mass = 0.05 mol × 84.01 g/mol = 4.2005 g
Weigh 4.2005 g of NaHCO₃ (about 4.20 g on a two-decimal balance), dissolve and make up to 250 mL.
The calculator prints 4.2005 g and the instruction "Dissolve and make up to 250 mL for 0.2 M". Dissolve the solid in less than 250 mL of water and then make up to the mark in a volumetric flask; adding 4.2 g to 250 mL of water gives slightly more than 250 mL of solution and so slightly less than 0.2 M.
Common mistakes
Assumptions and limitations
Keep learning
References
- International Union of Pure and Applied Chemistry. Compendium of Chemical Terminology (the 'Gold Book'). Source
- Bureau International des Poids et Mesures. The International System of Units (SI Brochure). 9th ed. Sèvres: BIPM; 2019. Source
- Sinko PJ, editor. Martin's Physical Pharmacy and Pharmaceutical Sciences. 8th ed. Philadelphia: Wolters Kluwer; 2023.
Common questions
The final volume of solution. Dissolve the solute in less solvent, then make up to the mark in a volumetric flask. Adding the solute to the full volume of solvent gives a slightly lower molarity.
Concentrated reagents are sold by volume but are not pure: conc. HCl is about 37% w/w with sg 1.19. Volume × sg gives grams of liquid, and × purity gives grams of HCl itself.
The MW of exactly what you weigh. CuSO₄·5H₂O is 249.68 g/mol, not the 159.61 g/mol of anhydrous CuSO₄ — using the wrong one gives an error of more than 50%.
1 M = 1,000 mM = 1,000,000 μM = 1,000,000,000 nM. Very small results are shown in scientific notation, e.g. 2.775e-7 M is 277.53 nM.
Prepared by the PharmaWallah student team. Verify results with your textbook or instructor. Report an error · Editorial policy