Chemkit

Molarity Calculator — Mass, Volume & Concentration

Molarity is the number of moles of dissolved solute per liter of solution (not per liter of solvent). It is the single most-used concentration unit in wet chemistry — every buffer recipe, titration, and stock preparation is a molarity calculation in disguise. This calculator solves the standard relation for any one of the four variables when the other three are known.

Molarity Calculator

Enter any three values and pick which to solve for. Sig figs default to the lowest-precision input.

Solve for

Equations used

Base relation and each rearrangement:

The equation

Molarity is defined as moles of solute divided by liters of solution:

M=nVM = \dfrac{n}{V}

and the number of moles of a compound with molecular (formula) weight MWMW is simply mass over MWMW:

n=mMWn = \dfrac{m}{MW}

Combining these gives the form the calculator uses:

M=mMW⋅VM = \dfrac{m}{MW \cdot V}

Rearranged for each variable:

  • Mass   m=M⋅MW⋅V\;m = M \cdot MW \cdot V
  • Molecular weight   MW=mM⋅V\;MW = \dfrac{m}{M \cdot V}
  • Volume   V=mMW⋅M\;V = \dfrac{m}{MW \cdot M}

Worked example

Prepare a 0.200 M NaCl solution in 500. mL of water. How much NaCl do you weigh?

Worked example — 0.200 M NaCl in 500. mL
M (target)
0.200 mol/L (3 sig figs)
MW (NaCl)
58.44 g/mol (PubChem CID 5234)
V
0.500 L (3 sig figs)
  1. Formula
  2. Substitute
  3. Note

    Rounded to three significant figures, weigh out 5.84 g of NaCl and dissolve it in water in a 500 mL volumetric flask, then fill to the mark. Fill to the volume of the final solution, not by adding 500 mL of water — see pitfalls below.

  4. Result

Common concentration ranges

ApplicationTypical range
Physiological saline (isotonic NaCl)0.154 M (0.9 % w/v)
Working PBS or TBS buffer10–100 mM
Enzyme kinetics substrates1 µM – 10 mM
Trace metals in environmental samples1 nM – 1 µM

Common pitfalls

  • Solution volume ≠ solvent volume. Molarity is defined per liter of final solution. Use a volumetric flask and top up to the mark; do not measure out water and add solute on top.
  • Temperature. Solution volume expands with temperature, so molarity drifts a little between 4 °C and 25 °C. For temperature-invariant work, prefer molality.
  • Hydrates. A “MW” from a bottle of CuSO₄·5H₂O is 249.68 g/mol, not 159.61 g/mol. The five waters count toward the mass you weigh.
  • Sig figs. Trailing zeros without a decimal point are ambiguous (500 vs 500.). This calculator counts the sig figs of the string you typed and uses the lowest count across your inputs. Override with the sig-figs selector if you know your instrument is more precise.

Practice problems

Attempt each on paper, then expand the worked solution to check your arithmetic and sig-figs.

Problem 1 Easy Prepare 2.00 L of 0.500 M NaCl. How much NaCl do you weigh?

Answer: 58.4 g NaCl

Solve for mass (m)
M (target)
0.500 mol/L
MW (NaCl)
58.44 g/mol
V
2.00 L
  1. Formula
  2. Substitute
  3. Result
Problem 2 Easy You dissolve 0.372 g of KCl in enough water to make 250. mL of solution. What is the molarity?

Answer: 0.0200 M (20.0 mM)

Solve for concentration (M)
m
0.372 g
MW (KCl)
74.55 g/mol
V
0.250 L
  1. Formula
  2. Substitute
  3. Result
Problem 3 Intermediate How many mL of a 0.150 M glucose stock do you draw to deliver 30.0 mg of glucose?

Answer: 1.11 mL

Solve for volume (V)
m (target)
0.0300 g (30.0 mg)
MW (D-glucose)
180.16 g/mol
M (stock)
0.150 mol/L
  1. Formula
  2. Substitute
  3. Result
Problem 4 Intermediate You dissolve 2.42 g of an unknown salt in 100.0 mL of water and a titration confirms the concentration is 0.100 M. What is the salt's molecular weight?

Answer: 242 g/mol

Solve for molecular weight (MW)
m
2.42 g
V
0.1000 L
M
0.100 mol/L
  1. Formula
  2. Substitute
  3. Note

    MW is the sig-fig-limiting output — 0.100 M and 0.1000 L give a three-sig-fig answer.

  4. Result
Problem 5 Hard You need 100.0 mL of a 0.0500 M CuSO₄ solution. Your bottle is labeled 'CuSO₄·5H₂O'. How much do you weigh?

Answer: 1.25 g of CuSO₄·5H₂O

Solve for mass (m) — hydrate
M (target)
0.0500 mol/L
MW (CuSO₄·5H₂O)
249.68 g/mol
V
0.1000 L
  1. Formula
  2. Substitute
  3. Note

    Use the full hydrate MW (249.68) — one mole of pentahydrate delivers one mole of CuSO₄. Using anhydrous MW (159.61) would under-weigh by about 36 %.

  4. Result

Frequently asked questions

What is molarity?
Molarity (M) is the amount of solute, in moles, dissolved per liter of solution. It is the most common concentration unit in analytical and inorganic chemistry: 1 M = 1 mol/L.
What is the difference between molarity and molality?
Molarity divides moles of solute by the total volume of the solution in liters. Molality (m) divides moles of solute by the mass of the solvent in kilograms. Molality does not change with temperature; molarity does, because solution volume expands with temperature.
How do I calculate molarity from mass?
Convert mass to moles by dividing by the molecular weight, then divide moles by the volume of the solution in liters: M = m / (MW × V). For 5.85 g NaCl (MW 58.44 g/mol) dissolved to make 0.500 L of solution, M = 5.85 / (58.44 × 0.500) = 0.200 M.
What units should I use?
The calculator accepts mass in g, mg, µg, or kg; volume in L, mL, or µL; and concentration in M, mM, µM, or nM. Internally every value is converted to grams, liters, and mol/L before the arithmetic, so any mixed unit input works as long as MW is entered in g/mol.

Sources

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