What Is Molarity & How Is It Calculated?

It is convenient for measurement purposes to know how much of a substance is dissolved in a given volume of solution; this is what chemists mean by "concentration." Molarity is the most common way of expressing concentration when working with solutions and chemical reactions. The reason is that reactants (elements or compounds) combine in whole-number ratios when their amounts are expressed in units called "moles." For example, 2 moles of hydrogen gas combine with 1 mole of oxygen gas to produce 2 moles of water, by the chemical reaction: 2H₂ + O₂ = 2H₂O.

A mole of a substance is defined as a specific number of atoms or molecules called the "Avogadro number," which is 6.022 × 10²³. The number is derived from international agreement, based on the number of atoms in exactly 12 grams (g) of the carbon isotope "C-12." The convenience this "counting unit," the Avogadro number, provides is seen when considering, for example, the weights of 1 mole each of oxygen, water and carbon dioxide, which are 16.00 g, 18.02 g and 44.01 g, respectively.

Molarity, or molar concentration (M), is defined as the number of moles of a substance, or "solute," dissolved in 1 liter of solution. Molarity is not to be confused with "molality," which is concentration expressed as moles of solute per kilogram of solvent. Examples will help clarify the concept of molarity and how it works.

Consider a problem that asks for the molarity of a solution containing 100 grams (g) of sodium chloride, NaCl, in 2.5 liters of solution. First, determine the "formula weight" of NaCl by adding together the "atomic weights" of its elements, Na and Cl, as follows:

22.99 + 35.45 = 58.44 g NaCl/mole.

Next, calculate of number of moles in 100 g of NaCl by dividing the weight of NaCl by its formula weight:

100 g NaCl ÷ [58.44 g NaCl/mole NaCl] = 1.71 moles NaCl.

Finally, calculate the molarity of the solution by dividing the number of moles of NaCl by the volume of the solution:

1.71 moles NaCl ÷ 2.5 liters = 0.684 M.

Consider a problem that asks for the weight of sodium sulfate, Na₂SO₄, required to prepare 250 milliliters (ml) of a 0.5 M solution. The first step is to calculate the number of moles of Na₂SO₄ required by multiplying the volume of the solution by the molarity:

0.25 liters × 0.5 moles Na₂SO₄/liter = 0.125 moles Na₂SO₄

Next, the formula weight of Na₂SO₄ is determined by adding together the atomic weights of its constituent atoms. A molecule of Na₂SO₄ contains 2 atoms of Na, 1 atom of S (sulfur) and 4 atoms of O (oxygen), therefore its formula weight is:

[22.99 × 2] + 32.07 + [16.00 × 4] = 45.98 + 32.07 + 64.00 = 142.1 g Na₂SO₄/mole

Finally, the weight of Na₂SO₄ required is calculated by multiplying the number of moles by the formula weight:

0.125 moles Na₂SO₄ × 142.1 g Na₂SO₄/mole Na₂SO₄ = 17.76 g Na₂SO₄.