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7.16: Concentrations: Molarity Equation and Calculations

  • Page ID
    213253
  • Learning Objectives
    • Calculate the molarity of a solution.

    As stated previously, chemists have defined several types of concentrations, which each use a different chemically-acceptable unit, or combination of units, to indicate the amount of solute that is dissolved in a given amount of solvent.  Three percent-based concentrations, mass percent, volume percent, and mass/volume percent, were defined and applied in Sections 7.13, 7.14, and 7.15, respectively.  The following paragraphs will present and apply the equation that is used to determine the molarity of a solution, which, in contrast to the previously-discussed concentrations, is not calculated as a percentage.

    Molarity Equation

    As shown below, the molarity of a solution is defined as the ratio of the molar amount of solute that is present in a solution, relative to the volume of the solution, as a whole.  Recall that the variable that is utilized to represent the molar quantity of a substance is "n."  Because, in contrast to the concentrations that have been discussed in the previous sections of this chapter, molarity is not expressed as a percentage, this proportion should not be multiplied by 100.

    \(\text{Molarity}\) = \( \dfrac{ \rm{n_{solute} \; (\rm{mol})}}{\rm{V_{solution} \; (\rm{L})}} \)

    As stated previously, mass percents and volume percents can be calculated using an alternative equation, in which the masses or volumes, respectively, of the solute and the solvent that are contained in a solution are added to obtain the mass or volume, respectively, of that solution, as a whole.  In contrast, mass/volume percents, which are most often calculated for solutions that are prepared by dissolving solid solutes in liquid solvents, cannot be reliably determined using an analogous equation.  Recall that, in order to create this type of solution, the solid solute particles must overcome the attractive forces that exist between the liquid solvent molecules, and that after the solute particles have dispersed throughout the solvent, the solvent molecules interact more strongly with the solvated solute particles than with other solvent molecules.  As a result of these solute-solvent interactions, the solvated solute particles occupy less space than they had prior to their solvation, which causes the volume of the solution, as a whole, to decrease, relative to the combined volumes of the individual solute and solvent.  Because the magnitude of this volumetric contraction varies based on the solute and solvent that are utilized to prepare a solution, calculating the mass/volume percent of a solution by adding the volumes of its components is prohibitively challenging.  Since, like mass/volume percents, molarities are primarily calculated for solutions that contain solid solutes and liquid solvents, only the equation that is shown above can be applied to reliably determine the molarity of a solution.

    The equation that is presented above can be modified to more heavily-emphasize the identity of the solute by writing the chemical formula of that substance as a subscript on the "Molarity" variable.  For example, the phrase "molarity of sodium chloride" can be symbolized as "MolarityNaCl."  However, because most word-processing programs and coding languages cannot represent information as a subscript within another subscript, only chemical formulas that do not contain numerical subscripts can be expressed using the notation that is shown above.  For example, consider the phrase "molarity of tin (IV) phosphite."  While the letters in the chemical formula for tin (IV) phosphite, Sn3(PO3)4, could be properly-formatted as subscripts, the numerical subscripts within this formula cannot be further subscripted, and the resultant notation, "MolaritySn3(PO3)4," is, therefore, not chemically-correct.  As a result, chemists developed an alternative notation, in which the chemical formula of the solute is enclosed in square brackets, to represent the molarity of a specific substance.  Because the chemical formula for the solute is typed using standard formatting, this notation, which will be incorporated into calculations that will be performed in Chapter 8, can be used to express the molarity of any chemical.  For example, the phrase "molarity of tin (IV) phosphite" can be symbolized as [Sn3(PO3)4].

    Molarity Calculations

    In order to be utilized in the equation that is shown above, the amount of solute that is contained in the solution must be expressed in moles.  While the volumetric quantities that are incorporated into the volume percent and mass/volume percent equations must be reported in milliliters, the molarity calculation that is shown above requires that the volume of the solution be provided in liters.  Therefore, if either of these measurements is reported using an alternative unit, its value would need to be converted to the appropriate unit prior to being incorporated into the molarity equation.  Finally, the chemical formula of each component must be written as the secondary unit on its associated numerical quantity.

    The equation that is shown above is simplified by dividing the molar amount of solute by the volume of the solution, as a whole.  During this process, no unit cancelation occurs, because the units that are present in the numerator and denominator, "mol" and "L," respectively, do not match one another.  Therefore, the unit that is used to report the molarity of a solution must contain both of these units, which are related to one another in a fractional format, "mol/L," and the chemical formula of the solute is written as the secondary unit on this calculated quantity.  Alternatively, the unit for molarity can be abbreviated as a capital "M," which should be italicized, M, when typed and underlined, M, if hand-written.

    Finally, because molarities are not defined as exact quantities, their values should be reported using the correct number of significant figures.

    For example, calculate the molarity of a 795 milliliter solution that is prepared by dissolving 1.64 moles of gold (III) chloride in water.

    In order to calculate the molarity of a solution, each substance that is referenced in the problem must first be classified as a solute or a solvent.  Because the indicator word "in" is present in the given statement, the chemical that is mentioned after this word, water, H2O, is the solvent in this solution, and the remaining substance, gold (III) chloride, AuCl3, is the solute, "by default."

    Before the molarity equation can be applied, the validity of the units that are associated with the given numerical values must be confirmed.  As stated above, the amount of solute that is contained in the solution must be expressed in moles, and the volume of the solution must be provided in liters.  Therefore, since the amount of solute is reported in moles, this quantity can be directly incorporated into the molarity equation.  However, the volume of the solution is not expressed in an acceptable unit.  Therefore, this value must be converted to liters, as shown below, before it can be utilized to solve the given problem.

    \( {\text {795}} {\cancel{\rm{mL} \; \rm{solution}}} \times\) \( \dfrac{\rm{L} \; \rm{solution}}{1,000 \; \cancel{\rm{mL} \; \rm{solution}}}\) = \( {\text {0.795}} \; \rm{L} \; \rm{solution}\) 

    Both values can now be incorporated into the molarity equation, as shown below.  No unit cancelation occurs during the simplification of this mathematical statement, because the units that are present in the numerator and denominator, "mol" and "L," respectively, do not match one another.  Therefore, the unit that is used to report the molarity of this solution must contain both of these units, which are related to one another in a fractional format, "mol/L."  Alternatively, the unit for molarity can be abbreviated as a capital "M", which should be italicized, M, when typed and underlined, M, if hand-written.  The chemical formula of the solute is written as the secondary unit on the calculated quantity, and applying the correct number of significant figures to this value results in the final answer that is shown below.

    \(\text{Molarity}\) = \( \dfrac{1.64 \; \rm{mol} \; \rm{AuCl_3}}{0.795 \; \rm{L} \; \rm{solution}}\)

    \(\text{Molarity}\) = \({2.062893... \rm{mol/L} \; \rm{AuCl_3}} ≈ {2.06 \; \rm{mol/L} \; \rm{AuCl_3}} ≈ {2.06 \; M \; \rm{AuCl_3}}\)

    Exercise \(\PageIndex{1}\)

    Calculate the molarity of a 1.43 liter solution that is prepared by dissolving 7.05 x 1024 molecules of potassium hydroxide in water.

    Answer
    In order to calculate the molarity of a solution, each substance that is referenced in the problem must first be classified as a solute or a solvent.  Because the indicator word "in" is present in the given statement, the chemical that is mentioned after this word, water, H2O, is the solvent in this solution, and the remaining substance, potassium hydroxide, KOH, is the solute, "by default."

    Before the molarity equation can be applied, the validity of the units that are associated with the given numerical values must be confirmed.  As stated above, the amount of solute that is contained in the solution must be expressed in moles, and the volume of the solution must be provided in liters.  Therefore, since the volume of the solution is reported in liters, this quantity can be directly incorporated into the molarity equation.  However, because the amount of solute is not expressed in an acceptable unit, the given value must be converted to moles before it can be utilized to solve the given problem.  The phrase "molecules of potassium hydroxide" indicates that the Avogadro's number equality that is shown below is required to convert the given amount of solute to a molar quantity.

    1 mol KOH = 6.02 × 1023 KOH molecules

    In order to completely eliminate the unit "molecules of potassium hydroxide," the equality that is written above must be utilized as a conversion factor.  When using a calculator, any quantity that is expressed in scientific notation should be offset by an additional set of parentheses, and applying the correct number of significant figures to the calculated quantity results in the numerical value that is shown below.

    \( {7.05 \times 10^{24} \cancel{\rm{molecules} \; \rm{KOH}}} \times\) \( \dfrac{1 \; \rm{mol} \; \rm{KOH}}{6.02 \times 10^{23} \cancel{\rm{molecules} \; \rm{KOH}}}\) = \( {\text {11.71096...}}\) \({\rm{mol} \; \rm{KOH}}\) ≈ \( {\text {11.7}}\) \({\rm{mol} \; \rm{KOH}}\)

    Both values can now be incorporated into the molarity equation, as shown below.  No unit cancelation occurs during the simplification of this mathematical statement, because the units that are present in the numerator and denominator, "mol" and "L," respectively, do not match one another.  Therefore, the unit that is used to report the molarity of this solution must contain both of these units, which are related to one another in a fractional format, "mol/L."  Alternatively, the unit for molarity can be abbreviated as a capital "M," which should be italicized, M, when typed and underlined, M, if hand-written.  The chemical formula of the solute is written as the secondary unit on the calculated quantity, and applying the correct number of significant figures to this value results in the final answer that is shown below.

    \(\text{Molarity}\) = \( \dfrac{11.7 \; \rm{mol} \; \rm{KOH}}{1.43 \; \rm{L} \; \rm{solution}}\)

    \(\text{Molarity}\) = \({8.181818... \rm{mol/L} \; \rm{KOH}} ≈ {8.18 \; \rm{mol/L} \; \rm{KOH}} ≈ {8.18 \; M \; \rm{KOH}}\)

    Exercise \(\PageIndex{2}\)

    Calculate the molarity of a 1.17 deciliter solution that is prepared by dissolving 31.96 grams of magnesium nitrate in water.

    Answer
    In order to calculate the molarity of a solution, each substance that is referenced in the problem must first be classified as a solute or a solvent.  Because the indicator word "in" is present in the given statement, the chemical that is mentioned after this word, water, H2O, is the solvent in this solution, and the remaining substance, magnesium nitrate, Mg(NO3)2, is the solute, "by default."

    Before the molarity equation can be applied, the validity of the units that are associated with the given numerical values must be confirmed.  As stated above, the amount of solute that is contained in the solution must be expressed in moles, and the volume of the solution must be provided in liters.  Therefore, neither the amount of solute nor the volume of the solution is expressed in an acceptable unit.   The given amount of solute must be converted to moles before it can be utilized to solve the given problem.  The phrase "grams of magnesium nitrate" indicates that the  molecular weight of magnesium nitrate, Mg(NO3)2, is required to convert the given amount of solute to a molar quantity.  The mass contribution calculations for magnesium nitrate, Mg(NO3)2, are shown below.  

    \( {\text {1}}\) \({\cancel{\rm{mol} \; \rm{Mg}}}\) × \( \dfrac{24.31 \; \rm{g} \; \rm{Mg}}{1\; \cancel{\rm{mol} \; \rm{Mg}}}\) = \( {\text {24.31}}\) \({\rm{g} \; \rm{Mg}}\)

    \( {\text {2}}\) \({\cancel{\rm{mol} \; \rm{N}}}\) × \( \dfrac{14.01 \; \rm{g} \; \rm{N}}{1\; \cancel{\rm{mol} \; \rm{N}}}\) = \( {\text {28.02}}\) \({\rm{g} \; \rm{N}}\)

    \( {\text {6}}\) \({\cancel{\rm{mol} \; \rm{O}}}\) × \( \dfrac{16.00 \; \rm{g} \; \rm{O}}{1\; \cancel{\rm{mol} \; \rm{O}}}\) = \( {\text {96.00}}\) \({\rm{g} \; \rm{O}}\)

    The numerical value of the molecular weight of magnesium nitrate, Mg(NO3)2, 148.33, is determined by adding the mass contributions that are shown above and can be reported using a molecular weight equality, as shown below,

    1 mol Mg(NO3)2 = 148.33 g Mg(NO3)2

    or formatted as a "hidden" conversion factor, 148.33 g/mol Mg(NO3)2.

    In order to completely eliminate the unit "grams of magnesium nitrate," the molecular weight of magnesium nitrate must be utilized as a conversion factor.  Applying the correct number of significant figures to the calculated quantity results in the numerical value that is shown below.

    \( {\text {31.96}} {\cancel{\rm{g} \; \rm{Mg(NO_3)_2}}} \times\) \( \dfrac{1 \; \rm{mol} \; \rm{Mg(NO_3)_2}}{148.33 \; \cancel{\rm{g} \; \rm{Mg(NO_3)_2}}}\) = \( {\text {0.2154655...}}\) \({\rm{mol} \; \rm{Mg(NO_3)_2}}\) ≈ \( {\text {0.22}}\) \({\rm{mol} \; \rm{Mg(NO_3)_2}}\)

    Furthermore, the volume of the solution must be converted to liters, as shown below, before it can be utilized to solve the given problem.

    \( {\text {1.17}} {\cancel{\rm{dL} \; \rm{solution}}} \times\) \( \dfrac{\rm{L} \; \rm{solution}}{10 \; \cancel{\rm{dL} \; \rm{solution}}}\) = \( {\text {0.117}} \; \rm{L} \; \rm{solution}\)  

    Both values can now be incorporated into the molarity equation, as shown below.  No unit cancelation occurs during the simplification of this mathematical statement, because the units that are present in the numerator and denominator, "mol" and "L," respectively, do not match one another.  Therefore, the unit that is used to report the molarity of this solution must contain both of these units, which are related to one another in a fractional format, "mol/L."  Alternatively, the unit for molarity can be abbreviated as a capital "M," which should be italicized, M, when typed and underlined, M, if hand-written.  The chemical formula of the solute is written as the secondary unit on the calculated quantity, and applying the correct number of significant figures to this value results in the final answer that is shown below.

    \(\text{Molarity}\) = \( \dfrac{0.22 \; \rm{mol} \; \rm{Mg(NO_3)_2}}{0.117 \; \rm{L} \; \rm{solution}}\)

    \(\text{Molarity}\) = \({1.88034... \rm{mol/L} \; \rm{Mg(NO_3)_2}} ≈ {1.9 \; \rm{mol/L} \; \rm{Mg(NO_3)_2}} ≈ {1.9 \; M \; \rm{Mg(NO_3)_2}}\)

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