What is the percent by mass concentration of a solution?

• Calculate the mass/volume percent 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.  The following paragraphs will present and apply the equation that is used to calculate a mass/volume percent, which is the final type of percent-based concentration that will be discussed in this chapter.

The mass/volume percent of a solution is defined as the ratio of the mass of solute that is present in a solution, relative to the volume of the solution, as a whole.  Because this type of concentration is expressed as a percentage, the indicated proportion must be multiplied by 100, as shown below.

\(\text{Mass/Volume Percent}\) = \( \dfrac{ \rm{m_{solute} \; (\rm{g})}}{\rm{V_{solution} \; (\rm{mL})}} \) × \({100}\)

As discussed in the previous two sections of this chapter, 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.  While mass percents are typically reported for solid- and liquid-phase solutions, and volume percents are usually determined for liquid- and gas-phase solutions, a mass/volume percent concentration is most often calculated for solutions that are specifically prepared by dissolving solid solutes in liquid solvents.  In order to create this type of solution, the solid solute particles must overcome the attractive forces that exist between the liquid solvent molecules, in order to move throughout and occupy the "empty" spaces that are temporarily created during the solvation process.  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 and, consequently, exist in closer physical proximity to those solute particles, relative to 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.  Therefore, only the equation that is shown above can be applied to reliably determine the mass/volume percent of a solution.

In order to be incorporated into the equation that is shown above, the mass of the solute must be expressed in grams, the volume of the solution must be provided in milliliters, and the chemical formula of each component must be written as the secondary unit on its associated numerical quantity.  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 mass/volume percent equation.

During the multiplication and division processes that are used to solve this equation, no unit cancelation occurs, because the units that are present in the numerator and denominator, "g" and "mL," respectively, do not match one another.  Therefore, the unit that results from the division of the indicated quantities is "g/mL," which is a unit that is typically utilized to report the density of a substance.  Because densities and mass/volume percent concentrations have unique definitions and are calculated using different equations, these measurements are distinctive quantities and, consequently, cannot be expressed using the same unit.  Therefore, the mass and volume units are eliminated during the simplification of the mass/volume percent equation, even though "g" and "mL" do not cancel, mathematically, and the calculated concentration is expressed as a percentage.  However, as stated previously, the quantity of solute that is present in a given solution can be expressed using three unique percent-based concentrations.  In order to distinguish a mass/volume percent, which is calculated by simplifying a mass-to-volume ratio, from the other percent-based concentrations, the unit in which a mass/volume percent concentration is reported is "% m/v," and the chemical formula of the solute is written as the secondary unit on this calculated quantity.

Finally, because mass/volume percents are not defined as exact quantities, their values should be reported using the correct number of significant figures.  However, "100" is an exact number and, therefore, does not impact the significance of the final reported concentration.

Calculate the mass/volume percent of a 762.5 milliliter solution that is prepared by dissolving 289.15 grams of calcium azide, Ca(N3)2, in water.

Before this equation can be applied, the validity of the units that are associated with the given numerical values must be confirmed.  As stated above, the mass of the solute must be expressed in grams, and the volume of the solution must be provided in milliliters.  Therefore, the given quantities are both expressed in the appropriate unit and can be directly incorporated into the mass/volume percent equation, as shown below.  The mass and volume units are eliminated during the simplification of this equation, even though "g" and "mL" do not cancel, mathematically, in order to avoid obtaining a density unit as a result of dividing the given quantities.  In order to distinguish a mass/volume percent, which is calculated by simplifying a mass-to-volume ratio, from the other percent-based concentrations, the unit in which the resultant concentration is reported is "% m/v Ca(N3)2."  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.

\(\text{Mass/Volume Percent}\) = \( \dfrac{289.15 \; \rm{g} \; \rm{Ca(N_3)_2}}{762.5 \; \rm{mL} \; \rm{solution}}\) × \({100}\)

\(\text{Mass/Volume Percent}\) = \({37.92131... \%\ \rm{m/v} \; \rm{Ca(N_3)_2}} ≈ {37.92 \%\ \rm{m/v} \; \rm{Ca(N_3)_2}}\)