7.14: Concentrations: Volume Percent


Learning Objectives
• Calculate the 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 equations that are used to calculate a volume percent, which is the second of three percent-based concentrations that will be discussed in this chapter.

Volume Percent Equations and Indicator Words

The volume percent of a solution is defined as the ratio of the volume of solute that is present in a solution, relative to the volume of the solution, as a whole.  Additionally, because this type of concentration, which is typically calculated for liquid- and gas-phase solutions, is expressed as a percentage, this proportion must be multiplied by 100, as shown below.

$$\text{Volume Percent}$$ = $$\dfrac{ \rm{V_{solute} \; (\rm{mL})}}{\rm{V_{solution} \; (\rm{mL})}}$$ × $${100}$$

Because a solution is comprised of both a solute and a solvent, the volume of a solution, as a whole, is equal to the sum of the volumes of the solute and the solvent that it contains.  Therefore, the following equation can also be used to calculate the volume percent of a solution.

$$\text{Volume Percent}$$ = $$\dfrac{ \rm{V_{solute} \; (\rm{mL})}}{\rm{V_{solute} \; (\rm{mL}) \; + \; V_{solvent} \; (\rm{mL})}}$$ × $${100}$$

While these two equations are equivalent to one another, the first equation should be applied to determine the volume percent of a solution if a numerical quantity is associated with the word "solution" in a given problem.  In contrast, the second equation should be utilized if the amount of solvent that is present in the solution is specified.

Volume Percent Calculations

All of the volumes that are incorporated into the equations that are shown above must be expressed in milliliters, and the chemical formula of each component must be written as the secondary unit on its associated numerical quantity.  Therefore, if the amount of solute, solvent, or solution is reported using an alternative unit, its value would need to be converted to milliliters prior to being incorporated into a volume percent equation.  In order to simplify the second equation, order of operations dictates that the addition of the solute and solvent volumes occurs first.  The mathematical statement that results is identical to the first equation that is presented above.

During the subsequent multiplication and division, the volume unit is canceled, because "mL" is present in the numerator and the denominator in the proportion that is being simplified.  Because all of the volume-based units are eliminated as a result of this cancelation, the calculated answer is expressed as a percentage.  However, as stated above, 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 volume percent, which is calculated by simplifying a volume-to-volume ratio, from the other percent-based concentrations, the unit in which a volume percent concentration is reported is "% v/v," and the chemical formula of the solute is written as the secondary unit on this calculated quantity.

Finally, because 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.

Exercise $$\PageIndex{1}$$

Calculate the volume percent of a solution that is prepared by mixing 519.2 milliliters of helium and 168.4 milliliters of molecular chlorine.

In order to calculate the volume percent of a solution, each substance that is referenced in the problem must first be classified as a solute or a solvent.  Because neither the indicator word "in" nor a solubility limit is present in the given statement, the relative amounts of helium, He, and molecular chlorine, Cl2, must be compared to determine which chemical is the solvent and which substance is the solute in this solution.  Helium, He, is the solvent in this solution, as it is the chemical that is present in the greatest amount, 519.2 milliliters, and molecular chlorine, Cl2, is the solute, as it is present in a lesser quantity, 168.4 milliliters.  Alternatively, because a solution can only contain one solvent, after identifying helium, He, as the solvent in this solution, molecular chlorine, Cl2, can be classified as the solute "by default."

Because the amount of helium, He, is specified in the given problem, the second equation that is presented above should be applied to determine the volume percent of this solution.  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, all of the volumes that are incorporated into the equations that are shown above must be reported in milliliters.  Therefore, the given quantities are both expressed in the appropriate unit and can be directly incorporated into the indicated equation, as shown below.  In order to simplify this equation, order of operations dictates that the addition of the solute and solvent volumes occurs first.  During the subsequent multiplication and division, the volume unit is canceled, because "mL" is present in the numerator and the denominator in the proportion that is being simplified.  Since all of the volume-based units are eliminated as a result of this cancelation, the unit in which the resultant volume percent concentration is reported is "% v/v Cl2," in order to distinguish a volume percent, which is calculated by simplifying a volume-to-volume ratio, from the other percent-based concentrations.  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.

$$\text{Volume Percent}$$ = $$\dfrac{168.4 \; \rm{mL} \; \rm{Cl_2}}{168.4 \; \rm{mL} \; \rm{Cl_2} \; + \; 519.2 \; \rm{mL} \; \rm{He}}$$ × $${100}$$

$$\text{Volume Percent}$$ = $$\dfrac{168.4 \; \cancel{\rm{mL}} \; \rm{Cl_2}}{687.6 \; \cancel{\rm{mL}} \; \rm{solution}}$$ × $${100}$$

$$\text{Volume Percent}$$ = $${24.49098... \%\ \rm{v/v} \; \rm{Cl_2}} ≈ {24.49 \%\ \rm{v/v} \; \rm{Cl_2}}$$

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