# 7.13: Concentrations: Mass Percent

- Page ID
- 222377

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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)- Calculate the mass percent of a solution.

As stated in the previous section of this chapter, chemists often utilize unsaturated solutions, which contain less than the maximum amount of solute that can dissolve in a specified amount of solvent, when performing chemical reactions. However, because the quantity of the solute that is contained in an unsaturated solution cannot be described using a solubility limit, which must, by definition, correspond to a maximum amount of solute, 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 mass percent, which is the first of three percent-based concentrations that will be discussed in this chapter.

## Mass Percent Equations and Indicator Words

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

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

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

\(\text{Mass Percent}\) = \( \dfrac{ \rm{m_{solute} \; (\rm{g})}}{\rm{m_{solute} \; (\rm{g}) \; + \; m_{solvent} \; (\rm{g})}} \) × \({100}\)

While these two equations are equivalent to one another, the first equation should be applied to determine the mass 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.

## Mass Percent Calculations

All of the masses that are incorporated into the equations that are shown above must be expressed in grams, 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 grams prior to being incorporated into a mass percent equation. In order to simplify the second equation, order of operations dictates that the addition of the solute and solvent masses occurs first. The mathematical statement that results is identical to the first equation that is presented above.

During the subsequent multiplication and division, the mass unit is canceled, because "g" is present in the numerator and the denominator in the proportion that is being simplified. Because all of the mass-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 mass percent, which is calculated by simplifying a *mass-to-mass* ratio, from the other percent-based concentrations, the unit in which a mass percent concentration is reported is "% m/m," and the chemical formula of the solute is written as the secondary unit on this calculated quantity.

Finally, because mass 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.

For example, calculate the mass percent of a 235 gram solution that is prepared by dissolving 96 grams of lithium acetate in water.

In order to calculate the mass percent 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, H

_{2}O, is the solvent in this solution, and the remaining substance, lithium acetate, LiC

_{2}H

_{3}O

_{2}, is the solute, "by default."

Because a numerical quantity is associated with the word "solution" in the given problem, the first equation that is presented above should be used to determine the mass 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 masses that are incorporated into the equations that are shown above must be reported in grams. Therefore, the given quantities are both expressed in the appropriate unit and can be directly incorporated into the indicated equation, as shown below. During the multiplication and division processes that are used to solve this equation, the mass unit is canceled, because "g" is present in the numerator and the denominator in the proportion that is being simplified. Since all of the mass-based units are eliminated as a result of this cancelation, the unit in which the resultant mass percent concentration is reported is "% m/m LiC_{2}H_{3}O_{2}," in order to distinguish a mass percent, which is calculated by simplifying a *mass-to-mass* 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.

\(\text{Mass Percent}\) = \( \dfrac{96 \; \cancel{\rm{g}} \; \rm{LiC_2H_3O_2}}{235 \; \cancel{\rm{g}} \; \rm{solution}}\) × \({100}\)

\(\text{Mass Percent}\) = \({40.85160... \%\ \rm{m/m} \; \rm{LiC_2H_3O_2}} ≈ {41 \%\ \rm{m/m} \; \rm{LiC_2H_3O_2}}\)

Calculate the mass percent of a solution that is prepared by mixing 132 grams of tin (II) sulfate and 850. grams of water.

**Answer**- In order to calculate the mass 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 tin (II) sulfate, SnSO
_{4}, and water, H_{2}O, must be compared to determine which chemical is the solvent and which substance is the solute in this solution. Water, H_{2}O, is the solvent in this solution, as it is the chemical that is present in the greatest amount, 850. grams, and tin (II) sulfate, SnSO_{4}, is the solute, as it is present in a lesser quantity, 132 grams. Alternatively, because a solution can only contain one solvent, after identifying water, H_{2}O, as the solvent in this solution, tin (II) sulfate, SnSO_{4}, can be classified as the solute "by default."

Because the amount of solvent, water, H_{2}O, is specified in the given problem, the second equation that is presented above should be applied to determine the mass 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 masses that are incorporated into the equations that are shown above must be reported in grams. Therefore, the given quantities are all 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 masses occurs first. During the subsequent multiplication and division, the mass unit is canceled, because "g" is present in the numerator and the denominator in the proportion that is being simplified. Since all of the mass-based units are eliminated as a result of this cancelation, the unit in which the resultant mass percent concentration is reported is "% m/m SnSO_{4}," in order to distinguish a mass percent, which is calculated by simplifying a*mass-to-mass*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.\(\text{Mass Percent}\) = \( \dfrac{132 \; \rm{g} \; \rm{SnSO_4}}{132 \; \rm{g} \; \rm{SnSO_4} \; + \; 850. \; \rm{g} \; \rm{H_2O}}\) × \({100}\)

\(\text{Mass Percent}\) = \( \dfrac{132 \; \cancel{\rm{g}} \; \rm{SnSO_4}}{982 \; \cancel{\rm{g}} \; \rm{solution}}\) × \({100}\)

\(\text{Mass Percent}\) = \({13.441955... \%\ \rm{m/m} \; \rm{SnSO_4}} ≈ {13.4 \%\ \rm{m/m} \; \rm{SnSO_4}}\)