Mass Percent Formula: Concept and Solved Examples

Mass Percent Formula: Sugar solution can be prepared by dissolving sugar crystals in water. The sweetness of the solution depends on the amount of sugar being dissolved. In other words, we can say the sweetness of the solution depends on the concentration of sugar in the water.  

In this article, mass percent formula, you will understand different methods of calculation of mass percent, i.e., mass by mass percentage- formula, mass by volume percentage and volume by volume percentage. You will also explore the formula of mass percentage, percent composition formula, and numerical mass percentage in detail.

What is the Concentration of a Solution?

In chemistry, concentration is the measure of how much of a given substance is mixed with another substance. This can apply to any sort of chemical mixture but most often is used in relation to solutions.

The concentration of a solution is defined as the amount of solute present per unit mass or per unit volume of the solution.

The concentration of a solution is expressed in several ways such as:

1. Mass by the mass percentage of solute in solution.
2. Mass by volume percentage of solute in solution.
3. Volume by volume percentage of solute in solution.

What is the Solution?

A solution is a homogeneous mixture of two or more substances. A solution essentially contains two components:

1. Solute 
2. Solvent

(i) Solute: The substance that is being dissolved or undergoes a change in its physical state is called a solute.

(ii) Solvent: The substance that dissolves the other substance and does not undergo a change in its physical state is called a solvent.

For example, when common salt is dissolved in water, common salt is a solute, and water is a solvent. Here, crystals of common salt undergo a physical state change from solid to fine particles attached to water molecules.

Mass by Mass Percentage Formula

In this measurement system, the concentration of a solution is expressed as the mass of solute in grams present per \(100\,{\rm{ g}}\) of the solution.

Thus, \({\rm{the}}\,{\rm{concentration}}\,{\rm{of}}\,{\rm{solute}}\,{\rm{in\, mass}}\,{\rm{by}}\,{\rm{mass}}\,{\rm{percentage}}{\mkern 1mu} {\rm{ = }}{\mkern 1mu} \frac{{{\rm{Mass}}\;{\rm{of}}\;{\rm{solute}}}}{{{\rm{Mass}}\;{\rm{of}}\;{\rm{solution}}}} \times 100\)

While expressing the concentration in terms of mass by mass percentage formula, both the mass of solute and that of the solution must be expressed in the same mass unit, viz. both in grams or both in kilograms, etc.

Example: If \(24{\rm{ g}}\) of glucose is dissolved in \(200\,{\rm{ g}}\) of solution, then

Mass by the mass percentage formula of glucose in solution \( = \frac{{{\rm{Mass\;of\;solute}}}}{{{\rm{Mass\;of\;solution}}}} \times 100\)

\(= \frac{{24}}{{200}} \times 100 = 12\% \)

Solved Example on Mass by Mass Percentage Formula

Q.1. A sugar syrup of mass \(214.2\,{\rm{ g}}\) contains \(34.2\,{\rm{ g}}\) of sugar. Calculate the concentration of sugar in the syrup.
Ans: Mass of solute \( = {\rm{ }}34.2\,{\rm{ g}}\)
Mass of the solution \( = {\rm{ }}214.2\,{\rm{ g}}\)
\(\therefore \) Mass percentage sugar in the syrup \( = \frac{{{\rm{Mass\;of\;solute}}}}{{{\rm{Mass\;of\;solution}}}} \times 100\)
\( = \frac{{34.2 \times 100}}{{214.2}}\)
\( = 15.96\;\% \)

Q.2. A solution is made by mixing \(32\, {\rm{ g}}\) methanol and \(230\, {\rm{ g}}\) ethanol at \(300\, {\rm{ K}}.\) What is the percentage by mass of methanol in the solution?
Ans: Here, Mass of methanol (solute) \( = {\rm{ }}32\,{\rm{ }}\,{\rm{g}}\)
Mass of ethanol (solvent) \( = {\rm{ }}230\,{\rm{ }}\,{\rm{g}}\)
So, the mass of solution \( = 32 + 230 = 262\;\,{\rm{g}}.\)
\(\therefore \) Percentage by mass of methanol in the solution
\( = \frac{{{\rm{Mass\;of\;solute}}}}{{{\rm{Mass\;of\;solution}}}} \times 100\)
\( = \frac{{32 \times 100}}{{262}}\)
\( = 12.21\,\% \)

Mass by Volume Percentage

In this measurement system, the concentration of a solution is expressed in terms of the mass of solute per \(100\) units of the volume of the solution. Thus,

The concentration of solute in mass by volume percentage \( = \frac{{{\rm{Mass\;of\;solute}}}}{{{\rm{Volume\;of\;solution\;\;}}}} \times 100\)

Depending upon the units of the volume of the solution, the unit of mass by volume percentage may be,

1. gram per millilitre (g/mL)
2. gram per litre (g/L)
3. kilogram per litre (kg/L)

For example, if \(10\,\,{\rm{g}}\) of sugar is dissolved in \(100\,\,{\rm{mL}}\) of the solution, then

Mass by volume percentage of sugar in solution \( = \frac{{{\rm{10\;g\;of\;sugar}}}}{{{\rm{100\;mL\;of\;solution\;\;}}}} \times 100\)

\( = \left( {\frac{{10}}{{100}}} \right)\left( {\frac{{\rm{g}}}{{{\rm{mL}}}}} \right) \times 100 = 10\,\% \;\left( {{\rm{w/v}}} \right){\rm{\;in\;g/mL}}\)

Solved Example On Mass by Volume Percentage

Q.1 If \({\rm{40}}\,{\rm{g}}\) of hydrochloric acid \(\left( {{\rm{HCl}}} \right)\) is present in \({\rm{100}}\,\,{\rm{mL}}\) of solution. Calculate the mass by volume percentage of \(\left( {{\rm{HCl}}} \right)\) in solution.
Ans: Mass by volume percentage of \(\left( {{\rm{HCl}}} \right)\) in solution \( = \frac{{{\rm{40\;g\;of\;HCl}}}}{{{\rm{100\;mL\;of\;solution}}}}\; \times 100\)
\( = \left( {\frac{{40}}{{100}}} \right)\left( {\frac{{\rm{g}}}{{{\rm{mL}}}}} \right) \times 100 = 40\% \;\left( {{\rm{w/v}}} \right)\)

Q.2. A \(200{\rm{ ml}}\) solution contains \(15{\rm{ g}}\) of solute. Find the concentration of a solution.
Ans: Mass by volume percentage \( = \frac{{{\rm{Mass\;of\;solute}}}}{{{\rm{volume\;of\;solution}}}} \times 100\)
\( = \frac{{15}}{{200}} \times 100 = 7.5\% \,\,\left( {{\rm{w/v}}} \right)\)

Volume by Volume Percentage

In this measurement system, the concentration is expressed in terms of the volume of solute per \(100\) units of volume of solution. Thus,

The concentration of solute in volume by volume percentage \( = \frac{{{\rm{Volume\;of\;solute}}}}{{{\rm{Volume\;of\;solution\;\;}}}} \times 100\)

For example, if \(10\,\,{\rm{mL}}\) of alcohol is present in \(100\,\,{\rm{mL}}\) solution, then

Volume by volume percentage of alcohol in solution \( = \frac{{{\rm{10\;mL\;of\;alcohol}}}}{{{\rm{100\;mL\;of\;solution\;\;}}}} \times 100 = 10\% \;\left( {{\rm{v/v}}} \right)\)

Volume by Volume Percentage Solved Example

Q.1. \(25\,\,{\rm{mL}}\) of ethanol is present in \(150\,\,{\rm{mL}}\) of its aqueous solution. What is the concentration of the solution?
Ans: \({\rm{Volume\;of\;ethanol  =  25\;mL}}\)
\({\rm{Volume\;of\;solution}} = {\rm{Volume\;of\;ethanol}} + {\rm{Volume\;of\;water}} = 150\,\,{\rm{\;mL}}\)
Volume by volume percentage of ethanol in solution \( = \frac{{{\rm{Volume\;of\;ethanol}}}}{{{\rm{Volume\;of\;solution\;\;}}}}\; \times 100\)
\( = \frac{{25}}{{150}} \times 100 = 16.67\,\% {\rm{\;}}\left( {{\rm{v/v}}} \right)\)

Calculation of Mass Percent from Density

Mass of the unit volume of the substance is called density.

Density \({\rm{\; = }}\frac{{{\rm{Mass}}}}{{{\rm{Volume}}}}\)

Q.1. The solubility of potassium chloride in water at \(20{\,^{\rm{o}}}{\rm{C}}\) is \(34.7{\rm{ g}}\) in \(100{\rm{ g}}\) of water. The density of the solution is \(1.3\,{\rm{g}}\,{\rm{m}}{{\rm{L}}^{{\rm{ – 1}}}}{\rm{.}}\) Calculate the concentration of potassium chloride in the solution in \({\rm{\% \;}}\,\left( {{\rm{w/v}}} \right){\rm{.}}\)
Ans: \(34.7\,{\rm{ g}}\) of potassium chloride is present in \(100\,{\rm{ g}}\) of water.
\({\rm{Mass\;of\;solution}} = 100\;\,{\rm{g}} + 34.7\,\;{\rm{g}} = 134.7\,\;{\rm{g}}\)
\(134.7\;{\rm{g\;of\;the\;solution\;contains}}\;34.7\;{\rm{g\;of\;potassium\;chloride}}{\rm{.}}\)
Density \(\; = \frac{{{\rm{Mass}}}}{{{\rm{Volume}}}}\)
\(1.3 = \frac{{134.7}}{{{\rm{Volume}}}}\)
\({\rm{Volume\;of\;}}134.7{\rm{\;g\;of\;solution}}\; = 103.62\;{\rm{mL}}\)
\(103.62\;{\rm{mL\;of\;the\;solution\;contains}}\;34.7\;{\rm{g\;of\;potassium\;chloride}}.\)
Therefore, mass of potassium chloride in \(100\;\,{\rm{mL}}\) of the solution \( = \frac{{34.7 \times 100}}{{103.62\;\;}} = 33.49\;\,{\rm{g}}\)
Concentration of the solution \( = 33.49\,\% \,\;\left( {{\rm{w/v}}} \right)\)

Mass Percent Formula for Titration

Titration is the method used to determine the unknown concentration of a solution. In this method, the titrant is prepared as a standard solution, and it is titrated against a solution of unknown concentration.

The mass percentage is calculated by the formula \({{\rm{M}}_{\rm{1}}}{{\rm{V}}_{\rm{1}}}{\rm{ =  }}{{\rm{M}}_{\rm{2}}}{{\rm{V}}_{\rm{2}}}{\rm{,}}\) where \({\rm{M}}\) is the molarity (the moles of a solute per litre volume of a solution) and V is the volume.

Summary

Understanding concentration terms is required in chemistry to prepare a variety of solutions. The article Mass Percent Formula will assist you in understanding mass by mass percentage formula, mass by volume percentage, and volume by volume percentage. You can use this knowledge to solve numerical problems involving solution concentration.

FAQs on Mass Percent Formula

Following are the frequently asked questions on the mass percent formula:

Q.1. What is the Mass Percent Formula?
Ans: Mass percentage \( = \frac{{{\rm{Mass\;of\;solute}}}}{{{\rm{Mass\;of\;solution}}}} \times 100\)

Q.2. What is the formula of mass by volume percentage?
Ans: Concentration of solute in mass by volume percentage \( = \frac{{{\rm{Mass\;of\;solute}}}}{{{\rm{Volume\;of\;solution}}}} \times 100\)

Q.3. What are the steps for finding the mass percentage?
Ans: First, find out the mass of solute and solution. Mass of solution is the sum of the masses of solute and solvent. Then substitute these values in the following equation and simplify.
Mass percentage \( = \frac{{{\rm{Mass\;of\;solute}}}}{{{\rm{Mass\;of\;solution}}}} \times 100\)

Q.4. How do you calculate percent by mass?
Ans: Percentage by mass is calculated by the following method.
Mass percentage \( = \frac{{{\rm{Mass\;of\;solute}}}}{{{\rm{Mass\;of\;solution}}}} \times 100\)
Mass of solution \( = \) Mass of solute \(+\) Mass of solvent

Q.5. What is the unit of percent by mass?
Ans: Percent by mass as no unit. Its value is represented by the percentage.

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