Percent concentration

Percentage concentrations are useful to show the direct comparison between a solvent and a solute. There are three ways of expressing concentration as a percentage.

1. Mass/mass percent concentration, (m/m)% \[
   \frac{\textrm{mass of solute}\left(g\right)}{\textrm{mass of solution (g) }}\times100
   \]

   Example: What is the percent by mass concentration of sodium chloride solution prepared by dissolving [latex]3.25\textrm{g}[/latex] of sodium chloride in [latex]42.50\textrm{g}[/latex] of water?

   Masses of the solute and the solvent are given, and we are asked to calculate the percent concentration of sodium chloride by mass:\[\begin{align*}
   = & \frac{\textrm{mass of solute}\left(g\right)}{\textrm{mass of solution}\left(g\right)}\times100\\
   = & \frac{3.25g}{42.50g+3.25g}\times100\\
   = & 7.1\%
   \end{align*}\] 
2. Volume/volume percent concentration, (v/v)% \[
   \frac{\textrm{volume of solute}\left(mL\right)}{\textrm{volume of solution}\left(mL\right)}\times100\]

   Example: If ethanol [latex]8.00mL[/latex] is dissolved in enough water to give [latex]80.00mL[/latex] of the solution, what is the percent by volume concentration of ethanol in the resulting solution?

   We are asked to calculate ethanol concentration in percent by volume. The volumes of the solute and the solution are given: \[\begin{align*}
   = & \frac{\textrm{volume of solute}\left(mL\right)}{\textrm{volume of solution}\left(mL\right)}\times100\\
   = & \frac{8.00mL}{80.00mL}\times100\\
   = & 10.0\%
   \end{align*}\]

    

3. Mass/volume percent concentration, (m/v)% \[
   \frac{\textrm{mass of solute}\left(g\right)}{\textrm{volume of solution}\left(mL\right)}\times100
   \]

   Example: If [latex]5.0g[/latex] of [latex]\ce{MgCl}_{2}[/latex] is dissolved in enough water to give [latex]200\textrm{mL}[/latex] of solution, calculate the concentration of [latex]\ce{MgCl}_{2}[/latex] in (m/v)%.

   We are asked to calculate the concentration of  [latex]\ce{MgCl}_{2}[/latex] in (m/v)%. The mass of the [latex]\ce{MgCl}_{2}[/latex] and the volume of the resulting solution are given: \[\begin{align*}
   = & \frac{\textrm{mass of solute}\left(g\right)}{\textrm{volume of solution}\left(mL\right)}\times100\\
   = & \frac{5.0g}{200mL}\times100\\
   = & 2.5\%
   \end{align*}\]

    

Parts Per Million and Parts Per Billion

Parts per million (ppm) and parts per billion (ppb) are typically used to describe trace quantities of solutes found within a solution. They represent the parts of an analyte found within one million or one billion parts of a medium respectively.

Parts can be in either mass or volume units. Both solute and solvent must have the same units. For example, one gram of solute is present in a million grams of solution, and one millilitre of benzene is present in a million millilitres of air. To illustrate: \[
1ppm=\frac{1mL}{1000000mL}\times1000000
\]

Commonly, however, analytes are found dissolved in water. As the density of water is extremely close to  [latex]1~gmL^{-1}[/latex], it is often appropriate to report [latex]1~ppm[/latex] as equal to [latex]1~mgL^{-1}[/latex] for aqueous solutions near room temperature.

Molarity (M)

Molarity is the number of solute moles present in one litre of a solution. This is the most common way to show the concentration of a solution within chemical laboratories thanks to its direct relationship to moles (which, as discussed in 5.2 ‘Molar Mass’, can be easily determined by mass). We can determine molarity by rearranging the following formula:

\[
n=cv
\]

Where, [latex]n[/latex] is the number of moles [latex]\left(\textrm{mol}\right)[/latex]
[latex]c[/latex] is the molarity of the solution [latex]\left(\textrm{molL}^{-1}\right)[/latex]
[latex]v[/latex] is the volume of the solution [latex]\left(\textrm{L}\right)[/latex]

This is one of the most important formulas in chemistry! It allows us to determine the amount of moles of a substance present within a given concentration and the volume of a substance. However, rearranging allows us to determine the moles required to create a desired concentration for a given volume. As such, a common version of the above equation utilised by chemists is as follows:

\[
c=\frac{n}{v}
\]

Figure 5.3.1: Simulation by PhET Interactive Simulations, University of Colorado Boulder, licensed under CC-BY-4.0 (https://phet.colorado.edu).

In combination with [latex]n=\frac{m}{M}[/latex], these equations allow chemists to determine the mass of a solute needed to achieve a specific concentration within a solution. These two equations represent the majority of undergraduate chemistry calculations – try to remember both off by hard! Let’s observe how we can find the mass of a compound needed to achieve a desired concentration.

Dilution of Solutions

Dilution is a process used to lower the concentration of the original solution by adding solvents. The concentrated solution is known as the stock solution. However, the amount of solute present in the solution remains constant before and after the dilution; only the volume changes. The following equation shows the relationship between concentration and volume before and after dilution. This equation is built upon the fact that the same amount of solute is present in the solution before (stock solution) and after (diluted solution) the dilution: \[
C_{1}V_{1}=C_{2}V_{2}
\]

Where,

[latex]C_{1}[/latex]= Concentration of the stock solution

[latex]V_{1}[/latex]= Volume of the stock solution

[latex]C_{2}[/latex]= Concentration of the diluted solution

[latex]V_{2}[/latex]= Volume of the diluted solution

Chemistry Is Everywhere: Preparing IV Solutions

In a hospital emergency room, a physician orders an intravenous (IV) delivery of 100 mL of 0.5% KCl for a patient suffering from hypokalemia (low potassium levels). Does an aide run to a supply cabinet and take out an IV bag containing this concentration of KCl (Figure 5.3.2)?

Not likely. It is more probable that the aide must make the proper solution from an IV bag of sterile solution and a more concentrated, sterile solution, called a stock solution, of KCl. The aide is expected to use a syringe to draw up some stock solution and inject it into the waiting IV bag and dilute it to the proper concentration. Thus the aide must perform a dilution calculation.

If the stock solution is 10.0% KCl and the final volume and concentration need to be 100 mL and 0.50%, respectively, then to determine how much stock solution to use:

[latex]\begin{array}{rcl} (10\%)V_1&=&(0.50\%)(100\text{ mL}) \\ V_1&=&5\text{ mL} \end{array}[/latex]

Of course, the addition of the stock solution affects the total volume of the diluted solution, but the final concentration is likely close enough, even for medical purposes.

Medical and pharmaceutical personnel are constantly dealing with dosages that require concentration measurements and dilutions. It is an important responsibility: calculating the wrong dose can be useless, harmful, or even fatal!

In most cases, only a certain maximum amount of solute can be dissolved in a given amount of solvent. This maximum amount is called the solubility of the solute. It is usually expressed in terms of the amount of solute that can dissolve in 100 g of the solvent at a given temperature. Table 5.3.1 “Solubilities of Some Ionic Compounds” lists the solubilities of some simple ionic compounds. These solubilities vary widely: NaCl can dissolve up to 31.6 g per 100 g of H₂O, while AgCl can dissolve only 0.00019 g per 100 g of H₂O.

When the maximum amount of solute has been dissolved in a given amount of solvent, we say that the solution is saturated with solute. When less than the maximum amount of solute is dissolved in a given amount of solute, the solution is unsaturated. These terms are also qualitative terms because each solute has its own solubility. A solution of 0.00019 g of AgCl per 100 g of H₂O may be saturated, but with so little solute dissolved, it is also rather dilute. A solution of 36.1 g of NaCl in 100 g of H₂O is also saturated but rather concentrated. Ideally, we need more precise ways of specifying the amount of solute in a solution.

In some circumstances, it is possible to dissolve more than the maximum amount of a solute in a solution. Usually, this happens by heating the solvent, dissolving more solute than would normally dissolve at regular temperatures, and letting the solution cool down slowly and carefully. Such solutions are called supersaturated solutions and are not stable; given an opportunity (such as dropping a crystal of solute in the solution), the excess solute will precipitate from the solution.

It should be obvious that some solutes dissolve in certain solvents but not others. NaCl, for example, dissolves in water but not in vegetable oil. Beeswax dissolves in liquid hexane but not water. What is it that makes a solute soluble in some solvents but not others?

The answer is intermolecular interactions. Intermolecular interactions include London dispersion forces, dipole-dipole interactions, and hydrogen bonding. From experimental studies, it has been determined that if the molecules of a solute experience the same intermolecular forces that the solvent does, the solute will likely dissolve in that solvent. So, NaCl — a very polar substance because it is composed of ions — dissolves in water, which is very polar, but not in oil, which is generally nonpolar. Nonpolar wax dissolves in nonpolar hexane but not in polar water. This concept leads to the general rule that “like dissolves like” for predicting whether a solute is soluble in a given solvent (see Figure 5.3.3). However, this is a general rule, not an absolute statement, so it must be applied with care.