5.8 Concentration Conversions
This tutorial outlines how to take a solubility (in g/100g) and convert it to other concentration units. The following Excel worksheet can be used to log our results as we work through these.
For this exercise we will determine various concentrations of a 136 g saturated aqueous NaCl solution at 25 °C.
We will be using the solubility of NaCl in water at 25 °C of 36g/100g as reported by Wikipedia.
We need to know the volume of the saturated solution (to get concentration such as molarity) which requires us to look up the density of the solution from other sources. Remember, densities are not additive.
Below is a table of aqueous sodium chloride solution densities (in kg L–1 or equivalently, g mL–1) at various concentrations and temperatures (borrowed from handymath.com).
Table: Densities (in kg/L) of NaCl in water at various concentrations and temperatures.
| Conc (wt%) | 0 °C | 10 °C | 25 °C | 40 °C | 60 °C | 80 °C | 100 °C |
|---|---|---|---|---|---|---|---|
| 1 | 1.0075 | 1.0071 | 1.0041 | 0.9991 | 0.99 | 0.9785 | 0.9651 |
| 2 | 1.0151 | 1.0144 | 1.0111 | 1.0059 | 0.9967 | 0.9852 | 0.9719 |
| 4 | 1.0304 | 1.0292 | 1.0253 | 1.0198 | 1.0103 | 0.9988 | 0.9855 |
| 8 | 1.0612 | 1.0591 | 1.0541 | 1.048 | 1.0381 | 1.0264 | 1.0134 |
| 12 | 1.0924 | 1.0895 | 1.0837 | 1.077 | 1.0667 | 1.0549 | 1.042 |
| 16 | 1.1242 | 1.1206 | 1.114 | 1.1069 | 1.0962 | 1.0842 | 1.0713 |
| 20 | 1.1566 | 1.1525 | 1.1453 | 1.1377 | 1.1268 | 1.1146 | 1.1017 |
| 24 | 1.19 | 1.1856 | 1.1778 | 1.1697 | 1.1584 | 1.1463 | 1.1331 |
| 26 | 1.2071 | 1.2025 | 1.1944 | 1.1861 | 1.1747 | 1.1626 | 1.1492 |
Plot
Figure 5.12: Density of aqueous NaCl solution various concentrations and temperatures.
To use this data, we need to know the concentration (in wt%) of a 36 g NaCl / 100 g water solution to choose the correct density at 25 °C. We will calculate it below.
Volume of Solution
1. Find the relevant masses of a saturated solution
We know the mass of the solution is 136 g (given by the problem). Given that the solubility of a saturated NaCl(aq) solution is 36g/100g, we know 36 g of NaCl is present and 100 g of H2O is present. Adding these together gives 136 g of solution (to the nearest whole number).
\[\begin{align*} m_{\mathrm{soln.}} &= m_{\mathrm{solute}} + m_{\mathrm{solvent}} \\[1.5ex] &= 36.0~\mathrm{g~NaCl} + 100~\mathrm{g~H_2O}\\[1.5ex] &= 136~\mathrm{g} \end{align*}\]
2. Find mass fraction
We now calculate the mass fraction of NaCl (to two decimal places).
\[\begin{align*} \omega_{\mathrm{solu.}} &= \dfrac{m_{\mathrm{solu.}}}{m_{\mathrm{soln.}}}\\[1.5ex] &= \dfrac{36.0~\mathrm{g}}{136.0~\mathrm{g}}\\[1.5ex] &= 0.26471\\[1.5ex] &= 0.26 \end{align*}\]
3. Find mass %
It is important to note that “% by mass” has a variety of names. “wt%” is “% by mass”. Find this to the nearest whole number.
\[\begin{align*} \mathrm{mass}~\% &= \omega_{\mathrm{solu.}} \times 100\%\\[1.5ex] &= 0.26471 \times 100\% \\[1.5ex] &= 26.471\%\\[1.5ex] &= 26\% \end{align*}\]
4. Find the density of solution
Cross-referencing the solution density table, we see that a saturated aqueous solution of NaCl with a wt% of 26 (from step 3) at 25 °C has a density of 1.1944 kg L–1.
We convert this density to g mL–1 as follows
\[\begin{align*} \dfrac{1.1944~\mathrm{kg}}{\mathrm{L}} \left ( \dfrac{10^3~\mathrm{g}}{\mathrm{kg}} \right ) \left ( \dfrac{\mathrm{L}}{10^3~\mathrm{mL}} \right ) = 1.1944~\mathrm{g~mL^{-1}} \end{align*}\]
5. Find the volume of solution
Using density and mass, we can find volume (to three decimal places).
\[\begin{align*} d_{\mathrm{soln.}} = \dfrac{m_{\mathrm{soln.}}}{V_{\mathrm{soln.}}} \longrightarrow V_{\mathrm{soln.}} &= \dfrac{m_{\mathrm{soln.}}}{d_{\mathrm{soln.}}} \\[1.5ex] &= \dfrac{136.0~\mathrm{g}}{\mathrm{1.1944~g~mL^{-1}}}\\[1.5ex] &= 113.865~\mathrm{mL} \left ( \dfrac{\mathrm{L}}{10^3~\mathrm{mL}} \right ) \\[1.5ex] &= 0.113865~\mathrm{L}\\[1.5ex] &= 0.114~\mathrm{L}\\[1.5ex] \end{align*}\]
We can now determine solubility in other concentration units. Below are the unrounded values calculated from above.
6. Organize data
| Property | Value |
|---|---|
| msolu | 36 g |
| msolv | 100 g |
| msoln | 136 g |
| mass fraction | 0.26471 |
| mass % | 26.471 |
| dsoln | 1.1944 g mL–1 |
| Vsoln | 0.113865 L |
Mass Concentration
Find mass concentration to two decimal places.
\[\begin{align*} \rho &= \dfrac{\mathrm{g~solute}}{\mathrm{L~solution}}\\[1.5ex] &= \dfrac{36.0~\mathrm{g}}{\mathrm{0.113865~\mathrm{L}}}\\[1.5ex] &= 316.1639~\mathrm{g~L^{-1}}\\[1.5ex] &= 316.16~\mathrm{g~L^{-1}} \end{align*}\]
Molarity
We can now determine the molar solubility (M ; mol L–1) of NaCl (at 25 °C). First convert 36.0 g NaCl to moles.
\[\begin{align*} n_{\mathrm{solu.}} &= 36.0~\mathrm{g~NaCl} \left (\dfrac{\mathrm{mol~NaCl}}{58.44~\mathrm{g~NaCl}} \right )\\[1.5ex] &= 0.6160~\mathrm{mol~NaCl} \end{align*}\]
We can now calculate the molar solubility (M) to two decimal places.
\[\begin{align*} M &= \dfrac{\mathrm{mol~solute}}{\mathrm{L~solution}}\\[1.5ex] &= \dfrac{0.6160~\mathrm{mol~NaCl}}{0.113865~\mathrm{L}}\\[1.5ex] &= 5.4101~\mathrm{mol~L^{-1}}\\[1.5ex] &= 5.41~\mathrm{mol~L^{-1}} \end{align*}\]
Molality
We use dimensional analysis to convert 100 g of water into 1 kg of water.
\[\begin{align*} m &= \dfrac{\mathrm{mol~solute}}{\mathrm{kg~solvent}}\\[1.5ex] &= \dfrac{0.616~\mathrm{mol~NaCl}}{100~\mathrm{g~water}} \left ( \dfrac{1000~\mathrm{g}}{1~\mathrm{kg}} \right ) \\[1.5ex] &= 6.16~\mathrm{mol~kg^{-1}} \end{align*}\]
Mole Fraction
To find mole fraction, we must obtain the solute and solvent amount in moles. We already know how much NaCl (in mol) we have. Let’s determine moles of water.
\[100~\mathrm{g~H_2O} \left ( \dfrac{\mathrm{mol}}{18.02~\mathrm{g}} \right ) = 5.549~\mathrm{mol~H_2O}\]
Now find the mole fraction (to two decimal places).
\[\begin{align*} \chi_{\mathrm{solu}} &= \dfrac{n_{\mathrm{solu.}}}{n_{\mathrm{soln.}}}\\[1.5ex] &= \dfrac{0.616~\mathrm{mol~NaCl}}{0.616~\mathrm{mol~NaCl} + 5.549~\mathrm{mol~water}}\\[1.5ex] &= 0.09992\\[1.5ex] &= 0.10 \end{align*}\]
Mole Percent
Multiply the mole fraction by 100% to obtain the mole % (to two decimal places).
\[\begin{align*} \mathrm{mol~\%} &= \chi_{\mathrm{solu.}}\times 100\% \\[1.5ex] &= 0.09992 \times 100\%\\[1.5ex] &= 9.992\%\\[1.5ex] &= 9.99\% \end{align*}\]
Mass-Volume Percent
The m/v% of the solution (in g mL–1 to two decimal places) can be found as follows.
\[\begin{align*} m/v~\% &= \dfrac{m_{\mathrm{solute~(g)}}}{V_{\mathrm{solution~(mL)}}} \times 100\% \\[1.5ex] &= \dfrac{36~\mathrm{g~NaCl}}{113.865~\mathrm{mL}} \times 100\% \\[1.5ex] &= 31.6164\%\\[1.5ex] &= 31.62\% \end{align*}\]
Parts per million (by mass)
The parts per million (ppm) concentration unit is reserved for very dilute solutions. However, we can still calculate the concentration (in ppm by mass) of a saturated aqueous NaCl solution.
\[\begin{align*} \mathrm{ppm} &= \dfrac{m_{\mathrm{solu.}}}{m_{\mathrm{soln.}}} \times 10^6 \\[1.5ex] &= \dfrac{36.0~\mathrm{g~NaCl}}{136~\mathrm{g~soln.}} \times 10^6 \\[1.5ex] &= 2.65\times 10^5~\mathrm{ppm~(by~mass)} \end{align*}\]
Parts per million (by volume)
The parts per million (ppm) concentration unit is reserved for very dilute solutions. However, we can still calculate the concentration (in ppm by volume) of a saturated aqueous NaCl solution.
First, determine the volume of 36.0 g NaCl by using the density of NaCl (dNaCl = 2.165 g cm–3).
\[\begin{align*} d_{\mathrm{NaCl}} &= \dfrac{m_{\mathrm{NaCl}}}{V_{\mathrm{NaCl}}} \longrightarrow \\[1.5ex] V_{\mathrm{NaCl}} &= \dfrac{m_{\mathrm{NaCl}}}{d_{\mathrm{NaCl}}} \\[1.5ex] &= \dfrac{36.0~\mathrm{g~NaCl}}{2.165~\mathrm{g~cm^{-3}}}\\[1.5ex] &= 16.628~\mathrm{cm^3} \\[1.5ex] &= 16.628~\mathrm{mL} \end{align*}\]
Recall that cm3 is a mL. Now solve for ppm by volume.
\[\begin{align*} \mathrm{ppm} &= \dfrac{V_{\mathrm{solu.}}}{V_{\mathrm{soln.}}} \times 10^6 \\[1.5ex] &= \dfrac{16.628~\mathrm{mL~NaCl}}{113.865~\mathrm{mL~soln.}} \times 10^6 \\[1.5ex] &= 1.46\times 10^5~\mathrm{ppm~(by~volume)} \end{align*}\]
Below are the (rounded) tabulated concentrations of a saturated aqueous solution of NaCl at 25 °C.
| Concentration | Unit | Value |
|---|---|---|
| g/100g | g/100g | 36 |
| mass fraction | unitless | 0.26 |
| mass percent | % | 26.47 |
| mass concentration | g L–1 | 316.16 |
| molarity | mol L–1 | 5.41 |
| molality | mol kg–1 | 6.16 |
| mole fraction | unitless | 0.10 |
| mole percent | % | 9.99 |
| mass-volume percent | % | 31.62 |
| ppm (by mass) | ppm | 2.65 × 105 |
| ppm (by volume) | ppm | 1.46 × 105 |
Practice
You fill a jug of seawater. We assume the seawater is a solution of NaCl and water. The solution has a density of 1.08 g mL–1 and a concentration of 0.600 M. What is the percent by mass and percent by volume (each to two decimal places) of the salt in the solution? NaCl has a density of 2.165 g cm–3.
Solution
Volume of solution
Assume a 100g solution.
\[\begin{align*} V_{\mathrm{soln}} &= 100~\mathrm{g~solution} \left ( \dfrac{1~\mathrm{mL~solution}}{1.08~\mathrm{g~solution}} \right ) \left (\dfrac{1~\mathrm{L}}{10^3~\mathrm{mL}} \right ) \\ &= 0.092592~\mathrm{L~solution} \end{align*}\]
Moles of solute
\[\begin{align*} n_{\mathrm{solute}} &= 0.600~M \times 0.092592~\mathrm{L~solution} = 0.055556~\mathrm{mol~NaCl} \end{align*}\]
Mass of solute
\[\begin{align*} m_{\mathrm{solute}} &= 0.055556~\mathrm{mol~NaCl} \left ( \dfrac{58.44~\mathrm{g~NaCl}}{1~\mathrm{mol~NaCl}} \right ) = 3.2467~\mathrm{g~NaCl} \end{align*}\]
Percent by mass
\[\begin{align*} \mathrm{mass~\%} &= \dfrac{m_{\mathrm{solute}}}{m_{\mathrm{solution}}} \times 100\% \\[1.5ex] &= \dfrac{3.2467~\mathrm{g~NaCl}}{100.0~\mathrm{g~solution}} \times 100\% \\[1.5ex] &= 3.25\% \end{align*}\]
Volume of solute
\[\begin{align*} V_{\mathrm{solute}} &= 3.2467~\mathrm{g~NaCl} \left ( \dfrac{\mathrm{mL}}{2.165~\mathrm{g~NaCl}} \right ) \left (\dfrac{1~\mathrm{L}}{10^3~\mathrm{mL}} \right ) \\[1.5ex] &= 1.4996\times 10^{-3}~\mathrm{L~NaCl} \end{align*}\]
Percent by volume
\[\begin{align*} \mathrm{vol~\%} &= \dfrac{V_{\mathrm{solute}}}{V_{\mathrm{solution}}} \times 100\% \\[1.5ex] &= \dfrac{1.4996\times 10^{-3}~\mathrm{L~NaCl}}{0.092592~\mathrm{L~solution}} \times 100\% \\[1.5ex] &= 1.62\% \end{align*}\]
Practice
A solution is 3.45% NaCl by mass. The solution has a density of 1.11 g mL–1. What is the volume (in L) of the solution?
Solution
Mass of solute
Assume a 100g solution.
\[\begin{align*} m_{\mathrm{solu}} &= \dfrac{\%~\mathrm{by~mass}}{100~\%} \times \mathrm{m_{soln}} \\ &= \dfrac{3.45\%}{100~\%} \times \mathrm{100~g} \\ &= 3.45~\mathrm{g} \end{align*}\]
Mass of solution
\[\begin{align*} m_{\mathrm{soln}} &= m_{\mathrm{solu}} + m_{\mathrm{solv}} \\ &= 3.45~\mathrm{g} + 100~\mathrm{g} \\ &= 103.45~\mathrm{g} \end{align*}\]
Volume of solution
\[\begin{align*} V_{\mathrm{soln}} &= 103.45~\mathrm{g~solution} \left ( \dfrac{1~\mathrm{mL~solution}}{1.11~\mathrm{g~solution}} \right ) \left (\dfrac{1~\mathrm{L}}{10^3~\mathrm{mL}} \right ) \\ &= 0.093~\mathrm{L~solution} \end{align*}\]