# Describing chemical solutions¶

The composition of a chemical solution can be specified using sets of various quantities (e.g. amount fractions, mass fractions, concentrations, molalities). In this document we develop the formulas to switch from one set to another.

The results are summarized in the Tables section, at the end of the document.

## Some definitions¶

A (chemical) solution is a mixture of one or more solutes dissolved in a solvent. 1

The amount of substance (symbol $$n$$) is the quantity that measures a number of elementary entities (like atoms, ions, molecules, etc.). 2 Its S.I. unit is the mole (symbol $$\text{mol}$$). An amount of substance of one mole contains $$6.02214076\times10^{23}$$ elements (this is the Avogadro number). 3

Let’s consider a solution of $$p$$ solutes identified by the indices $$1$$ through $$p$$, in a solvent identified by the index $$0$$. The quantities related to the whole solutions have the subscript $$\text{s}$$.

We will need the following quantities
• $$n_i$$ is the amount $$i$$ in the solution (unit $$\text{mol}$$),

• $$M_i$$ is the mass of substance $$i$$ in the solution (unit $$\text{kg}$$),

• $$V_i$$ is the volume of substance $$i$$ in the solution (unit $$\text{m}^3$$).

Again, in these, $$i$$ can be either $$0$$ (for the solvent), $$1$$ to $$p$$ (for the solutes) or $$\text{s}$$ (for the whole solution).

The composition of the solution, in terms of substance $$i\in [0, p]$$, can be expressed using

• Amount fraction (a.k.a. mole fraction), symbol $$x$$, is the ratio of the amount of solute to the amount of solution. 4 It has no unit.

(1)$x_i = \frac{n_i}{n_\text{s}}.$

As, by definition, $$n_\text{s}=\sum_{i=0}^p n_i$$, the amount fractions obey

(2)$\sum_{i=0}^p x_i = 1.$
• mass fraction (a.k.a. percent weight or %wt when expressed with a denominator of 100), symbol $$w$$, is the ratio of the mass of solute to the mass of solution. 5 It has no unit.

(3)$w_i = \frac{M_i}{M_\text{s}}.$

As, by definition, $$M_\text{s}=\sum_{i=0}^p M_i$$, the amount fractions obey

$\sum_{i=0}^p w_i = 1.$
• concentration (a.k.a. molarity), symbol $$c$$, is the amount of solute per volume of solution. 6 Its S.I. unit is $$\text{mol.m}^{-3}$$

(4)$c_i = \frac{n_i}{V_\text{s}}.$
• molality, symbol $$m$$, is the amount of solute per mass of solvent (not per mass of solution!). 7 Its S.I. unit is $$\text{mol.kg}^{-1}$$

(5)$m_i = \frac{n_i}{M_0}.$

To convert from one set of expressions to another, we will need the molar masses and volumes of the constituents.

• The molar mass of a substance $$i$$, symbol $$\mathcal{M}_i$$, is the mass of one mole of said substance, unit $$\text{kg.mol}^{-1}$$. As a consequence, the mass of substance $$i$$ in the solution is

$M_i = n_i\mathcal{M}_i.$

The molar mass of a solution is computed as the sum of the molar masses of its constituents weighted by their amount fractions

(6)$\mathcal{M}_\text{s} = \sum_{i=0}^p x_i\mathcal{M}_i.$
• The molar volume of a substance $$i$$, symbol $$\mathcal{V}_i$$, is the volume of one mole of said substance, unit $$\text{m}^3\text{mol}^{-1}$$. As a consequence, the volume of substance $$i$$ in the solution is

$V_i = n_i\mathcal{V}_i.$

For some solutions, the molar volume of a solution can be computed as the sum of the molar masses of its constituents weighted by their amount fractions

(7)$\mathcal{V}_\text{s} = \sum_{i=0}^p x_i\mathcal{V}_i.$

A solution for which this relation is true is called an ideal solution.

## Some conversions¶

In this section we establish the formulas to convert from one set of quantities to another.

### From the amount fractions¶

Let the amount fractions $$x_0$$ through $$x_p$$ be known.

The molar mass of the solution is, by its definition (6),

$\mathcal{M}_\text{s} = \sum_{i=0}^p x_i\mathcal{M}_i.$

The molar volume of the solution is by its definition (7),

$\mathcal{V}_\text{s} = \sum_{i=0}^p x_i\mathcal{V}_i.$

Using these,

the mass fraction of the solute $$i$$ derives from its definition (3),

(8)$w_i = \frac{M_i}{M_\text{s}} = \frac{n_i \mathcal{M}_i}{n_\text{s}\mathcal{M}_\text{s}} = x_i \frac{\mathcal{M}_i}{\mathcal{M}_\text{s}},$

the concentration of the solute $$i$$ derives from its definition (4),

(9)$c_i = \frac{n_i}{V_\text{s}} = \frac{n_i}{n_\text{s}\mathcal{V}_\text{s}} = \frac{x_i}{\mathcal{V}_\text{s}},$

and the molality of the solute $$i$$ derives from its definition (5),

$m_i = \frac{n_i}{M_0} = \frac{n_i}{n_0\mathcal{M}_0} = \frac{n_i}{n_\text{s}} \frac{n_\text{s}}{n_0} \frac{1}{\mathcal{M}_0} = \frac{x_i}{x_0} \frac{1}{\mathcal{M}_0}.$

### From the mass fractions¶

Let the mass fractions $$w_0$$ through $$w_p$$ be known.

The amount fraction of the solute $$i$$ is given by inverting (8)

(10)$x_i = w_i \frac{\mathcal{M}_\text{s}}{\mathcal{M}_i}.$

The molar mass of the solution is obtained by injecting (10) into (2)

$\sum_{i=0}^p w_i \frac{\mathcal{M}_\text{s}}{\mathcal{M}_i} = 1,$

which yields

$\mathcal{M}_\text{s} = \frac{1}{\sum_{i=0}^p \frac{w_i}{\mathcal{M}_i}}.$

We get the molar volume of the solution by injecting (10) into its definition (7)

$\mathcal{V}_\text{s} = \mathcal{M}_\text{s} \sum_{i=0}^p \frac{w_i \mathcal{V}_i}{\mathcal{M}_i}.$

The concentration of the solute $$i$$ derives from its definition (4)

$c_i = \frac{n_i}{n_\text{s}\mathcal{V}_\text{s}} \frac{\mathcal{M}_i\mathcal{M}_\text{s}}{\mathcal{M}_i\mathcal{M}_\text{s}} = w_i\frac{\mathcal{M}_\text{s}}{\mathcal{V}_\text{s}\mathcal{M}_i}.$

The molality of the solute $$i$$ derives from its definition (5)

$m_i = \frac{n_i}{M_0} = \frac{n_i\mathcal{M}_i}{M_0\mathcal{M}_i} = \frac{M_iM_\text{s}}{M_0\mathcal{M}_iM_\text{s}} = \frac{w_i}{w_0}\frac{1}{\mathcal{M}_i}.$

### From the concentrations¶

Let the concentrations $$c_0$$ through $$c_p$$ be known.

The amount fraction derives from its definition (1)

(11)$x_i = \frac{n_i}{n_\text{s}}\frac{\mathcal{V}_\text{s}}{\mathcal{V}_\text{s}} = c_i\mathcal{V}_\text{s}.$

The molar mass of the solution is obtain by injecting this expression in (7)

$\mathcal{M}_\text{s} = \sum_{i=0}^p x_i\mathcal{M}_i = \mathcal{V}_\text{s}\sum_{i=0}^p c_i\mathcal{M}_i.$

We get the molar volume of the solution by injecting (11) in (2)

$\sum_{i=0}^p c_i\mathcal{V}_\text{s} = 1,$

from which we obtain

$\mathcal{V}_\text{s} = \frac{1}{\sum_{i=0}^p c_i}.$

The mass fraction derives fron its definition (3)

$w_i = \frac{M_i}{M_\text{s}} = \frac{n_i\mathcal{M}_i}{n_\text{s}\mathcal{M}_\text{s}}\frac{\mathcal{V}_\text{s}}{\mathcal{V}_\text{s}} = x_i\frac{\mathcal{M}_i}{\mathcal{M}_\text{s}}. = c_i\frac{\mathcal{M}_i\mathcal{V}_\text{s}}{\mathcal{M}_\text{s}}.$

The molality derives from its definition (5)

$m_i = \frac{n_i}{M_0}\frac{n_0V_\text{s}}{n_0V_\text{s}} = \frac{c_i}{c_0}\frac{n_0}{M_0}.$

### From the molalities¶

Let the molalities $$m_0$$ through $$m_p$$ be known.

If we sum the molalities, by their definitions, we have

$\sum_{j=0}^p m_j = \sum_{j=0}^p \frac{n_j}{M_0} = \frac{n_\text{s}}{n_0\mathcal{M}_0} = \frac{1}{x_0\mathcal{M}_0}.$

So the amount fraction of the solvent is

(12)$x_0 = \frac{1}{\sum_{j=0}^p m_j\mathcal{M}_0}.$

The expressions for the solutes derive from their definitions (1)

$x_i = \frac{n_i}{n_\text{s}}\frac{n_0\mathcal{M}_0}{M_0} = m_i x_0 \mathcal{M}_0 = \frac{m_i}{\sum_{j=0}^p m_j}, \qquad i \neq 0.$

From these we derive the molar mass of the solution

$\mathcal{M}_\text{s} = \sum_{i=0}^p \frac{m_i}{\sum_{j=0}^p m_j} \mathcal{M}_i,$

and the molar volume of the solution

$\mathcal{V}_\text{s} = \sum_{i=0}^p \frac{m_i}{\sum_{j=0}^p m_j} \mathcal{V}_i.$

Combining (8) and (12), we obtain the mass fraction of the solvent

$w_0 = x_0\frac{\mathcal{M}_0}{\mathcal{M}_\text{s}} = \frac{1}{\mathcal{M}_\text{s}\sum_{j=0}^p m_j}.$

Similarly for the solutes

$w_i = x_i\frac{\mathcal{M}_i}{\mathcal{M}_\text{s}} = \frac{m_i}{\sum_{j=0}^p m_j}\frac{\mathcal{M}_i}{\mathcal{M}_\text{s}}, \qquad i \neq 0.$

Using (9), we can compute the concentration of the solute

$c_0=\frac{x_0}{\mathcal{V}_\text{s}} = \frac{1}{\mathcal{M}_0\mathcal{V}_\text{s}\sum_{j=0}^p m_j}.$

Similarily for the solutes

$c_i=\frac{x_i}{\mathcal{V}_\text{s}} = \frac{m_i}{\mathcal{V}_\text{s}\sum_{j=0}^p m_j}, \qquad i \neq 0.$

## Tables¶

### Molar mass and molar volume of the solution¶

Property of the solution

as a function of the

amount fractions

mass fractions

concentrations

molalities

Molar mass $$\mathcal{M}_\text{s}$$

$\sum_{i=0}^p x_i\mathcal{M}_i$
$\frac{1}{\sum_{i=0}^p \frac{w_i}{\mathcal{M}_i}}$
$\mathcal{V}_\text{s}\sum_{i=0}^p c_i\mathcal{M}_i$
$\frac{\sum_{i=0}^p m_i\mathcal{M}_i}{\sum_{j=0}^p m_j}$

Molar volume $$\mathcal{V}_\text{s}$$

$\sum_{i=0}^p x_i\mathcal{V}_i$
$\mathcal{M}_\text{s}\sum_{i=0}^p\frac{w_i\mathcal{V}_i}{\mathcal{M}_i}$
$\frac{1}{\sum_{i=0}^p c_i}$
$\frac{\sum_{i=0}^p m_i\mathcal{V}_i}{\sum_{j=0}^p m_j}$

### Conversions between representations¶

Quantity

expressed in terms of

amount fractions

mass fractions

concentrations

molalities

amount fractions $$x$$

$x_i=w_i \frac{\mathcal{M}_\text{s}}{\mathcal{M}_i}$
$x_i=c_i\mathcal{V}_\text{s}.$
$x_0=\frac{1}{\mathcal{M}_0\sum_{j=0}^p m_j}$
$x_i=\frac{m_i}{\sum_{j=0}^p m_j}$

mass fraction $$w$$

$w_i=\frac{\mathcal{M}_i}{\mathcal{M}_\text{s}}$
$w_i=c_i\frac{\mathcal{M}_i\mathcal{V}_\text{s}}{\mathcal{M}_\text{s}}$
$w_0=\frac{1}{\mathcal{M}_\text{s}\sum_{j=0}^p m_j}$
$w_i=\frac{m_i}{\sum_{j=0}^p m_j}\frac{\mathcal{M}_i}{\mathcal{M}_\text{s}}$

concentration $$c$$

$c_i=\frac{x_i}{\mathcal{V}_\text{s}}$
$c_i=w_i\frac{\mathcal{M}_\text{s}}{\mathcal{V}_\text{s}\mathcal{M}_i}$
$c_0=\frac{1}{\mathcal{M}_0\mathcal{V}_\text{s}\sum_{j=0}^p m_j}$
$c_i=\frac{m_i}{\mathcal{V}_\text{s}\sum_{j=0}^p m_j}$

molality $$m$$

$m_i=\frac{x_i}{x_0} \frac{1}{\mathcal{M}_0}$
$m_i=\frac{w_i}{w_0}\frac{1}{\mathcal{M}_i}$
$m_i=\frac{c_i}{c_0}\frac{1}{\mathcal{M}_0}$