# Molar concentration

"Molarity" redirects here. It is not to be confused with Molality.

In chemistry, the molar concentration, $c_i$ is defined as the amount of a constituent $n_i$ (usually measured in moles – hence the name) divided by the volume of the mixture $V$:[1]

$c_i = \frac {n_i}{V}$

It is also called molarity, amount-of-substance concentration, amount concentration, substance concentration, or simply concentration. Whereas mole fraction is a ratio of moles to moles, molar concentration is a ratio of moles to volume. The volume $V$ in the definition $c_i = n_i/V$ refers to the volume of the solution, not the volume of the solvent. One litre of a solution usually contains either slightly more or slightly less than 1 liter of solvent because the process of dissolution causes volume of liquid to increase or decrease (discussed further at volume fraction).

The reciprocal quantity represents the dilution (volume) which can appear in Ostwald's law of dilution.

## Units and notation

In addition to the notation $c_i$ there is a notation using brackets and the formula of a compound like [A]. This notation is encountered especially in equilibrium constants and reaction quotients.

The SI unit is mol/m3. However, more commonly the unit mol/L is used. A solution of concentration 1 mol/L is also denoted as "1 molar" (1M). In many publication styles, the "M" symbol (as well as mM, µM, and so on) is like the degree (°) and percent (%) symbols in that it is closed up to the number, whereas most unit symbols (for example, cm, mm, L, mL, g, kg, s) take an intervening space. Some styles deprecate the M, mM, µM notation and replace it with mol/L, mmol/L, or µmol/L.

1 mol/L = 1 mol/dm3 = 1 mol dm−3 = 1 M = 1000 mol/m3.

An SI prefix is often used to denote concentrations. Commonly used units are listed in the table hereafter:

Name Abbreviation Concentration Concentration (SI unit)
millimolar mM 10−3 mol/dm3 100 mol/m3
micromolar μM 10−6 mol/dm3 10−3 mol/m3
nanomolar nM 10−9 mol/dm3 10−6 mol/m3
picomolar pM 10−12 mol/dm3 10−9 mol/m3
femtomolar fM 10−15 mol/dm3 10−12 mol/m3
attomolar aM 10−18 mol/dm3 10−15 mol/m3
zeptomolar zM 10−21 mol/dm3 10−18 mol/m3
yoctomolar yM[2] 10−24 mol/dm3
(1 particle per 1.6 L)
10−21 mol/m3

## Related quantities

### Number concentration

Main article: Number concentration

The conversion to number concentration $C_i$ is given by:

$C_i = c_i \cdot N_{\rm A}$

where $N_{\rm A}$ is the Avogadro constant, approximately 6.022×1023 mol−1.

### Mass concentration

The conversion to mass concentration $\rho_i$ is given by:

$\rho_i = c_i \cdot M_i$

where $M_i$ is the molar mass of constituent $i$.

### Mole fraction

Main article: Mole fraction

The conversion to mole fraction $x_i$ is given by:

$x_i = c_i \cdot \frac{M}{\rho} = c_i \cdot \frac{\sum_i x_i M_i}{\rho}$
$x_i= c_i \cdot \frac{\sum x_j M_j}{\rho - c_i M_i}$

where $M$ is the average molar mass of the solution, $\rho$ is the density of the solution and j is the index of other solutes.

A simpler relation can be obtained by considering the total molar concentration namely the sum of molar concentrations of all the components of the mixture.

$x_i = \frac{c_i}{c} = \frac{c_i}{\sum c_i}$

### Mass fraction

The conversion to mass fraction $w_i$ is given by:

$w_i = c_i \cdot \frac{M_i}{\rho}$

### Molality

Main article: Molality

The conversion to molality (for binary mixtures) is:

$b_2 = \frac{{c_2}}{{\rho - c_2 \cdot M_2}} \,$

where the solute is assigned the subscript 2.

For solutions with more than one solute, the conversion is:

$b_i = \frac{{c_i}}{{\rho - \sum c_i \cdot M_i}} \,$

## Properties

### Sum of molar concentrations – normalizing relation

The sum of molar concentrations gives the total molar concentration, namely the density of the mixture divided by the molar mass of the mixture or by another name the reciprocal of the molar volume of the mixture. In an ionic solution ionic strength is proportional to the sum of molar concentration of salts.

### Sum of products molar concentrations-partial molar volumes

The sum of products between these quantities equals one.

$\sum_i c_i \cdot \bar{V_i} = 1$

### Dependence on volume

Molar concentration depends on the variation of the volume of the solution due mainly to thermal expansion. On small intervals of temperature the dependence is :

$c_i = \frac {{c_{i,T_0}}}{{(1 + \alpha \cdot \Delta T)}}$

where $c_{i,T_0}$ is the molar concentration at a reference temperature, $\alpha$ is the thermal expansion coefficient of the mixture.

## Spatial variation and diffusion

Molar and mass concentration have different values in space where diffusion happens.

## Examples

Example 1: Consider 11.6 g of NaCl dissolved in 100 g of water. The final mass concentration $\rho$(NaCl) will be:

$\rho$(NaCl) = 11.6 g / (11.6 g + 100 g) = 0.104 g/g = 10.4 %

The density of such a solution is 1.07 g/mL, thus its volume will be:

$V$ = (11.6 g + 100 g) / (1.07 g/mL) = 104.3 mL

The molar concentration of NaCl in the solution is therefore:

$c$(NaCl) = (11.6 g / 58 g/mol) / 104.3 mL = 0.00192 mol/mL = 1.92 mol/L

Here, 58 g/mol is the molar mass of NaCl.

Example 2: Another typical task in chemistry is the preparation of 100 mL (= 0.1 L) of a 2 mol/L solution of NaCl in water. The mass of salt needed is:

$m$(NaCl) = 2 mol/L x 0.1 L x 58 g/mol = 11.6 g

To create the solution, 11.6 g NaCl are placed in a volumetric flask, dissolved in some water, then followed by the addition of more water until the total volume reaches 100 mL.

Example 3: The density of water is approximately 1000 g/L and its molar mass is 18.02 g/mol (or 1/18.02=0.055 mol/g). Therefore, the molar concentration of water is:

$c$(H2O) = 1000 g/L / (18.02 g/mol) = 55.5 mol/L

Likewise, the concentration of solid hydrogen (molar mass = 2.02 g/mol) is:

$c$(H2) = 88 g/L / (2.02 g/mol) = 43.7 mol/L

The concentration of pure osmium tetroxide (molar mass = 254.23 g/mol) is:

$c$(OsO4) = 5.1 kg/L / (254.23 g/mol) = 20.1 mol/L.

Example 4: A typical protein in bacteria, such as E. coli, may have about 60 copies, and the volume of a bacterium is about $10^{-15}$ L. Thus, the number concentration $C$ is:

$C$ = 60 / (10−15 L)= 6×1016 L−1

The molar concentration is:

$c = C / N_A$ = 6×1016 L−1 / (6×1023 mol−1) = 10−7 mol/L = 100 nmol/L
Reference ranges for blood tests, sorted by molar concentration.

If the concentration refers to original chemical formula in solution, the molar concentration is sometimes called formal concentration. For example, if a sodium carbonate solution has a formal concentration of $c$(Na2CO3) = 1 mol/L, the molar concentrations are $c$(Na+) = 2 mol/L and $c$(CO32-) = 1 mol/L because the salt dissociates into these ions.

## References

1. ^ IUPAC, Compendium of Chemical Terminology, 2nd ed. (the "Gold Book") (1997). Online corrected version:  (2006–) "amount concentration, c".
2. ^ David Bradley. "How low can you go? The Y to Y".