Mole fraction is used very frequently in the construction of phase diagrams. It has a number of advantages:

• it is not temperature dependent (as is molar concentration) and does not require knowledge of the densities of the phase(s) involved
• a mixture of known mole fraction can be prepared by weighing off the appropriate masses of the constituents
• the measure is symmetric: in the mole fractions x = 0.1 and x = 0.9, the roles of 'solvent' and 'solute' are reversed.
• In a mixture of ideal gases, the mole fraction can be expressed as the ratio of partial pressure to total pressure of the mixture
• In a ternary mixture one can express mole fractions of a component as functions of other components mole fraction and binary mole ratios:

  ${\displaystyle {\begin{aligned}x_{1}&={\frac {1-x_{2}}{1+{\frac {x_{3}}{x_{1}}}}}\\[2pt]x_{3}&={\frac {1-x_{2}}{1+{\frac {x_{1}}{x_{3}}}}}\end{aligned}}}$ 

Differential quotients can be formed at constant ratios like those above:

${\displaystyle \left({\frac {\partial x_{1}}{\partial x_{2}}}\right)_{\frac {x_{1}}{x_{3}}}=-{\frac {x_{1}}{1-x_{2}}}}$ 

or

${\displaystyle \left({\frac {\partial x_{3}}{\partial x_{2}}}\right)_{\frac {x_{1}}{x_{3}}}=-{\frac {x_{3}}{1-x_{2}}}}$ 

The ratios X, Y, and Z of mole fractions can be written for ternary and multicomponent systems:

${\displaystyle {\begin{aligned}X&={\frac {x_{3}}{x_{1}+x_{3}}}\\[2pt]Y&={\frac {x_{3}}{x_{2}+x_{3}}}\\[2pt]Z&={\frac {x_{2}}{x_{1}+x_{2}}}\end{aligned}}}$ 

These can be used for solving PDEs like:

${\displaystyle \left({\frac {\partial \mu _{2}}{\partial n_{1}}}\right)_{n_{2},n_{3}}=\left({\frac {\partial \mu _{1}}{\partial n_{2}}}\right)_{n_{1},n_{3}}}$ 

or

${\displaystyle \left({\frac {\partial \mu _{2}}{\partial n_{1}}}\right)_{n_{2},n_{3},n_{4},\ldots ,n_{i}}=\left({\frac {\partial \mu _{1}}{\partial n_{2}}}\right)_{n_{1},n_{3},n_{4},\ldots ,n_{i}}}$ 

This equality can be rearranged to have differential quotient of mole amounts or fractions on one side.

${\displaystyle \left({\frac {\partial \mu _{2}}{\partial \mu _{1}}}\right)_{n_{2},n_{3}}=-\left({\frac {\partial n_{1}}{\partial n_{2}}}\right)_{\mu _{1},n_{3}}=-\left({\frac {\partial x_{1}}{\partial x_{2}}}\right)_{\mu _{1},n_{3}}}$ 

or

${\displaystyle \left({\frac {\partial \mu _{2}}{\partial \mu _{1}}}\right)_{n_{2},n_{3},n_{4},\ldots ,n_{i}}=-\left({\frac {\partial n_{1}}{\partial n_{2}}}\right)_{\mu _{1},n_{2},n_{4},\ldots ,n_{i}}}$ 

Mole amounts can be eliminated by forming ratios:

${\displaystyle \left({\frac {\partial n_{1}}{\partial n_{2}}}\right)_{n_{3}}=\left({\frac {\partial {\frac {n_{1}}{n_{3}}}}{\partial {\frac {n_{2}}{n_{3}}}}}\right)_{n_{3}}=\left({\frac {\partial {\frac {x_{1}}{x_{3}}}}{\partial {\frac {x_{2}}{x_{3}}}}}\right)_{n_{3}}}$ 

Thus the ratio of chemical potentials becomes:

${\displaystyle \left({\frac {\partial \mu _{2}}{\partial \mu _{1}}}\right)_{\frac {n_{2}}{n_{3}}}=-\left({\frac {\partial {\frac {x_{1}}{x_{3}}}}{\partial {\frac {x_{2}}{x_{3}}}}}\right)_{\mu _{1}}}$ 

Similarly the ratio for the multicomponents system becomes

${\displaystyle \left({\frac {\partial \mu _{2}}{\partial \mu _{1}}}\right)_{{\frac {n_{2}}{n_{3}}},{\frac {n_{3}}{n_{4}}},\ldots ,{\frac {n_{i-1}}{n_{i}}}}=-\left({\frac {\partial {\frac {x_{1}}{x_{3}}}}{\partial {\frac {x_{2}}{x_{3}}}}}\right)_{\mu _{1},{\frac {n_{3}}{n_{4}}},\ldots ,{\frac {n_{i-1}}{n_{i}}}}}$ 

The mass fraction w_i can be calculated using the formula

${\displaystyle w_{i}=x_{i}{\frac {M_{i}}{\bar {M}}}=x_{i}{\frac {M_{i}}{\sum _{j}x_{j}M_{j}}}}$ 

where M_i is the molar mass of the component i and M̄ is the average molar mass of the mixture.

The mixing of two pure components can be expressed introducing the amount or molar mixing ratio of them ${\displaystyle r_{n}={\frac {n_{2}}{n_{1}}}}$ . Then the mole fractions of the components will be:

${\displaystyle {\begin{aligned}x_{1}&={\frac {1}{1+r_{n}}}\\[2pt]x_{2}&={\frac {r_{n}}{1+r_{n}}}\end{aligned}}}$ 

The amount ratio equals the ratio of mole fractions of components:

${\displaystyle {\frac {n_{2}}{n_{1}}}={\frac {x_{2}}{x_{1}}}}$ 

due to division of both numerator and denominator by the sum of molar amounts of components. This property has consequences for representations of phase diagrams using, for instance, ternary plots.

Mixing binary mixtures with a common component gives a ternary mixture with certain mixing ratios between the three components. These mixing ratios from the ternary and the corresponding mole fractions of the ternary mixture x₁₍₁₂₃₎, x₂₍₁₂₃₎, x₃₍₁₂₃₎ can be expressed as a function of several mixing ratios involved, the mixing ratios between the components of the binary mixtures and the mixing ratio of the binary mixtures to form the ternary one.

${\displaystyle x_{1(123)}={\frac {n_{(12)}x_{1(12)}+n_{13}x_{1(13)}}{n_{(12)}+n_{(13)}}}}$ 

Multiplying mole fraction by 100 gives the mole percentage, also referred as amount/amount percent [abbreviated as (n/n)% or mol %].

The conversion to and from mass concentration ρ_i is given by:

${\displaystyle {\begin{aligned}x_{i}&={\frac {\rho _{i}}{\rho }}{\frac {\bar {M}}{M_{i}}}\\[3pt]\Leftrightarrow \rho _{i}&=x_{i}\rho {\frac {M_{i}}{\bar {M}}}\end{aligned}}}$ 

where M̄ is the average molar mass of the mixture.

The conversion to molar concentration c_i is given by:

${\displaystyle {\begin{aligned}c_{i}&=x_{i}c\\[3pt]&={\frac {x_{i}\rho }{\bar {M}}}={\frac {x_{i}\rho }{\sum _{j}x_{j}M_{j}}}\end{aligned}}}$ 

where M̄ is the average molar mass of the solution, c is the total molar concentration and ρ is the density of the solution.

The mole fraction can be calculated from the masses m_i and molar masses M_i of the components:

${\displaystyle x_{i}={\frac {\frac {m_{i}}{M_{i}}}{\sum _{j}{\frac {m_{j}}{M_{j}}}}}}$ 

In a spatially non-uniform mixture, the mole fraction gradient triggers the phenomenon of diffusion.