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Mole fraction

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(Redirected from Mole percent) Proportion of a constituent in a mixture
mole fraction
Other namesmolar fraction, amount fraction, amount-of-substance fraction
Common symbolsx
SI unit1
Other unitsmol/mol

In chemistry, the mole fraction or molar fraction, also called mole proportion or molar proportion, is a quantity defined as the ratio between the amount of a constituent substance, ni (expressed in unit of moles, symbol mol), and the total amount of all constituents in a mixture, ntot (also expressed in moles):

x i = n i n t o t {\displaystyle x_{i}={\frac {n_{i}}{n_{\mathrm {tot} }}}}

It is denoted xi (lowercase Roman letter x), sometimes χi (lowercase Greek letter chi). (For mixtures of gases, the letter y is recommended.)

It is a dimensionless quantity with dimension of N / N {\displaystyle {\mathsf {N}}/{\mathsf {N}}} and dimensionless unit of moles per mole (mol/mol or mol ⋅ mol) or simply 1; metric prefixes may also be used (e.g., nmol/mol for 10). When expressed in percent, it is known as the mole percent or molar percentage (unit symbol %, sometimes "mol%", equivalent to cmol/mol for 10). The mole fraction is called amount fraction by the International Union of Pure and Applied Chemistry (IUPAC) and amount-of-substance fraction by the U.S. National Institute of Standards and Technology (NIST). This nomenclature is part of the International System of Quantities (ISQ), as standardized in ISO 80000-9, which deprecates "mole fraction" based on the unacceptability of mixing information with units when expressing the values of quantities.

The sum of all the mole fractions in a mixture is equal to 1:

i = 1 N n i = n t o t ;   i = 1 N x i = 1 {\displaystyle \sum _{i=1}^{N}n_{i}=n_{\mathrm {tot} };\ \sum _{i=1}^{N}x_{i}=1}

Mole fraction is numerically identical to the number fraction, which is defined as the number of particles (molecules) of a constituent Ni divided by the total number of all molecules Ntot. Whereas mole fraction is a ratio of amounts to amounts (in units of moles per moles), molar concentration is a quotient of amount to volume (in units of moles per litre). Other ways of expressing the composition of a mixture as a dimensionless quantity are mass fraction and volume fraction are others.

Properties

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:
    x 1 = 1 x 2 1 + x 3 x 1 x 3 = 1 x 2 1 + x 1 x 3 {\displaystyle {\begin{aligned}x_{1}&={\frac {1-x_{2}}{1+{\frac {x_{3}}{x_{1}}}}}\\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:

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

or

( x 3 x 2 ) x 1 x 3 = x 3 1 x 2 {\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:

X = x 3 x 1 + x 3 Y = x 3 x 2 + x 3 Z = x 2 x 1 + x 2 {\displaystyle {\begin{aligned}X&={\frac {x_{3}}{x_{1}+x_{3}}}\\Y&={\frac {x_{3}}{x_{2}+x_{3}}}\\Z&={\frac {x_{2}}{x_{1}+x_{2}}}\end{aligned}}}

These can be used for solving PDEs like:

( μ 2 n 1 ) n 2 , n 3 = ( μ 1 n 2 ) n 1 , n 3 {\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

( μ 2 n 1 ) n 2 , n 3 , n 4 , , n i = ( μ 1 n 2 ) n 1 , n 3 , n 4 , , n i {\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.

( μ 2 μ 1 ) n 2 , n 3 = ( n 1 n 2 ) μ 1 , n 3 = ( x 1 x 2 ) μ 1 , n 3 {\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

( μ 2 μ 1 ) n 2 , n 3 , n 4 , , n i = ( n 1 n 2 ) μ 1 , n 2 , n 4 , , n i {\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:

( n 1 n 2 ) n 3 = ( n 1 n 3 n 2 n 3 ) n 3 = ( x 1 x 3 x 2 x 3 ) n 3 {\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:

( μ 2 μ 1 ) n 2 n 3 = ( x 1 x 3 x 2 x 3 ) μ 1 {\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

( μ 2 μ 1 ) n 2 n 3 , n 3 n 4 , , n i 1 n i = ( x 1 x 3 x 2 x 3 ) μ 1 , n 3 n 4 , , n i 1 n i {\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}}}}}

Related quantities

Mass fraction

The mass fraction wi can be calculated using the formula

w i = x i M i M ¯ = x i M i j x j M j {\displaystyle w_{i}=x_{i}{\frac {M_{i}}{\bar {M}}}=x_{i}{\frac {M_{i}}{\sum _{j}x_{j}M_{j}}}}

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

Molar mixing ratio

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

x 1 = 1 1 + r n x 2 = r n 1 + r n {\displaystyle {\begin{aligned}x_{1}&={\frac {1}{1+r_{n}}}\\x_{2}&={\frac {r_{n}}{1+r_{n}}}\end{aligned}}}

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

n 2 n 1 = x 2 x 1 {\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 to form ternary mixtures

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.

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

Mole percentage

Multiplying mole fraction by 100 gives the mole percentage, also referred as amount/amount percent .

Mass concentration

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

x i = ρ i ρ M ¯ M i ρ i = x i ρ M i M ¯ {\displaystyle {\begin{aligned}x_{i}&={\frac {\rho _{i}}{\rho }}{\frac {\bar {M}}{M_{i}}}\\\Leftrightarrow \rho _{i}&=x_{i}\rho {\frac {M_{i}}{\bar {M}}}\end{aligned}}}

where is the average molar mass of the mixture.

Molar concentration

The conversion to molar concentration ci is given by:

c i = x i c = x i ρ M ¯ = x i ρ j x j M j {\displaystyle {\begin{aligned}c_{i}&=x_{i}c\\&={\frac {x_{i}\rho }{\bar {M}}}={\frac {x_{i}\rho }{\sum _{j}x_{j}M_{j}}}\end{aligned}}}

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

Mass and molar mass

The mole fraction can be calculated from the masses mi and molar masses Mi of the components:

x i = m i M i j m j M j {\displaystyle x_{i}={\frac {\frac {m_{i}}{M_{i}}}{\sum _{j}{\frac {m_{j}}{M_{j}}}}}}

Spatial variation and gradient

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

References

  1. ^ IUPAC, Compendium of Chemical Terminology, 2nd ed. (the "Gold Book") (1997). Online corrected version: (2006–) "amount fraction". doi:10.1351/goldbook.A00296
  2. Zumdahl, Steven S. (2008). Chemistry (8th ed.). Cengage Learning. p. 201. ISBN 978-0-547-12532-9.
  3. Rickard, James N.; Spencer, George M.; Bodner, Lyman H. (2010). Chemistry: Structure and Dynamics (5th ed.). Hoboken, N.J.: Wiley. p. 357. ISBN 978-0-470-58711-9.
  4. ^ "ISO 80000-9:2019 Quantities and units — Part 9: Physical chemistry and molecular physics". ISO. 2013-08-20. Retrieved 2023-08-29.
  5. "SI Brochure". BIPM. Retrieved 2023-08-29.
  6. ^ Thompson, A.; Taylor, B. N. (2 July 2009). "The NIST Guide for the use of the International System of Units". National Institute of Standards and Technology. Retrieved 5 July 2014.
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