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Mole Fraction

14th Sep 2019 @ 11 min read

Analytical Chemistry

The mole is the fundamental SI unit for measuring the amount of substance. One mole comprises of approximately 6.022 × 1023 elementary units.

1 \,\text{mol}\approx 6.022 \times 10^{23}

The mole is a frequently used unit in chemistry. Chemical reactions are always balanced using moles of the reactant and the product. The concentration of a solution involves the mole of a solute. Some examples are molar concentration or molarity, molality, mole fraction, molar density. The mole fraction is another way of expressing the concentration.

Definition, Symbols, and Formula

The mole fraction of a substance in the mixture is the ratio of the mole of the substance in the mixture to the total mole of the mixture. Since it is the ratio of moles to moles, it is a dimensionless quantity. The mole fraction is sometimes called the amount fraction.

For solutions and liquid mixtures, the symbol x is used to denote and for a gaseous component, the symbol y is used to denote it.

For a mixture of the ith component,

x_i=\frac{n_i}{n_\text{mix}}

Here, xi is the molar fraction of the ith component, ni is the mole of the ith component in the mixture, and nmix is the total mole of the mixture. But we also know, the total mole of a mixture is the sum of the mole of each component in the mixture.

n_\text{mix}=\sum{n_i}

From the above two equations,

x_i=\frac{n_i}{\sum{n_i}}

Note: The sum of the mole fraction of each component is always equal to one.

x_1+x_2+x_3+...=\sum{x_i}=1

Example 1: Air

Statement: A sample of air contains 39.05 mol of nitrogen, 10.47 mol of oxygen, 0.45 mol of argon, and 0.03 mol of argon.

The atmosphere of Earth
Figure 1: The Atmosphere of Earth
[Image Source: The Massachusetts Institute of Technology]

Solution: Let nN2, nO2, nAr, and nCO2 be moles of nitrogen, oxygen, argon, and carbon dioxide.

The mole fraction of nitrogen, oxygen, argon, and carbon dioxide in the given of air be yN2, yO2, yAr, and yCO2.

y_{\text{N}_2} &=\frac{n_{\text{N}_2}}{n_{\text{N}_2} + n_{\text{O}_2} + n_\text{Ar} + n_{\text{CO}_2}}\\ &=\frac{39.05}{39.05+10.47+0.45+0.03}\\ &=\frac{39.05}{50.00}\\ &=0.781\,0 y_{\text{O}_2} &=\frac{n_{\text{O}_2}}{n_{\text{N}_2} + n_{\text{O}_2} + n_\text{Ar} + n_{\text{CO}_2}}\\ &=\frac{10.47}{39.05+10.47+0.45+0.03}\\ &=\frac{10.47}{50.00}\\ &=0.209\,4 y_\text{Ar} &=\frac{n_\text{Ar}}{n_{\text{N}_2} + n_{\text{O}_2} + n_\text{Ar} + n_{\text{CO}_2}}\\ &=\frac{0.45}{39.05+10.47+0.45+0.03}\\ &=\frac{0.45}{50.00}\\ &=0.009\,0 y_{\text{CO}_2} &=\frac{n_{\text{CO}_2}}{n_{\text{N}_2} + n_{\text{O}_2} + n_\text{Ar} + n_{\text{CO}_2}}\\ &=\frac{0.03}{39.05+10.47+0.45+0.03}\\ &=\frac{0.03}{50.00}\\ &=0.000\,6

Thus, the sample of air consists of 0.781 0 of nitrogen, 0.209 4 of oxygen, 0.009 0 of argon, and 0.000 6 of carbon dioxide.

In a solution, the mole fraction of a solute is the ratio of the mole of the solute to the total mole of the solution.

n_\text{solute}=\frac{n_\text{solute}}{n_\text{solution}}

Example 2: Sucrose

Statement: Sucrose, aka table sugar or granulated sugar, is a disaccharide. It is made up of glucose and fructose and has a molecular formula of C12H22O11. Sucrose has a high solubility in water. Consider a solution of two tablespoon (40 g) of sucrose in 100 g of water.

Sucrose Solution
Figure 2: Sucrose Solution
[Image Source: Wikimedia]

The molar mass of sucrose is 342 g mol−1 and of water, 18 g mol−1.

Structure of Sucrose
Figure 3: Structure of Sucrose

Solution: The solution contains 40 g of sucrose and 100 g of water. The respective mole of sugar and water are as follows:

n_{\text{C}_{12}\text{H}_{22}\text{O}_{11}}=\frac{40}{342}=0.117\,\text{mol} n_{\text{H}_2\text{O}}=\frac{100}{18}=5.56\,\text{mol}

Let xC12H22O11 and xH2O be the mole fraction of sucrose and water in the solution.

x_{\text{C}_{12}\text{H}_{22}\text{O}_{11}} &=\frac{n_{\text{C}_{12}\text{H}_{22}\text{O}_{11}}}{n_{\text{C}_{12}\text{H}_{22}\text{O}_{11}}+n_{\text{H}_2\text{O}}}\\ &=\frac{0.117}{0.117+5.56}\\&=0.021 x_{\text{H}_2\text{O}} &=\frac{n_{\text{H}_2\text{O}}}{n_{\text{C}_{12}\text{H}_{22}\text{O}_{11}}+n_{\text{H}_2\text{O}}}\\ &=\frac{5.56}{0.117+5.56}\\ &=0.979

Thus, the solution has 0.021 of sugar and 0.979 of water.

Note: The mole fraction is an intensive property. So, it would not vary from sample to sample. It is also temperature and pressure independent.

Molar Percentage or Mole Per cent

The molar percentage or mole per cent is the same as mole fraction when expressed in the percentage.

 \text{mol \%} &=\text{mole fraction} \times 100\\ &=\frac{n_i}{n_\text{mix}} \times 100

Example 3: Iron (II) Sulphate

Statement: Ferrous sulphate has the molecular formula FeSO4. Its solution is typically greenish in colour. A solution of FeSO4 contains 20 g of FeSO4 and 100 g of H2O. The molar mass of FeSO4 is 152 g mol−1 and of H2O, 18 g mol−1.

Iron (II) Sulphate Solution
Figure 4: Iron (II) Sulphate Solution
[Image Source: Wikimedia]

Solution: The gram of ferrous sulphate and water in the solution is 20 g and 100 g. The respective mole of ferrous sulphate and water is as follows:

n_{\text{FeSO}_4}=\frac{20}{152}=0.132\,\text{mol} n_{\text{H}_2\text{O}}=\frac{100}{18}=5.56\,\text{mol}

The mole per cent of ferrous sulphate and water is as follows:

\text{mol \% of}\,\text{FeSO}_4 &=\frac{n_{\text{FeSO}_4}}{n_{\text{FeSO}_4}+n_{\text{H}_2\text{O}}} \times 100 \\ &=\frac{0.132}{0.132+5.56} \times 100 \\ &=\frac{0.132}{5.692} \times 100 \\ &=2.3\,\% \text{mol \% of}\,\text{H}_2\text{O} &=\frac{n_{\text{H}_2\text{O}}}{n_{\text{FeSO}_4}+n_{\text{H}_2\text{O}}} \times 100 \\ &=\frac{5.56}{0.132+5.56} \times 100 \\ &=\frac{5.56}{5.692} \times 100 \\ &=97.7\,\%

Thus, the solution is 2.3 % FeSO4 by the mole.

Relationship with Other Quantities

The mole fraction is a way to express the composition of a mixture. Other quantities are molarity, molality, mass fraction. The relationship between them can be established.

Molar Concentration or Molarity

The molar concentration or molarity is the mole of a solute per unit volume of the solution. Typically volume is expressed in litre (L) or decimetre cube (dm−3).

C_i=\frac{n_i}{V}

Converting the mole fraction to the molar concentration or molarity,

C_i &=\frac{n_i}{V}\\ &=\frac{n_i}{V} \times \frac{\sum{n_i}}{\sum{n_i}}\\ &=\frac{n_i}{\sum{n_i}} \times \frac{\sum{n_i}}{V}\\ &=x_i C

Here, C=\frac{\sum{n_i}}{V}is the total molar concentration.

The above equation relates the mole fraction, molar concentration, and total molar concentration.

C_i &=x_i \times \frac{\sum{n_i}}{V}\\ &=x_i \times \frac{\sum{m_i}}{\sum{m_i}} \times \frac{\sum{n_i}}{V} \\ &=x_i \times \frac{\sum{n_i}}{\sum{m_i}} \times \frac{\sum{m_i}}{V} \\ &=x_i \times \frac{1}{M} \times \rho \\ &=x_i \times \frac{\rho}{M}

Here, \rho=\frac{\sum{m_i}}{V} is the density of the solution and M=\frac{\sum{m_i}}{\sum{n_i}} is the average molar mass of the solution. But we also know M=\sum{x_i M_i}

C_i=\frac{x_i \rho}{\sum{x_i M_i}}

The above equation relates the mole fraction, molar concentration, and density.

Mass Concentration

The mass concentration is the ratio the mass of a solute to the volume of the solution.

\rho_i=\frac{m_i}{V}

Converting the mole fraction into the mass concentration,

\rho_i &=\frac{m_i}{V}\\ &=\frac{n_i M_i}{V}\\ &=\frac{n_i M_i}{V} \times \frac{\sum{n_i}}{\sum{n_i}}\\ &=\frac{n_i}{\sum{n_i}} \times M_i \times \frac{\sum{n_i}}{V}\\ &=x_i M_i C

Here, Mi is the molar mass of the solute and C is the total molar concentration.

The mole fraction can also be expressed in terms of the density of a solution.

\rho_i &=x_i M_i \times \frac{\sum{n_i}}{V}\\ &=x_i M_i \times \frac{\sum{m_i}}{\sum{m_i}} \times \frac{\sum{n_i}}{V} \\ &=x_i M_i \times \frac{\sum{n_i}}{\sum{m_i}} \times \frac{\sum{m_i}}{V} \\ &=x_i M_i \times \frac{1}{M} \times \rho \\ &=x_i \rho \times \frac{M_i}{M}\\ &=x_i \rho \times \frac{M_i}{\sum{x_i M_i}}

Here, ρ is the density of the solution, M is the average molar mass of the solution.

Molality

The molality is the mole of a solute per unit mass of the solvent. The molality is denoted by b and has the unit mol kg−1.

b_i=\frac{n_i}{m_\text{solvent}}

Consider a single-solute solution, where b is the molality, x is the mole fraction of the solute.

b &=\frac{n_\text{solute}}{m_\text{solvent}} \\ &=\frac{n_\text{solute}}{n_\text{solute}+n_\text{solvent}} \times \frac{n_\text{solute}+n_\text{solvent}}{m_\text{solvent}} \\ &=x \times \frac{n_\text{solute}+n_\text{solvent}}{M_0 n_\text{solvent}} \\ &=x \times \frac{1}{M_0 (1-x)} \\ &=\frac{x}{M_0(1-x)}

Here, M0 is the molar mass of the solvent.

For a single-solvent solution having j solutes,

x_i &=\frac{n_i}{\sum{n_i}}\\ &=\frac{n_i}{n_\text{solvent}+\sum{n_j}} \\ &=\frac{n_i}{\frac{m_\text{solvent}}{M_0}+\sum{n_j}} \\ &=\frac{(n_i \big/ m_\text{solvent})M_0}{1+M_0 \frac{\sum{n_j}}{m_\text{solvent}}} \\ &=\frac{b_i M_0}{1+M_0 \sum{b_j}}

Mass Fraction

The mass fraction is a very similar quantity. It is the ratio of the mass of a solute to the total mass of the solution.

w_i=\frac{m_i}{\sum{m_i}} w_i &=\frac{m_i}{\sum{m_i}}\\ &=\frac{n_i M_i}{\sum{n_i M_i}}\\ &=\frac{n_i M_i}{\sum{n_i}} \times \frac{\sum{n_i}}{\sum{n_i M_i}}\\ &={x_i M_i} \times \frac{1}{\sum{\frac{n_i}{\sum{n_i}} M_i}}\\ &=\frac{x_i M_i}{\sum{x_i M_i}}

Here, Mi is the molar mass of the ith solute.

Mixing Ratio

The mixing ratio of two pure substances is the ratio of the mole of a substance to the other.

r=\frac{n_2}{n_1}

Expressing r in terms of x1,

r &=\frac{n_2}{n_1} \\ &=\frac{(n_2+n_1)-n_1}{n_1} \\ &=\frac{n_2+n_1}{n_1}-1 \\ &=\frac{1}{x_1}-1

Expressing r in terms of x2,

r &=\frac{n_2}{n_1}\\ &=\frac{n_2}{(n_1+n_2)-n_2}\\ &=\frac{1}{(n_1+n_2)\big/n_2-1}\\ &=\frac{1}{1\big/ x_2-1}\\ &=\frac{x_2}{1-x_2}

Solved Problem

Problem 1: Ferric Solution

Statement: A ferric chloride solution contain 40 g of ferric chloride and 60 g of water. The molar mass of ferric chloride is 162.2 g mol−1 and of water, 18.0 g mol−1.

Ferric Chloride Solution
Figure 5: Ferric Chloride Solution
[Image Source: All Chemical]

Solution: Converting the gram to the mole,

n_{\text{H}_2\text{O}} &=\frac{m_{\text{H}_2\text{O}}}{M_{\text{H}_2\text{O}}}\\ &=\frac{60}{18.0}\\ &=3.33\,\text{mol}

The mole fraction of the ferric chloride solution and water are as follows:

x_{\text{FeCl}_3} &=\frac{n_{\text{FeCl}_3}}{n_{\text{FeCl}_3}+n_{\text{H}_2\text{O}}} \\ &=\frac{0.247}{0.247+3.33} \\ &=0.069 x_{\text{H}_2\text{O}} &=\frac{n_{\text{H}_2\text{O}}}{n_{\text{FeCl}_3}+n_{\text{H}_2\text{O}}} \\ &=\frac{3.33}{0.247+3.33} \\ &=0.931

Thus, the solution contains 6.9 % of ferric chloride.

Problem 2: Exhaust from a Factory

Statement: A sample of flue gas from a factory comprise of 20 g of carbon dioxide, 10 g of water vapour, 31 g of oxygen, and 48 g of nitrogen. The molar mass of carbon dioxide, water vapour, oxygen, and nitrogen is 44 g mol−1, 18 g mol−1, 32 g mol−1, and 28 g mol−1.

A Chimney of the Factory
Figure 6: A Chimney of the Factory
[Image Source: Pixabay]

Solution: Converting the gram to the mole,

n_{\text{CO}_2} =\frac{20}{44} =0.454\,\text{mol} n_{\text{H}_2\text{O}} =\frac{10}{18} =0.556\,\text{mol} n_{\text{O}_2} =\frac{31}{32} =0.969\,\text{mol} n_{\text{N}_2} =\frac{48}{28} =1.71\,\text{mol}

The mole fraction of these gases are as follows:

y_{\text{CO}_2} &=\frac{n_{\text{CO}_2}}{n_{\text{CO}_2}+ n_{\text{H}_2\text{O}}+n_{\text{O}_2}+ n_{\text{N}_2}} \\ &=\frac{0.454}{0.454+0.556+0.969+1.71} \\ &=0.12 y_{\text{H}_2\text{O}} &=\frac{n_{\text{H}_2\text{O}}}{n_{\text{CO}_2}+ n_{\text{H}_2\text{O}}+n_{\text{O}_2}+ n_{\text{N}_2}} \\ &=\frac{0.556}{0.454+0.556+0.969+1.71} \\ &=0.15 y_{\text{O}_2} &=\frac{n_{\text{O}_2}}{n_{\text{CO}_2}+ n_{\text{H}_2\text{O}}+n_{\text{O}_2}+ n_{\text{N}_2}} \\ &=\frac{0.969}{0.454+0.556+0.969+1.71} \\ &=0.26 y_{\text{N}_2} &=\frac{n_{\text{N}_2}}{n_{\text{CO}_2}+ n_{\text{H}_2\text{O}}+n_{\text{O}_2}+ n_{\text{N}_2}} \\ &=\frac{1.71}{0.454+0.556+0.969+1.71} \\ &=0.47

Problem 3: o-Xylene in Diethyl Ether

Statement: When two methyl groups are attached to a benzene ring at an ortho position, o-xylene is formed. o-Xylene is an aromatic compound having the formula C6H4(CH3)2. It is soluble in dimethyl ether, one of the most common solvents in chemical industries. If 2.3 mol of o-xylene is dissolved in 12.1 mol of dimethyl ether, estimate the mole percent of each.

Solution: Let nxy and ndme be the mole of o-xylene and dimethyl ether.

n_\text{xy} =2.3 \,\text{mol} &&n_\text{dme} =12.1 \,\text{mol}

Let xxy and xdme be the mole fraction of o-xylene and dimethyl ether.

x_\text{xy} &=\frac{n_\text{xy}}{n_\text{xy}+n_\text{dme}}\\ &=\frac{2.3}{12.1+2.3}\\ &=\frac{2.3}{14.4}\\ &=0.160 x_\text{dme} &=\frac{n_\text{dme}}{n_\text{xy}+n_\text{dme}}\\ &=\frac{12.1}{12.1+2.3}\\ &=\frac{12.1}{14.4}\\ &=0.840

Thus, the mixture has 16 % o-xylene and 84 % dimethyl ether.

Practice Problem

Find the mole fraction in the following problems.

Problem 1: Sodium Chloride

Sodium chloride is a common salt having formula NaCl. It is highly soluble in water. A beaker contains 100 g of water, and 23.4 g of sodium chloride is dissolved in it. The molar mass of sodium chloride is 58.44 g mol−1 and of water, 18.02 g mol−1.

Problem 2: Indigo Dye

Indigo dye is a naturally occurring dye having a dense blue colour. It is widely used for dyeing of textiles. The molar mass of indigo dye is 262.27 g mol−1 and of water, 18.02 g mol−1. A industrial vessel has 0.1 g of indigo per 100 g of water.

Indigo Dye
Figure 7: Indigo Dye

Problem 3: Water Gas Shift Reaction

The water-gas shift reaction is a famous industrial reaction. Carbon monoxide reacts with steam to form carbon dioxide and hydrogen.

\text{CO}+\text{H}_2\text{O} \longrightarrow \text{CO}_2+\text{H}_2

An industrial pipe has a flow rate of 40 mol s−1 of carbon monoxide, 40 mol s−1 of steam, 17 mol s−1 of carbon dioxide, and 23 mol s−1 of hydrogen.

  1. 067 3 of sodium chloride and 0.933 7 of water
  2. 007 % of indigo dye and 99.993 % of water
  3. 33 of carbon monoxide, 0.33 of steam, 0.15 of carbon dioxide, and 0.19

Summary

  • The mole fraction of a component in the mixture is the ratio of the mole of the component to the total mole of the mixture. It is denoted by x for liquids and solids or y for gases.
  • x_i=\frac{n_i}{\sum{n_i}}
  • For solution,
  • x_i=\frac{n_\text{solute}}{n_\text{solution}}
  • The mole fraction is an intensive property. So, it would not vary from sample to sample for a given uniform mixture. It is also independent of temperature and pressure.
  • The sum of each mole fraction in a mixture is always equal to one.
  • \sum{x_i}=1

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