What Is a Mole Fraction? The Complete Guide

In this up growing world, science is that particular fact which is taking the door opened for the civilization and humanity, although it has its own pros and cons. Among all the science subjects chemistry is one of the leading subject holding a great industrial outweigh. While comprehend chemistry there might be a little portion of it will rise up, that is the concept of mole fraction or molar fraction. It is nothing but the unit of the amount of the components present in a solution. But in order to know about molar fraction or mole fraction we have to know some other terms that are inter related with mole fraction.

Avogadro’s Law and Mole Concept

So the basic law of Avogadro is at the same temperature and pressure, equal volumes of all gasses contain equal number of molecules.

  • Application of Avogadro’s law:
  1. All elementary gases except inert gasses are diatomic.
  2. Molecular mass of a gas = 2 x its vapor density i.e. M=2D
  3. At STP 6.023 x 1023 molecules of any gas occupies 22.4L
  • AVOGADRO NUMBER: (N0 or NA) = 6.023 x 1023mol-1
  • The number of molecules in 1 cm3 of gas in STP is equal to LOSCHIMDT NUMBER which is 2.66 x 1019
  • The reciprocal of Avogadro number is known as AVOGRAM.
  • Mole is the amount of any substance what can be solid, liquid or gas that contains 6.023 x 1023 numbers of elementary entities (atoms, molecules, ions, protons, electrons, etc.)

Atomic, Molecular and Formula Mass

ATOMIC MASS: it is defined as the ratio of mass of an atom of an element and 1/12th the mass of one carbon atom (C12). It is called as the relative mass and its unit is a.m.u or u

1 amu = actual mass of one atom of C12/12

=   12/6.023 x 1023/12

=   1.65 x 10-24

Average atomic mass: atomic mass of isotope x percentage of abundance/100

MOLECULAR MASS: sum of atomic masses of all elements present in a molecule is called molecular mass or the mass of one molecule relative to 1/12th the mass of one carbon atom is called its molecular mass.

FORMULA MASS: sum of all atomic masses of all the atoms present the formula of the compound. (ionic compound does not have any discrete by mass)

EQUIVALENT MASS: it is the number of parts by mass of an element that combines with or is displaced from a compound by 1.008 parts of hydrogen, 8parts of oxygen and 35.5 parts of chlorine by mass.

Atomic mass, molecular mass, formula mass and equivalent mass expressed in gram is called gram atom, gram molecule, gram formula and gram equivalent respectively.

EXAMPLE:

1 gram atom of sodium = 23g

1 gram molecule of Oxygen = 32gm

1 gram formula of sodium chloride = 58.5 gm

1 gram equivalent of magnesium = 12 gm

1 mol of element = 1 gram atom of element

1 mol of compound = 1 gram molecule of compound

I.e. 44 g of CO2 = 1 mol CO2

1 mol of an ion = 1 gram ion of an ion

1 mol of a gas = 22.4 L of gas in STP

EXAMPLE:

Calculate the total number of molecules and oxygen atoms and neutrons in 4.4g of CO2.

No. of atoms in CO2 = 4.4/44 = 0.1

No. of molecules = 0.1 x 6.023 x 1023 = 6.023 x 1022

No. of oxygen atoms = no. of molecules x 2 = 12.044 x 1022 = 1.20 x 1023

Number of 0n1 = number of molecules x (6 + 2 x 8)

= 1.32 x 1024 [6C12 8O16]

Dalton’s atomic theory

  • Matter consists of smallest indivisible particles that can take part in chemical reaction.
  • All the atoms of an element have same mass and same properties but atoms of different elements have different masses and different properties.
  • Atoms of different elements combine in a simple whole number whole number ratio form compound in a chemical reaction.
  • Atoms are neither created nor destroyed in a chemical reaction

Modern View of Atom

  • An atom is divisible into other similar particles which are known as subatomic particles. It can also combine in non whole number ratio as in the case of non stoichiometric compounds like Fe93O
  • Atoms of same element also differ in mass and mass related properties as in the cases of isotopes.
  • A chemical reaction involves rearrangement of atoms.

Methods for determination of the equivalent mass

Hydrogen displacement method:

  • A known mass of the element is changed into oxide directly. The mass of the oxide is noted. Mass of oxygen combined = mass of oxide – mass of element

Metal to metal displacement method:

  • More active metal can displace less active metal from its salts’ solution. The mass of the displaced metal bear the same ratio as their equivalent weights.

Double decomposition method:

  • The mass of the compound reacted and the mass of the product formed are in the ratio of their equivalent masses.
  • The equivalent mass of the compound is the sum of the equivalent masses of its radicals.
  • The equivalent mass of a radical is equal to the formula mass of the radical divided by its charge

                                                           AB + CD  ——> AD + CB

By electrolysis:

  • According to faraday’s second law of electrolysis when the same quantity of electricity flows thorough solutions of different electrolytes equivalent amount of substances deposited at the different electrodes.

Weight of X deposited/weight of Y deposited = eq. wt. of X/eq. wt. of Y =w1/w2 = E1/E2

Where w1 and w2 are deposited wt. Of metals at electrodes and E1 and E2 are equivalent weights respectively.

Conversion method:

  • When one compound of metal is converted into another compound of the same metal

E.g.       Metal carbonate   ——->  metal oxide + CO2

Then,

Weight of compound 1/weight of compound 2

=eq. wt of metal + eq. wt. of anion of compound 1/ eq. wt of metal + eq. wt. of anion of compound 2

So basically mole fraction is somehow inter related with the concentration of solutions. So the terms of concentrations is needed to be known.

CONCENTRATIONS OF SOLUTIONS

The concentration of a solution reflects the relative proportion of solute (B) and solvent (A) present in the solution. The various concentration terms are

  • Percentage by mass (%w/W)

%w/W = mass of solute/mass of solution x 100

10% H2SO4 w/W means 10 g of pure H2SO4 is present in 100g of H2SO4 solution.

  • Percentage (% w/v)

% w/v =mass of solute/volume of solution x 100

10% w/V H2SO4 solution means 10g of pure H2SO4 is present in 100 mL of H2SO4                      solution.

  • Percentage by volume (%v/V)

The concentration term is defined for a solution of liquid into liquid

%v/V = volume of solute/volume of solution x 100

42.8% v/V ethanol solution means that 42.8 ml pure ethanol is present in 100mL of ethanol solution.

Mole Fraction

In the field of chemistry mole fraction or molar fraction is the unit of the amount of a components divided by the whole amount of components present in a solution. Suppose a solution of A and B contains nA moles of A and nB moles of B then

Mole fraction of A would be  xA= nA/ nA+ nB

Mole fraction of B would be  xB= nB/ nA+ nB

                                                                                          XA + XB = 1

This same equation can be presented by 100 in denominator is called the molar proportion or molar percentage.

Amount fraction is also called as molar fraction. It is totally similar to the number fraction, which defines the amount of molecules in a component. Divided by the whole number of all the molecules. There is new term amount of substance fraction is used instead of mole fraction by the national institute of standards and technology because this term does not contain the unit mole. Meanwhile molar concentration represents the quotient of moles over volume and the mole fraction is the ratio of moles. The mole fraction basically is a procedure to express the structure of a mixture of its quantity which is dimensionless.

There are some great advantages people can get using mole fraction are

  • A solution which has a known mole fraction can be created by weighing off the accurate masses of the components.
  • The measurement of mole fraction will be symmetric if the mole fractions are x= 0.1 and x=0.9 then the characters of solvent and solute will be reserved.
  • Mole fraction does not depend on the temperature. And also does not require the knowledge of density of phases that is involved.
  • The mole fraction can be presentable as the partial pressure of the total pressure of he solution.

Molarity

it is defined as the number of moles of solute dissolved in one liter of solution.

M= no. of moles of solute/ volume of solution In liter

Suppose wB gram of solute dissolved in 1 liters solution and molecular mass of solute is MB                                 g/mol.

M= wB / MB= 10 x d(g/mL) x PC(w/W) ÷ MB

= moles of solute

Unit : mol/liter

Normality

It is the number of gram equivalents of solute dissolved in 1 liter of solution.

N = Wb/Eb = 10 x d(g/mL) xPC(w/W)÷ Eb

Molarity

It is defined as the number of moles of solute present in 1 kg of solvent.

M = number of moles I solute/mass of solvent In kg

Unit: mol/kg

Parts Per Million (ppm)

It is defined as the quantity of the solute in grams present in 106 grams of the solution.

Ppm = mass of solute/mass of solution x 106

Atmospheric pollution in cities is also expressed in ppm volume. It refers to the volume of the pollutant in 106 units of volume. 10 ppm of SO2 in air means 10mL of SO2 is present in 106 mL of air.

Hardness of water is also expressed in ppm. Parts by mass of hardness causing salt equivalent to CaCO3 present in 106 parts by mass of the hard water sample

Some important relations

  1. Mole fraction and molality

m= xB/1- xB x 1000/MA = xB/1- xB x 55.56

xB = mole fraction of solute

MA = molar mass of solvent

  1. Molarity, molality and density of solution

m= 1000 x M/1000 x d-M x MB

MB = molar mass of solute.

  1. Mormality = nf x Molarity

Molarity of the mixture of the two solutions of same solute and solvent

Mmixture= M1V1 + M2V2/ V1+ V2

Here M1 and V1 are the molarity and volume of first solution and M2 and V2 are the molarity and volume of second solution.

If dilution is carried out = V2 = V1+V2

Normality of the mixture of two solutions of same solute and solvent

Nmixture= N1V1 + N2V2/ V1+V2

Laws of Chemical Combination

Law of conservation of mass:

In all physical and chemical changes the total mass of the system remains unchanged

  • It is not applicable to nuclear reactions

Law of definite proportion:

A pure compound always contains the same elements combined in the same ratio of their mases.

Law of multiple proportion:

If two elements combined together to form more than one compounds then the masses of one element that combined with a fixed mass of the other element are in the ratio of simple whole numbers.

  • This law in not valid for the higher homologues of hydrocarbons.

Gay lussac’s law of gaseous volumes

When different gases combined to form new compounds then the volumes of the reactants and the gaseous products are in the simple whole number of ratio.

Stoichiometry, limiting Reagent and Percentage Composition

Stoichiometry:

It deals with the calculations of the quantities of reactants and products involved in balanced chemical reaction

Limiting reagent:

The reacteant which is consumed first and determines the amount of the product formed in a chemical reaction is called limiting reagent.  The reactant which is not completely consumed in a chemical reaction is called excess reagent.

Percentage composition:

Percentage composition of a compound is a relative mass of each of the constituent element in 100 parts of it. The percentage of the compound is readily calculated from the formula of the compound.

Let the molecular mass of compound is M and W the mass of the element present in a molecule.

Final lines

So this all about the tactics of concentration of various solutions. This methods of processing the fact about inner constructive particles of a component is very useful to apply in the applications of chemistry in higher education. Among of all these mole fraction or molar fraction is a very important point to be noted down as it is now entered the syllabus of JEE and NEET examinations. So if you are finding trouble to know about mole fraction or something other related about the concentration of solution then then the methods mentioned above should be followed.

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