# Mole Fraction Class XI

It is the ratio of a number of moles of a particular component to the total number of moles of the solution. If a substance ‘A’ dissolves in substance ‘B’ and their number of moles are nA and nB respectively; then the mole fractions of A and B are given as Mole fraction of A, AVERAGE ATOMIC WEIGHT
For elements, which have atoms with different relative masses (isotopes) the atomic mass is taken as the weighted mean of the atomic masses.
General formula for Average atomic weight = For example, chlorine contains two isotopes 35 amu, 37 amu. The relative abundance of these two is in the ratio of 3: 1. Thus the atomic mass of chlorine is the average of different relative masses. Therefore atomic mass of chlorine is equal to MINIMUM MOLECULAR WEIGHT: (MMW)
It is the molecular weight of a compound shown by presence of minimum number of atoms [i.e. for monomer = 1, Dimer = 2, Trimer = 3, Tetramer = 4] e.g. Insulin contains 3.4% S, find its MMW.
100g Insulin contains 3.4g S (sulphur). Assuming it be monomer [as nothing specified] one sulphur atom 32 AMU (atomic mass), We can write, 3.4 g S is contained by 100g insulin 32g S is contained by CARAT
This is unit to measure purity of gold eg: 24 carat = 100% Pure gold 18 carat = 75% Pure gold
DETERMINATION OF MOLECULAR FORMULAE
The molecular formula of a compound may be defined as the formula which gives the actual number of atoms of various elements present in the molecule of the compound. For example, the molecular formula of the compound glucose can be represented as C6H12O6. A molecular of glucose contains six atoms of carbon, twelve atoms of hydrogen and six atoms of oxygen.
In order to find out molecular formula of a compound, the first step is to determine its empirical formula from the percentage composition.
The empirical formula of a compound may be defined as the formula which gives the simplest whole number ratio of atoms of the various elements present in the molecule of the compound. The empirical formula of the compound glucose (C6H12O6) is which shows that C, 1: H and O are present in the simplest ratio 1:2: 1.
Empirical formula mass of a substance is equal to the sum of atomic masses of all the atoms in the empirical formula of the substance.
The molecular formula is a whole number multiple of the empirical formula. Thus Molecular formula = (Empirical formula) × n where n = 1, 2, 3, ... Steps for writing the empirical formula
The percentage of the elements in the compound is determined by suitable methods and from the data collected, the empirical formula is determined by the following steps:
(i) Divide the percentage of each element by its atomic mass. This will give the relative number of moles of various elements present in the compound.
(ii) Divide the quotients obtained in the above step by the smallest of them so as to get a simple ratio of moles of various elements.
(iii) Multiply the figures, so obtained by a suitable integer if necessary in order to obtain a whole number ratio.
(iv) Finally, write down the symbols of the various elements side by side and put the above number as the subscripts to the lower right-hand corner of each symbol. This will represent the empirical formula of the compound.
Steps for writing the molecular formula
(i) Calculate the empirical formula as described above.
(ii) Find out the empirical formula mass by adding the atomic masses of all the atoms present in the empirical formula of the compound.
(iii) Divide the molecular mass (determined experimentally by some suitable method) by the empirical formula mass and find out the value of n.
Vapour density
Sometimes in numerical, the molecular mass of a volatile substance is not given, instead, vapour density is given. Vapour density can be defined as Vapour density Factor-Label Method
The method is similar to mole method with a difference that this method involves the use of a factor to calculate the required amount.
For. eg. For a general reaction aA+bB→cC+dD, If the amount of any component is given & another component is required the method can be applied as. Weight of C produced This method is most useful to solve sequential reactions or to solve problems involving percentage yield [ discussed later in Chapter]
For eg. For a hypothetical sequential reaction.
2A→B+C; 3B+2D→2E+C; E+2F→ G
If gm of “A” is taken & mixed with excess of D & F then wt. of “G” produced will be wt. of “G” where MA & Ma are molecular masses of “A” & “G” respectively
In the case of limiting reagent (to be discussed later), the amount of liming reagent will be used for solving using factor label.
In the case of % purity or % yield (to be discussed later another factor is multiplied in addition to the existing factors (check % yield section).

Principal of Atom Conservation (POAC)
In a chemical reaction, atoms are conserved, so moles of atoms shall also be conserved. This is known as principal of atomic conservation. This principal is helpful in solving problems of nearly all stoichiometric calculations e.g.

KClO3(s)→ KCl(s)+O2(g)

Applying POAC for K atoms
Moles of K atoms in KClO3 = Moles of K atoms in KCI
Since one mole of KClO3contains 1 mol of K atom. Similarly, 1 mol of KCI contains one mole of K atoms. (Mass volume relationship of reactant and product)
In this way applying POAC we can break the chemical equation into a number arithmetic equations without balancing the chemical equation. Moreover number of reactions and their sequence from reactants to products are not required. It is important to note that POAC can be applied for the atoms which remain conserved in a chemical reaction.
Please note that the balance reaction is essential is both Mole method of solving and Factor Label Method. Only while using POAC (which would be applicable only in certain cases), balance reaction is not required.
You can use POAC for all atoms in the compound only IF all the reactants and products are known.
Another method for solving stoichiometric problems is by using equivalent concept. This method also does not require balance reaction. However, this would be dealt separately in Equivalent Concept covered in Redox Reactions.
Illustration 13: How many grams of oxygen (O2) are required to completely react with 0.200 g of hydrogen (H2) to yield water (H2O)? Also, calculate the amount water formed. (At.Mass H = 1; 0 = 16).
Solution: The balanced equation for the reaction is Illustration 14: What volume of oxygen at N.T.P. can be produced by 6.125 g of potassium chlorate according to the reaction 2KCIO3 → 2KCI + 3O2
Solution: The given chemical equation is:

2KCIO3 →      2KCI +    3O2

2 mole        2 moles    3 mole

2 × 122.5 g   2×22.4 L at N.T.P Illustration 15 : What mass of zinc is required to produce hydrogen by reaction with HCl which is enough to produce 4 mol of ammonia according to the reactions.

Zn + 2HCl → ZnCl2 + H2

3H2 + N2 → 2NH3

Solution : The given equations are  Try deriving the above relation and also get the relation between mole fraction and molarity Molarity (M) and Molality (m) for Pure Substances:
1. Water: Let the sample of water has 1000 ml
Mass of water = 1000 gm [density of water = 1gm/ML.] Parts per million (ppm) →Amount of solute (in g) with 106 g solvent
Parts per billion (ppb) → amount of solute (in g) with  g solvent
Concentration of Oleum
Industrial preparation of is often done by adding water to commercially available oleum

( a mixture of H2SO4 & SO3 ).
The concentration of oleum in such cases is expressed as X% oleum, which can be defined as 100 gm of this oleum is capable of making X gms of H2SO4 when X – 100 gm of water is added into the oleum solution.
Can you guess from the definition what could be the minimum value of X while representing the concentration of Oleum? For eg. 100 g oleum means, 100 gm of oleum on addition of 9 gm of water will give 109 gm of pure
The concentration term also give an idea of the composition of H2SO4 & SO3 (“Combined SO3 ” & free SO3 ”) in the oleum.

For the above example, the composition can be calculated as follows:
The amount of H2O required gives the amount of “free SO3” present in the oleum since H2O will react with “free SO3” to give pure H2SO4 .
∴ for 109% oleum 100 gm of oleum has that much “free ” which requires 9 gm of H2O for the following reaction.
SO3+ H2O→H2SO4
Since 1/2 mole of H2O are required therefore moles of SO3 will also be 1/2
SOpresent in 100 gm of oleum =1/2 mole = 40 gm
∴ H2SO4in 100 gm of oleum = 60 gm
∴ wt. wise composition of oleum = 60 gm H2SO4, 40 gm SO3
Can you guess what will happen if we add more water than required into oleum? And what will happen when we add less water than required.
As the value of “X” increases as represented by X% oleum it is observed that % of in the mixture also increases. Can you use this info to calculate what could be the maximum value of “X” in the % of Oleum.

Therefore, the empirical formula is Na2CO3.
Illustration 8 : A compound has the following composition:
Mg = 9.76%, S = 13.01%, O = 26.01%, = 51.22%.
What is its empirical formula? [Mg=24, S=32, O=16, H=1]
Solution : = 2.846 = 7
Hence, the empirical formula is MgSO4. 7H2O..
Illustration 9: What is the simplest formula of the compound which has the following percentage composition. Carbon 80%, Hydrogen 20%. If the molecular mass is 30, calculate its molecular formula.

Solution: Molecular formula = Empirical formula × n =C2H4 × 2 =C4H8O2 .
Illustration 11: An organic compound on analysis the following data: C = 57.82%, H = 3.6%, and the rest is oxygen. Its vapour density is 83. Find its empirical and molecular formula.
Solution: Calculation of empirical formula: To calculate empirical formula: STOICHIOMETRY
The word ‘stoichmetry’ is derived from two Greek words stoicheion (meaning element) and metron (meaning measure). Stoichiometry, thus, deals with the calculation of masses (sometimes volumes also) of the reactants and the products involved in a chemical reaction. Before understanding how to calculate the amounts of reactants required or those produced in a chemical reaction, let us study what information is available from the balanced chemical equation for a given reaction. Let us consider the combustion of methane. A balanced equation for this reaction is as given below: The above balance reaction gives the following gives information
(a) For every 1 mole of CH4, 2 mole O2 will be required to produce 1 mole of CO2 and 2 moles of H2O-Mole - Mole relation.
(b) For every 16 gms of CH4 64 gms of O2 will be required to produce 44 gms of CO2 and 36 gms of H2O.
(c) Ratio of moles of CO2: H2O at any time = 1: 2
(d) There will be no change in total mass of all reactants and products at any time for any chemical reaction.
(e) For the above reaction only, there will be no change in a total number of moles of all reactants and products. In order to solve the problems based on chemical calculations the following steps, in general, are quite helpful.
(i) Write the balanced chemical equation.
(ii) Write the atomic mass/molecular mass/moles/molar volumes of the species involved in calculations.
(iii) Calculate the result by applying unitary method.
There is no law of conservation of moles.
Apart from the usual Mole method used for solving stoichiometric problems, sometimes Factor Label Method and POAC method are also used depending upon the type of problem.
Students are advised to have an understanding of the utility of each of the method for solving stoichiometric problems.

Illustration 16: 1.20 gm of a sample of Na2CO3 and K2CO3 was dissolved in water to form 100 ml of a solution. 20 ml of this solution required 40 ml of 0.1 M HCl for complete neutralisation. Calculate the weight of Na2COin the mixture.
If another 20 ml of this solution is treated with an excess of BaCl2 what will be the weight of the precipitate? Limiting Reagent
In many situations, one of the reactants is present in excess, therefore, some of this reactant is left over on completion of the reaction. For example, consider the combustion of hydrogen.
2H2 (g) + O2 (g) → 2H2O (g)
Suppose that 2 moles of H2 and 2 moles of O2 are available of reaction. It follows from the equation that only 1 mole of O2 is required for complete combustion of 2 moles of H2; 1 mole of O2 will, therefore, be left over on completion of the reaction. The amount of the product obtained is determined by the amount of the reactant that is completely consumed in the reaction. This reactant is called the limiting reagent. Thus, limiting reagent may be defined as the reactant which is completely consumed during the reaction.
In the above example is the limiting reagent. The amount of formed will, therefore, be determined by the amount of. Since 2 moles of are taken, it will form 2 moles of on combustion.
The best method to identify limiting reagent is by dividing given moles of each reactant by their stoichiometric coefficient, the one with least ratio is limiting reagent. It is particularly useful when number of reactants are more than two.]
Illustration 17: How much magnesium sulphide can be obtained from 2.00 g of magnesium and 2. 00 g of sulphur by the reaction Mg + S→ MgS? Which is the limiting reagent?
Calculate the amount of the reactants which remains unreacted.
Solution: First of all each of the masses is expressed in moles: From the equation, Mg + S MgS, it follows that one mole of Mg reacts with one mole of S. We are given more moles of Mg than of S. Therefore, Mg is in excess and some of it will remain unreacted when the reaction is over. S is the limiting reagent and will control the amount of product. From the equation, we note that one mole of S gives one mole of MgS, so 0.0624 moles of S will react with 0.0624 moles of Mg to form 0.0624 mole of MgS.
Molar mass of MgS = 56.4 g
Mass of MgS formed = 0.062456.4 g = 3.52gof MgS
Moles of Mg left unreacted = 0.0824 – 0.0624 moles of Mg
= 0.0200 moles of Mg
Mass of Mg left unreacted = moles of Mg × molar mass of Mg
= 0.0200 × 24.3 g of Mg
= 0.486 g of Mg

Calculation Involving Percent Yield
In general, when a reaction is carried out in the laboratory we do not obtain actually the theoretical amount of the product. The amount of the product that is actually obtained is called the actual yield. Knowing the actual yield and theoretical yield, the percentage yield can be calculated as % yield The actual amount of any limiting reagent consumed in such incomplete reactions is given by [% yield given moles of limiting reagent] [For reversible reactions]. For irreversible reaction with % yield less than 100, the reactants are converted to product (desired) and waste.
Can you now guess what factor should be added while solving the problem using factor label method if the % yield or % purity is also mentioned in the numerical?
CONCENTRATION TERMS
Many chemical reactions occur in the solution state and
hence stoichiometric problems may sometimes involve different concentration terms of solutions to give an idea of the amount of solute and solvent present in the solutions. There are various ways to represent the concentration of the solution as mentioned below.
In a solution, the designation of solute and solvent is often a matter of convenience, however, many times the one present in smaller quantity is termed as a solute.Also, a solution may have more than one solute but solvent cannot be more than one.
All the concentration terms given below are applicable only when there is a homogenous solution or when solute completely dissolves in the solvent.
Method of Expressing Concentration of Solution
For each of the concentration terms, classify them which type of ratio are they (w/w or w/v or v/v)? Also, guess whether they will be temperature dependent or not?