Stoichiometry & Stoichiometric Calculations and Concentration Terms - Notes | Study Chemistry Class 11 - NEET

Stoichiometry

Stoichiometry is the calculation of the quantities of reactants and products involved in a chemical reaction.
Following methods can be used for solving problems:
(a) Mole Method (For Balance reaction)
(b) POAC method (Balancing not required but common sense, use it with slight care)
(c) Equivalent Concept

➢ Concept of Limiting Reagent

Limiting Reagent: It is a very important concept in chemical calculation. It refers to reactant which is present in minimum stoichiometry quantity for a chemical reaction.
Concept of Limiting Reagent

It is the reactant consumed fully in a chemical reaction. So all calculations related to various products or in the sequence of reactions are made on the basis of limiting reagent.
It comes into the picture when a reaction involves two or more reactants. For solving any such reaction, the first step is to calculate L.R.

Calculation of Limiting Reagent:
(a) By calculating the required amount by the equation and comparing it with the given amount. [Useful when only two reactants are there].

(b) By calculating the amount of any one product taking each reactant one by one irrespective of other reactants. The one giving the least product is limiting reagent.

(c) Divide given moles of each reactant by their stoichiometric coefficient, the one with the least ratio is the limiting reagent. [Useful when the number of reactants are more than two or two.

➢ Percentage Yield

The percentage yield of product is:

 
The actual amount of any limiting reagent consumed in such incomplete reactions is given by:
% yield × given moles of limiting reagent [For reversible reactions].

Example 1. A compound which contains one atom of X and two atoms of Y for each three atoms of Z is made by mixing 5 gm of X, 1.15 × 10²³ atoms of Y and 0.03 mole of Z atoms. Given that only 4.40 gm of compound results. Calculate the atomic weight of Y if atomic weight of X and Z are 60 and 80 respectively. 
Solution. Moles of x = 5/60 = 1/12 = 0.083

Moles of z = 0.03
x + 2y + 3z → xy₂z₃
For limiting reagent, 0.083/1 = 0.083
0.19/2 = 0.095 , 0.03/3 = 0.01
Hence z is limiting reagent
wt of xy₂z₃ = 4.4 gm = moles × molecular wt.
Moles of xy₂z₃ = 1/3 × 0.03 = 0.01
300 + 2m = 440
⇒ 2m = 440 - 300 ⇒ m =70  

➢ POAC Rule

POAC is the simple mass conservation:
KClO₃ → KClO₂
Apply the POAC on K.
moles of K in KClO₃ = moles of K in KCl
1 × moles of KClO₃ = 1 × moles of KCl
moles of KClO₃ = moles of KCl
Apply POAC on O
moles O in KClO₃ = moles of O in O₂
3 × moles of KClO₃ = 2 × moles of O₂
Structure of KCLO₃

Example 2. In the gravimetric determination of phosphorous, an aqueous solution of dihydrogen phosphate ion (H₂PO₄^-) is treated with a mix of ammonium & magnesium ions to precipitate magnesium ammonium phosphate MgNH₄ PO₄.6H₂O. This is heated and decomposed to magnesium Pyrophosphate, Mg₂P₂O₇ which is weighted. A solution of H₂PO₄^- yielded 1.054 gm of Mg₂P₂O₇ what weight of NaH₂PO₄ was present originally.
NaH₂PO₄ → Mg₂P₂O₇ apply POAC on P
Solution. Let wt of NaH₂PO₄ = w gm
moles of P in NaH2PO4 = moles of P in Mg₂P₂O₇
w/120 × 1 = 1.054/232 × 2
w = 1.054 x 120/232 × 2 = 1.09 gm

Dulong's and Petit's Law
Atomic weight × specific heat (cal/gm°C) ∝ ≌ 6.4
Gives approximate atomic weight and is applicable for metals only.
Take care of units of specific heat.

Example 3. 7.5 mL of a hydrocarbon gas was exploded with an excess of oxygen. On cooling, it was found to have undergone a contraction of 15 mL. If the vapour density of the hydrocarbon is 14, determine its molecular formula. (C = 12, H = 1) 

Solution. 

on cooling the volume contraction = 15 ml
i.e The volume of H₂O (g) = 15 ml
V.D of hydrocarbon = 14
Molecular wt. of C_xH_x = 28
12x + y = 28 ...(1)
From reaction

12 x + 4 = 28
12 x = 24
x = 2
Hence Hydrocarbon is C₂H₄.

Example 4. Calculate the weight of FeO produced from 2g VO and 5.75 g of Fe₂O₃. Also report the limiting reagent.
VO + Fe₂O₃ → FeO + V₂O₅
Solution. Balancing the given equatio_n
2VO + Fe₂O₃ → 6FeO + V₂O₅
Mole before reaction = 2/67 = 5.75/160
0.0298 = 0.0359 = 0 = 0
Mole after reaction 

Mole of FeO formed = 0.0359 × 2
Weight of FeO formed = 0.0359 × 2 × 72
= 5.17 g
The limiting reagent is one which is used completely, i.e. Fe₂O₃ here.

➢ Try Yourself!

Q.1. What is the number of moles of Fe(OH)₃ (s) that can be produced by allowing 1 mole of Fe₂S₃, 2 moles of H₂O and 3 moles of O₂ to react?
2Fe₂S₃ (s) + 6H₂O(l) + 3O₂(g ) → 4Fe (OH )₃(s) + 6S(s)
Ans. 1.34 moles

Q.2. In a process for producing acetic acid, oxygen gas is bubbled into acetaldehyde containing manganese (II) acetate (catalyst) under pressure at 60ºC.
2CH₃CHO + O₂ →2CH₃COOH
In a laboratory test of this reaction, 20 g of CH₃CHO and 10 g of O₂ were put into a reaction vessel. 
(a) How many grams of CH₃COOH can be produced? 
(b) How many grams of the excess reactant remain after the reaction is complete?
Ans. 
(a) 27.27 g
(b) 2.73 g

➢ Concentration of Solution

The concentration terms are divided into two types.
1. Temperature dependant- the concentration terms that have a volume term, are affected by change in temperature. Example- Molarity, Normality etc
2. Temperature independant - the concentration terms that do not have a volume term and only mass/moles do not depend upon temperature. Example- Molality, mole fraction etc.

The concentration of a solution can be expressed in any of the following ways:
(a) % by Weight: Amount of solute dissolved in 100 gm of solution.
4.9% H₂SO₄ by weight: 100 gm of solution contains 4.9 gm of H₂SO₄.
(b) % by Volume: Volume of solute dissolved in 100 ml of solution.
x% H₂SO₄ by volume: 100 ml of solution contains x ml H₂SO₄
(c) % Weight by Volume: Weight of solute present in 100 ml of solution.
(d) % Volume by Weight: Volume of solute present in 100 gm of solution.

➢ Concentration Terms

• Molarity (M): No. of moles of solute present in 1000 ml of solution.
  Molarity (M) = moles of solute/volume of solution (lit)
  M = m.moles of solute/volume of solution (ml)
• Molality (m): No. of moles of solute present in 1000 gm of solvent.
  m = moles of solute/wt. of solvent in kg  
  m = m.moles of solute/wt. of solvent in gm  
• Normality (N): No of gm. equivalents of solute present in 1000 ml of solution.
  
  
  Number of gram equivalent  of solute  = Mass of solute in gram/equivalent weight of solute
• Equivalent weight of solute (E) = Molar mass of solute / Valence factor
  (i) Valence factor for base = acidity of base
  (ii) Valence factor for acid = basicity of acid
  (iii) Valence factor for element = valency
  Thus, if one gram equivalent of a solute is present in one liter of the solution, the concentration of the solution is said to be one normal.

The relationship between Normality and Molarity can be expressed as:
► Molarity × Molecular mass = Strength of solution (g/L)  
► Normality × Equivalent mass = Normality of the solution (g/L)  
► Molarity × Molecular mass = Normality × Equivalent mass     
Normality = n × Molarity

• Formality (f): It is the number of formula mass in grams present per litre of solution. In case formula mass is equal to molecular mass, formality is equal to molarity. Like molarity and normality, the formality is also dependent on temperature. It is used for ionic compounds in which there is no existence of a molecule. The mole of ionic compounds is called formole and molarity as formality.
  
  
  
  Where,
  w = weight of solute,
  f = formula weight of solute
  V = volume of solution
  n_f = no. of gram formula weight
  
• Mole Fraction: The mole fraction of a particular component in a solution is defined as the number of moles of that component per mole of solution. If a solution has n_A mole of A & n_B mole of B.
  
  And, X_A + X_B = 1
• Parts per million (ppm):
   
• Volume strength of H₂O₂: The strength of H₂O₂ is represented as 10V, 20V, 30V etc.
  20V H₂O₂ means one litre of this sample of H₂O₂ on decomposition gives 20 lit. of O₂ gas at S.T.P.
  Decomposition of H₂O₂ is given as:
  
  1 mole = 1/2 × 22.4 lt O₂ at S.T.P.
  ⇒ 34 g = 11.2 lt O₂ at S.T.P.
  To obtain 11.2 litre O₂ at S.T.P., at least 34 gm H₂O₂ must be decomposed, and for 20 lt O₂, we should decompose at least 34/112 × 20 gm H₂O₂.
  1 lt solution of H₂O₂ contains 34/112 × 20  gm H₂O₂
  1 lt solution of H₂O₂ contains 34/112 × 20/17 equivalents of H₂O₂ 
  (E H₂O₂ = M/2 = 34/2 = 17)
  Normality of H₂O₂ = 34/112 × 20/17
  Normality of H₂O₂ (N)
  
  Another Method:
  
  From law of equivalence we have:
  gm eq. of O₂ = gm eq. of H₂O₂
  gm eq. of O₂ = moles × n factor of O₂, = 20/22.4 x 4 = 20/5.6
  gm. eq. of H₂O₂ = 20/5.6
  and the volume of H₂O₂ is 1 lit.
  this means 1 lit of H₂O₂ has 20/5.6 gm eq.
  i.e. Normality N = 20/5.6
  Normality of H₂O₂  
  
  Molarity of H₂O₂ (M) 
  
  Strength (in g/l): Denoted by S.
  Strength = molarity × mol. wt.
  = molarity × 34
  strength = Normality × Eq. weight.
  = Normality × 17

► Strength of Oleum 
Oleum is SO₃ dissolved in 100% H₂SO₄. Sometimes, oleum is reported as more than 100% by weight, say y% (where y > 100). This means that (y - 100) grams of water, when added to 100 g of given oleum sample, will combine with all the free SO₃ in the oleum to give 100% sulphuric acid.Structure of Oleum

► Relationship between Molarity, Molality & Density of Solution 
Let the molarity of solution be 'M', molality be 'm' and the density of solution be d gm/m.
Molarity implies that there are M moles of solute in 1000 ml of solution.
wt. of solution = density × volume
= 1000 d gm wt of solute = MM₁
where M₁ is the molecular wt of solute.
wt of solvent = (1000 d - MM₁) gm.
(1000 d - MM₁) gm of solvent contains M moles of solute.
1000 gm of solvent have

 =  Molality no. of moles of solute present in 1000 gm of solvent
 
= Molality on simplifying


► Relationship between Molality & Mole Fraction 
Consider a binary solution consisting of two components A (Solute) and B (Solvent).
Let x_A & x_B are the mole fraction of A & B respectively.

If molality of solution be m then

 =

where M_B is the molecular wt of the solvent B

Example 5. A bottle labeled with "12V H₂O₂" contains 700 ml solution. If a student mix 300 ml water in it what is the g/ liter strength & normality and volume strength of the final solution.
Solution. N = 12/5.6
Meq. of H₂O₂ = 12/5.6 x 100
let the normality of H₂O₂ on dilution be N.
Meq. before dilution = Meq. after dilution
N × 1000 = 12/5.6 x 100 N = 12/5.6 × 7/10 = 1.5 M = 1.5/2
strength gm/lit = 1.5/2 x 34 = 25.5
volume strength = N × 5.6 =  8.4 V 

Example 6. Calculate the percentage of free SO₃ in oleum (considered as a solution of SO₃ in H₂SO₄) that is labeled '109% H₂SO₄ '. 
Solution. '109% H₂SO₄' refers to the total mass of pure H₂SO₄, i.e., 109 g that will be formed when 100 g of oleum is diluted by 9 g of H₂O which (H₂O) combines with all the free SO₃ present in oleum to form H₂SO₄.
H₂O + SO₃ → H₂SO₄
1 mole of H₂O combines with 1 mole of SO₃
or 18 g of H₂O combines with 80 g of SO₃
or 9 g of H₂O combines with 40 g of SO₃.
Thus, 100 g of oleum contains 40 g of SO₃ or oleum contains 40% of free SO₃.

Example 7. A 62% by mass of an aqueous solution of an acid has a specific gravity of 1.8. This solution is diluted such that the specific gravity of the solution became 1.2. Find the % by wt of acid in the new solution.
Solution. Density = 1.8
Volume of solution = Let x gm water be added in solution then
d = 0.6 x = 100
⇒ x = 166.67
Mass of new solution = 100 + 166.67 = 266.67
266.67 gm solution contains 62 gm of acid
% by mass = 23.24 %

Example 8. An aqueous solution is 1.33 molal in methanol. Determine the mole fraction of methanol & H₂O.
Solution. 


⇒ x_A = 0.02394 x_B, x_A x_B = 1
⇒ 1.02394 x_B = 1

 = 0.98, x_A = 0.02

Second Method: Let wt. of solvent = 1000 gm, molality = 1.33

= moles of solute
mole fraction of solute = ,

mole fraction of solute = 0.02

mole fraction of solvent = 1 - 0.02 = 0.98

Example 9. The density of 3 M solution of sodium thiosulphate (Na₂S₂O₃) is 1.25 g/mL. Calculate
(i) amount of sodium thiosulphate
(ii) mole fraction of sodium thiosulphate
(iii) molality of Na  and S₂O₃^2- ions
Solution.
(i) Let us consider one litre of sodium thiosulphate solution
wt. of the solution = density × volume (mL)
= 1.25 × 1000 = 1250 g.
wt. of Na₂S₂O₃ present in 1 L of the solution
= molarity × mol. wt.
= 3 × 158 = 474 g.
wt. % of Na₂S₂O₃ = 37.92%

(ii) Mass of 1 litre solution = 1.25 × 1000 g = 1250 g
[∵ density = 1.25 g/l]
Mole fraction of Na₂S₂O₃ = Number of moles of Na₂S₂O₃/Total number of moles
Moles of water = 1250 – 158 × 3/18 = 43.1
Mole fraction of Na₂S₂O₃ = 3/3 + 43.1 = 0.065

(iii) 1 mole of sodium thiosulphate (Na₂S₂O₃) yields 2 moles
of Na⁺ and 1 mole of S₂O₃^2-
Molality of Na₂S₂O₃ = 3 × 1000/776 = 3.87
Molality of Na⁺ = 3.87 × 2 = 7.74 m
Molality of S₂O^2-₃ = 3.87 m

Example 10. A solution of NaCl 0.5% by wt. If the density of the solution is 0.997 g/ml, calculate
(a) The molality
(b) Molarity
(c) Normality
(d) Mole fraction of the solute
Solution. 
Number of moles of NaCl
= 0.5/58.5 = 0.00854

(a) By definition,

= 100/0.997 = 100.3
(b) Now Molarity

(c) Normality

(d) To calculate mole fraction of the solute
No. of moles of water in 99.5 g = 99.5/18 = 5.5277
Moles of NaCl = 0.5/58.5 = 8.547 x 10^-3
X_H2O = 1 - X_NaCI = 0.9984

_{Example 11. Find the molarity and molality of a 15% solution of} H₂SO_{4 (density of} H₂SO_{4 = 1.020 g cm− 3) (Atomic mass: H = 1, O = 16, S = 32 amu).}
Solution. 15% of a solution of H₂SO₄ means 15 g of H₂SO₄ are present in 100 g of the solution i.e.
Mass of H₂SO₄ dissolved = 15 g
Mass of the solution = 100 g
Density of the solution = 1.02 g/cm³ (given)
Calculation of molality:
Mass of solution = 100 g
Mass of H₂SO₄ = 15g
Mass of water (solvent) = 100 – 15 = 85 g
Mol. mass of H₂SO₄ = 98
∴ 15 g H₂SO₄ = 15/98 = 0.153 moles
Thus 85 g of the solvent contain 0.153 moles 100 g of the solvent containing
= 0.153/85 × 1000 = 1.8
Hence the molality of H₂SO₄ solution = 1.8 m
Calculation of molarity:
15 g of H₂SO₄ = 0.153 moles
Vol. of solution = W t. of solution/Density of solution = 100/1.02 = 98.04 cm³
Thus 98.04 cm³ of solution contain H₂SO₄ = 0.153 moles

Tip: Practice makes a man perfect. Practice as many questions as you can on this topic. Make handwritten short notes for the formulas!