Table of contents  
Stoichiometry 
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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
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.
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 × 10^{23} 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 → xy_{2}z_{3}
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 xy_{2}z_{3} = 4.4 gm = moles × molecular wt.
Moles of xy_{2}z_{3} = 1/3 × 0.03 = 0.01
300 + 2m = 440
⇒ 2m = 440  300 ⇒ m =70
POAC is the simple mass conservation:
KClO_{3} → KClO_{2}
Apply the POAC on K.
moles of K in KClO_{3} = moles of K in KCl
1 × moles of KClO_{3} = 1 × moles of KCl
moles of KClO_{3} = moles of KCl
Apply POAC on O
moles O in KClO_{3 }= moles of O in O_{2}
3 × moles of KClO_{3} = 2 × moles of O_{2}
Structure of KCLO_{3}
Example 2. In the gravimetric determination of phosphorous, an aqueous solution of dihydrogen phosphate ion (H_{2}PO_{4}^{}) is treated with a mix of ammonium & magnesium ions to precipitate magnesium ammonium phosphate MgNH_{4} PO_{4}.6H_{2}O. This is heated and decomposed to magnesium Pyrophosphate, Mg_{2}P_{2}O_{7} which is weighted. A solution of H_{2}PO_{4}^{} yielded 1.054 gm of Mg_{2}P_{2}O_{7} what weight of NaH_{2}PO_{4} was present originally.
NaH_{2}PO_{4} → Mg_{2}P_{2}O_{7 }apply POAC on P
Solution. Let wt of NaH_{2}PO_{4} = w gm
moles of P in NaH2PO4 = moles of P in Mg_{2}P_{2}O_{7}
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 H_{2}O (g) = 15 ml
V.D of hydrocarbon = 14
Molecular wt. of C_{x}H_{x} = 28
12x + y = 28 ...(1)
From reaction
12 x + 4 = 28
12 x = 24
x = 2
Hence Hydrocarbon is C_{2}H_{4}.
Example 4. Calculate the weight of FeO produced from 2g VO and 5.75 g of Fe_{2}O_{3}. Also report the limiting reagent.
VO + Fe_{2}O_{3} → FeO + V_{2}O_{5}
Solution. Balancing the given equatio_{n}
2VO + Fe_{2}O_{3} → 6FeO + V_{2}O_{5}
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. Fe_{2}O_{3} here.
Q.1. What is the number of moles of Fe(OH)_{3} (s) that can be produced by allowing 1 mole of Fe_{2}S_{3}, 2 moles of H_{2}O and 3 moles of O_{2} to react?
2Fe_{2}S_{3} (s) + 6H_{2}O(l) + 3O_{2}(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.
2CH_{3}CHO + O_{2} →2CH_{3}COOH
In a laboratory test of this reaction, 20 g of CH_{3}CHO and 10 g of O_{2} were put into a reaction vessel.
(a) How many grams of CH_{3}COOH 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
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% H_{2}SO_{4} by weight: 100 gm of solution contains 4.9 gm of H_{2}SO_{4}.
(b) % by Volume: Volume of solute dissolved in 100 ml of solution.
x% H_{2}SO_{4} by volume: 100 ml of solution contains x ml H_{2}SO_{4}
(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.
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
► Strength of Oleum
Oleum is SO_{3} dissolved in 100% H_{2}SO_{4}. 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 SO_{3} 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 = MM_{1}
where M_{1} is the molecular wt of solute.
wt of solvent = (1000 d  MM_{1}) gm.
(1000 d  MM_{1}) 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 H_{2}O_{2}" 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 H_{2}O_{2} = 12/5.6 x 100
let the normality of H_{2}O_{2} 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 SO_{3} in oleum (considered as a solution of SO_{3} in H_{2}SO_{4}) that is labeled '109% H_{2}SO_{4 }'.
Solution. '109% H_{2}SO_{4}' refers to the total mass of pure H_{2}SO_{4}, i.e., 109 g that will be formed when 100 g of oleum is diluted by 9 g of H_{2}O which (H_{2}O) combines with all the free SO_{3} present in oleum to form H_{2}SO_{4}.
H_{2}O + SO_{3} → H_{2}SO_{4}
1 mole of H_{2}O combines with 1 mole of SO_{3}
or 18 g of H_{2}O combines with 80 g of SO_{3}
or 9 g of H_{2}O combines with 40 g of SO_{3}.
Thus, 100 g of oleum contains 40 g of SO_{3} or oleum contains 40% of free SO_{3}.
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 & H_{2}O.
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 (Na_{2}S_{2}O_{3}) is 1.25 g/mL. Calculate
(i) amount of sodium thiosulphate
(ii) mole fraction of sodium thiosulphate
(iii) molality of Na^{ } and S_{2}O_{3}^{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 Na_{2}S_{2}O_{3} present in 1 L of the solution
= molarity × mol. wt.
= 3 × 158 = 474 g.
wt. % of Na_{2}S_{2}O_{3} = 37.92%
(ii) Mass of 1 litre solution = 1.25 × 1000 g = 1250 g
[∵ density = 1.25 g/l]
Mole fraction of Na_{2}S_{2}O_{3} = Number of moles of Na_{2}S_{2}O_{3}/Total number of moles
Moles of water = 1250 – 158 × 3/18 = 43.1
Mole fraction of Na_{2}S_{2}O_{3} = 3/3 + 43.1 = 0.065
(iii) 1 mole of sodium thiosulphate (Na_{2}S_{2}O_{3}) yields 2 moles
of Na^{+} and 1 mole of S_{2}O_{3}^{2}
Molality of Na_{2}S_{2}O_{3} = 3 × 1000/776 = 3.87
Molality of Na^{+} = 3.87 × 2 = 7.74 m
Molality of S_{2}O^{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_{H}_{2O} = 1  X_{NaCI} = 0.9984
_{Example 11. Find the molarity and molality of a 15% solution of }H_{2}SO_{4}_{ }_{(density of }H_{2}SO_{4 }_{= 1.020 g cm− 3) (Atomic mass: H = 1, O = 16, S = 32 amu).}
Solution. 15% of a solution of H_{2}SO_{4} means 15 g of H_{2}SO_{4} are present in 100 g of the solution i.e.
Mass of H_{2}SO_{4} dissolved = 15 g
Mass of the solution = 100 g
Density of the solution = 1.02 g/cm^{3} (given)
Calculation of molality:
Mass of solution = 100 g
Mass of H_{2}SO_{4} = 15g
Mass of water (solvent) = 100 – 15 = 85 g
Mol. mass of H_{2}SO_{4} = 98
∴ 15 g H_{2}SO_{4} = 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 H_{2}SO_{4} solution = 1.8 m
Calculation of molarity:
15 g of H_{2}SO_{4} = 0.153 moles
Vol. of solution = W t. of solution/Density of solution = 100/1.02 = 98.04 cm^{3}
Thus 98.04 cm^{3 }of solution contain H_{2}SO_{4} = 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!
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