OLEUM AND ITS PERCENTAGE(%) LABBELING

(1) Oleum can be represented by the formula ySO₃.H₂O where y is the total molar sulphur trioxide content .the value of y can be varied to different Oleum sample.

(2) Oleum also be expressed  as H₂SO₄.xSO3 where x is molar free suphur trioxide.

(3) Oleum is the solution of of sulphur trioxide in sulphuric acid , it is also known as  fuming sulphuric acid or Pyrosulphuric acid (H₂S₂O₇=H₂SO₄+SO₃).

(4) Oleum sample contain two type of SO₃.

(i) Free SO3:- It is that SO₃ which combined with water water to give H₂SO₄ .

(ii) Combined SO3:- The SO₃ which is present in H₂SO₄ itself is known as combined SO₃ and it does not react with water.

PERCENTAGE(%) LABELLING OF OLEUM :

 OLEUM concentration represented by (100+X) like 104.5%, 109%, 113.4% ,118% etc.

" (100+X)% labelling of Oleum defined as  a number which is greater than 100, it mean (X ) is  the amount of water in gm required to destroyed free SO₃  completely, present in 100 gm  of oleum sample"

For example:  109% oleum sample means that 9 gm water required to destroyed the free SO3 completely , present in 100 gm of Oleum Sulphate sample. 

CALCULATIONS OF FREE SO₃ IN (X%) OLEUM SAMPLE :

Given - X %  labelled 100 gm Oleum, it means  (X-100) gm water required to destroyed all free SO₃ present in 100 gm Oleum.
                                    W_H2O= (X-100) gm

BY STOICHIOMETRIC CALCULATION:

Weight of SO₃ = nSO3 X Molecular wt SO₃

Weight of H₂SO₄ = nH₂SO₄ X Molecular wt of H₂SO₄

 Thus % (Free) SO₃ present in 100 gm 109% labelled Oleum Sample is

Calculation of maximum % labelling of Oleum sample: 

NOTE-For maximum possible labbelling amount of SO₃ is 100 gm, It means 100 gm Oleum sample contains 100 gm SO₃ only and zero( 0 )gm H₂SO₄ ,hence maximum labbelling possible is 122.5 %.

EXAMPLE (1): Calculate the % of free SO₃ in an Oleum sample that is labelled as 118 % ?

SOLUTION:  

EXAMPLE (2): If the percentage free SO₃ in an Oleum sample is 20% then label the sample of Oleum in term of percentage H₂SO₄.?

SOLUTION: 

EXAMPLE(3): Two sample of Oleum are labelled as 109% and 115%,what is the difference between weight of free SO₃ in these samples ?.

SOLUTION:  Given % labelling (X)=109%  and 115% ,find difference between weight of free SO₃ in these samples ?. 

 Difference between weight of free SO₃ in these samples are= 66.66-44= 26.67 gm

EXAMPLE(4): What is the %SO₃ in Oleum sample that is labelled as 104.5% H₂SO₄ ?.

SOLUTION: Given % labelling (X) =104.5%, Find %( Free) SO₃=?

EXAMPLE(5): 9 gm water is added into Oleum sample labelled as 112% H₂SO₄  then the amount of free SO₃ remaining in the solution is ? (STP=1atm and 273K).

SOLUTION: Initial free moles of SO₃=  =2/3 moles

Moles of water that combined with free moles of SO₃ =9/18=1/2 moles

Moles of free SO₃ left 2/3-1/2=1/6 moles

Volume of free SO₃ at STP=1/6X22.4=3.73 L

EXAMPLE(6): Find out the % labelling of oleum Sulphate in which mole fraction of SO₃ is 0.2 ?.

SOLUTION: We know mole mass fraction percentage is equal to % labelling.

EXAMPLE (7): Find out the % labelling of new oleum sample obtained by mixing of 4.5 gm of water in 100 gm of 109% labelled oleum sample ?.

SOLUTION: Wt of SO3 in Original Oleum

The amount of free SO₃ destroyed by 4.5 gm water is added 

The amount of free SO₃ destroyed by 4.5 gm water is = x80=20 gm

Wt of left SO3 =40-20=20 gm

               Given Wt of (Free) SO₃= 20 % Find % labelling (X)

             % labeling(X) = 104.5 %

EXAMPLE(8): 100 gm of 120% labelled  Oleum is diluted with 15 gm of water. determined the new % labelling of Oleum ?.

SOLUTION: :   Wt of SO₃ in Original Oleum

The amount of free SO3 destroyed by 15 gm water is added

                  % labelling(X) = 105 %

The amount of free SO3 destroyed by 4.5 gm water=15/18 x 80=66.66 gm

Wt of left SO3 =88.88-66.66=22.22 gm

Given Wt of (Free) SO₃= 22.22 % Find % labelling (X)

% labelling(X) = 105 %

EXAMPLE(9): Calculate amount of total H₂SO₄ when 100 gm 109% labelled Oleum sample is completely destroyed by water ?.
SOLUTION: the amount H₂SO₄

Originally 109% 100 gm Oleum sample contains 40 gm free SO₃ and 60 gram H₂SO₄

Hence total Wt of H₂SO₄ =60 gm +49 gm =109 gm

EXAMPLE(10): 25 gm of Oleum sample required 2 gm of water ,find out the % labelling of sample .

SOLUTION: :  25 gm oleum required 2 gm water

                       1 gm require …………. 2/25 gm water

                      100 gm require ……..2/25x100=8 gm

                  Hence % labelling is 108%

EXAMPLE(11): A mixture is prepared by mixing of 20 gm SO₃ in 30 gm of H₂SO₄ .

(I)            Find the mole fraction of SO₃ .

(II)          Determine % labelling of Oleum sample.

SOLUTION:

 (i)    Total wt of Oleum is 20 gm SO₃+ 30 gm H₂SO₄

(ii)

Given Wt of (Free) SO₃= 40 % Find % labelling (X)

                                         % labelling(X) = 109%

EXAMPLE(12): What  volume of 1M NaOH (in ml)will required to react completely with 100 gm of Oleum which is 109 % labelled ?.

SOLUTION: We know that 109% Oleum sample contains 40 gm SO3 and 60 gm H₂SO₄

                     (E _wt= SO₃=80/2=40 gm and E _wt  H₂SO₄ =98/2=49 gm)

At equivalent point 

      No of equivalent of SO₃ + No of equivalent H₂SO₄ = no of equivalent of NaOH

EXAMPLE (13) : 0.5 gm of fume H₂SO₄ (Oleum ) is diluted with water, this solution is completely neutralised by 26.7 ml of 0.4 N NaOH. Find the percentage free SO3 in sample solution.?

SOLUTION: Given total wt of Oleum sample is 0.5 gm, let x gm SO3 and (0.5-x) H₂SO₄  

                             (E _wt= SO₃=80/2=40 gm and E _wt H₂SO₄=98/2=49 gm)

At equivalent point    

 

 No of equivalent of SO₃ + No of equivalent H₂SO₄ = no of equivalent of NaOH

EXAMPLE (14): A mixture of H₂CO₃ liquid and CO₂ gas is labelled  as Oleum sample . 50 gm such mixture contains  22% CO₂ , find out  the % labelling of such mixture.

SOLUTION:

                                  Given 22% CO2

 % labeling(X) of CO₂ = 109%

EXAMPLE (15): Calculate how much H2SO4 will be obtained from 400 gm of Oleum sample having labelling 104.5%?

SOLUTION: 104.5 % labelled means 100 Oleum sample required 4.5 gm water to completely destroyed free SO3 present in 100 gm sample

100 gm Oleum sample require 4.5 gm water to destroyed all free SO3

Weight of SO₃ in destroyed by 18 gm water is =18/18x80= 80 gm 

Weight of H₂SO₄ = 400-80= 320 gm present in 400 gm Oleum sample

Weight of H₂SO₄ newly formed is =18/18x98= 98 gm

Total Weight H₂SO₄ =320+98=418 gm

Related Questions:

(1) Calculate the amount of 80 % pure NaOH sample required to react with 21.3 gm Chlorine in hot condition.

(2) When 1.25 gm sample of Chalk is strongly heated. 0.44 gm of CO2 is produced. Calculate % purity of Chalk sample?

(3) The Mass of 80% pure H2SO4 required to completely neutralize 106 gm of Na2CO3?

(4) When dilute HCl is added to 5.73 gm of contaminated CaCO3, 2.49 gm of CO2 is obtained. Find % purity of CaCO3 sample?

(5) 200 gm of CaCO3 sample on heating produces 11.2 liters of CO2 (g) at STP. Calculate % purity of CaCO3 sample?

(6) A sample of impure Iron pyrite (FeS2) when 13.9 gm heated then it produces iron (iii) oxide (Fe2O3) and Sulphur dioxide (SO2). If 8.02 gm Iron (iii) oxide is obtained, what was the % purity of given sample (original-FeS2)?

(7) Galena (an ore) is partially oxidized by passing air through it at high temperature. After some time, the passage of air is stopped, but the heating is continued in a closed furnance such that the contents undergo self-reduction. The weight (in kg) of Pb produced per kg of O2 consumed is______ . (Atomic weights in g mol–1 : O = 16, S = 32, Pb = 207). (JEE-Advanced 2018 Paper-2)

(8) P and Q are two elements which from P2Q3 and PQ2. if 0.15 mole of P2Q3 weighs 15.9 gm and 0.15 mole of PQ2 weighs 9.3 gm, then, what weights of P and Q?

(9) 367.5 gm KClO3 (122.5) when heated then calculate. 

(1) How many gram of O2 is produced? 

(2) How Many litre of O2 is produced at STP?

(10) How many moles O2 are required to needed produced 5 moles of Fe2O3.

(11) How many grams of Fe2O3 formed by heating of 18 gm FeO with O2.

(12) If 4 gm of bleaching powder dissolved to give 100 ml solution 25 ml of it react with excess of CH3COOH and KI .The Iodine liberation required 10 ml of 0.125 N Hypo solutions. Calculate % of available chlorine in the sample.

(13) 25 ml of household bleach solution was mixed with 30 ml of 0.50 M KI and 10 ml of 4.0N acetic acid. In this titration of the liberated iodine 48 ml of 0.25 N Na2S2O3 was used to reach the end point. The Molarity of household bleach solution is?