# A sample of Oleum is labelled 118% . The percentage of free SO3 in the sample is

Avinash
8 years ago
Oleum, also known as fuming sulfuric acid, is a solution of various compositions ofsulfur trioxideinsulfuric acid, or sometimes more specifically todisulfuric acid(also known as pyrosulfuric acid).

Its not Oleum 118%,its 118% of H2SO4 labelled oleum.

Oleums can be described by the formulaySO3.H2O where y is the total molar sulfur trioxide content. The value ofycan be varied, to include different oleums. They can also be described by the formula H2SO4.xSO3where x is now defined as the molar free sulfur trioxide content. Oleum is generally assayed according to the free SO3content by weight. It can also be expressed as a percentage of sulfuric acid strength; for oleum concentrations, that would be over 100%. For example, 10% oleum can also be expressed as H2SO4.0.13611SO3,1.0225SO3.H2O or 102.25% sulfuric acid. The conversion between% acid and% oleum is:% acid = 100 + 18/80 *% oleum
Calculation of free SO3 in 118% H2SO4 labelled oleum
H2O + SO3 → H2SO4
As 18g of water combines with 80g of SO3
100g of oleum contains 80g of SO3 or 80% free SO3
Kartikey Sharma
19 Points
8 years ago
I am not getting it . Pls tell in simpler terms i.e in 118g H2SO4   100 gwater  & 18g H2SOis there or anything else.
Avinash
8 years ago
Dont get confused. I’ll try to make it simpler

sulfur trioxide+ sulfuric acid → oleum
SO3(g) +H2SO4(aq) → H2S2O7(l)

oleum + water →sulfuric acid
H2S2O7(l) +H2O(l) → 2H2SO4(aq)

Look at hte above reactions, Oleum issulfur trioxide in sulfuric acid
this oleum H2S2O7 can be described asH2SO4.xSO3 where x is now defined as the molar free sulfur trioxide content.Now here x=1. So we have to calculate the free content of SO3. which is thepercentage of sulfuric acid strength

Strength of oleum is 100+amt. of H2O.

no. of moles of SO3 = no.of moles of H2O

wt of SO3 / 80 = ((%Strength of oleum -100))/18

% free SO3 = wt of SO3 *100/100.
kowshal
13 Points
6 years ago
118% oleum means...100gm sample and 18gm is the water requried to combine so3 in the sample....1mole of so3 =80 gms..1mol of H20=18gm...1mol of H20 combines 1mol of so3...then 18gm of water combines 80gm of oleum..this is called free oleum..remaining 20%H2so4
Dhruv
20 Points
6 years ago
After addition of h2o in solution SO3 reacts with H2O to form H2SO4 . So final amount of H2SO4 is 118%
Tan
23 Points
6 years ago
Given percent is 118 % The formula to find free so3 is (x-100 )×80 ----------- 18Wer x is the given % So the ans will be 118 - 100 / 18 ×80
Shalini sharma
11 Points
5 years ago
2NaOH+H2SO4->NA2SO4+2H2OUse 109 g of H2SO4, because 109% oleum means 109g H2SO4 is formed.M1V1/2=M2V21×V=109×1000×2/98V=2224.5mL
Aryan Kumar
13 Points
5 years ago
See, x% labelled oleum means 100 g of such oleum on dilution will give x gm of H2SO4Wt of H2O added=(x-100)gmMoles of H2O added=(x-100)/18.Moles of SO3present in oleum=(x-100)/18 .Wt of SO3 present in oleum=(x-100)/18*80gm.% of free SO3 in oleum=(x-100)/18*80%Now, for your question,%of SO3= 18/18*80%=80%In the above,80 is the molecular wt of SO3.SO3+H2O-->H2SO4.So one mole of h2o reacts with 1 mole of SO3 to form one mole of H2SO4.
rohith
16 Points
4 years ago

Dont get confused. I’ll try to make it simpler

sulfur trioxide+ sulfuric acid → oleum
SO3(g) +H2SO4(aq) → H2S2O7(l)

oleum + water →sulfuric acid
H2S2O7(l) +H2O(l) → 2H2SO4(aq)

Look at hte above reactions, Oleum issulfur trioxide in sulfuric acid
this oleum H2S2O7 can be described asH2SO4.xSO3 where x is now defined as the molar free sulfur trioxide content.Now here x=1. So we have to calculate the free content of SO3. which is thepercentage of sulfuric acid strength

Strength of oleum is 100+amt. of H2O.

no. of moles of SO3 = no.of moles of H2O

wt of SO3 / 80 = ((%Strength of oleum -100))/18

% free SO3 = wt of SO3 *100/100.
rohith
16 Points
4 years ago
Given percent is 118 % The formula to find free so3 is (x-100 )×80 ----------- 18Wer x is the given % So the ans will be 118 - 100 / 18 ×80

Maahi
15 Points
4 years ago
Formula to find free so3 is 80*(x-100)/18
Where x is amount of oleum in labelled sample
It is formula to calculate percentage of free so3
3 years ago
Hello student

Oleum sample always consist of SO3 and H2SO4
The labelling of the sample denotes the amount of H2SO4 in the sample after water is added to it. This is due to the fact that free SO3 reacts with water to form H2SO4.
SO3 + H2O → H2SO4

Let us assume 100 g of sample
Let the mass of free SO3 = x g
Hence, the mass of free H2SO4 = (100 – x) g

when water is added to the sample, one mole of SO3 forms one mole of H2SO4
hence, mass of additional H2SO4 formed = x/80 * 98       ( as molar mass of SO3 = 80 g; molar mass of H2SO4 = 98 g)
= 1.225x
Hence the total mass of H2SO4 in the new sample = (100 – x) + 1.225x
= 100 + 0.225x
Now, In the given question Oleum is labelled as 118%
i.e. 100g of oleum forms 118g of pure H2SO4
Hence,
100 + 0.225x = 118
or, 0.225x = 18
or, x = 80 g
Hence percentage of free SO3 in oleum = 80/100 * 100 = 80%

Hope it helps
Regards,
Kushagra
Yash Chourasiya
3 years ago
Dear Student

118%oleum means100g of oleum requires18g ofH2​Oto form118g ofH2​SO4​.
SO3​ + H2​O → H2​SO4
1mol 1mol
(80g) (18g)

18gH2​O≡ 80gSO3​
∴Percentage of freeSO3​ = (18/18)​×80
=80%