Oil is the most important mineral for Azerbaijan. From chemist’s point of view it is a mixture of a — Organic Chemistry Chemistry Question
Determination of water in oil
Oil is the most important mineral for Azerbaijan. From chemist’s point of view it is a mixture of a great number of substances of different nature – both organic and inorganic. At all stages of petroleum refining, it is important to control the content of inorganic impurities, including water, in oil. One of the popular chemical methods for the determination of water in various organic matrices is named Karl Fischer titration. The method is based on the reaction described by R. Bunsen:
I2 + SO2 + H2O → HI + H2SO4
Indicate the oxidant and reductant in the reaction.
Model Answer
Reductant – sulfur dioxide, oxidizer – iodine.
Put the coefficients in this reaction. Sum of the least integer coefficients is
A) 14 B) 9 C) 7 D) 10
Model Answer
C) 7
Reagent which was proposed by the German chemist Karl Fischer is a mixture of pyridine, sulfur dioxide, iodine, and methanol. In this case, the following reactions occur during titration:
SO2 + CH3OH + H2O + I2 → 2 HI + CH3OSO3H
Py + HI → PyH+I–
Py + CH3OSO3H → PyH+ CH3OSO3 –
What is the role of pyridine in the composition of Fischer reagent?
Model Answer
As a base, pyridine binds to acids that are formed during the process (HI and H2SO4) and neutralize them.
What substance(s) could be used instead of pyridine?
A) Imidazole B) Pyrrol C) Hydrazine D) Butylamine
Model Answer
The substance must have basic properties – A) Imidazole, C) Hydrazine, D) Butylamine.
However, the above composition of titrant may lead to various side reactions. Write the possible reactions taking place between the components of the Fischer reagent and the following substances:
a) aldehydes, RC(O)H,
b) ketones, RC(O)R,
c) mercaptans, RSH,
d) organic peroxides, ROOH.
Indicate if an overestimation or underestimation of the water content is observed in each case.
Model Answer
Reactions of ketones and aldehydes with methanol lead to ketals and acetals. The result is overstated, since water is released:
RCH=O + 2 CH3OH → RCH(OCH3)2 + H2O.
Reactions of aldehydes with sulfur dioxide and base give sulfite aldehyde derivative. The result is understated, because water is absorbed.
RCH=O + SO2 + H2O + Py → [RCH(OH)SO3 –]PyH+.
Iodine reacts with mercaptans. The result is overstated, since iodine is consumed:
2 RSH + I2 → RSSR + 2 HI.
Reaction of hydrogen iodide with peroxides:
ROOH + 2HI → I2 + ROH + H2O.
Hydroperoxides produce equivalent amounts of iodine and water. The Fischer titration is free from interference. If some other strong oxidizing agents (elemental bromine, chlorine) are present, excess SO2 is passed through the sample. This reduces these substances to chloride and bromide respectively, which no longer interfere. Other peroxides (percarbonate or diacylperoxide) react according to the following equation at different rates:
R–CO–O–O–CO–R + 2 HI → 2 RCOOH + I2.
In this case, determination of water is performed at low temperatures (up to –60°C), so that any possible side reactions can be «frozen».
Fischer reagent is to be standardized prior to use, i.e. it is necessary to set amount of water corresponding to one volume of a titrant. The reagent was prepared by the following procedure: 49 g of iodine were dissolved in 158 g of pyridine, then 38.5 g of liquid sulfur dioxide was introduced while cooling. Thereafter, mixture was diluted up with methanol to 1 dm3.
a) Using the above data, calculate the theoretical titre of the Fischer reagent (in mg/mL).
b) Calculate the practical titre (in mg/mL) of Fischer reagent, if to reach the endpoint for the titration of 5 g of a mixture of methanol and water (water content of 1% by volume), 19 cm3 of titrant was spent.
c) Why do the results obtained differ?
d) Calculate the water content (in %) in the sample of sour oil, if the content of mercaptan in sulfur recalculation is 1 wt.% and titration of 1.00 g sample of sour oil dissolved in methanol required 7.5 cm3 of titrant.
Model Answer
a) First, calculate the amounts of substance of iodine and sulphur dioxide.
n(I2) = 49 / (2⋅127) = 0.193 mol, n(SO2) = 38.5 / 64 = 0.6 mol.
Iodine is completely consumed. One molecule of iodine reacts with one molecule of sulphur oxide, so the theoretical titre is (in mg/cm3):
m(H2O) = 0.193 / 1000 ⋅ 18 = 3.5 mg / cm3.
b) The practical titre is (use the density of methanol (0.7918) as the density of mixture):
m(H2O) = (5 / 0.7918) ⋅ 0.01 / 19 = 3.3 mg / cm3.
c) The results obtained differ because of the presence of water in the reactants, but also because of water vapor in the air.
d) The amount of iodine that reacted with mercaptans:
n(I2) = 0.01 / 32 / 2 mol.
Multiplying this value by the molar mass of water, the value of overstatement can be calculated (in milligrams):
m(H2O) = n(I2) M(H2O) = 2.8 mg.
The total mass of determined water is therefore:
m(H2O) = 7.5 × 3.3 = 24.8 mg.
The mass of water in the sample is:
m(H2O) = 24.8 – 2.8 = 22.0 mg.
Water content (mass %) in the sample is:
w(H2O) = 22.0 mg / 1.00 g ⋅ 100 % = 2.20 %.
The modern version of water determination in different samples is coulometric Karl Fischer titration. During the titration iodine is generated electrochemically and the water content is defined as the total amount of current passing through the coulometer. Titration is stopped when the generation of iodide stops in the system.
a) Using Faraday's law, calculate the mass fraction of water in the oil sample (10.0 g), if 375.3 coulombs passed through a coulometer.
b) 1.000 g portion of sugar was dissolved in 15 cm3 of methanol and chloroform mixture. Calculate the molar and mass fractions of water and sugar, if for a coulometric titration of sugar a total charge of 567.2 coulombs was required, while 31.1 coulombs were required to titrate solvent.
Model Answer
a) In oxidation of iodine two electrons are involved. Then, using the Faraday equation water content (mass %) is calculated:
m(H2O) = M Q / (nF) = 18 × 375.3 / (2 × ⋅96500) = 0.0035 g.
w(H2O) = 0.035 / 10 ⋅ 100% = 0.35 %.
b) The charge passed during the titration of the sugar sample is:
Q = 567.2 – 31.1 = 536.1 C.
Then, water mass is 0.050 g (from Faraday law), and the content (mass %) is 5.0 %. The mole fraction of water is (sucrose – С12Н22О11):
x(H2O) = (0.05 / 18) / (0.05 / 18 + 0.95 / 342) * 100 % = 50 %.