Calculation of the % yield
Mr of SnCl2.2H2O = 229.9350
Mr of SnCl2 = 189.616
Mass of SnCl2.2H2O weighed = 3.0020g
Mass of SnCl2 in 3.0020g of SnCl2.2H2O = x 3.0020g
= 2.4756g of SnCl2
Therefore, this mass (2.4756g) of SnCl2 reacted with the 3.2950g of iodine to give SnCl2I2.
Equation 1: SnCl2 + I2 ------------> SnCl2I2
Number of moles of SnCl2 =
= 1.306 x 10 -2 moles of SnCl2
From equation (1) ;
1 mole of SnCl2 produces 1 mole of SnCl2I2
Therefore,
1.306 x 10 -2 moles of SnCl2 produced 1.306 x 10 -2 moles of SnCl2I2
Equation 2: 2SnCl2I2 -----------------> SnCl4 + SnI4
From equation (2);
2 moles SnCl2I2 produces 1 mole of SnI4
Therefore,
1.306 x 10 -2 moles of SnCl2I2 should produce x 1.306 x 10 -2 moles of SnI4
= 6.53 x 10 -3 moles of SnI4
Mass of 6.53 x 10 -3 moles of SnI4 ( Theoretical yield)
Mr of SnI4 = 626.3286
Mass = (6.53 x 10 -3) X 626.3286
= 4.0899g
Actual yield = 2.0850g
% yield = X 100
= 51 %
Percentage of Iodine in Tin (IV) Iodide
Procedure:-
-
Approximately 0.25g of SnI4 was weighed in a glass stoppered flask, 30cm3 of concentrated HCl and a few cm3 of chloroform were added to it.
-
The solution was then titrated with a standard solution of potassium iodate ( containing about 1.25g in 250 cm3) until on vigorously shaking of the flask, the violet coloration in the organic layer just disappeared ( i.e the aqueous part becomes pale yellow).
- The experiment was repeated twice in order to obtain accurate titre values.
Mass of sample used in first titration = 0.2514g
Mass of sample used in second titration = 0.2523g
Titration Table
Answers to questions
-
1st equation
IO3- + 5I- + 6H+ ----------> 3I2 + 3H2O
2nd equation
IO3- + 2I2 + 6H+ + 5Cl- ----------> 5ICl + 3H2O
Proving that KIO3 is equivalent to 2I-
We multiply the 1st equation by 2 and the 2nd equation by 3. We get:-
2IO3- + 10I- + 12H+ ----------> 6I2 + 6H2O....(1)
3IO3- + 6I2 + 18H+ + 15Cl- ----------> 15 ICl + 9H2O...(2)
Combining the two equations, we get:-
IO3- + 2I- + 6H+ + 5Cl- ----------> 5ICl + 3H2O
From this equation, we conclude that KIO3 is equivalent to 2I-
- Calculating % of iodine in the sample
Average mass of sample used =
= 0.25185g
Average volume of potassium iodate used =
= 34.95 cm3
250cm3 of KIO3 solution contains 1.25g of potassium iodate.
Therefore,
34.95cm3 of KIO3 solution contains X 34.95 g of potassium iodate
= 0.17475g of potassium iodate
Number of moles of KIO3 in 0.17475g of KIO3
Mr of KIO3 = 214.0011
Number of moles =
= 8.166 x 10 -4 moles
Equation:
IO3- + 2I- + 6H+ + 5Cl- ----------> 5ICl + 3H2O
From the equation,
1 mole of IO3- reacts with 2 moles of iodide
Therefore,
8.166 x 10 -4 moles of IO3- reacted with 2 x (8.166 x 10 -4 moles) of iodide.
=1.6332 x 10-3 moles of iodide.
Mass of iodide present in the sample
Ar of iodine = 126.9045
Mass of iodide= Ar x number of moles
= 126.9045 x 1.6332 x 10-3
= 0.20726g
Percentage of iodine in the sample
Mass of sample= 0.25185g
Mass of iodide= 0.20726g
% of iodine in sample = x 100
=82.30%