[ckretai]

I'm a bit confused with my calculation for finding the iodine number of an oil sample. My 0.10g oil sample was mixed with 5mL of cyclohexane, and then 7.00mL of Wijs reagent (ICl) was added.

After an hour, I added 7mL of 10% KI to form molecular iodine from the unreacted ICl:
ICl + KI -> KCl + I₂

Then I titrated with 0.0500M sodium thiosulfate. The volume of Na₂S₂O₃ needed to reach the end point of my titration was 11.6mL, or 0.00058 moles of Na₂S₂O₃.
In my control titration (no lipid present), 24.7mL of Na₂S₂O₃ = 0.001235 moles were required to reach the endpoint.

There was 1 mole of I₂ reacting with every 2 moles of Na₂S₂O₃ beacuse:
I₂ + 2S₂O₃^2- -> 2I^- + S₄O₆^2-

So that means:
0.00058/2 = 0.00029 moles I₂ in the oil solution.
0.001235/2 = 0.0006175 moles I₂ in the control titration.

The difference between the unreacted iodine in the control and the sample is the amount of iodine that actually reacted with the lipid.
0.0006175 - 0.00029 = 0.0003275 moles.
Converted to grams, using the molecular mass of I₂, 253.809g/mol:
(0.0003275)(253.809)=0.08312g of I₂.

I'm not really sure where to go from here. I know that the iodine number is the grams of iodine consumed by 100g of the oil. I'm almost there, right?

[Borek]

I just skimmed your calculations, logic looks good, haven't checked the numbers.

Now it is about simple proportions: if 0.1 g of an oil consumes 0.08312 g of iodine, how much would 100 g of the oil consume?

[ckretai]

I just skimmed your calculations, logic looks good, haven't checked the numbers.

Now it is about simple proportions: if 0.1 g of an oil consumes 0.08312 g of iodine, how much would 100 g of the oil consume?

Oh, so I am right there! 0.1g x 1000 = 100g, so 0.08312g x 1000 = 83.12g is my iodine number.