Question

Thyroxine \(\left(\mathrm{C}_{15} \mathrm{H}_{11} \mathrm{I}_{4} \mathrm{NO}_{4}\right)\) is a hormone synthesized by the thyroid gland and used to control many metabolic functions in the body. A physiologist determines the mass percent of thyroxine in a thyroid extract by igniting \(0.4332 \mathrm{~g}\) of extract with sodium carbonate, which converts the iodine to iodide. The iodide is dissolved in water, and bromine and hydrochloric acid are added, which convert the iodide to iodate. (a) How many moles of iodate form per mole of thyroxine? (b) Excess bromine is boiled off and more iodide is added, which reacts as shown in the following equation: $$ \mathrm{IO}_{3}^{-}(a q)+6 \mathrm{H}^{+}(a q)+5 \mathrm{I}^{-}(a q) \longrightarrow 3 \mathrm{I}_{2}(a q)+3 \mathrm{H}_{2} \mathrm{O}(l) $$ How many moles of iodine are produced per mole of thyroxine? (Hint: Be sure to balance the charges as well as the atoms.) What are the oxidizing and reducing agents in the reaction? (c) The iodine reacts completely with \(17.23 \mathrm{~mL}\) of \(0.1000 \mathrm{M}\) thiosulfate as shown in the following unbalanced equation: $$ \mathrm{I}_{2}(a q)+\mathrm{S}_{2} \mathrm{O}_{3}^{2-}(a q) \longrightarrow \mathrm{I}^{-}(a q)+\mathrm{S}_{4} \mathrm{O}_{6}^{2-}(a q) $$ What is the mass percent of thyroxine in the thyroid extract?

Short answer

Number of moles of iodate per mole of thyroxine: 4. Number of moles of iodine per mole of thyroxine: 12. Mass percent of thyroxine: 12.90%.

Step-by-step solution

Determine the number of moles of iodate formed per mole of thyroxine

Thyroxine \(\mathrm{C_{15}H_{11}I_{4}NO_{4}}\) contains 4 iodine atoms (I). When converted to iodate \(\mathrm{IO_3^-}\), each mole of thyroxine will produce 4 moles of iodate. Therefore, the number of moles of iodate formed per mole of thyroxine is 4.

Determine the number of moles of iodine produced per mole of thyroxine

For the reaction given: \(\mathrm{IO_3^-}(aq) + 6 \mathrm{H^+}(aq) + 5 \mathrm{I^-}(aq) \longrightarrow 3 \mathrm{I_2}(aq) + 3 \mathrm{H_2O}(l)\), the ratio of iodate (\(\mathrm{IO_3^-}\)) to iodine (\(\mathrm{I_2}\)) is 1:3. Since 4 moles of thyroxine produce 4 moles of iodate, these 4 moles of iodate will produce 4 x 3 = 12 moles of iodine. Therefore, the number of moles of iodine produced per mole of thyroxine is 12.

Identify the oxidizing and reducing agents

In the reaction: \(\mathrm{IO_3^-}(aq) + 6 \mathrm{H^+}(aq) + 5 \mathrm{I^-}(aq) \longrightarrow 3 \mathrm{I_2}(aq) + 3 \mathrm{H_2O}(l)\), \(\mathrm{IO_3^-}\) is the oxidizing agent because it gains electrons (is reduced) and \(\mathrm{I^-}\) is the reducing agent because it loses electrons (is oxidized).

Calculate the moles of \(\mathrm{I_2}\)

Given that the reaction of iodine with thiosulfate is: \(\mathrm{I_2}(aq) + \mathrm{S_2O_3^{2-}}(aq) \longrightarrow \mathrm{I^-}(aq) + \mathrm{S_4O_6^{2-}}(aq)\), first calculate the moles of thiosulfate used. Moles of \(\mathrm{S_2O_3^{2-}}\) = Volume \(\mathrm{(L)}\) x Molarity \(\mathrm{(M)}\) = 0.01723 L x 0.1000 M = 0.001723 moles. Since the ratio of \(\mathrm{I_2}\) to \(\mathrm{S_2O_3^{2-}}\) is 1:2, moles of \(\mathrm{I_2}\) = 0.001723 / 2 = 0.0008615 moles.

Calculate the mass percent of thyroxine in the extract

Given that 1 mole of thyroxine produces 12 moles of iodine, the number of moles of thyroxine in the sample is \(\mathrm{0.0008615/12 = 7.179 \times 10^{-5}}\) moles. The molar mass of thyroxine is \(\mathrm{C_{15}H_{11}I_{4}NO_{4}}\) = 777.87 g/mol. Therefore, the mass of thyroxine is \(\mathrm{7.179 \times 10^{-5} \times 777.87 = 0.0559 g}\). The mass percent of thyroxine in the extract is \(\mathrm{(0.0559 / 0.4332) \times 100 = 12.90\%}\).

Key concepts

mass percent calculation

Calculating mass percent involves determining how much of a particular component is present in relation to the total mass of a sample. Here's how to approach it:
Suppose you need to find the mass percent of thyroxine in a thyroid extract.
You start with the extracted sample's initial mass and determine the mass of thyroxine present.
Next, you use the formula:
Mass Percent = \[ \frac{\text{Mass of Component}}{\text{Total Mass of Sample}} \times 100 \]
By substituting the known values, you can find the mass percent of thyroxine in your sample.

iodate formation

Iodate formation is a process where iodide ions are converted into iodate ions.
In the given problem, after the thyroid extract is ignited with sodium carbonate, iodide is formed.
This iodide is then treated with bromine and hydrochloric acid to form iodate.
The relevant chemical conversion is:
\[ \text{Iodide (I}^-\text{) to Iodate (IO}^-_3\text{)} \]
This conversion is crucial as it enables the subsequent steps in the analysis, such as the formation of iodine and eventually, the quantification of thyroxine.

oxidizing and reducing agents

In redox reactions, it's essential to identify the substances that gain or lose electrons.
Oxidizing agents gain electrons and get reduced, while reducing agents lose electrons and get oxidized.
Looking at the reaction:
\[ \text{IO}_3^-(aq) + 6 \text{H}^+(aq) + 5 \text{I}^-(aq) \rightarrow 3 \text{I}_2(aq) + 3 \text{H}_2 \text{O}(l) \]
In this case, \text{IO}_3^- is the oxidizing agent because it is reduced, and \text{I}^- is the reducing agent because it is oxidized.

thiosulfate titration

Thiosulfate titration is a quantitative method to determine the concentration of a reactant.
In the provided problem, iodine produced in the reaction is titrated with thiosulfate.
The reaction is:
\[ \text{I}_2(aq) + \text{S}_2\text{O}_3^{2-}(aq) \rightarrow \text{I}^-(aq) + \text{S}_4\text{O}_6^{2-}(aq) \]
Here, you first calculate moles of thiosulfate used. Then, using the stoichiometric ratios, you determine the moles of iodine.
Finally, these moles help deduce the mass and mass percent of thyroxine in the thyroid extract.