The reaction between fructose and the periodate ion is described by the following equation.

C₆H₁₂O₆(aq) + 5 IO₄^-(aq) $\rightarrow$ 5 IO₃^-(aq) + 5 HCO₂H(aq) + H₂CO(aq)
Calculate the molarity of a fructose solution if 22.35 mL of 1.050 M IO₄^- is required to consume 12.56 mL of this solution.

Question

The reaction between fructose and the periodate ion is described by the following equation.

C₆H₁₂O₆(aq) + 5 IO₄^-(aq)

\rightarrow

5 IO₃^-(aq) + 5 HCO₂H(aq) + H₂CO(aq)
Calculate the molarity of a fructose solution if 22.35 mL of 1.050 M IO₄^- is required to consume 12.56 mL of this solution.

Quizplus · Accepted Answer

The stoichiometry of the reaction shows that 1 mole of fructose reacts with 5 moles of periodate ion. Using the molarity and volume of the periodate solution, we can calculate the moles of periodate: $1.050 \, \text{M} \times 0.02235 \, \text{L} = 0.0234675 \, \text{moles}$. Since 5 moles of periodate react with 1 mole of fructose, the moles of fructose are $0.0234675 \, \text{moles} / 5 = 0.0046935 \, \text{moles}$. The molarity of the fructose solution is then $0.0046935 \, \text{moles} / 0.01256 \, \text{L} = 0.3735 \, \text{M}$, which falls between 0.25 M and 0.5 M.

The Reaction Between Fructose and the Periodate Ion Is Described →\rightarrow→

The Reaction Between Fructose and the Periodate Ion Is Described $\rightarrow$