Open-chain fructose, $\mathrm{CH}{}_2\mathrm{OH}-\mathrm{CO}-\mathrm{CH}(\mathrm{OH})-\mathrm{CH}(\mathrm{OH})-\mathrm{CH}(\mathrm{OH})-\mathrm{CH}{}_2\mathrm{OH}$, is oxidised completely by excess $\mathrm{HIO}{}_4$. Count every cleavage, including the cleavage of the alpha-hydroxy acid formed from the ketone carbon. The number of moles of $\mathrm{HIO}{}_4$ consumed and of $\mathrm{CO}{}_2$ formed per mole of fructose are respectively

The chain has five C-C bonds, each between carbons carrying oxygen functions, and each is cleaved once, so 5 mol HIO4 are used. Cleavage of C1-C2 (alpha-hydroxy ketone) gives HCHO and turns C2 into a carboxyl group, and this alpha-hydroxy acid is then cleaved at C2-C3, releasing C2 as $\mathrm{CO}{}_2$. C1 and C6 give HCHO and C3, C4, C5 give HCOOH, so 1 mol of $\mathrm{CO}{}_2$ is formed.