Oxidation of monosaccharides

Question

When 20 grams of glucose solution react with ammoniacal copper(I) chloride solution (Fehling’s reagent) 14.3 grams of Cu2O are precipitated. What is the mass percentage of this solution?

Answer 1

Not sure you have this exactly right. It is my impression that Fehling's reagent contains copper(II) chloride, or copper(II) sulfate, or some other salt of copper(II). The reaction with glucose depends on the reduction of Cu²⁺ to Cu¹⁺, so you wouldn't want to have copper(I) to begin with. If we go with that caveat, then the simplified reaction can be represented as

glucose + Fehlings (2Cu²⁺⁾ ==> oxidized product of glucose + Cu₂O (red ppt)

The bottom line is that 1 mole glucose produces 1 mole Cu₂O

moles Cu₂O produced = 14.3 g x 1 mole/143 g = 0.1 moles

moles glucose present = 0.1 moles

mass glucose present = 0.1 moles x 180 g/mole = 18 grams glucose present

mass percentage = 18 g/20 g (x100%) = 90%