Determining the Effect of Different Sugar Concentrations on Benedict's Solution
PLANNING:
Hypothesis and prediction:
My hypothesis is that the sugar solution with the highest
concentration will turn the precipitate the darkest colour. This will
be the 10% sugar solution. This is because the more amount of sugar it
contains the more it will reduce the Copper. It will be a much darker
precipitate compared to the other two.
Background theory:
Benedict's solution is an aqueous solution of Copper (II) Sulphate,
Sodium carbonate and Sodium citrate. It is an alkaline solution used
to test for the presence of aldehyde groups (RCHO). The reducing sugar
(Glucose) reduces the copper (II) Sulphate to Copper (I) oxide. The
colour of the precipitate varies dependent on the strength of the
reducing sugar present. The colour can vary from blue to red-brick:
indicating a high concentration of sugar. Glucose contains an aldehyde
group, so it is able to reduce the Benedict's solution and form a
precipitate. An aldehyde contains the general formula; RCHO, where the
R represents Hydrogen. They are formed from partial oxidation of
primary alcohols. An aldeyde is formed due to cabonyl groups; these
contain an Oxygen atom joined by a double bond to carbon. If the
carbonyl is joined to a hydrogen atom, then the compound is an
aldehyde. Glucose is a monosaccharide. Monosaccharides can take the
form of linear or ring structures. The carbonyl on the carbon 1
supplies the electron which joins the carbon 1 to carbon 5. When this
bond breaks, there are extra electrons which are then used to reduce
other molecules. The heating thus breaks these bonds, so then the free
electrons reduce the Copper (II) Sulphate to Copper (I) oxide in the
Benedict's solution. The reason why the Benedict's solution is readily
reduced is because it has a high PH, hence alkaline, whereas the
Carboxyl group in the Glucose gives it acidic properties. It splits
open the ring structure, consequently releasing the electrons which
are accepted by the Copper. The Copper (II) ions act as a mild
If this experiment were designed to determine the amount of Fructose in a solution, describe what, if anything, would need to change in the reaction? Explain why there would or would not need to be changes. (5
CL-, as the ions of H+ and OH- react to form H2O. These spectator ions
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