Question

I did an experiment known as "Identification of an unknown aldehyde and unknown ketone."

First I added 0.2 mL aldehyde, 1 mL ethanol, 49 drops of water, 2 more drops ethanol, and 4 drops of phenylhydrazone in a flask. I put the solution in ice and crystals formed. My end product weighed 0.105 g. Please calculate the percent yield of the phenylhydrazone reaction assuming the aldehyde is the limiting reagent.

Aldehyde= 4-methylbenzaldehyde (1.019 g/ mL = density)

Answer 1

The reaction of formation of Phenylhydrazone derivative of given aldehyde is as follows:

From the reaction it is clear that 1 mol aldehyde + 1 mole phenylhydrazine gives 1 mole of hydrazone derivative.

In terms of molar mass(MW) we can say that, 120 g of an aldehyde gives 210 g of hydrazone derivative.

i.e. 120 g aldehyde 210 g of Hydrazone derivative.

In our experiment we used 0.2 mL of aldehyde whose density is 1.019 g/mL.

We know that, density = mass / volume

mass = density x volume

0.2 mL is the volume of aldehyde used which has density 1.019 g/mL. Put these values in above equation and find out the mass of aldehyde used

mass of aldehyde = density of aldehyde x volume of aldehyde

mass of aldehyde = 1.019 x 0.2

mass of aldehyde = 0.2038 g.

i.e we have used 0.2038 g of aldehyde.

Recall the equation (1),

120 g of aldehyde 210 g of phenylhydrazone derivative.

Let us find out the expected(Theoretically calculated weight) weight of phenylhydrazone from 0.2038 g of a starting aldehyde

120 g aldehyde 210 g of phenylhydrazone

0.2038 g aldehyde (say) 'Mcal' phenylhydrazone

So, Mcal = (210 x 0.2038) / 120............(Just a cross multiplication method used)

Mcal = 0.3567 g phenylhydrazone

But practically observed yield of hydrazone (say) Mobs = 0.105 g phenylhydrazone

Now, %yield = (Mobs / Mcal ) x 100

So, % yield of phenylhydrazone derivative = (0.105 / 0.3567) x 100

% yield of phenylhydrazone derivative = 29.44 %

The % yield of phenylhydrazone in the raection assuming aldehyde as limiting reagent is 29.44 %.