Question

Please help! (These are the instructions I had)

1)         Preparation of K₂[Cu(C₂O₄)₂] 2H₂O

CuSO₄•5H₂O + 2K₂C₂O₄•H₂O → K₂[Cu(C₂O₄)₂]•2H₂O + K₂SO₄ + 5H₂O

Heat a solution of 2.5g of potassium oxalate monohydrate in 12.5mL of water contained in a 50mL beaker (approx. 90°C). Heat 1.55g of copper sulfate pentahydrate in 3mL of water to about 90°C and add it rapidly, with vigorous stirring to the hot potassium oxalate solution. Set on the bench and allow to cool to room temperature then cool the mixture by setting the beaker in an ice bath for 15 to 30min.

Weigh a filter paper then place it in a Buchner funnel atop a filter flask. Wet it slightly with distilled water with the vacuum on, then suction-filter the blue crystals. Try to keep the crystals in the middle of the filter paper and use a rubber scraper to remove the remaining crystals from the beaker. Add ~2mL of cold water to your beaker to try to remove the remaining crystals and wash the crystals on the filter paper. Then wash your crystals successively with 2mL of absolute ethanol, and finally 2mL of acetone. Remove the filter paper with a forceps, place on a watch glass, and allow to air dry. Weigh the air-dried material and store it in a vial. You should obtain about 1.75g of product.

2)         Preparation of K₃[Al(C₂O₄)₃] 3H₂O

2 Al + 6 KOH + 6 H₂C₂O₄•2H₂O → 2 K₃[Al(C₂O₄)₃]•3H₂O + 3 H₂ + 12 H₂O

Place .25g of aluminum powder in a 50mL beaker and cover with 2.5mL of hot water. Add 6mL of 5M KOH solution in small portions to regulate the vigorous evolution of hydrogen. If you have any residual metal after the gas is evolved, heat the liquid to almost boiling in order to dissolve any residual metal. Maintain the heating and add a solution of 3.75g of oxalic acid in 27mL of water in small portions. During the neutralization, hydrated alumina will precipitate, but it will re-dissolve at the end of the addition after gentle boiling. You may need to gently boil for 10 minutes. Add 12 mL of ethanol and cool on the bench until near room temperature, then cool the solution in an ice bath. If oily material separates, stir the solution and scratch the sides of the beaker with your glass rod to induce crystallization. Suction-filter the product using a Buchner funnel and suction flask and wash with a 2mL portion of absolute ethanol. Air dry the product, weigh it, and store in a stoppered bottle. You should obtain about 2.75g of product.

My initial mass for the K₂[Cu(C₂O₄)₂] 2H₂O paper was .52 g and the final was 2.21 g

My initial mass for the K₃[Al(C₂O₄)₃] 3H₂O paper was .50 g and the final was .99 g

Please show calculations for theoretical and percent yield

Answer 1

To solve this question we need to make limiting reactant calculations, in your experiments 1 of the reactants has a bigger ammount than the theoretically needed

For the K₂[Cu(C₂O₄)₂] 2H₂O (Molecular weight is 353.75 )

The statement says you need 1 mole of CuSO4•5H2O (Molecular weight 249.55) and 2 moles of K₂C₂O₄•H₂O (Molecular weight 184.2).

In your experiment you have:

2.5 grams of K₂C₂O₄•H₂O, we must calculate the number of moles with the formula moles = weight / molecular weight

moles = 2.5 / 184.2 = 0.0135722 moles

and you have 1.55g of CuSO4•5H2O, therefore we have 0.0062111 moles

Now we can find the limiting reactant.

1 mole of CuSO4•5H2O reacts with 2 moles of K₂C₂O₄•H₂O

we have 0.006211 moles of Cupper compound. To find the ammount we need of oxalate we make the next analysis:

$\frac{2molesoxalate}{1moleCuppercompound} * 0.006211 moles of Cupper compound$

We need 0.01242 moles of Oxalate to consume totally the cupper compound, we can see that we have 0.01357 moles, more than the ammount needed.

We can say that the CuSO4•5H2O is the limiting reactant.

The original reaction is:

CuSO4•5H2O + K₂C₂O₄•H₂O $\rightarrow$ K₂[Cu(C₂O₄)₂] 2H₂O

We can see that 1 moles of Cupper compound produces 1 mole of K₂[Cu(C₂O₄)₂] 2H₂O according to the stoichiometry. Since the copper compound is the limiting reactant we can say that all the moles of copper compound will produce the compound we are looking for. So if the reaction is completed totally we should get

0.006211 moles of K₂[Cu(C₂O₄)₂] 2H₂O

to find the weight we must apply the formula weight = moles * molecular weight

Weight (grams) = 0.006211 * 353.75 = 2.1971 grams

This is the theoretical yield, this is the ammount of product that you would get if you had perfect experimental conditions. In real life you hardly would get this because some product remains in the flask or reaction is not completed in 100%.

From the results you report in your experiment

The product you got is the final weight of the paper minus the initial weight of paper

2.21 grams - 0.52 grams = 1.69 grams

That´s the product you got:

You got 1.69 grams out of 2.19 grams possible.

Percent yield = Actual yield / Theoretical yield * 100

Percent yield = 1.69 / 2.19 = 0.7717 *100 = 77.17 %

For the Second Reaction.

We have to follow the same steps that before.

We have 0.25 gr of Al (MW 27) or 0.009259 moles of Al

We have 6 ml of 5M KOH, Molarity is moles / Volume. moles is Molarity * Volume = 0.006 * 5 = 0.03 moles

We also have 3.75 grams of oxalyc acid H₂C₂O₄•2H₂O (MW 126) , or 0.02976 moles

Now we have to make calculations to find the limiting reactant. Let´s make calculations for Al

2 Al + 6KOH + 6 H₂C₂O₄•2H₂O $\rightarrow$ 2 K₃[Al(C₂O₄)₃]•3H₂O + 3H₂ + 12 H₂O

2 moles of Al need 6 moles of KOH

So we have 0.009259 moles of Al, to find the moles of KOH we need to react this compound we apply the next equation:

$\frac{6molesKOH}{2molesAl} * 0.009259 moles of Al$

If we solve this we find that we need 0.02777 moles of KOH to make the Al to react completely

We have 0.03 moles of KOH, we can see that KOH is in excess

From the reaction we can see that 2 moles of Al need 6 moles H₂C₂O₄•2H₂O (similar proportions like KOH and Al)

So we repeat the operation:

$\frac{6molesOxalycAcid}{2molesAl} * 0.009259 moles of Al$

We need 0.0277 moles of oxalyc acid, we really have 0.02976 moles. Again we see we have more moles than we theoretically need.

Because of this we can safely say that the Al is the limiting reactant.

From the original reaction:

2 Al + 6KOH + 6 H₂C₂O₄•2H₂O $\rightarrow$ 2 K₃[Al(C₂O₄)₃]•3H₂O + 3H₂ + 12 H₂O

We can see that 2 moles of Al produces 2 moles of K₃[Al(C₂O₄)₃]•3H₂O, since the coefficients are the same it is true to say that 1 mole of Al produce 1 mole of K₃[Al(C₂O₄)₃]•3H₂O

So we have 0.009259 moles of Al, according to stoichiometry we should be able to produce 0.009259 moles of K₃[Al(C₂O₄)₃]•3H₂O (the molecular weight of this compound is 462.28)

The Theoretical weight you should get is 0.009259 * 462.28 = 4.28 grams

That is your theoretical yield.

From your experiment the product you got is:

0.99 - 0.5 = 0.49 grams

Percent yield = theoretical yield / actual yield = 0.49 / 4.28 grams = 0.1144 *100 = 11.44 %

From this last experiment you got a very low percent yield, I guess that some aluminum powder didn´t dissolve correctly or some stayed in the beaker unscratched.