Purpose

The purpose of this investigation is to explore the percent yield of the precipitate in the reaction

Introduction

For known amounts of reactants, theoretical amounts of products can be calculated in a chemical reaction or process. Calculated amounts of products are called theoretical yield. In these calculations, the limiting reactant is the limiting factor for the theoretical yields of all products.

However, in a reaction to prepare a compound, you may get less than the theoretical yield, because of incomplete reactions or loss. The amount recovered divided by the theoretical yield gives a percent yield (% yield) or actual yield.

The above reaction will produce a solid precipitate of  and an aqueous solution of.  These products will be separated by filtration and the precipitate will be collected in precipitate will be collected in filter paper.  It will dry overnight and then be weighed.  The mass obtained will then be compared to the theoretical yield, in order to determine the percentage yield of the reaction.

Materials

Filter Paper
Small Filter
2 Graduated Cylinders
Erlenmeyer flask
50mL Beaker
Scoopula
100mL Volumetric Flask
Weigh Boat
Mass Balance
Distilled Water

Procedure

  • Make 100mL of 0.1M  by placing 1.110g (±0.001g) of  into a 100mL volumetric flask, and dissolving it with distilled water.  Use only a little distilled water at first to dissolve the solid calcium chloride, then, fill the flask up to the etched line.
  • Make 100mL of 0.1M  by placing 3.81g (±0.001g) of  into a 100mL volumetric flask, and dissolving it with distilled water.  Use only a little distilled water at first to dissolve the solid sodium phosphate, then, fill the flask up to the etched line.
  • Prepare the filter paper by folding it into quarters, then opening one side in order to obtain a cone shape. Measure the mass of the dry filter paper.

  • Place filter paper into filter.  Then, place the filter into the Erlenmeyer flask.
  • Measure 10.00mL of
  • Measure 10.00mL of
  • Mix the 10mL of  and 10mL of  into a beaker.  Stir for approximately 5 seconds.
  • Pour this solution into the prepared filter.
  • Allow all of the liquid to drain into the Erlenmeyer flask.
  • Remove the filter paper from the filter.  Place it onto a glass plate and allow the filter paper with the precipitate to dry overnight in a closed environment.
  • Measure and record the mass of the dried filter paper with precipitate.
  • Repeat steps 3-11 five times.

Observations

Table 1: Qualitative Observations

Trial NumberVolume of

CaCl2(aq)

Volume of

Na3Po4 (aq)

Mass of Dry Filter Paper

(+/- .001g)

Mass of Filter Paper  + Precipitate

(+/- .001g)

110.00mL10.00mL0.77g0.95g
210.00mL10.00mL0.78g0.96g
310.00mL10.00mL0.77g0.98g
410.00mL10.00mL0.77g0.96g
510.00mL10.00mL0.77g0.95g

Mass of  used to make 100mL of 0.1M  solution CaCl2(ag) solution= 1.110 (+/- .001g)

Mass of  used to make 100mL of 0.1M  solution Na3PO4 solution= 3.810 (+/- 0.001g)

Calculations

Solutions Calculations to produce 100mL of 0.1M solutions

Theoretical Yield

The reactants were made in 100mL quantities.  These need to be scaled down in order to find the mass of each reactant present in 10mL extracts of each reactant.


Trial NumberMass of Dry Filter Paper (+/- .001g)Mass of Filter Paper  + Precipitate (+/- .001g)Mass of Ca3(PO4)2 (s) (+/- .002g)
10.770g0.950g0.180g
20.780g0.960g0.180g
30.770g0.980g0.210g
40.770g0.960g0.190g
50.770g0.950g0.180g

Conclusion & Evaluation

This percent yield of  from the reaction was found to be 183%.  This number implies that there was significant error in this experiment, as the amount of precipitate formed was approximately double the expected amount.

This skewed result could have been caused by certain external factors.  Firstly, because the precipitate was left to dry overnight, it could have accumulated dust, as well as reacted with other particles in the air to increase its weight.  Thus, the product is impure.  Precautions were made in order to avoid this, by placing the wet filter sheets into a closed environment.  However, it was not completely sealed, thus allowing for this error.  This factor can be eliminated in the future by placing the filter sheets in a completely sealed environment to dry overnight.

Also, this could have been due to impurities in the reactants.  If the solid calcium chloride and sodium phosphate were not completely decomposed in the distilled water this would cause the actual yield in the reaction to be higher than the theoretical yield of the reaction.   More specifically, if the limiting reactant was not completely decomposed, there would be excess of the other reagent remaining in the product.  This would add to the weight of the product, resulting in a high percentage yield.  In the future, the solutions should be mixed and allowed to sit for a few hours, to ensure that all of the solid particles are fully decomposed.

There is also uncertainty associated with the instruments used in this experiment.  If the things measured were a slightly more than should be this could have affected the concentrations of the solutions, and thus caused a larger reaction than expected.  To avoid this in the future, more accurate equipment should be used.  For example, pipettes should be used to measure the 10mL of solution instead of graduated cylinders.

Thus, this experiment found the percent yield of the reaction to be 183%.  This suggests that there were errors associated with the procedure as the amount of precipitate formed was almost double the expected value.  The errors discussed address the reasons as to why this occurred, and solutions were provided to avoid these errors in the future and increase the accuracy of the experiment.

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1 Comment on "Percentage Yield Lab Answers"

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T. MOSEKI
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what if the percentage yield is 13%?

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