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Percent Yield of a Precipitate

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The purpose of the experiment is to determine the percent yield of the precipitate; by performing double displacement reaction between solutions of two different compounds.


First of all when making a solution of two different compounds; there will many variables that can be considered during the experiment. However, the variables are controlled variables.

Controlled Variables

∙amount of water that will be dissolved with the compound (amount of water until the compound is dissolved)

∙temperature inside the laboratory (constant)

∙air pressure (constant)

∙mass of reactant used (mass is less that 3g)

When we consider a double displacement reaction; we will need to think about the solubility rules. In our experiment we consider the rule that “Most nitrate (NO3-) salts are soluble.” And “Most sulfate are soluble. Notable exceptions are BaSO4……” By this rules we can check the solutions that we will use during the experiment that whether it is soluble or not. Also, we can check after the reaction, which of the compound is the precipitate. Hence, before the experiment it is possible to predict whether the reaction will include precipitate by the solubility rules. In this case the precipitate will be BaSO4.


In this experiment the reaction equation will be:

CuSO4?H2O + Ba(NO3)2 →Ba SO4+ Cu(NO3)2

In Lab#6 that we have done in Dec 15 2003; we have found out that the mole for hydrate in copper (II) sulfate is 5. Thus, by this experiment I will continue to use the answer that I have found in Lab#6. So by using the knowledge from Lab#6 the reaction equation will be:

CuSO45H2O + Ba(NO3)2 →BaSO4+ Cu(NO3)2

We can measure the mass of copper (Ⅱ) sulfate and the mass of the barium nitrate. Thus, we can calculate the theoretical yield of barium sulfate. By this, it is possible to find out the percent yield of barium sulfate; if we found the experimental yield.

The percent yield of barium sulfate= experimental yield of barium sulfate

theoretical yield of barium sulfate


filter paper

copper(Ⅱ) sulfate

barium nitrate

glass stirring rod

beaker (250ml)

two volumetric flasks (50ml)


plastic tray


i. Take out the solutes of Cu(II)SO4 and Ba(NO3)2 and measure them approximately


ii. Insert each of the solution inside a volumetric flask and mix these solutes into water and stir it with a glass stirring rod until it dissolves.

iii.After it dissolved; mix the two solutions into a beaker.

iv.Make sure there is a precipitate inside the beaker.

v.Before you filter the solution that is made in the beaker; measure the mass of the filter paper.

vi.Now filter the solution contained inside the beaker.

vii.Leave it to filter for a while.

viii.After it has filtered; place it in the oven for the night.

ix.After it has dried up; measure the mass of filter paper.

Data table

Table1- the change of the state and the mass of Cu(II)SO4 and Ba(NO3)2

State (before inserted in water)State(after inserted in water)Mass (+0.002g)

Cu (II)SO4

Light blue, white crystal, Blue, transparent2.945g

Ba(NO3)2White, powderedWhite, translucent3.000g

Table 2- the change of the mass and the state of BaSo4 (the precipitate)

StateMass with filter paper

BaSo4Light blue solid, powered

With no smell5.273g

Table 3- the change of the mass of filter paper


Filter Paper2.495g

Data Analysis

First of all, since we have found out the mass of Cu(II)SO4 and Ba(NO3)2; we will need to find out the mole for these compound.

2.945g Cu(II)SO4∙ 5H2O 1 mol Cu(II)SO4∙ 5H2O

249.71g Cu(II)SO4∙ 5H2O

= 0.01179……mol Cu(II)SO4∙ 5H2O(required)

3.000g Ba(NO3)2 1 mol Ba(NO3)2

261.36g Ba(NO3)2

= 0.01147……mol Ba(NO3)2 (required)

Next, we will need to find out which of the compound is the limited reactant.

To do that, we will need to multiple the mole that we got in Cu(II)SO4∙ 5H2O with the mole ratio between Cu(II)SO4∙ 5H2O and Ba(NO3)2.

0.01179……mol Cu(II)SO4∙ 5H2O 1 mol Ba(NO3)2

1 mol Cu(II)SO4∙ 5H2O

= 0.01179……mol Ba(NO3)2 (needed)

If we compare the numbers of moles that are required and needed for Ba(NO3)2 ; we do not have enough. Hence, Ba(NO3)2 is the limited reactant; and, Cu(II)SO4∙ 5H2O is in excess. Now, since we know that Ba(NO3)2 is the limited reactant, we can find out the theoretical yield of the precipitate; BaSO4.

0.01147……mol Ba(NO3)2 1 mol BaSO4 233.4g BaSO4

1 mol Ba(NO3)21 mol BaSO4

=2.67906…… g BaSO4

To finally find out the percent yield of BaSO4; we will need to calculate the experimental yield of BaSO4. To do that we will need to find out the difference between the mass of the filter paper and filter paper + BaSO4.

5.273g – 2.495g


Finally, we can find the percent yield of BaSO4 by dividing experimental yield with theoretical yield.

2.778g BaSO4 100

2.67906…… g BaSO4

= 103.69295……%

≈ 103.7 %


As said in the purpose; we have done this experiment to find out the percent yield of the precipitate. By the calculation that was done in the data analysis; the percent yield of BaSO4 is 103.7%.

Also, we can find out the percent error in this experiment.

Percent error = |accepted value – experimental value| / accepted value 100

= |2.67906……g – 2.778g| / 2.67906……g 100

= 3.693%


The percent yield in this experiment is 103.7%, and the percent error is 3.693% that relatively low; thus, it is a pretty accurate percentage. However, since the percent yield and the percent error is not completely perfect there are some errors that have to be considered.

During the transfer process of the two solutions into a beaker to mix it; some of the solution might have been left inside the container. This might have caused some error.

During the stirring process of the two solutions inside the beaker with glass stirring rod; some liquid were stuck on the rod. Because some of the liquid was taken out by the rod; it could of caused some error.

During the filtration; it is possible that I did not rinse off enough of the precipitate and some of the compound was left behind that should have been filtered. This could have caused some errors in the mass of the precipitate.

When we transferred the compound from the plastic tray that we used to measure the compound; some of the compound might have been left on the plastic tray. This might have caused a difference in the masses of the compound.

When we finally mixed the two solutions into a beaker, some of the compound that was inside a volumetric flask had some reminder on the side of the flask. This might of caused some errors in the reaction.

When we mixed the two solutions; immediately we filtered the solution. It is possible that the reaction was not finished, and this might have caused same error.

It is possible that when we poured water to the compounds and stirred it; it might have not been fully dissolved.

When the ionic compounds were weighted, it was exposed to the air. This might have reacted with water that exists in the atmosphere; and, for a hydrate. Consequently, this would increase the weight.

Take your time to do the transfer process and the stirring process carefully as possible during the lab.

Make sure to remember to rinse off the solution or the compound that might have been left over in the filter funnel, plastic tray or in the volumetric flask.

Also, I should have rinsed off the precipitate that was on the filter paper more carefully during the experiment. So there was only the precipitate that was on the filter paper.

We will need more time to actually be sure that the reaction is finished.

Always add the excess water during the dissolving process and the stirring process; because, we donot know how much water is necessary to dissolve the compound.

Shorten the time of exposing the ionic compounds; to avoid from the compounds reacting with water to form hydrate.

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