By Aneeqa H., Ariel T., Carrie F., Fabiha N.

  • The PowerPoint presentation for this experiment can be downloaded here.

Introduction




The objective of this experiment is to determine the amount of chloride ions present in a given solution of Sodium Chloride using Silver Nitrate as a reagent. The solubility of Silver Chloride is very low in water at room temperature. The obtained precipitate can be filtered and washed free of other impurities and the mass of chloride present can be estimated gravimetrically (by determining mass).

The chloride ion is formed when the element Chlorine picks up one electron to form a negatively charged ion (Cl−). A chloride is a chemical compound that results from the combination of chlorine gas with a metal. An example is Sodium Chloride (NaCl).

Figure 1 - 3D image of the atomic structure of Silver Chloride (AgCl), a chloride.
Figure 1 - 3D image of the atomic structure of Silver Chloride (AgCl), a chloride.

Sodium Chloride (NaCl - Also known as Table Salt) is a compound that is widely used as a condiment and a food preservative. It is mass-produced through the evapouration of seawater and other sources such as brine wells. Sodium Chloride is also used in the production of the element Chlorine through the electrolysis of NaCl dissolved in H2O. This process is represented by the following equation:

2 NaCl + H2O → Cl2 + H2 + 2 NaOH
Silver Nitrate (AgNO3 - Also known as Lunar Caustic) is a soluble compound which is produced by dissolving Ag in a solution of Nitric Acid and evapourating the solution. This process is represented by the following equation:

4 Ag (s) + 6 HNO3 (aq) → 4 AgNO3 (aq) + 3 H2O (l) + NO (g) + NO2 (g)

Gravimetric analysis is the set of procedures to determine the quantity of a substance present in the mass of a solid.
This usually involes seven steps:

1. Drying and measuring the masses of samples to be analysed
2. Dissolving the sample in distilled water.
3. Precipitating the substance by adding a reagent.
4. Separating the precipitate from the solution by filtration.
5. Washing the precipitate free of impurities.
6. Drying the precipitate to obtain mass.
7. From the mass and known composition of the precipitate, the amount of the original ion can be determined.
8. For this experiment, 0.2 g of NaCl and excess amounts of AgNO3 will be used:

The double displacement reaction taking place in the experiment is represented by the following chemical equation:
NaCl + AgNO3 → AgCl + NaNO3

Mole to Mole ratio 1 : 1 : 1 : 1

The theoretical yield of Cl ions is as follows:

Molar Mass of NaCl = 58.44 g/mol
Given NaCl = 0.2 g

Molar mass of AgNO3 = 143.318 g/mol
Let x = expected yield of AgCl
external image 1-1.jpg


AgCl:
external image 1-2.jpg

% by mass of Cl: 0.25%

Thus, the amount of Cl in 0.49g of AgCl is:


(0.25) (0.49) =
0.1212 g

To gravimetrically measure the amount of chloride ions present in 0.1212 g of Cl, we must first determine how many moles of Cl is in 0.1212 g. The formula for this is:
external image 1-3.jpg

∴ There are 0.00341897 moles of chloride ions.

Now that the number of moles of chloride ions has been determined, to find the exact number of chloride ions present, we simply multiply the number of moles with Avogadro's Number:


(0.00341897) (6x10^23) = 2.0589 x 10^21


Therefore, in this experiment, the number of chloride ions hypothesized to be present is 2.0589 x 10^21.


Problem: What is the amount of chloride ions present in 0.2 g of Sodium Chloride (NaCl)?

Hypothesis: The number of chloride ions present in 0.2 g of NaCl is approximately 2.06 * 10^21.


Materials




The following materials were used in the experiment.

  • Erlenmeyer Flask (2)
  • Beaker (1)
  • Funnel (1)
  • Ashless Filter Paper (1)
  • Paper Clips (4)
  • Balance
  • 0.2 g of Sodium Chloride (NaCl)
  • 3 g of Silver Nitrate (aqueous) (AgNO3)
  • Distilled Water
  • Dropper
  • Test Tubes (2)
  • Bunsen Burner
  • Crucible and lid
  • Crucible tongs
  • Retort Stand
  • Ring Clamp
  • Clay Triangle
  • Safety Goggles
  • Spatula
  • Stirring Rod
  • Graduated Cylinder


Procedure




Formation of the Precipitate

1. The mass of the filter paper was measured using the balance.
2. Sodium Chloride (NaCl) was carefully added onto the filter paper using a spatula until the difference in mass between the filter paper with NaCl and the filter paper alone was equal to the desired mass of NaCl (0.02 g).
3. The correct amount of NaCl was put into an Erlenmeyer flask.
4. Distilled water was added to NaCl to fill up half of the test tube. The test tube was shaken to help NaCl dissolve faster in distilled water. This resulted in an aqueous solution of NaCl.
5. The mass of the graduated cylinder was measured and recorded.
6. Aqueous Silver Nitrate (AgNO3) was dropped into the graduated cylinder using a dropper, until the difference between the mass of the graduated cylinder with AgNO3 (aq) and the graduated cylinder was equal to the desired amount of AgNO3 (aq) (0.3 g).

*Note that if the mass of AgNO3 (aq) exceeds the desired amount, it will not have an effect on the chemical reaction of the experiment since AgNO3 (aq) is not the limiting reagent.

7. AgNO3 (aq) was poured into a test tube.
8. AgNO3 (aq) was poured onto NaCl in the Erlenmeyer flask. A white precipitate (AgCl) was formed.
9. The precipitate was left until it completely settled down.
10. A few drops of AgNO3 (aq) was dropped into the solution to see if further precipitation occurred.

Filtration of the Solution Containing the Precipitate

11. The filter paper was attached on to the top of a funnel using paper clips.
12. The funnel with the filter paper was placed on top of an Erlenmeyer flask.
13. The solution containing the precipitate was poured onto the filter paper. The precipitate remained on the filter paper as the aqueous solution of NaNO3 passed through the paper and dropped into the Erlenmeyer flask.
14. Distilled water was periodically poured onto the filter paper so that all the NaNO3 solution passed though the filter paper, leaving only AgCl.
15. After all of the NaNO3 solution drained through the paper, some final drops were collected in a test tube before they fell in the Erlenmeyer flask.
16. A few drops of NaCl (aq) were added to the solution using a dropper to sense the presence of Ag + ions.
17. The AgCl precipitate was left until all the NaNO3 drained through.

DSC02990.JPG
Figure 2 - Precipitate (AgCl) on the filter paper after filtration


Measurement of the Mass of AgCl by Drying Filter Paper

18. After filtration, the filter paper with the precipitate was left until it dried completely.
19. The mass of dried filter paper with the precipitate was measured and recorded. The difference between the initial mass of the filter paper and the mass of the filter paper with the precipitate is the mass of AgCl.

Measurement of Mass of AgCl by Burning the Ashless Filter Paper

20. The retort stand, ring clamp, clay triangle and the Bunsen burner were set up.
21. The mass of the crucible and lid were measured and recorded.
22. The filter paper was carefully folded with the precipitate inside, and placed in the crucible.
23. The crucible was placed on the clay triangle, and the lid was set slightly off-centre.
24. The Bunsen burner was lit to heat the crucible with a gentle flame.
25. Gradually, the intensity of the flame was increased. This was continued until no more of the ashless filter paper was left.
26. The lid was heated above the flame using crucible tongs.
27. The crucible, lid and its contents were allowed to cool.

The mass of the crucible with contents (AgCl ) and lid was measured. The difference of this mass and the initial mass of the crucible and lid is the mass of AgCl.



Safety Precautions




1. Wear safety goggles throughout the entire experiment.
2. Ensure that all the containers are clean and free of unwanted substances.
3. Long hair should be tied back and loose sleeves should be rolled up.
4. Dispose of all chemical waste appropriately.
5. Make sure the lab bench is clean and empty before beginning the lab.
6. Avoid physical contact with Silver Nitrate as it is both toxic and corrosive.
7. Avoiding excess physical contact with and breathing in power chemicals such as Silver Chloride.
8. The Bunsen burner flame should not be left attended. It should always be shut off when not in use.
9. The retort stand should be firmly attached to its base. All clamps being used should be securely fastened to the ring stand.
10. The hot crucible must be handled with care and should only be moved using tongs


Observations




Quantitative Observations

Data recorded from the experiment is displayed in the table below:


Object (s)
Mass (g)
Calculation
Formation of the Precipitate
Filter paper
1.04 g

Filter paper with NaCl
1.24 g

NaCl
0.2 g
(1.24 g - 1.0 g)
Graduated cylinder
22.48 g

Graduated cylinder with AgNO3
27.08 g

AgNO3
3.4 g
(27.48 g - 22.48 g)
Measurement of the mass of AgCl by drying the filter paper
Filter paper with AgCl
1.43 g

AgCl
0.39 g
(1.43 g - 1.04 g)
Measurement of the mass of AgCl by burning the ashless filter paper
Crucible and lid
32.13 g

Crucible, lid and contents (AgCl)
32.64 g

AgCl
0.51 g
(32.64 g - 32.13 g)

Qualitative Observations

Aqueous Silver Nitrate (AgNO3) is a clear solution. Aqueous Sodium Chloride (NaCl) is also a clear solution.

When aqueous Silver Nitrate (AgNO3) was added to aqueous Sodium Chloride (NaCl), a precipitate was formed immediately. The precipitate was a powder and milky white in colour.

The precipitate was insoluble, which confirms our previous knowledge that AgCl has a low solubility in water.

When the crucible was heated with the contents inside, the ashless filter paper was burnt. As it burnt, the filter paper turned to carbon dioxide and was released into the air. Gray coloured contents were left behind in the crucible.


Calculations




Already known information:
(Please refer to the calculations done in Introduction)

Molar mass of Cl = 35.45 g/mol

Percentage Composition of AgCl
75% Ag
25% Cl

Calculation of Amount of Chloride Ions Present in 0.2 g of NaCl based on Results Obtained by Drying Filter Paper
Mass of AgCl = 0.39 g (This is obtained from the difference of the mass of the filter paper with AgCl and the filter paper.)
(1.43 g - 1.04 g) = 0.39 g
Mass of the Cl Ions Present in the Sample of AgCl
= 25 % of 0.39 g
= 0.25 * 0.39 g
= 0.0975 g
Moles of Cl ions
To determine the number of moles of Cl ions present in the sample, the given mass of Cl must be divided by the molar mass of Cl.

= Given Mass of Cl
Molar mass of Cl
= 0.0975 g
35.45 g/mol

= 0.00275 mol
Number of Cl ions
To determine the number of Cl ions present in the sample, the number of moles of Cl ions must be multiplied by Avogadro’s number.

= (moles of Cl) * (Avogadro’s number)
= (0.00275 mol) * (6.022 *10^23 atoms/mol)
= 0.01656 * 10^23 atoms
= 1.656 * 10^21 atoms

The number of Chloride ions present in 0.2 g of NaCl is 1.656 * 10^21 (based on the results obtained by drying the filter paper).

Calculation of the Amount of Cl Ions present in 0.2 g of NaCl Based on the Results Obtained by Burning Ashless Filter Paper

Mass of AgCl = 0.51 g (This is obtained from the difference of the mass of the crucible with lid and contents and crucible with lid.)
= (32.64 g - 32.13 g)

Mass of the chloride ions present in the sample of AgCl

= 25 % of 0.51 g
= 0.25 * 0.51 g
= 0.1275 g

Moles of Cl Ions
To determine the number of moles of Cl ions present in the sample, the given mass of Cl must be divided by the molar mass of Cl.

= Given Mass of Cl
Molar mass of Cl
= 0.1275 g
35.45 g/mol

= 0.00360 g


Number of Cl Ions
To determine the number of Cl ions present in the sample, the number of moles of Cl ions must be multiplied by Avogadro’s number.

= (moles of Cl) * (Avogadro’s number)
= (0.00360 mol) * (6.022 *10^23 atoms/mol)
= 0.02167 * 10^23 atoms
= 2.167 * 10^21 atoms

The number of Chloride ions present in 0.2 g of NaCl is 2.167 * 10^21 (based on the results obtained by burning the ashless filter paper)

Conclusion




Two results were obtained in this experiment. The first was obtained from taking the mass of Silver Chloride (AgCl) by drying the filter paper. The second was obtained from taking the mass of AgCl by burning the ashless filter paper.

The amount of Chloride ions present in 0.2 g of Sodium Chloride is 1.656 * 10^21 (based on the results obtained from drying the filter paper).

The amount of Chloride ions present in 0.2 g of Sodium Chloride is 2.167 * 10^21 (based on the results obtained from burning the ashless filter paper).

The amount of chloride ions present in NaCl (reactant) will be the same as AgCl (product) due to the Law of Conservation of Mass. No chloride ions were lost or gained during the reaction, theoretically. This only stands true if all of the NaCl is used up in the reaction, making it the limiting reagent. Thus AgNO3, the other reactant, was the excess reagent. The double displacement reaction that took place between NaCl and AgNO3, resulted in the products: AgCl and NaNO3. AgCl has a low solubility in water at room temperature. It was used to gravimetrically estimate the amount of chloride ions present in the initial NaCl.

Theoretically, the number of Chloride ions present in 0.2 g of NaCl is approximately 2.06 *10^21. The results obtained were close to the predicted number of Chlorine ions, although they were not exact due to experimental error, proving that the gravimetric methods used were correct. These errors will be explained in detail in “Sources of Error”.


Discussion




1. How did the Law of Conservation of Mass help predict the amount of Cl in AgCl?

The Law of Conservation of Mass states that the mass of the reactants will be the mass of the products in any given chemical reaction. Mass is neither gained nor lost.
In the experiment, 0.2 g of NaCl was used. Through simple mathematics, we can determine the mass of chloride ions present in that amount:

Let x = the ratio between the mass of NaCl given and the molar mass of NaCl.
external image 1-4.jpg

So, to find the mass of Cl present in 0.2g of NaCl, we simply multiply the ratio by the molar mass of Cl:

external image 1-5.jpg
With the Law of Conservation of Mass, it can be concluded that no mass of Cl would be lost in the chemical reaction. As such, the mass of Cl in AgCl would also be 0.1212 g.

2. Why was it necessary to wash off all the impurities from the AgCl precipitate.

It was necessary to wash off all the impurities form the AgCl precipitate because otherwise the mass of the AgCl would be compromised, and the end number of chloride ions present would be inaccurate.

3. Calculate the percentage yield of AgCl. Why was the yield not 100%?

The percentage yield of AgCl can be calculated with the following equation:
external image 1-3.jpg

Therefore, the percentage yield was 80%.

It is theoretically impossible to get a yield of 100%. This is because there will always be possible sources of error (see: "Sources of Error").

external image 1-4.jpg
Therefore, the percentage yield was 104%.

It is impossible to get a percentage yield of 104% as this states that mass has been added during the process of the reaction, which contradicts the Law of Conservation of Mass. There will always be possible sources of error causing the percentage yield to be more or less than 100% (see: "Sources of Error").

4. Why does NaCl dissolve in water?

The reason that NaCl dissolves in water can be explained using the concepts of lattice energy and hydration energy. Lattice energy is the energy that is given out when a bond is formed so the structure is stable. The more stable a structure is, the more lattice energy is released during the bonding.

NaCl is a crystal structure. Crystal structures are formed such that each ion is surrounded by a cloud of ions of the opposite charge, making it extremely stable, and requiring greater lattice energy to be released. Crystals tend to have extremely high melting points as the compound will always try to be at its most stable state, and the only way to make it change is if the new state offered is more stable.


However, NaCl dissolves readily in H2O. As no outside energy is added, it's determined that the energy to dissolve it must come from the H2O.
H2O molecules have a dipole moment, meaning that each end of the molecule has an opposite charge. When NaCl is added to water, hydration takes place, which is when an ion is surrounded by H2O molecules.


Figure 3 - An example of a hydration shell. A Sodium ion is surrounded by H2O molecules.
Figure 3 - An example of a hydration shell. A Sodium ion is surrounded by H2O molecules.

As NaCl is an ionic compound, Na has a highly positive charge, while Cl has a highly negative charge. However, for an ion to be completely stable, it needs to be surrounded by a cloud of ions of the opposite charge, and when NaCl is in H2O, the H2O molecules provide that stability.
As this stability is greater than the stability that would be provided from a since NaCl molecule, the ions separate and go through hydration. The Na ions are surrounded be the negative end of the H2O molecule, where the Cl ions are surrounded by the positive one. This is known as the hydration shell. The energy used to separate and further stabilize the Na and Cl ions is known as hydration energy. This energy is greater than the lattice energy released when forming NaCl which is why it was able to separate the molecule into two ions.

Therefore, NaCl dissolves in water because that provides the Na and Cl ions with more stability.

5. Why doesn't AgCl dissolve in water?

AgCl doesn't dissolve in H2O because the lattice energy released during the formation of the compound is much stronger than the hydration energy that the hydration shell of H2O molecules would provide should the Ag and Cl ions separate.

The AgCl molecule is more stable as a solid precipitate than it would be dissolved and hydrated.

6. Why doesn’t an excess amount of Silver Nitrate effect the chemical reaction in the experiment?

This experiment is to determine the amount of chloride ions present in a given solution of NaCl, using AgNO3 as a reagent. According to the law of conservation of mass, the amount of chloride ions present in the reactants will be the same that of the products. Thus, in the double displacement reaction that is taking place in the experiment, all of the NaCl must be used up, making it the limiting reagent. Since it is the limiting reagent, it limits the amount of the other reactant (AgNO3) used, and it also limits the amount of products.

Therefore, if there’s an excess of AgNO3 in the reaction, it will remain as it is and will not alter the results of the experiment.

7. What are some properties of AgCl?

Silver Chloride (AgCl) is a white crystalline solid that may change colour upon prolonged exposure to light because it is light sensitive. First it becomes a purple colour, then black with prolonged exposure to light (especially ultraviolet light). The change of colour occurs when AgCl undergoes decomposition into Ag and Cl. It is a dry powder that does not absorb moisture from the air. Silver Chloride has very low solubility in water (0.00089 g per Litre at 10°C). However, it is soluble in dilute Ammonia, Sodium Thiosulfate, Sulfuric Acid, and Potassium Bromide solutions. It has a molar mass of 143.32 g/mol. AgCl melts at 455
°C and boils as 1550°C.

8. Why were the contents in the crucible grayish in colour after heating?

The contents that were heated in the crucible were AgCl and ashless filter paper. The ashless filter paper turned into carbon dioxide, upon heating, and was released into the air. Therefore, the remaining content was AgCl. AgCl is room temperature is a milky white coloured powder. Upon heating, AgCl turns into silver (Ag) and Chlorine (Cl) which is signaled by the grayish colouration of the sample. This explains why the contents in the crucible were grayish in colour.

9. What type of chemical reaction is taking place in this experiment?

A double displacement reaction is taking place in this experiment.
In a double displacement reaction, parts of two compounds switch places to form new compounds. This can be represented by this equation, where AB and CD are the reactants:

AB + CD → AD + CB

In the double displacement reaction that is taking place in the experiment, silver nitrate (AgNO3) combines with sodium chloride (NaCl) to form two new compounds: silver chloride (AgCl) and sodium nitrate (NaNO3), because the sodium and silver switched places.

AgNO3 (aq) + NaCl (aq) ---> AgCl (s) + NaNO3 (aq)


10.
Calculate the Percentage Error of the experiment.


Percentage error = (Difference of Actual and Theoretical Yield / Theoretical Yield) * 100%

Percentage error based on the results obtained from drying the filter paper:

Percentage Error
= (0.404 * 10^21 / 2.06 *10^21) *100%
= 0.1961 * 100%
= 19.61 %

Percentage error based on the results obtained from drying the filter paper is 19.61%.

Percentage error based on the results obtained from burning the ashless filter paper:

Percentage Error
= (0.107 * 10^21 / 2.06 *10^21) *100%
= 0.0519 * 100%
= 5.19 %

Percentage error based on the results obtained from burning the ashless filter paper is 5.19 %.



Sources of Experimental Error




As with all scientific experiments, a certain degree of experimental error is always present which could potentially yield inaccurate results. Due to the many steps that it took to complete this experiment, the sources of error were numerous. First, the reading on the electrical balance was observed to change constantly. This may have been due to slight air currents, or perhaps due to the fact that the contents being weighted were so light. Since the contents weighted were very light, it was difficult to achieve an accurate reading, even when the materials were measured multiple times.

When pouring the Na2NO3 and AgCl solution into the filter, some small amounts of AgCl were still stuck in the Erlenmeyer flask after numerous attempts to get the remaining AgCl out. This caused a slight alteration in the results since not all of the AgCl produced was included in the final mass causing the percentage yield to be inaccurate.


In the results obtained from burning the ashless filter paper, the mass of the AgCl weighted is greater than the expected result because of some of the ashless filter paper could not be completely burned away.

After heating the crucible and lid, the crucible was placed on the counter to allow for cooling. This caused it to pick up unwanted particles that may have been on the counter. These particles may have added to the total mass of the lid and crucible. Also, the crucible was contaminated with residues from previous experiments that could not be removed by any means prior to conducting the experiment. Due to limited resources, such a crucible was required to be used in the experiment. This may have altered the results of the experiment.



Suggestions


1. Use of better quality ashless filter paper that will completely burn away without leaving any unwanted residue.
2. Use of clean crucible and lid.
3. Minimum transfer of the samples from container to container.