Gravimetric_Estimation_Chloride_Ions

Jonah Chevrier, Nick Jiang, Ushhud Khalid, Philip Van-Lane
===**__The actual project can be found here. Don't write anything on it, we'll do that on Sunday (since rough draft's due), just write your info and stuff here...__**===


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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, washed free of other impurities and the mass of chloride present can be estimated gravimetrically (by determining mass ).
 * Objective:**


 * Links**

[|Project Outline]

//Reagent// - A substance used in a chemical reaction to detect, measure, examine, or produce other substances. (I believe reagent is similar to, if not is, a reactant.)
 * Definitions**

Gravimetric analysis can be described as obtaining a precipitate from a solution and removing any impurities from it, in order to find its net mass. This is used in this experiment, the objective of which is to measure the number of chloride ions in a solution of sodium chloride, using silver nitrate as the limiting reagent. When sodium chloride is added to silver nitrate is added, a double displacement reaction which results in silver chloride and sodium nitrate. The reaction occurs as follows: AgNO3(aq) + NaCl(aq) --> AgCl(s) + NaNO3(aq). A precipitate is produced (the silver chloride) in the solution of sodium nitrate. The precipitate can be extracted and washed free of impurities. The extract is weighed to give a theoretical mass.
 * Introduction**

The following formula was used to derive the mass of a molecule: Mass (g) / Molecular Mass (g/mol) = Mol quantity of the mass of the molecule Multiply that quotient by //Avogadro's number// to get the amount of atoms/ions present in the molecule.

Avogadro's number is a constant which represents the number of atoms in a gram of atoms, or the number of molecules in a gram of molecules, of a substance; it is defined by the quantity "mol."

To start the experiment, a specific amount of NaCl and AgNO3 must be used to find the amount of Chloride ions in the solution.

To find the amount of NaCl needed in a 100mL solution of distilled water, the following formula was used: 100mL * (1L/1000mL) *(0.2 mol / 1L) = 0.02 mol is the concentration of NaCl needed in 100mL of water 0.02mol * Molecular Mass of Sodium Chloride = Mass of NaCl needed in 100mL of water 0.02mol * 58.44277g/mol = 1.1688554g of NaCl needed

To find the amount of AgNO3 needed in a 100mL solution of distilled water, the following formula was used: 100mL * (1L/1000mL) *(0.1 mol / 1L) = 0.01 mol is the concentration of AgNO3 needed in 100mL of water 0.01mol * Molecular Mass of Silver Nitrate = Mass of AgNO3 needed in 100mL of water 0.02mol * 169.88g/mol = 1.6988g of AgNO3 needed

//(10/05/08 - 9:16 p.m., Ushhud)//

//(10/05/08 - 9:17 p.m., Ushhud)//
 * Materials**
 * Safety Goggles
 * 200mL of distilled water
 * ~1.1688554g of NaCl
 * ~1.6988g of AgNO3
 * 2 test tubes
 * Bunsen burner
 * Test tube tongs

(MUST BE IN PAST, PASSIVE TENSE!)
 * Procedure**
 * To obtain 10mL of 0.2M NaCl solution:**
 * 1) Took ~1.1688544g of NaCl and dissolved it in less than 100mL of distilled water.
 * 2) Added enough water to reach 100mL mark.
 * 3) Took 10mL of that 100mL solution to get 10mL of 0.2M NaCl.


 * To obtain 10mL of 0.1M AgNO3 solution:**
 * 1) Took ~1.6988g of AgNO3 and dissolved it in less than 100mL of distilled water.
 * 2) Added enough water to reach 100mL mark.
 * 3) Took 10mL of that 100mL solution to get 10mL of 0.1M AgNO3.

//(10/05/08 - 9:18 p.m., Ushhud)//
 * Experiment Procedure:**
 * 1) 10mL of 0.1M AgNO3(aq) was added to a test tube.
 * 2) 10mL of 0.2M NaCl was added to the test tube containing AgNO3. A reaction occured which resulted in the formation of a precipitate.
 * 3) The bunsen burner tube was attached to the gas nozzle and was carefully turned on.
 * 4) The test tube was clasped with the tongs and was carefully heated.
 * 5) Once the sodium nitrate evaporated, the bunsen burner was turned off; the silver chloride was allowed to cool.
 * 6) The difference of masses of the test tube with silver chloride as well as an empty one was found to get the mass of silver chloride, in grams.
 * 7) Using the molecular mass of silver chloride, the moles of silver chloride was found.
 * 8) Multiplying that amount by Avogadro's number gives the amount of molecules of silver chloride present in the precipitate.
 * 9) Since the ratio of silver to chlorine in silver chloride is 1:1, the amount of chloride ions present was that quantity found in step 8.


 * Safety Precautions**
 * Safety goggles were worn at all times
 * The work space was free of clutter
 * No chemicals or apparatus were tasted or smelled; if necessary, the wafting method was used
 * All long hair was tied back and all loose clothing was tied or removed
 * When the test tube was being heated, it was pointed away from any person
 * Made sure that the bunsen burner was far enough away from the operator when it was being used

//(10/04/08 - 10:20 p.m., Ushhud)//

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 * Observations**


 * Data Table**


 * Calculations**


 * Result**


 * Conclusion**

AgNO3(aq) + NaCl(aq) --> AgCl(s) + NaNO3(aq) Since the molecule ratio between AgNO3(aq) + NaCl(aq) is 1 : 1, NaCl was the limiting reagent as there was a set amount of salt to begin with. The double displacement formed a solid silver chloride as a precipitate due to silver chloride having a low solubility at room temperature. The silver chloride was then isolated from sodium nitrate and was weighted to be 0.3191grams. The number of moles was found though the equation of n = (Mass/Molecular mass) and was found to be 0.002226475 mol. The number of molecules was calculated from the equation of n = (# of molecules) / (Avogadro's number). The number of molecules was found to be 1.3407843432 * 10^21 and since the mole ratio is 1 to 1, the amount of chlorine ions was identical to the aforementioned figure. In terms of percentage yield, the figure came to be approximately 111.32%. The number was found by subtracting the actual mass from the theoretical mass, dividing by the theoreical mass and multiply it by 100. Usually the number would be below 100% since it’s impossible to obtain more than the theoretical mass or the maximum mass. The reason why the percentage went over 100% was that carbon particles still remained after the burning process and had added weight to the entire compound. This partly explains the percentage error of 11.32% found through this equation. **
 * The amount of chlorine ions present in the solution of sodium chloride and silver nitrate is   1.3407843432 * 10^21 molecules. This figure was calculated by first combining aqueous liquids of sodium chloride and silver nitrate to form sodium nitrate and silver chloride through double displacement.

** //(11/30/08 - 5:50 p.m., Nick)//
 * | [ (Mass of Theoretical Yield - Mass of Actual Yield ) / (Mass of Theoretical Yield) ] * 100% |
 * Suggested Modifications of the Experiment**

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 * Errors**