Analysis of Silver in an Alloy Free Essay Example

The purpose of this lab was to calculate the percent of silver in an alloy using gravimetric analysis. Through the procedure mentioned in this report, the percent of silver in an alloy of a U.S. Mint dime made before 1965 was 90.2% � 0.05%.

Introduction:

The alloy used for this experiment was a U.S. Mint ten-cent dime alloyed as silver and copper if made before the year 1965, after which mints began using copper and nickel instead. First, the dime piece was dissolved in nitric acid and the silver was precipitated as silver chloride.

The precipitate was then filtered, washed, dried and weighed. To obtain the percent of silver in the alloy, the mass of silver chloride was compared to the mass of the original sample. Because the results are based on the mass of a product, the procedure is determined as gravimetric analysis.

Nitric acid (HNO3) is required as an “oxidizing agent” because silver and copper are very non-reactive metals and cannot dissolve in hydrochloric acid or sulfuric acid.

The nitrate ion (NO3�) oxidized Ag(s) to Ag+(aq) and Cu(s) to Cu2+ (aq). As the nitric acid was reduced, the product became nitrogen monoxide (NO), a colorless gas that instantly reacts with oxygen in the air to produce nitrogen dioxide (NO2), an orange-brown gas. The following demonstrate the half reactions mentioned:

• Ag(s) –> Ag+(aq) + e-
• Cu(s) –> Cu2+ (aq) + 2e-
• 4 H+(aq) + NO3- + 3e- –> NO(g) + 2H2O(l)

To separate the silver ions and the copper ions, the silver was precipitated as silver chloride (AgCl) by adding dissolved sodium chloride (NaCl) to the solution, which also created the product of copper (II) chloride (CuCl2).

Theoretically, because copper (II) chloride is soluble in water and silver chloride is insoluble in water, the addition of chloride ions to the solution from the sodium chloride precipitated all the silver ions and none of the copper ions. The silver chloride precipitate was filtered from this solution. The following demonstrates the reaction for the precipitation mentioned:

Procedure:

Day 1

• The mass of a piece of dime made before 1965 (silver-copper alloy) was measured using an analytical balance and recorded.
• The dime/alloy was placed in a clean, group labeled 100-mL beaker.
• While under a fume hood, a 10 mL graduated cylinder was used to pour 10 mL of 6M nitric acid into the 100-mL beaker containing the alloy.
• To make sure the alloy had totally dissolved, the beaker containing the solution was placed on a hot plate
• After calculating two times the amount of sodium chloride necessary to precipitate the silver in the alloy sample (assuming that the sample is 100% silver), the amount of sodium chloride was massed out to � 0.001 on a sensitive balance and dissolved in 25 mL of distilled water.
• The precipitate was made by slowly adding the sodium chloride solution to the dissolved silver and stirring with a stirring rod.
• The beaker was then covered with plastic wrap to keep dust out of the solution and left to stand over night in a drying oven so that the precipitate particles could grow larger and filtered easier.

Day 2

• A B�chner Funnel, a B�chner Flask with a filter paper, and a vacuum aspirator was assembled and used to filter the solution.
• After the water was turned on and the vacuum aspirator was activated, distilled water from a wash bottle was sprayed to create a seal between the funnel and filter paper.
• The solution was poured into the funnel with the vacuum aspirator on and the wash bottle was used to clean the sides of funnel and the beaker being poured. The intention was to isolate the precipitate on the filter.
• To stop the vacuum aspirator, the rubber tubing connect the flask to the sink was removed before turning off the water to prevent the possibility of the filtrate being pushed up through the funnel backwards.
• Our filter apparatus was then disassembled and the funnel containing the precipitate in the filter paper was placed in a drying oven and left overnight.

Day 3

• The filter paper containing the precipitate was procured from the funnel using tweezers.
• The mass of the filter with the precipitate was weighed on an analytical balance and recorded.

Observations:

• Dime
• Filter
• Filter + AgCl
• NaCl
• Mass (g) � 0.001
• 0.309
• 0.067
• 0.435
• 0.332
• During Day 1 while the alloy was dissolving, the nitrate turned the color of turquoise and brown gas is let off. Through this process of dissolving, the toxic gases nitrogen monoxide (NO) and nitrogen dioxide (NO2) are evolved, hence the necessity for the mixture of the acid and alloy to take place under a fume hood.

________________________________________________________________________

Calculations:

Amount of NaCl necessary to precipitate the silver x2

Mass of Ag (dime*) x 1 mol Ag x 1 mol NaCl x 58.443g NaCl

107.868g Ag 1 mol Ag 1 mol NaCl

Assuming that the dime is 100% silver

0.309g Ag x 1 mol Ag x 1 mol NaCl x 58.443g NaCl = (0.167 � 0.001)g NaCl

107.868g Ag 1 mol Ag 1 mol NaCl

2(0.167) = (0.334 � 0.002)g NaCl*

*0.332 � 0.001g NaCl was used in the experiment

Mass of AgCl

(Mass of Filter + AgCl) – Mass of Filter

0.435g – 0.067g = (0.368 � 0.002)g AgCl

Mass of Ag

Mass of AgCl x Molar Mass Ag = %Uncertainty = Uncertainty x 100

Molar Mass AgCl Value

1.368g AgCl x 107.87g Ag =(0.277 � 0.001)gAg (0.5%)%Uncertainty = 0.002 x 100 = 0.5%

143.32g AgCl 0.368

Percent Silver in Dime

Mass Ag x 100 =

Mass Dime

0.277g Ag x 100 = (89.6 � 0.7)% 0.309g Dime

%UncertaintyDime =Uncertainty = 0.001 x 100 = 0.3%

Value 0.309

%Uncertainty%Ag in Dime = %UncertaintymassAg + %UncertaintyDime =0.5% + 0.3% = 0.8%

Therefore, 0.8% of 89.6% = 0.008 x 89.6 = 0.7%

Conclusions, Discussions, and Evaluations:

By means of the experiment procedure, the mass of the dried precipitate (AgCl) and filter paper was obtained. Next, the mass of AgCl precipitate and eventually the mass of silver itself was calculated. Through dividing the acquired mass of silver by the original mass of the dime, the percent of silver within the dime was achieved.

However, not all of the silver from the original sample was and could be accounted for. Because the funnel from Day 2’s procedure was not covered in plastic wrap before being left in the drying oven overnight, dust particles possibly could have collected in the filter in addition to the precipitate. This would offset the measurement of the mass of precipitate on Day 3. Furthermore, during Day 2’s procedure, there was the risk that the precipitate had “peptized”, forming minute particles that could pass through the filter due to the use of the wash bottle to wash the precipitate into the funnel with the dilute nitric acid solution.

This problem is solved if measures such as 2 mL of 6M HNO3 had been added to the 150mL of distilled water in the wash bottle. The outcome would not have shifted greatly, but it would have avoided the possibility allowing all of the particles to pass through. Additionally, it is my opinion that if the lab spanned a time slot such as 2 hours to complete the lab and the written report, the quality of the results would have more reputability because it was all done at once.

Why is a twofold excess of chloride added to precipitate the silver?

• A twofold excess of chloride is added to the precipitate to insure that all silver ions are precipitated. The NaCl and CuCl2 is dissolved in water and is therefore not an obstructing factor in the experiment.

If the funnel containing the silver chloride is not cool when its mass is determined, will the calculated percent silver be too high or too low? Why?

• If the funnel is not cool when the mass is determined, the convection currents made from the heat within the scale would lift the funnel to make the mass appear lighter ergo the calculated percent would be lower than the actual.

Calculate the mass of silver in your original sample. Using the mass and the current price of silver calculate how much the silver in this experiment was worth.

• The current price of silver is $0.56 per 1 gram. My mass of my original sample of silver was (0.309 � 0.001) g. The actual amount of silver in the sample is 0.309 x .896 = (0.277 � 0.001)g. The silver was worth (0.277x.56) = 0.16 cents.

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