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Analysis of Silver in an Alloy

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Analysis of Silver in an Alloy Mark Andrew Garner Gareth Smith January 28 - February 1, 2008 Aim and Outcome: 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) ...read more.


* 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 ...read more.


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. 1. Why is a twofold excess of chloride added to precipitate the silver? a. 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. 2. 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? a. 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. 3. 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. a. 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. Garner 4 ...read more.

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