1. A sliver of metal has been extracted from the head of a victim of a crime. Yo
ID: 488200 • Letter: 1
Question
1. A sliver of metal has been extracted from the head of a victim of a crime. You subject it to analysis by ICP and find it contains 95% tin, 2% copper, 2% bismuth, and 1% antimony. At the scene of the crime you catalog a number of metal items. Among them are a baseball bat, candlestick, a length of chain, a frying pan, and an antique sword. Can you identify the item most likely used as the weapon? Can you identify the likely alloy of the metal sliver?
2. There has been a disaster in the stockroom. The labels of several bottles of copper(II) says have been obliterated. With the aid of the chemical inventory you have narrowed the possibilities to copper (II) chloride, copper (II) nitrate, and copper (II) sulfate pentahydrate. Outline in reasonable detail, how you might identify each sample so you can restore the proper labels?
Explanation / Answer
1. Baseball bat is mainly made of aluminium based alloys
Swords should be made of copper of bronze materials (primarily copper)
A chain should be made of some alloy of iron.
Frying pan should be of stainless steel (Fe, C, Mn)
Candlestick can be made of an alloy known as Pewter (mainly tin and other metals like copper, bismuth, antimony) So this matches with that of the ICP analysis of the metal piece.
2. A pinch of all 3 salts can be taken and marked A, B and C and aqueous solution should be made by adding equal volumes of water.
Freshly prepared AgNO3 should be added to each one of them.
Copper Chloride will give white ppt CuCl2 + 2AgNO3 -------------> Cu(NO3)2 + 2AgCl (white ppt)
Copper nitrate will have no reaction
Copper sulphate will also give white ppt Cu(SO4)2 + AgNO3 ------------> Ag2SO4 (white ppt) + Cu(NO3)2
So, now we know which is Copper (II) nitrate and we can separate it out (Lets say it is C)
Next we should add an aquesous solution of Barium Chloride to each of A and B (CuCl2 and CuSO4)
No reaction with CuCl2
But we will get white ppt with CuSO4 BaCl2 + CuSO4 --------------> BaSO4 (white ppt) + CuCl2
So again, now we know which is CuSO4
Thus the remaining salt is CuCl2
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