Christian Lecce
Mr. Ribarich
Wednesday, February 20st, 2013
Purpose
To determine the mass of copper formed when excess aluminum is reacted with a given mass of a copper salt (Copper Chloride dihydrate), and the mole-to-mole ratio between the reactant and the product of a chemical reaction.
Apparatus
* 150ml beaker * Stirring rod * Ruler * Hotplate * Tweezers * 50ml graduated cylinder
Materials
* Copper (II) chloride dehydrate * Aluminum foil
Procedure
1) Determine the mass of a clean 150ml beaker.
2) Add 2g of copper (II) chloride dehydrate
3) Add 50ml of water and stir to dissolve the compound
4) Fold a 6cm x 10cm strip of aluminum foil lengthwise twice, then curl the folded strip into a circle so it fits on its edge in the beaker
5) Heat the solution with the aluminum until the solution loses its blue colour and turns brown with copper.
6) Once no more reaction can occur with the aluminum, remove the excess with out removing any copper
7) Heat the content until all of the liquid evaporates and there is no more smoke emitting from the beaker.
8) Once the beaker has nothing but copper in it, weigh it.
9) Subtract the total weight by the original weight of the beaker to get the weight of just the copper
Observations
1) The theoretical yield comes to 0.7435g of copper
3CuCl2-2H2O + 2Al -> 3Cu + 2AlCl3 + 6H2O Mass of CuCl2-2H2O | Mass of Clean Beaker | Mass of Copper created | Mole-to-Mole ratio | 2g | 113.5g | 0.8g | 1 : 1 (hydrate to copper) |
2g of CuCl2-2H2O
CuCl2-2H2O = 170.55 g/mol
2 / 170.55 = 0.0117mol
1 : 1 0.0117 x
0.0117 moles of Cu
Cu = 63.55g/mol
63.55 x 0.0117 = 0.7435g
2) The actual yield of the reaction is 0.8
0.8