In an experiment, a student isolated 6.356 g of pure copper from an initial sample of 9.902 g of copper chloride. Determine the empirical formula of this copper chloride compound. Show your work.
Copper chloride here can be copper (II) chloride (CuCl2) or copper (I) chloride (CuCl). The balanced chemical equation for the decomposition of the copper chlorides: CuCl2 = Cu + Cl2 2CuCl = 2Cu + Cl2 Determine the number of moles of 6.356 g pure Cu. Note: 1 mole Cu = 63.5463 grams Cu 6.356 grams Cu x (1 mole Cu/63.5463 grams Cu) = 0.1000 mole Cu Determine the moles CuCl2 and CuCl that will produce this mole Cu. CuCl2: 1 mole CuCl2 produces 1 mole Cu 0.1000 mole Cu x (1 mole CuCl2/1 mole Cu) = 0.1000 mole CuCl2 CuCl: 2 moles CuCl produce 2 moles Cu 0.1000 mole Cu x (2 mole CuCl/1 mole Cu) = 0.1000 mole CuCl Convert these moles CuCl2 and CuCl to grams. Note: 1 mole CuCl2 = 134.4527 grams CuCl2 0.1000 mole CuCl2 x (134.4527 grams CuCl2/1 mole CuCl2) = 13.44527 grams CuCl2 Note: 1 mole CuCl = 98.9995 grams CuCl 0.1000 mole CuCl x (98.9995 grams CuCl/1 mole CuCl) = 9.89995 grams CuCl --> nearer to the amount of initial reagent Therefore, the empirical formula of the compound is CuCl or copper (I) chloride. That's it!n_n
Cu: 6.356 g / 63.54 g/mol = 0.1000 mol Cl: (9.902-6.356) g / 35.453 = 0.1000 mol => ratio 1:1 or formula CuCl.
