Principles of chemistry

Question

a 0.858 sample of copper was reacted with oxygen to form 1.074g of copper oxide. determine the empirical formula of the copper oxide

Answer 1

From the conservation of mass, we can find the mass of oxygen in the sample.

1.074 g of product - 0.858 g Cu = 0.216 g oxygen

moles of oxygen = 0.216 g x 1 mole oxygen/15.99 g = 0.0135 mols oxygen in the sample

moles copper = 0.858 g x 1 mol copper/63.55 g = 0.0135 mols copper in the sample

Since the copper and oxygen are present in a 1:1 mole ratio, the empirical formula = CuO

Answer 2

Given Cu + O₂ --> Cu_xO_X (of unknown subscripts)

Given the molar masses for Copper (63.546 g/mol) and Oxygen (O - 15.999 g/mol)

Your prime possibilities (most common oxidation states of copper) are

4Cu + O₂ --> 2Cu₂O (Copper (I) oxide)

or

2Cu + O₂ --> 2CuO (Copper (II) oxide)

Convert mass of copper to moles: (0.858 g Cu) (1 mol Cu / 63.546 g Cu) = 0.0135 mol Cu

Let's assume Copper (II) Oxide. Since the reaction will balance with a 1:1 molar ratio of Cu to the CuO

0.0135 mol Cu oxide were formed

Given the Cu oxide mass formed of 1.074 g

1.074 g Cu oxide / 0.0135 mol Cu oxide = 79.54 g/mol of the oxide

Check to see if this is CuO = 62.546 + 15.999 = 79.54 g/mol - so CuO or Copper (II) oxide

If we had assumed Copper (I) oxide (Cu₂O) the molar ratio would be different.

(0.0135 mol Cu)(2 mol Cu₂O/4 mol Cu) = 0.00675 mol

1.074 g / 0.00675 mol Cu₂O = 159.11g/mol

Check molecular mass of Cu₂O 2(62.546) + 15.999 = 141.091 so this is NOT Cu₂O