Question

The Law of Definite proportions implies that in a pure compound elements combine by a simple ratio by mass. Berthollet argued against this by showing for oxides of copper that many stoichiometries were possible. In such an experiment 50.0g of copper is oxidised at 400oC and the products are cooled, collected and found to weigh 58.7g. Assuming the products are just a mixture of Cu2O and CuO, approximately what percentage by mass is CuO ? (give to the nearest 5%, e.g. 5, 10, 15, 20 etc

Answer 1

The mass of copper is given to be 50 g.

The molecular mass of copper is 63.5 amu.

So, the no of moles of copper present in 50 g is

N = 50/63.5 = 0.7874.

The mass of oxygen present in the product is

M = 58.7 - 50 = 8.7 g

So, the number of moles of oxygen atom is

n = 8.7/16 = 0.544.

Let x moles of Vu combine with x moles of oxygen tonform CuO.

Let 2y moles of Cu combined with y moles of oxygen.

Then,

x+2y = 0.7874

x+y = 0.544

Subtracting the second equation from the first,

Y = 0.7874-0.544 = 0.2434

X = 0.544-0.2434 = 0.3006

So, 0.3006 moles of CuO is formed.

Mass of CuO formed = 0.3006*79.545 = 23.91 g

By percent, this is

% = 100*23.91/58.7 = 40.7 = 40%