Question

In an experiment, 4.90 g of copper oxide was obtained from 3.92 g of copper. In another experiment, 4.55 g of copper oxide gave, on reduction, 3.64 g of copper. Show with the help of calculations that these figures verify the law of constant proportions.

Answer 1

To solve this problem, we need to work out the proportion of copper and oxygen in the two reactions.

1st chemical reaction:

Therefore, mass of oxygen(X) = mass of copper oxide – mass of copper

Mass of oxygen = 4.9 – 3.92 = 0.98 g

In the above reaction, the ratio of mass of copper to the mass of oxygen is 3.92 : 0.98 = 4:1

2nd chemical reaction:

CuO + H₂  → Cu + H₂O

4.55g   Xg    3.64g  Yg

Molecular mass of copper oxide = 75.5g

Molecular mass of hydrogen = 2g

Molecular mass of copper = 63.5g

Molecular mass of water = 18g

Molecular mass of oxygen = 16g

Mass of hydrogen = 2/63.5 × 3.64 = 0.11g

Mass of water = 18/75.5 × 4.55 = 1.08g

Mass of oxygen in water = 16/18 × 1.08 = 0.96g

In the above reaction, the ratio of mass of copper to the mass of oxygen is 3.64 : 0.96 = 4:1
Ratio of copper to oxygen in both the reactions is 4:1. This illustrates and verifies the law of constant proportions.