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.
To work out the proportion of copper and oxygen in the two reactions.
1st chemical reaction:
$4Cu + O_2 → 2Cu_2O$
3.92 g X g 4.90 g
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:
$CuO + H_2 → Cu + H_2O$
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.
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