# 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 - Science

Short Note

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.

#### Solution

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.

Concept: Laws of Chemical Combination - Law of Constant Proportions or Law of Definite Proportions
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#### APPEARS IN

Lakhmir Singh Class 9 Chemistry - Science Part 2
Chapter 3 Atoms and Molecules
Hots Questions | Q 77 | Page 131