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

Explanation:

According to experiment 1:

Mass of copper oxide = 4.90 g

Mass of copper = 3.92 g

Hence, the percentage of mass of Copper = $3.92 \times \frac{100}{4.90}=80 \%$

According to experiment 2:

Mass of copper oxide = 4.65 g

Mass of copper used = 3.64 g

Hence, the mass percentage of Copper = $3.64 \times \frac{100}{4.65}=78.279 \%$

From the above results, it can be observed that the percentage of copper in both the samples are nearly the same.  

Hence, the law of definite proportion is proved.

Answer 2

Answer:

According to experiment 1:

Mass of copper oxide = 4.90 g

Mass of copper = 3.92 g

Hence, the percentage of mass of Copper = 3.92 \times \frac{100}{4.90}=80 \%3.92×

4.90

100

=80%

According to experiment 2:

Mass of copper oxide = 4.65 g

Mass of copper used = 3.64 g

Hence, the mass percentage of Copper = 3.64 \times \frac{100}{4.65}=78.279 \%3.64×

4.65

100

=78.279%

From the above results, it can be observed that the percentage of copper in both the samples are nearly the same.

Hence, the law of definite proportion is proved.