(a) Define the Law of constant proportion by giving
one example.

(b) In an experiment, 1.288 g of copper oxide was obtained from 1.03 g of Cu. In another experiment, 3.672 g of copper oxide gave a reduction of 2.938 g of copper. Which law of chemical combination can be derived by this example?

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Solution

(a) The Law of constant proportion states that a chemical compound always contains exactly the same proportion of elements by mass.
For example, pure water obtained from different sources such as River, well, springs always contains hydrogen and oxygen together in the ratio of 1:8 by mass.

(b) First case:
Copper taken = 1.03 g
Copper oxide formed = 1.288 g
Oxygen present = Mass of copper oxide - Mass of copper
= 1.288 - 1.03 = 0.258 g
Hence the ratio of the mass of copper and oxygen in the first case will be:
Mass of copper: Mass of oxygen
= 1.03 : 0.258
= 3.99 : 1
= 4 : 1

Second case:
Copper oxide taken = 3.672 g
Copper produced = 2.938 g
Oxygen present = Mass of copper oxide - Mass of copper
= 3.672 - 2.938 = 0.734 g
Hence the ratio of the mass of copper and oxygen in the first case will be:
Mass of copper: Mass of oxygen
= 2.938: 0.734 = 4 : 1
From the above calculations we can see that the ratio (or proportion) of copper and oxygen elements in the two samples of copper oxide compound is the same 4:1. So, the given figures verify the law of constant proportions.

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Similar questions

Q. In an experiment, 1.288 g of copper oxide was obtained from 1.03 g of copper. In another experiment, 3.672 g of copper oxide gave, on reduction, 2.938 g of copper. Show that these figures verify the law of constant proportions.

In a experiment, 4.90g of copper oxide was obtained from 3.92g of copper. In another experiment, 4.55g copper oxide gave on reduction 3.64g of copper. Show that these figures verify the law of constant proportion.

Q. 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.

Q. In an experiment

1.288 g

of copper oxide was obtained from

1.03 g

of copper and

0.258 g

of oxygen. Calculate the ratio of copper and oxygen in the sample?

2.16

grams of Cu, on reaction with

H N O_{3}

, followed by ignition of the nitrate, gave

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g of copper oxide. In another experiment

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g of copper oxide, upon reaction with hydrogen, gave

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(a) Define the Law of constant proportion by giving one example. (b) In an experiment, 1.288 g of copper oxide was obtained from 1.03 g of Cu. In another experiment, 3.672 g of copper oxide gave a reduction of 2.938 g of copper. Which law of chemical combination can be derived by this example?

(a) Define the Law of constant proportion by giving
one example.

(b) In an experiment, 1.288 g of copper oxide was obtained from 1.03 g of Cu. In another experiment, 3.672 g of copper oxide gave a reduction of 2.938 g of copper. Which law of chemical combination can be derived by this example?