Question

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. ​

Answer 1

Solution :

In order to solve this problem we've to calculate the ratio of copper and oxygen in the two samples of the given copper oxide compound.

a) In the first experiment :

• Mass of copper = 1.03 g

• Mass of copper oxide = 1.288 g

★ Mass of oxygen = Mass of copper oxide - Mass of copper

→ Mass of oxygen = 1.288 - 1.03

→ Mass of oxygen = 0.258 g

In the first sample of copper oxide compound :

★ Mass of copper : Mass of oxygen

→ 1.03 : 0.258

→ $\sf\dfrac{1.03}{0.258}$ : 1

→ 3.99 : 1

→ 4 : 1

b) In the second experiment :

• Mass of copper = 2.938 g

• Mass of copper oxide = 3.672 g

★ Mass of oxygen = Mass of copper oxide - Mass of copper

→ Mass of oxygen = 3.672 - 2.938

→ Mass of oxygen = 0.734 g

In the second sample of copper oxide compound :

★ Mass of copper : Mass of oxygen

→ 2.938 : 0.734

→ $\sf\dfrac{2.938}{0.734}$ : 1

→ 4 : 1

From the above calculations, we can see that the ratio of copper & oxygen elements in both the samples of copper oxide compound is the same 4:1. So, given figures verify the law of constant proportions.

Answer 2

SoluTion :-

In order to solve this problem we've to calculate the ratio of copper and oxygen in the two samples of the given copper oxide compound.

i) In the first experiment :

• Mass of copper = 1.03 g
• Mass of copper oxide = 1.288 g

We know that,

★ Mass of oxygen = Mass of copper oxide - Mass of copper

So,

→ Mass of oxygen = 1.288 - 1.03

= 0.258 g.

In the first sample of copper oxide compound :

We know that,

★ Mass of copper : Mass of oxygen

$= \: \sf1.03 : 0.258$

$= \: \sf\dfrac{1.03}{0.258}$

$= \sf 3.99 : 1$

$= \sf4 : 1$

ii) In the second experiment :

• Mass of copper = 2.938 g
• Mass of copper oxide = 3.672 g

We know that,

★ Mass of oxygen = Mass of copper oxide - Mass of copper

So,

→ Mass of oxygen = 3.672 - 2.938

= 0.734 g.

In the second sample of copper oxide compound :

We know that,

★ Mass of copper : Mass of oxygen

$= \sf2.938 : 0.734$

$= \sf\dfrac{2.938}{0.734}$

$= \sf4 : 1$