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.

Explanation

In order to solve this problem we have to calculate the ratio (or proportion) of copper and oxygen in the two samples of copper oxide compound. Now,
(a) In the first experiment:
Mass of copper `= 1.03 g "….."(1)`
And, Mass of copper oxide `= 1.288 g`
So, Mass of oxygen = Mass of copper oxide - Mass of copper
`= 1.288-1.03"......"(2)`
Now, in the first sample of copper oxide compound:
Mass of copper: Mass of oxygen `= 1.03 : 0.258`
`= (1.03)/(0.258) : 1`
`= 3.99: 1`
`= 4 : 1"......."(3)`
(b) In the second experiment:
Mass of copper = 2.938 g`"......"(4)`
And, Mass of copper oxide `= 3.672 - 2.938`
`= 0.734 g "........"(5)`
Now, in the second sample of copper oxide compound:
Mass of copper : Mass of oxygen `= 2.938 : 0.734`
`= (2.938)/(0.734) : 1`
`= 4 : 1"........"(6)`
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.