In an experiment 1.288g of copper oxide was obtained from 1.03g of copper. In another experiment 3.672g of copper oxide obtain from the 2.938g of copper. Show that these figures verify the law of constant proportion.

Dear Student,

First experiment
Copper oxide = 1.288 g
Copper left = 1.03 g
Oxygen present = 1.288 - 1.03 = 0.258 g
 
Percentage of oxygen in CuO = (0.258 × 100) / 1.288 = 19.98 ~ 20 %
 
Second experiment
Copper oxide = 3.672 g
Copper left = 2.938 g
Oxygen present = 3.672 - 2.938 = 0.734 g
 
Percentage of oxygen in CuO = (0.734 × 100) / 3.672 = 20.03 ~ 20 %

So, it is verified that these figures verify the law of constant proportion.

Best wishes!

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