Weight of copper oxide obtained by treating 2.16 g of metallic copper with nitric acid and subsequent ignition was 2·7g. In another experiment, 1 · 15 g of copper oxide on reduction yielded 0·92g of copper. Show that the result illustrates the law of definite proportion.

Solution

In the first experiment:
Weight of copper = 2.16 g
Weight of copper oxide obtained on ignition = 2.70 g

therefore

Percentage of copper in copper oxide

equals space fraction numerator 2.16 over denominator 2.70 end fraction cross times 100 space equals space 80 percent sign

therefore Percentage space of space oxygen space in space copper space oxide space equals 100 minus 80 space equals 20 percent sign

In the second experiment.
Weight of copper oxide = 1.15 g

therefore Weight space of space copper space left space after space reduction space equals space 0.92 space straight g therefore space Percentage space of space copper space in space copper space oxide space equals space fraction numerator 0.92 over denominator 1.15 end fraction cross times 100 space equals space 80 percent sign

therefore Percentage space of space oxygen space in space copper space oxide space space equals space 100 space minus 80 space equals space 20 percent sign

Thus, the data illustrates the law of definite porportion as in both the experiments the percentage of copper and oxygen are same.

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