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Weight of copper oxide obtained by heating `2.16 g` of metallic copper with `HNO_(3)` and subsequent ingnition was `2.70 g`. In another experiment, `1.15 g` of copper oxide on reduction yielded `0.92 g` of copper. Show that the results illustrate the law of definite proportions. |
Answer» In the first experiment Weight of copper `= 2.16 g` Weight of copper oxide `= 2.70 g` Weight of oxygen `= 2.70 - 2.16 = 0.54 g` Percentage of copper `= ("Weight of copper")/("Weight of copper oxide")xx100=((2.16g))/((2.70g))xx100=80%` Percentage of oxygen `= ("Weight of oxygen")/("Weight of copper oxide")xx100=((0.54g))/((2.70g))xx100=20%` In the second experiment Weight of copper oxide `= 1.15 g` Weight of copper `= 0.92 g` , Weight of oxygen `= 0.23 g` Percentage of copper `= ("Weight of copper")/("Weight of copper oxide")xx100=((0.92g))/((1.15g))xx100=80%` Percentage of oxygen `= ("Weight of oxygen")/("Weight of copper oxide")xx100=((0.23g))/((1.15g))xx100=20%` Since the ratio by weights of copper and oxygen in the two compounds remains the same, the law of definite composition is illustrated. |
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