Two experiments were conducted: 1.375 g of copper (II) oxide, CuO2, was reduced by heating at high pressure with excess hydrogen to yield 1.098 g of copper Cu. 1.179 g of copper Cu was dissolved in nitric acid to yield copper (II) nitrate, which was converted to 1.476 g copper (II) oxide, CuO2, on ignition. Show that the results illustrate the law of definite proportions by calculating the percentage of oxygen in copper (II) oxide in experiments 1 and 2.

Question

King Jefferson

Engineering

29 December, 07:49

Two experiments were conducted: 1.375 g of copper (II) oxide, CuO2, was reduced by heating at high pressure with excess hydrogen to yield 1.098 g of copper Cu. 1.179 g of copper Cu was dissolved in nitric acid to yield copper (II) nitrate, which was converted to 1.476 g copper (II) oxide, CuO2, on ignition. Show that the results illustrate the law of definite proportions by calculating the percentage of oxygen in copper (II) oxide in experiments 1 and 2.

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America · Answer 1

Mass of oxygen present = mass of copper (ll) oxide - mass of copper residue = 1.375 - 1.098 = 0.277

Percentage of oxygen = 0.277/1.375 * 100 = 20.14% for experiment 1

For experiment 2

Mass of oxygen present = 1.476 - 1.179 = 0.297

Percentage by mass of oxygen present in the copper (ll) oxide = 0.297/1.476 * 100 = 20.12%

It will be noticed that though the source of the copper ll oxide was different, they still contain the same proportion of oxygen which verify the law of definite proportion which states that all pure sample of the same chemical compound contain similar elements combined in the same proportion by mass.