Final answer:
The mass percentage of copper in Cu(OH)₄SO₄ is 28.67%, and in Cu(OH)₂SO₄, it is 43.65%. These calculations are based on the molar masses of copper and the compounds, demonstrating the importance of the patina in protecting the Statue of Liberty from further corrosion.
Step-by-step explanation:
Mass Percentage of Copper in Compounds
The Statue of Liberty has turned green due to a patina formed by the compounds Cu(OH)₄SO₄ and Cu(OH)₂SO₄. To find the mass percentage of copper in each compound, we can use the formula: Mass percentage of Cu = (Mass of Cu in the compound / Molar mass of the compound) × 100%.
For Cu(OH)4SO4, the molar mass of Cu is 63.55 g/mol, and the molar mass of the entire compound is 63.55 g/mol (Cu) + 64.00 g/mol (O₄) + 1.01 g/mol (H₄) + 32.07 g/mol (S) + 16.00 g/mol (O) = 221.63 g/mol. The mass percentage of Cu is ((63.55 / 221.63) × 100%) = 28.67%.
For Cu(OH)2SO4, the molar mass of the compound is 63.55 g/mol (Cu) + 32.00 g/mol (O₂) + 2.02 g/mol (H₂) + 32.07 g/mol (S) + 16.00 g/mol (O) = 145.64 g/mol. The mass percentage of Cu is ((63.55 / 145.64) × 100%) = 43.65%.
The oxidation of copper and the formation of these compounds are part of corrosion and result in the blue-green color of the patina, which also serves as a protective layer against further corrosion, a process known as passivation.