4.Вычислите массовые доли элементов в соединении: CuCO₃

Mr(CuCO₃) = 64 + 12 + 16 × 3 = 64 + 12 + 48 = 124 а.е.м.

1. ω(Cu) = (Ar(Cu) / Mr(CuCO₃)) × 100% = (64 / 124) × 100% ≈ 51.61%
2. ω(C) = (Ar(C) / Mr(CuCO₃)) × 100% = (12 / 124) × 100% ≈ 9.68%
3. ω(O) = (3 × Ar(O) / Mr(CuCO₃)) × 100% = (3 × 16 / 124) × 100% = (48 / 124) × 100% ≈ 38.71%

Ответ: ω(Cu) ≈ 51.61%, ω(C) ≈ 9.68%, ω(O) ≈ 38.71%