Question

Arranged the following compounds according to increasing copper content: Cu₂O, CuCO_3, Cu₂SO_4, Cu(MnO₄)₂

Answer 1

M(Cu) = 63.54 g/mol

1. M(Cu₂O) = 143.09 g/mol

Proportion:

143.09 – 100 %

$63.54 \times 2$ – x

$x = \frac{63.54 \times 2 \times 100}{143.09} = 88.81 \; \%$

2. M(CuCO₃) = 123.55 g/mol

Proportion:

123.55 – 100 %

63.54 – x

$x = \frac{63.54 \times 100}{123.55} = 51.43 \; \%$

3. M(Cu₂SO₄) = 223.15 g/mol

Proportion:

223.15 – 100 %

$63.54 \times 2$ – x

$x = \frac{63.54 \times 2 \times 100}{223.15} = 56.95 \; \%$

4. M(Cu(MnO₄)₂) = 301.41 g/mol

Proportion:

301.41 – 100 %

63.54 – x

$x = \frac{63.54 \times 100}{301.41} = 21.08 \; \%$

Answer: Cu(MnO₄)₂ → CuCO₃ → Cu₂SO₄ → Cu₂O