Question

A 0. 50 g mixture of Cu2O and contains 0. 425 g of Cu. Mass of CuO (in g) in mixture is K, find value of 10 K in the mixture? (Cu = 63.5 u, O = 16 u) (Mark answer to nearest integer)

Answer 1

2

Answer 2

Let the masses be:

• Mass of CuO = $y$ g
• Mass of Cu₂O = $0.50 - y$ g

Step 1. Find the mass fraction of Cu in each compound

For Cu₂O (formula weight = $2(63.5)+16 = 143$ g/mol):

$$\text{Fraction of Cu} = \frac{2(63.5)}{143} = \frac{127}{143}$$

For CuO (formula weight = $63.5+16 = 79.5$ g/mol):

$$\text{Fraction of Cu} = \frac{63.5}{79.5} = \frac{127}{159}$$

Step 2. Write the equation for total Cu mass

The total Cu mass is given as 0.425 g:

$$\frac{127}{143}(0.50 - y) + \frac{127}{159}y = 0.425$$

Divide through by 127 (present in both terms):

$$\frac{0.50 - y}{143} + \frac{y}{159} = \frac{0.425}{127}$$

Multiply through by $143 \times 159$ to clear denominators:

$$159(0.50 - y) + 143y = \frac{0.425 \times 143 \times 159}{127}$$

Simplify the left‐side:

$$79.5 - 159y + 143y = 79.5 - 16y$$

Thus,

$$79.5 - 16y = \frac{0.425 \times 22737}{127} \quad \text{(since } 143 \times 159 = 22737\text{)}$$

Evaluating the right side:

$$\frac{0.425 \times 22737}{127} \approx 76.11$$

So,

$$79.5 - 16y \approx 76.11 \quad \Longrightarrow \quad 16y \approx 79.5 - 76.11 = 3.39$$ $$y \approx \frac{3.39}{16} \approx 0.2119 \text{ g}$$

Step 3. Compute the required value

The mass of CuO in the mixture is $K = y \approx 0.2119$ g. Then:

$$10K \approx 10 \times 0.2119 \approx 2.119 \quad \text{g}$$

Rounded to the nearest integer, $10K \approx 2$.

Summary

• Explanation (minimal):

  Let mass of CuO = $y$ g and Cu₂O = $0.50 - y$ g. Write copper content equation $\frac{127}{143}(0.50-y) + \frac{127}{159}y = 0.425$, solve for $y$, then compute $10y$.