Question

When a sample of copper oxide is heated in the presence of propane gas, c3h8, three products produce: Co2, H2o and Cu. Using the results below identify the formula of the oxide as copper 1 oxide or copper 11 oxide. Explain you answer by showing all calculations and discussing any laws necessary to support your answer.
Table looks like this:
Items: Masses:
Mass of empty test tube 20.15
Mass of test tube and copper(?) oxide 22.23
Mass of test tube and copper 21.76

Answer 1

Task:

When a sample of copper oxide is heated in the presence of propane gas, c3h8, three products produce: Co2, H2o and Cu. Using the results below identify the formula of the oxide as copper 1 oxide or copper 11 oxide. Explain you answer by showing all calculations and discussing any laws necessary to support your answer.

Table looks like this:

Items: Masses:

Mass of empty test tube 20.15

Mass of test tube and copper(?) oxide 22.23

Mass of test tube and copper 21.76

Solution:

The chemical equation for this reaction is

$$10 \mathrm{Cu}_{x} \mathrm{O} + \mathrm{C}_{3} \mathrm{H}_{8} = 10 \mathrm{x} \mathrm{Cu} + 3 \mathrm{CO}_{2} + 4 \mathrm{H}_{2} \mathrm{O}$$

x = 1 for copper (II) oxide

x = 2 for copper (I) oxide

From the data in the table we can find

The amount of copper is

$$n (\mathrm{mol}) = \frac{m (g)}{M W (g / \mathrm{mol})}$$ $$n (\mathrm{Cu}) = 1.61 / 63.5 = 2.53 \cdot 10^{-2} \mathrm{mol}$$

According to the chemical equation $n(\mathrm{Cu}_{x}\mathrm{O}) = n(\mathrm{Cu}) / x$

$$n \left(\mathrm{Cu}_{x} \mathrm{O}\right) = 2.53 \cdot 10^{-2} / x \ (\mathrm{mol})$$

From the other hand the amount of copper oxide is

$$n (\mathrm{Cu}_{x} \mathrm{O}) = \frac{m (\mathrm{Cu}_{x} \mathrm{O})}{M W (\mathrm{Cu}_{x} \mathrm{O})} = 2.08 / (63.5x + 16) \ \mathrm{mol}$$

We can write

$$n (\mathrm{Cu}) / x = \frac{m (\mathrm{Cu}_{x} \mathrm{O})}{M W (\mathrm{Cu}_{x} \mathrm{O})}$$ $$2.53 \cdot 10^{-2} / x = 2.08 / (63.5x + 16)$$ $$2.53 \cdot 10^{-2} \cdot (63.5x + 16) = 2.08x$$ $$2.53 \cdot 10^{-2} \cdot 63.5x + 2.53 \cdot 10^{-2} \cdot 16 = 2.08x$$ $$1.6x + 0.40 = 2.08x$$ $$2.08x - 1.6x = 0.40$$ $$0.48x = 0.40$$ $$x = 0.83 \approx 1 \ (\text{it's CuO})$$

Answer: It was copper (II) oxide (CuO)