Chemistry 5070/41 — May/June 2014

Walkthrough

Copper reacts with diatomic oxygen gas ($\text{O}_2$) to form copper(II) oxide ($\text{CuO}$).
Balancing the atoms on both sides requires $2$ copper atoms reacting with $1$ oxygen molecule to produce $2$ formula units of copper(II) oxide:

$$2\text{Cu} + \text{O}_2 \rightarrow 2\text{CuO}$$

Key Takeaways

• Copper(II) oxide has the formula $\text{CuO}$ because copper has a $+2$ oxidation state and oxide has a $-2$ charge ($\text{Cu}^{2+}$ and $\text{O}^{2-}$).
• Oxygen in air exists as diatomic molecules, $\text{O}_2$.

Common Mistakes

• Writing oxygen as atomic $\text{O}$ instead of $\text{O}_2$.
• Writing incorrect formulae such as $\text{Cu}_2\text{O}$ or $\text{CuO}_2$.

Things to Be Careful About

• Ensure both the copper and oxygen atoms balance on both sides.

Walkthrough

Copper is a reddish-brown/pink metal. When it is heated in air, it oxidises to copper(II) oxide ($\text{CuO}$), which is a black solid.

Key Takeaways

• $\text{CuO}$ is black.
• $\text{Cu}_2\text{O}$ is red/brown, but copper(II) oxide formed in excess air is black.

Common Mistakes

• Confusing copper metal (pink/brown) or hydrated copper(II) salts (blue) with copper(II) oxide (black).

Things to Be Careful About

• Give a clear, unambiguous colour term: black.

Walkthrough

Looking at Fig. 1.2, the scale has major divisions every $20\text{ cm}^3$ and subdivisions of $2\text{ cm}^3$ (since there are 10 subdivisions between 60 and 80, each small division is $\frac{20}{10} = 2\text{ cm}^3$).
The edge of the plunger aligns exactly one small division to the left of 70, which is $72\text{ cm}^3$.

Key Takeaways

• Determine the value of each subdivision on a scale before reading the value.
• Read the position corresponding to the flat leading face of the plunger.

Common Mistakes

• Miscounting the subdivisions (e.g. assuming each mark represents $1\text{ cm}^3$ instead of $2\text{ cm}^3$).

Things to Be Careful About

• Ensure you read from the correct side of the scale (the gas volume is between 0 and the plunger face).

Walkthrough

Clean, dry air is approximately $78\%$ nitrogen, $21\%$ oxygen, and $1\%$ argon/other gases. When oxygen is removed by reaction with copper, nitrogen remains as the unreactive major component (making up about four-fifths of the remaining gas).

Key Takeaways

• Nitrogen is the main component of air (approx. $78\%$) and is unreactive under these conditions.

Common Mistakes

• Naming noble gases or carbon dioxide, which are only present in trace amounts compared to nitrogen.

Things to Be Careful About

• The question asks to 'name' the gas, so write 'nitrogen' (or $\text{N}_2$).

Walkthrough

The initial volume of air in syringe A was $90\text{ cm}^3$.
The volume of gas remaining after complete reaction was $72\text{ cm}^3$.
The decrease in volume corresponds to the volume of oxygen gas consumed by the copper:

$$90\text{ cm}^3 - 72\text{ cm}^3 = 18\text{ cm}^3$$

Key Takeaways

• The contraction in air volume during this experiment equals the volume of oxygen that reacted.

Common Mistakes

• Using $100\text{ cm}^3$ as the starting volume instead of reading the stem value ($90\text{ cm}^3$).

Things to Be Careful About

• Ensure error carried forward (ecf) is applied if an incorrect value was recorded in part (b)(i).

Walkthrough

To find the number of moles of a gas from its volume at room temperature and pressure (r.t.p.):

$$\text{moles} = \frac{\text{volume of gas in } \text{cm}^3}{\text{molar gas volume in } \text{cm}^3}$$

Given that $1\text{ mole}$ occupies $24\,000\text{ cm}^3$:

$$\text{moles of } \text{O}_2 = \frac{18}{24\,000} = 0.00075\text{ moles}$$

(or $7.5 \times 10^{-4}\text{ moles}$).

Key Takeaways

• Number of moles of gas at r.t.p. = $\frac{V\text{ (in }\text{cm}^3\text{)}}{24\,000}$ or $\frac{V\text{ (in }\text{dm}^3\text{)}}{24}$.

Common Mistakes

• Dividing by $24$ directly without converting the volume from $\text{cm}^3$ to $\text{dm}^3$.

Things to Be Careful About

• Count the leading zeros carefully in decimal notation ($0.00075$).

Walkthrough

1. Use the stoichiometric ratio from the balanced equation $2\text{Cu} + \text{O}_2 \rightarrow 2\text{CuO}$:
   $1\text{ mole of } \text{O}_2$ reacts with $2\text{ moles of } \text{Cu}$.

$$\text{moles of } \text{Cu} = 2 \times 0.00075\text{ mol} = 0.0015\text{ mol}$$

1. Calculate the mass of copper using $\text{mass} = \text{moles} \times A_r$:

$$\text{mass} = 0.0015\text{ mol} \times 64\text{ g / mol} = 0.096\text{ g}$$

Key Takeaways

• Always use the mole ratio from the balanced chemical equation before calculating reacting masses.
• $\text{mass} = \text{moles} \times A_r$ (or $M_r$).

Common Mistakes

• Forgetting the $2:1$ molar ratio and assuming $\text{moles of Cu} = \text{moles of } \text{O}_2$.

Things to Be Careful About

• Ensure correct relative atomic mass ($A_r = 64$) is used for $\text{Cu}$.