Questions 22-24 refer to the following information. A group of students was asked to recover Cu(s) f — Stoichiometry Chemistry Question

STEPS:

1. Understand the relationship between the dissolved ions and the precipitate:
When excess sodium phosphate ($\text{Na}_3\text{PO}_4$) is added to the copper solution, all of the dissolved copper ions ($\text{Cu}^{2+}$) precipitate out completely as solid copper phosphate ($\text{Cu}_3(\text{PO}_4)_2$). Looking closely at the chemical formula of the precipitate, $\text{Cu}_3(\text{PO}_4)_2$, each formula unit contains exactly 3 copper ions ($\text{Cu}^{2+}$). Therefore, the mole ratio is:
$$\frac{3\text{ mol Cu}^{2+}}{1\text{ mol Cu}_3(\text{PO}_4)_2}$$

2. Calculate the moles of copper phosphate precipitate recovered:
You are given that $3.8\text{ g}$ of solid $\text{Cu}_3(\text{PO}_4)_2$ was recovered and its molar mass is $381\text{ g/mol}$.
* Use the mass-to-mole formula:
$$\text{Moles of Cu}_3(\text{PO}_4)_2 = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{3.8\text{ g}}{381\text{ g/mol}} \approx \mathbf{0.010\text{ mol}}$$
*(Note: On Section I of the AP Exam, calculators are not allowed. This calculation is designed to be easily estimated: $\frac{3.8}{381}$ is almost exactly $\frac{1}{100}$, or $0.010\text{ mol}$).*

3. Determine the moles of $\text{Cu}^{2+}$ in the original solution:
* Using the $3:1$ stoichiometric ratio, multiply the moles of precipitate by 3 to find the moles of copper ions originally present:
$$\text{Moles of Cu}^{2+} = 0.010\text{ mol Cu}_3(\text{PO}_4)_2 \times \frac{3\text{ mol Cu}^{2+}}{1\text{ mol Cu}_3(\text{PO}_4)_2} = \mathbf{0.030\text{ mol Cu}^{2+}}$$

4. Calculate the molar concentration (molarity) of $\text{Cu}^{2+}$:
The volume of the original solution sample was $100.0\text{ mL}$.
* Convert the volume to liters:
$$100.0\text{ mL} = 0.1000\text{ L}$$
* Solve for molarity ($M = \frac{\text{moles}}{\text{volume in L}}$):
$$[\text{Cu}^{2+}] = \frac{0.030\text{ mol}}{0.1000\text{ L}} = \mathbf{0.30\text{ M}}$$
* This aligns with Option B.

*

WHY_OTHERS_WRONG:

• Option A is incorrect (0.10 M): A student would arrive at this value if they forgot to apply the $3:1$ stoichiometric ratio between $\text{Cu}^{2+}$ and $\text{Cu}_3(\text{PO}_4)_2$. If you assume a $1:1$ ratio, the calculation yields $\frac{0.010\text{ mol}}{0.100\text{ L}} = 0.10\text{ M}$.
• Option C is incorrect (1.0 M): This value is too high by a factor of roughly 3. It typically represents a combined mistake of neglecting the stoichiometry ratio and making a decimal error during the volume conversion (such as using $0.01\text{ L}$ instead of $0.1\text{ L}$).
• Option D is incorrect (3.0 M): A student might choose this if they correctly multiplied the moles of precipitate by 3 to get $0.030\text{ mol}$ of $\text{Cu}^{2+}$ but then incorrectly divided by $0.01\text{ L}$ instead of $0.1\text{ L}$ during the molarity calculation, or forgot to divide by the volume altogether and misplaced the decimal point.