A sample of a solid labeled as NaCl may be impure. A student analyzes the sample and determines that — Stoichiometry Chemistry Question

STEPS:

1. Understand how the mass percent of an element is calculated:
The mass percent of chlorine in any chloride salt is determined by the ratio of the molar mass of chlorine to the total molar mass of the compound:
$$\text{Mass } \% \text{ of Cl} = \left( \frac{\text{Molar Mass of Cl}}{\text{Molar Mass of Compound}} \right) \times 100\%$$

2. Analyze the effect of impurities on mass percent:
* Pure $\text{NaCl}(s)$ is $61\%$ chlorine by mass.
* The analyzed sample has a chlorine mass percent of $75\%$.
* For an impurity to increase the overall mass percentage of chlorine above the baseline of $61\%$, the impuring substance must contain a higher mass percent of chlorine than pure $\text{NaCl}$ ($>61\%$).

3. Compare the potential impurities by examining cation molar masses:
Since we are comparing binary salts with a $1:1$ stoichiometric ratio of cation to chloride anion ($\text{Cl} \approx 35.5\text{ g/mol}$), the mass percent of chlorine is inversely related to the molar mass of the cation:
$$\text{Molar Mass of Cations: } \text{Li}^+ (6.94\text{ g/mol}) < \text{Na}^+ (22.99\text{ g/mol}) < \text{K}^+ (39.10\text{ g/mol})$$
* A lighter cation means a smaller total molar mass of the compound, which yields a higher percentage of chlorine by mass.
* A heavier cation means a larger total molar mass of the compound, which yields a lower percentage of chlorine by mass.

4. Calculate/estimate the chlorine mass percent of the options:
* For $\text{LiCl}$: Since lithium is significantly lighter than sodium, the total molar mass of $\text{LiCl}$ is much lower ($\approx 42.4\text{ g/mol}$). Its chlorine mass percentage is:
$$\text{Mass } \% \text{ of Cl in LiCl} = \frac{35.5\text{ g/mol}}{42.4\text{ g/mol}} \times 100\% \approx \mathbf{84\%}$$
* Because $84\% > 61\%$, mixing any amount of $\text{LiCl}$ with $\text{NaCl}$ will pull the average chlorine mass percent upward. A mixture of the two can logically result in an overall value of $75\%$ chlorine by mass.

5. Conclude:
The presence of $\text{LiCl}(s)$ is the only scenario consistent with raising the mass percent of chlorine to $75\%$, confirming Option D as the correct answer.

*

WHY_OTHERS_WRONG:

* Option A is incorrect: A sample containing only pure $\text{NaCl}(s)$ must have a constant composition of exactly $61\%$ chlorine by mass, as dictated by the Law of Definite Proportions.
* Option B is incorrect: Sodium iodide ($\text{NaI}$) contains no chlorine at all ($0\%$ chlorine by mass). Adding a substance with no chlorine to the sample would dilute the chlorine content, dropping the overall mass percentage below $61\%$.
* Option C is incorrect: Potassium ($\text{K} \approx 39.10\text{ g/mol}$) is heavier than sodium ($\text{Na} \approx 22.99\text{ g/mol}$). Consequently, the molar mass of $\text{KCl}$ is larger ($\approx 74.6\text{ g/mol}$), meaning its chlorine mass percentage is lower than that of $\text{NaCl}$:
$$\text{Mass } \% \text{ of Cl in KCl} = \frac{35.5\text{ g/mol}}{74.6\text{ g/mol}} \times 100\% \approx \mathbf{48\%}$$
Adding $\text{KCl}$ ($48\%$) to $\text{NaCl}$ ($61\%$) would pull the overall chlorine mass percentage downward, making it mathematically impossible to reach $75\%$.