[VISUAL] Sample | Mass of Carbon | Mass of Hydrogen A | 60. g | 12 g B | 72 g | 12 g C | 84 g | 10. — Stoichiometry Chemistry Question

STEPS:

1. Determine the empirical formula of propene ($\text{C}_3\text{H}_6$):
An empirical formula represents the simplest whole-number ratio of the atoms of each element in a compound. For propene, the molecular formula is $\text{C}_3\text{H}_6$. Dividing both subscripts by their greatest common divisor (3) gives the empirical formula of $\text{CH}_2$. This means we are looking for a hydrocarbon sample that has a $1:2$ molar ratio of carbon ($\text{C}$) to hydrogen ($\text{H}$).
2. Recall the molar masses of Carbon and Hydrogen:
To convert the given masses of carbon and hydrogen into moles, we use their approximate molar masses from the periodic table:
* $\text{Molar mass of Carbon (C)} \approx 12\text{ g/mol}$
* $\text{Molar mass of Hydrogen (H)} \approx 1\text{ g/mol}$
3. Calculate the moles of Carbon and Hydrogen in each sample:
* For Sample A:
* $\text{Moles of C} = \frac{60.\text{ g}}{12\text{ g/mol}} = 5.0\text{ mol}$
* $\text{Moles of H} = \frac{12\text{ g}}{1\text{ g/mol}} = 12\text{ mol}$
* Molar ratio (C:H) = $5:12$ (Empirical formula: $\text{C}_5\text{H}_{12}$)
* For Sample B:
* $\text{Moles of C} = \frac{72\text{ g}}{12\text{ g/mol}} = 6.0\text{ mol}$
* $\text{Moles of H} = \frac{12\text{ g}}{1\text{ g/mol}} = 12\text{ mol}$
* Molar ratio (C:H) = $6:12$, which simplifies directly to $1:2$ (Empirical formula: $\text{CH}_2$).
* For Sample C:
* $\text{Moles of C} = \frac{84\text{ g}}{12\text{ g/mol}} = 7.0\text{ mol}$
* $\text{Moles of H} = \frac{10.\text{ g}}{1\text{ g/mol}} = 10.\text{ mol}$
* Molar ratio (C:H) = $7:10$ (Empirical formula: $\text{C}_7\text{H}_{10}$)
* For Sample D:
* $\text{Moles of C} = \frac{90.\text{ g}}{12\text{ g/mol}} = 7.5\text{ mol}$
* $\text{Moles of H} = \frac{10.\text{ g}}{1\text{ g/mol}} = 10.\text{ mol}$
* Molar ratio (C:H) = $7.5:10$, which simplifies to $3:4$ (Empirical formula: $\text{C}_3\text{H}_4$).
4. Identify the matching sample:
Sample B features a $1:2$ molar ratio of carbon to hydrogen, yielding the empirical formula $\text{CH}_2$. This matches the empirical formula of propene, confirming Option B as the correct answer.

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WHY_OTHERS_WRONG:

• Option A is incorrect (Sample A): This sample has an empirical formula of $\text{C}_5\text{H}_{12}$ (pentane), which is a $5:12$ ratio. A student might choose this if they mistakenly compared the mass values directly ($60\text{ g} : 12\text{ g} = 5:1$) or forgot to convert the masses into moles.
• Option C is incorrect (Sample C): This sample has an empirical formula of $\text{C}_7\text{H}_{10}$ ($7:10$ ratio) rather than $\text{CH}_2$.
• Option D is incorrect (Sample D): This sample has an empirical formula of $\text{C}_3\text{H}_4$ (propyne), representing a $3:4$ molar ratio.