A sample of a compound that contains only the elements C, H, and N is completely burned in O2 to pro — Stoichiometry Chemistry Question

STEPS:

1. Analyze the combustion reaction: The original compound contains only carbon (C), hydrogen (H), and nitrogen (N). When it is completely burned in excess oxygen ($\text{O}_2$), all of the carbon in the compound is oxidized to carbon dioxide ($\text{CO}_2$), and all of the hydrogen is oxidized to water ($\text{H}_2\text{O}$).
2. Calculate the moles of carbon (C) in the compound sample:
* Find the moles of $\text{CO}_2$ produced using its molar mass ($\approx 44.0\text{ g/mol}$):
$$n_{\text{CO}_2} = \frac{44.0\text{ g of }\text{CO}_2}{44.0\text{ g/mol}} = \mathbf{1.0\text{ mol of }\text{CO}_2}$$
* Since each molecule of $\text{CO}_2$ contains exactly one carbon atom, the moles of carbon atoms in the original compound must equal the moles of $\text{CO}_2$ produced:
$$n_{\text{C}} = n_{\text{CO}_2} = \mathbf{1.0\text{ mol of C}}$$
3. Calculate the moles of hydrogen (H) in the compound sample:
* Find the moles of $\text{H}_2\text{O}$ produced using its molar mass ($\approx 18.0\text{ g/mol}$):
$$n_{\text{H}_2\text{O}} = \frac{45.0\text{ g of }\text{H}_2\text{O}}{18.0\text{ g/mol}} = \mathbf{2.5\text{ mol of }\text{H}_2\text{O}}$$
* Since each molecule of $\text{H}_2\text{O}$ contains exactly two hydrogen atoms, the moles of hydrogen atoms in the original compound must be twice the moles of water produced:
$$n_{\text{H}} = 2 \times n_{\text{H}_2\text{O}} = 2 \times 2.5\text{ mol} = \mathbf{5.0\text{ mol of H}}$$
4. Determine the molar ratio of Carbon to Hydrogen (C:H):
* The simplest mole-to-mole ratio of carbon to hydrogen in the starting compound is:
$$\text{Ratio (C : H)} = 1.0 : 5.0$$
5. Evaluate the options for a matching C:H ratio:
* An empirical formula represents the simplest whole-number ratio of elements in a compound. Therefore, any possible empirical formula for this compound must exhibit a carbon-to-hydrogen ratio of exactly 1:5.
* Examining the options, only $\text{CH}_5\text{N}$ (Option B) has a C:H ratio of 1:5. This matches the official answer key.

*

WHY_OTHERS_WRONG:

• Option A is incorrect: In $\text{CH}_2\text{N}$, the carbon-to-hydrogen ratio is 1:2. If this were the correct formula, burning the compound would yield a 1:1 mole ratio of $\text{CO}_2$ to $\text{H}_2\text{O}$ (for example, $1.0\text{ mol of CO}_2$ and $1.0\text{ mol of H}_2\text{O}$, which corresponds to only $18.0\text{ g of H}_2\text{O}$ being produced rather than $45.0\text{ g}$).
• Option C is incorrect: In $\text{C}_2\text{H}_5\text{N}$, the carbon-to-hydrogen ratio is 2:5 (or 1:2.5). Burning a compound with this ratio would yield a $2:2.5$ (or $4:5$) mole ratio of $\text{CO}_2$ to $\text{H}_2\text{O}$, which is structurally different from our determined $1.0 : 2.5$ mole ratio of $\text{CO}_2$ to $\text{H}_2\text{O}$.
• Option D is incorrect: In $\text{C}_3\text{H}_3\text{N}_2$, the carbon-to-hydrogen ratio is 3:3 (or 1:1). Burning this compound would yield a $3:1.5$ (or $2:1$) mole ratio of $\text{CO}_2$ to $\text{H}_2\text{O}$, which does not align with the experimental results.