Determined the simplest of empincal formala of a Compound that Contains \( 82.2 \% \) of Nitrogen and \( 17.8 \% \) of hydrogen by mass- if the molecular weight of this compound was found rxperimentally to be 17.04 g , what is che molecalar formula of Compound (H-1.00g, \( N=14.0 \mathrm{~g} \) )

To determine the molecular formula of the compound containing \( 82.2\% \) nitrogen (N) and \( 17.8\% \) hydrogen (H) by mass, with a given molecular weight of \( 17.04 \) g/mol, follow these steps: ### 1. **Assume a 100 g Sample** This simplifies the calculations by directly converting percentages to grams. - **Nitrogen (N):** \( 82.2\% \) of 100 g = **82.2 g** - **Hydrogen (H):** \( 17.8\% \) of 100 g = **17.8 g** ### 2. **Calculate Moles of Each Element** Use the atomic weights provided: - **Nitrogen (N):** \( 14.0 \) g/mol - **Hydrogen (H):** \( 1.00 \) g/mol \[ \text{Moles of N} = \frac{82.2 \text{ g}}{14.0 \text{ g/mol}} \approx 5.871 \text{ mol} \] \[ \text{Moles of H} = \frac{17.8 \text{ g}}{1.00 \text{ g/mol}} = 17.8 \text{ mol} \] ### 3. **Determine the Simplest Mole Ratio** Divide each by the smallest number of moles to find the ratio: \[ \text{Ratio of N} = \frac{5.871}{5.871} = 1 \] \[ \text{Ratio of H} = \frac{17.8}{5.871} \approx 3.034 \approx 3 \] The simplest whole-number ratio of N to H is **1:3**. ### 4. **Establish the Empirical Formula** Based on the ratio, the empirical formula is: \[ \text{Empirical Formula} = \text{NH}_3 \] ### 5. **Calculate the Empirical Formula Weight** \[ \text{Weight of NH}_3 = 14.0 \text{ g (N)} + 3 \times 1.00 \text{ g (H)} = 17.0 \text{ g/mol} \] ### 6. **Determine the Molecular Formula** Compare the empirical formula weight to the given molecular weight: \[ n = \frac{\text{Molecular Weight}}{\text{Empirical Formula Weight}} = \frac{17.04 \text{ g/mol}}{17.0 \text{ g/mol}} \approx 1 \] Since \( n \approx 1 \), the molecular formula is the same as the empirical formula. ### **Final Molecular Formula** \[ \boxed{\text{NH}_3} \]